MBBS and BDS entrance Chemistry: NEET-UG questions on Nepal's MBBS syllabus: Inorganic Chemistry

2016 · NEET-UG (India) · 4 marks

Consider the molecules $\mathrm{CH_4}$ , $\mathrm{NH_3}$ and $\mathrm{H_2O}$ . Which of the given statements is false? (a) The H–C–H bond angle in $\mathrm{CH_4}$ , the H–N–H bond angle in $\mathrm{NH_3}$ , and the H–O–H bond angle in $\mathrm{H_2O}$ are all greater than $90^\circ$ . (b) The H–O–H bond angle in $\mathrm{H_2O}$ is larger than the H–C–H bond angle in $\mathrm{CH_4}$ . (c) The H–O–H bond angle in $\mathrm{H_2O}$ is smaller than the H–N–H bond angle in $\mathrm{NH_3}$ . (d) The H–C–H bond angle in $\mathrm{CH_4}$ is larger than the H–N–H bond angle in $\mathrm{NH_3}$ .

Answer

(b) The H–O–H bond angle in $\mathrm{H_2O}$ is larger than the H–C–H bond angle in $\mathrm{CH_4}$ .

All three molecules are $sp^3$ but lone pairs squeeze the bond angle: $\mathrm{CH_4}$ ( $109.5^\circ$ , no lone pair) > $\mathrm{NH_3}$ ( $107^\circ$ , one lone pair) > $\mathrm{H_2O}$ ( $104.5^\circ$ , two lone pairs). So H–O–H is smaller, not larger, than H–C–H, which makes (b) the false statement. The other three statements are true.

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2016 · NEET-UG (India) · 4 marks

Among the following, the correct order of acidity is: (a) $\mathrm{HClO_3<HClO_4<HClO_2<HClO}$ (b) $\mathrm{HClO<HClO_2<HClO_3<HClO_4}$ (c) $\mathrm{HClO_2<HClO<HClO_3<HClO_4}$ (d) $\mathrm{HClO_4<HClO_2<HClO<HClO_3}$

Answer

(b) $\mathrm{HClO<HClO_2<HClO_3<HClO_4}$

Acidity of the chlorine oxoacids rises with the number of oxygen atoms (oxidation state of Cl: +1, +3, +5, +7). More O atoms disperse the negative charge of the conjugate base ( $\mathrm{ClO^-}$ to $\mathrm{ClO_4^-}$ ) better, so the anion is more stable and the acid stronger. $\mathrm{HClO_4}$ is the strongest.

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2016 · NEET-UG (India) · 4 marks

In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (a) $\mathrm{Al^{3+}<Mg^{2+}<Na^{+}<F^{-}}$ (increasing ionic size) (b) $\mathrm{B<C<N<O}$ (increasing first ionisation enthalpy) (c) $\mathrm{I<Br<Cl<F}$ (increasing electron gain enthalpy) (d) $\mathrm{Li<Na<K<Rb}$ (increasing metallic radius)

Answer

(b) $\mathrm{B<C<N<O}$ (increasing first ionisation enthalpy)

Nitrogen has a half-filled stable $2p^3$ configuration, so its first ionisation enthalpy is higher than oxygen's: the actual order is $\mathrm{B<C<O<N}$ . Hence (b) does not agree with the property (this is the NTA key answer). (a) is right because for isoelectronic ions size falls as nuclear charge rises, and (d) is right because metallic radius increases down group 1. Note that option (c) as printed here ( $\mathrm{I<Br<Cl<F}$ ) looks like a misprint of the original paper's $\mathrm{I<Br<F<Cl}$ , which is correct since Cl has a more negative electron gain enthalpy than F.

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2016 · NEET-UG (India) · 4 marks

Which of the following statements is false? (a) $\mathrm{Mg^{2+}}$ ions form a complex with ATP. (b) $\mathrm{Ca^{2+}}$ ions are important in blood clotting. (c) $\mathrm{Ca^{2+}}$ ions are not important in maintaining the regular beating of the heart. (d) $\mathrm{Mg^{2+}}$ ions are important in the green parts of plants.

Answer

(c) $\mathrm{Ca^{2+}}$ ions are not important in maintaining the regular beating of the heart.

$\mathrm{Ca^{2+}}$ (with $\mathrm{Na^+}$ and $\mathrm{K^+}$ ) is essential for the rhythmic beating of the heart and for muscle contraction, so (c) is false. $\mathrm{Mg^{2+}}$ complexes with ATP and sits at the centre of chlorophyll, and $\mathrm{Ca^{2+}}$ is needed for blood clotting, so the other statements are true.

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2016 · NEET-UG (India) · 4 marks

Which one of the following statements is correct when $\mathrm{SO_2}$ is passed through acidified $\mathrm{K_2Cr_2O_7}$ solution? (a) The solution turns blue. (b) The solution is decolourized. (c) $\mathrm{SO_2}$ is reduced. (d) Green $\mathrm{Cr_2(SO_4)_3}$ is formed.

Answer

(d) Green $\mathrm{Cr_2(SO_4)_3}$ is formed.

$\mathrm{SO_2}$ is a reducing agent. In acid it reduces orange $\mathrm{Cr_2O_7^{2-}}$ to green $\mathrm{Cr^{3+}}$ : $\mathrm{Cr_2O_7^{2-}+3SO_2+2H^+\rightarrow2Cr^{3+}+3SO_4^{2-}+H_2O}$ , giving green $\mathrm{Cr_2(SO_4)_3}$ . The solution turns green (not blue or colourless), and $\mathrm{SO_2}$ is oxidised, not reduced.

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2016 · NEET-UG (India) · 4 marks

The electronic configurations of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are: (a) $[\mathrm{Xe}]4f^7 6s^2$ , $[\mathrm{Xe}]4f^8 6s^2$ and $[\mathrm{Xe}]4f^8 5d^1 6s^2$ (b) $[\mathrm{Xe}]4f^6 5d^1 6s^2$ , $[\mathrm{Xe}]4f^7 5d^1 6s^2$ and $[\mathrm{Xe}]4f^9 6s^2$ (c) $[\mathrm{Xe}]4f^6 5d^1 6s^2$ , $[\mathrm{Xe}]4f^7 5d^1 6s^2$ and $[\mathrm{Xe}]4f^8 5d^1 6s^2$ (d) $[\mathrm{Xe}]4f^7 6s^2$ , $[\mathrm{Xe}]4f^7 5d^1 6s^2$ and $[\mathrm{Xe}]4f^9 6s^2$

Answer

(d) $[\mathrm{Xe}]4f^7 6s^2$ , $[\mathrm{Xe}]4f^7 5d^1 6s^2$ and $[\mathrm{Xe}]4f^9 6s^2$

Eu ( $Z=63$ ) has the half-filled $4f^7$ with $6s^2$ . Gd ( $Z=64$ ) keeps the stable half-filled $4f^7$ and puts the extra electron in $5d$ : $4f^75d^16s^2$ . Tb ( $Z=65$ ) returns to $4f^9 6s^2$ . The other options place the electrons wrongly for Eu, Gd or Tb.

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2016 · NEET-UG (India) · 4 marks

When copper is heated with conc. $\mathrm{HNO_3}$ it produces: (a) $\mathrm{Cu(NO_3)_2}$ and $\mathrm{NO_2}$ (b) $\mathrm{Cu(NO_3)_2}$ and $\mathrm{NO}$ (c) $\mathrm{Cu(NO_3)_2}$ , $\mathrm{NO}$ and $\mathrm{NO_2}$ (d) $\mathrm{Cu(NO_3)_2}$ and $\mathrm{N_2O}$

Answer

(a) $\mathrm{Cu(NO_3)_2}$ and $\mathrm{NO_2}$

With concentrated $\mathrm{HNO_3}$ : $\mathrm{Cu+4HNO_3\rightarrow Cu(NO_3)_2+2NO_2+2H_2O}$ . NO is obtained with dilute (about 30%) nitric acid, and $\mathrm{N_2O}$ with very dilute acid and more active metals.

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2016 · NEET-UG (India) · 4 marks

Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (a) $\mathrm{I_2>Br_2>Cl_2>F_2}$ (b) $\mathrm{Cl_2>Br_2>F_2>I_2}$ (c) $\mathrm{Br_2>I_2>F_2>Cl_2}$ (d) $\mathrm{F_2>Cl_2>Br_2>I_2}$

Answer

(b) $\mathrm{Cl_2>Br_2>F_2>I_2}$

Bond enthalpy falls from $\mathrm{Cl_2}$ (about 242 kJ/mol) to $\mathrm{Br_2}$ (193) to $\mathrm{I_2}$ (151) as the atoms get larger. Fluorine is the exception: $\mathrm{F_2}$ (about 158) is lower than $\mathrm{Cl_2}$ because of strong lone pair–lone pair repulsion in the small F–F bond. So (d) is a tempting but wrong order.

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2016 · NEET-UG (India) · 4 marks

Predict the correct order among the following: (a) lone pair – lone pair > lone pair – bond pair > bond pair – bond pair (b) lone pair – lone pair > bond pair – bond pair > lone pair – bond pair (c) bond pair – bond pair > lone pair – bond pair > lone pair – lone pair (d) lone pair – bond pair > bond pair – bond pair > lone pair – lone pair

Answer

(a) lone pair – lone pair > lone pair – bond pair > bond pair – bond pair

A lone pair is held by only one nucleus, so it spreads out and occupies more space than a bonded pair. VSEPR theory therefore gives repulsion order lp–lp > lp–bp > bp–bp. This is why bond angles shrink as lone pairs are added, as in $\mathrm{CH_4}$ , $\mathrm{NH_3}$ and $\mathrm{H_2O}$ .

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2016 · NEET-UG (India) · 4 marks

The product obtained as a result of a reaction of nitrogen with $\mathrm{CaC_2}$ is: (a) $\mathrm{Ca(CN)_2}$ (b) $\mathrm{CaCN}$ (c) $\mathrm{CaCN_3}$ (d) $\mathrm{Ca_2CN}$

Answer

(a) $\mathrm{Ca(CN)_2}$

Heating calcium carbide in nitrogen gives calcium cyanamide, $\mathrm{CaC_2+N_2\rightarrow CaNCN\,(CaCN_2)+C}$ , which contains the cyanamide ion $\mathrm{NCN^{2-}}$ . The official key accepts option (a), the closest of the four formulas, which is a cyanide-type calcium compound with $\mathrm{CN}$ groups. The other options are not valid formulas.

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2016 · NEET-UG (India) · 4 marks

Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option. Column I: (a) $\mathrm{XeF_6}$ , (b) $\mathrm{XeO_3}$ , (c) $\mathrm{XeOF_4}$ , (d) $\mathrm{XeF_4}$ . Column II: (i) distorted octahedral, (ii) square planar, (iii) pyramidal, (iv) square pyramidal. Code: (a) (b) (c) (d) (a) (i) (iii) (iv) (ii) (b) (i) (ii) (iv) (iii) (c) (iv) (iii) (i) (ii) (d) (iv) (i) (ii) (iii)

Answer

(a) (a)–(i), (b)–(iii), (c)–(iv), (d)–(ii)

$\mathrm{XeF_6}$ : $sp^3d^3$ , distorted octahedral. $\mathrm{XeO_3}$ : $sp^3$ , pyramidal. $\mathrm{XeOF_4}$ : $sp^3d^2$ , square pyramidal. $\mathrm{XeF_4}$ : $sp^3d^2$ , square planar. This sequence (i), (iii), (iv), (ii) matches the first code.

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2016 · NEET-UG (India) · 4 marks

Which of the following has the longest C–O bond length? (Free C–O bond length in CO is 1.128 Å.) (a) $\mathrm{Ni(CO)_4}$ (b) $[\mathrm{Co(CO)_4}]^{-}$ (c) $[\mathrm{Fe(CO)_4}]^{2-}$ (d) $[\mathrm{Mn(CO)_6}]^{+}$

Answer

(c) $[\mathrm{Fe(CO)_4}]^{2-}$

Metal-to-CO back-donation fills the $\pi^*$ orbital of CO, weakening and lengthening the C–O bond. The more negative the charge on the metal, the stronger the back-bonding. $[\mathrm{Fe(CO)_4}]^{2-}$ has the highest negative charge, so it has the longest C–O bond. $[\mathrm{Mn(CO)_6}]^+$ is positive and has the shortest.

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2016 · NEET-UG (India) · 4 marks

Which is the correct statement for the given acids? (a) Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid. (b) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid. (c) Both are triprotic acids. (d) Both are diprotic acids.

Answer

(b) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.

Phosphinic acid is $\mathrm{H_3PO_2}$ , with only one P–OH group, so it is monobasic. Phosphonic acid is $\mathrm{H_3PO_3}$ , with two P–OH groups, so it is dibasic. The P–H hydrogens are not ionisable.

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2016 · NEET-UG (India) · 4 marks

Which one of the following compounds shows the presence of intramolecular hydrogen bond? (a) $\mathrm{H_2O_2}$ (b) HCN (c) Cellulose (d) Concentrated acetic acid

Answer

(c) Cellulose

In cellulose, the many OH groups on the same chain can hydrogen bond internally, and this intrachain bonding stabilises its structure (along with interchain bonds). $\mathrm{H_2O_2}$ , HCN and concentrated acetic acid (dimer) show only intermolecular hydrogen bonding.

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2016 · NEET-UG (India) · 4 marks

Boric acid is an acid because its molecule (a) contains replaceable $\mathrm{H^+}$ ion (b) gives up a proton (c) accepts $\mathrm{OH^-}$ from water releasing proton (d) combines with proton from water molecule

Answer

(c) accepts $\mathrm{OH^-}$ from water releasing proton

$\mathrm{B(OH)_3}$ has an empty p-orbital on boron, so it is a Lewis acid, not a proton donor: $\mathrm{B(OH)_3+H_2O\rightarrow[B(OH)_4]^-+H^+}$ . Its H atoms are not directly ionisable, so (a) and (b) are wrong.

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2016 · NEET-UG (India) · 4 marks

$\mathrm{AlF_3}$ is soluble in HF only in presence of KF. It is due to the formation of (a) $\mathrm{K_3[AlF_3H_3]}$ (b) $\mathrm{K_3[AlF_6]}$ (c) $\mathrm{AlH_3}$ (d) $\mathrm{K[AlF_3H]}$

Answer

(b) $\mathrm{K_3[AlF_6]}$

$\mathrm{AlF_3}$ is largely ionic/polymeric and insoluble in HF. With KF, extra $\mathrm{F^-}$ ions coordinate to $\mathrm{Al^{3+}}$ to give the soluble complex $\mathrm{AlF_3+3KF\rightarrow K_3[AlF_6]}$ . The hydride-containing species in the other options cannot form from fluoride.

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2016 · NEET-UG (India) · 4 marks

The suspension of slaked lime in water is known as (a) limewater (b) quicklime (c) milk of lime (d) aqueous solution of slaked lime

Answer

(c) milk of lime

Slaked lime is $\mathrm{Ca(OH)_2}$ . A suspension of it in water is milk of lime; the clear filtered saturated solution is limewater. Quicklime is $\mathrm{CaO}$ .

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2016 · NEET-UG (India) · 4 marks

The hybridizations of atomic orbitals of nitrogen in $\mathrm{NO_2^+}$ , $\mathrm{NO_3^-}$ and $\mathrm{NH_4^+}$ respectively are (a) $sp,\ sp^3$ and $sp^2$ (b) $sp^2,\ sp^3$ and $sp$ (c) $sp,\ sp^2$ and $sp^3$ (d) $sp^2,\ sp$ and $sp^3$

Answer

(c) $sp,\ sp^2$ and $sp^3$

$\mathrm{NO_2^+}$ : 2 bond pairs, no lone pair, linear, $sp$ . $\mathrm{NO_3^-}$ : 3 bond pairs, trigonal planar, $sp^2$ . $\mathrm{NH_4^+}$ : 4 bond pairs, tetrahedral, $sp^3$ .

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2016 · NEET-UG (India) · 4 marks

Which of the following fluoro-compounds is most likely to behave as a Lewis base? (a) $\mathrm{BF_3}$ (b) $\mathrm{PF_3}$ (c) $\mathrm{CF_4}$ (d) $\mathrm{SiF_4}$

Answer

(b) $\mathrm{PF_3}$

$\mathrm{PF_3}$ has a lone pair on phosphorus that it can donate, so it acts as a Lewis base (as a ligand, like CO). $\mathrm{BF_3}$ is electron-deficient (Lewis acid); $\mathrm{CF_4}$ and $\mathrm{SiF_4}$ have no lone pair on the central atom, and $\mathrm{SiF_4}$ is a Lewis acid.

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2016 · NEET-UG (India) · 4 marks

Which of the following pairs of ions is isoelectronic and isostructural? (a) $\mathrm{CO_3^{2-},\ NO_3^-}$ (b) $\mathrm{ClO_3^-,\ CO_3^{2-}}$ (c) $\mathrm{SO_3^{2-},\ NO_3^-}$ (d) $\mathrm{ClO_3^-,\ SO_3^{2-}}$

Answer

(d) $\mathrm{ClO_3^-,\ SO_3^{2-}}$

Both have 42 electrons and 3 bond pairs plus 1 lone pair on the central atom, so both are pyramidal. $\mathrm{CO_3^{2-}}$ and $\mathrm{NO_3^-}$ are isoelectronic and planar but the other pairs mix different electron counts or shapes.

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2016 · NEET-UG (India) · 4 marks

In context with beryllium, which one of the following statements is incorrect? (a) It is rendered passive by nitric acid. (b) It forms $\mathrm{Be_2C}$ . (c) Its salts rarely hydrolyze. (d) Its hydride is electron-deficient and polymeric.

Answer

(c) Its salts rarely hydrolyze.

Because of its very small size and high charge density, $\mathrm{Be^{2+}}$ salts are extensively hydrolysed, so the statement is incorrect. The others are true: Be is passivated by $\mathrm{HNO_3}$ , forms $\mathrm{Be_2C}$ , and $\mathrm{BeH_2}$ is electron-deficient and polymeric.

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2016 · NEET-UG (India) · 4 marks

Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour? (a) $\mathrm{Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O}$ (b) $\mathrm{3S+2H_2SO_4\rightarrow 3SO_2+2H_2O}$ (c) $\mathrm{C+2H_2SO_4\rightarrow CO_2+2SO_2+2H_2O}$ (d) $\mathrm{CaF_2+H_2SO_4\rightarrow CaSO_4+2HF}$

Answer

(d) $\mathrm{CaF_2+H_2SO_4\rightarrow CaSO_4+2HF}$

Here no oxidation state changes; it is a simple acid displacement of the volatile acid HF. In (a), (b) and (c) sulphur goes from $+6$ in $\mathrm{H_2SO_4}$ to $+4$ in $\mathrm{SO_2}$ , so the acid oxidises Cu, S and C.

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2016 · NEET-UG (India) · 4 marks

Which of the following pairs of d-orbitals will have electron density along the axes? (a) $d_{z^2},\ d_{xz}$ (b) $d_{xz},\ d_{yz}$ (c) $d_{z^2},\ d_{x^2-y^2}$ (d) $d_{xy},\ d_{x^2-y^2}$

Answer

(c) $d_{z^2},\ d_{x^2-y^2}$

These two ( $e_g$ ) orbitals have lobes pointing along the axes. $d_{xy}$ , $d_{xz}$ and $d_{yz}$ ( $t_{2g}$ ) have lobes between the axes, so options containing them are wrong.

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2016 · NEET-UG (India) · 4 marks

The correct geometry and hybridization for $\mathrm{XeF_4}$ are (a) octahedral, $sp^3d^2$ (b) trigonal bipyramidal, $sp^3d$ (c) planar triangle, $sp^3d^3$ (d) square planar, $sp^3d^2$

Answer

(d) square planar, $sp^3d^2$

$\mathrm{XeF_4}$ has 4 bond pairs and 2 lone pairs on Xe, so 6 electron pairs: octahedral electron geometry, $sp^3d^2$ . The two lone pairs sit trans to each other, giving a square planar shape, so (a) is wrong about molecular geometry.

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2016 · NEET-UG (India) · 4 marks

Among the following, which one is a wrong statement? (a) $\mathrm{PH_5}$ and $\mathrm{BiCl_5}$ do not exist. (b) $p\pi$ - $d\pi$ bonds are present in $\mathrm{SO_2}$ . (c) $\mathrm{SeF_4}$ and $\mathrm{CH_4}$ have same shape. (d) $\mathrm{I_3^+}$ has bent geometry.

Answer

(c) $\mathrm{SeF_4}$ and $\mathrm{CH_4}$ have same shape.

$\mathrm{SeF_4}$ has 4 bond pairs and 1 lone pair (see-saw), while $\mathrm{CH_4}$ is tetrahedral, so the statement is wrong. The others hold: $\mathrm{PH_5}$ and $\mathrm{BiCl_5}$ do not exist, $\mathrm{SO_2}$ has $p\pi$ - $d\pi$ bonding, and $\mathrm{I_3^+}$ is bent.

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2016 · NEET-UG (India) · 4 marks

The correct increasing order of trans-effect of the following species is (a) $\mathrm{NH_3>CN^->Br^->C_6H_5^-}$ (b) $\mathrm{CN^->C_6H_5^->Br^->NH_3}$ (c) $\mathrm{Br^->CN^->NH_3>C_6H_5^-}$ (d) $\mathrm{CN^->Br^->C_6H_5^->NH_3}$

Answer

(b) $\mathrm{CN^->C_6H_5^->Br^->NH_3}$

The trans-effect series runs roughly $\mathrm{CN^-}>\mathrm{C_6H_5^-}>\mathrm{Br^-}>\mathrm{NH_3}$ in decreasing strength (strong $\pi$ -acceptors and sigma donors are strongest, $\mathrm{NH_3}$ weakest of these). This is the ordering the key accepts, although the question wording says "increasing".

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2016 · NEET-UG (India) · 4 marks

Which one of the following statements related to lanthanons is incorrect? (a) Europium shows +2 oxidation state. (b) The basicity decreases as the ionic radius decreases from Pr to Lu. (c) All the lanthanons are much more reactive than aluminium. (d) Ce(+4) solutions are widely used as oxidizing agent in volumetric analysis.

Answer

(c) All the lanthanons are much more reactive than aluminium.

This is the statement marked incorrect in the answer key; the other three are true. $\mathrm{Eu}$ shows $+2$ (stable $4f^7$ ), basicity of $\mathrm{Ln(OH)_3}$ falls with the lanthanoid contraction, and $\mathrm{Ce^{4+}}$ is a common volumetric oxidant.

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2016 · NEET-UG (India) · 4 marks

Jahn-Teller effect is not observed in high spin complexes of (a) $d^7$ (b) $d^8$ (c) $d^4$ (d) $d^9$

Answer

(b) $d^8$

Jahn-Teller distortion needs unequal occupation of degenerate orbitals. High-spin $d^8$ is $t_{2g}^6e_g^2$ (symmetric), so no distortion. $d^4$ ( $t_{2g}^3e_g^1$ ) and $d^9$ ( $t_{2g}^6e_g^3$ ) have uneven $e_g$ occupation and distort strongly; $d^7$ is uneven in $t_{2g}$ and distorts weakly.

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2015 · NEET-UG (India) · 4 marks

In which of the following pairs, both the species are not isostructural? (a) $\mathrm{XeF_4,\ XeO_4}$ (b) $\mathrm{SiCl_4,\ PCl_4^+}$ (c) diamond, silicon carbide (d) $\mathrm{NH_3,\ PH_3}$

Answer

(a) $\mathrm{XeF_4,\ XeO_4}$

$\mathrm{XeF_4}$ is square planar ( $sp^3d^2$ , two lone pairs) whereas $\mathrm{XeO_4}$ is tetrahedral ( $sp^3$ ), so they are not isostructural. $\mathrm{SiCl_4}$ / $\mathrm{PCl_4^+}$ are both tetrahedral, diamond/SiC are both tetrahedral networks, and $\mathrm{NH_3}$ / $\mathrm{PH_3}$ are both pyramidal.

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2015 · NEET-UG (India) · 4 marks

On heating which of the following releases $\mathrm{CO_2}$ most easily? (a) $\mathrm{CaCO_3}$ (b) $\mathrm{K_2CO_3}$ (c) $\mathrm{Na_2CO_3}$ (d) $\mathrm{MgCO_3}$

Answer

(d) $\mathrm{MgCO_3}$

Thermal stability of carbonates rises down group 2 and with larger cation size. $\mathrm{Mg^{2+}}$ is the smallest and most polarising, so $\mathrm{MgCO_3}$ decomposes most easily. $\mathrm{K_2CO_3}$ and $\mathrm{Na_2CO_3}$ are very stable.

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2015 · NEET-UG (India) · 4 marks

Number of possible isomers for the complex $[\mathrm{Co(en)_2Cl_2}]\mathrm{Cl}$ will be: (en = ethylenediamine) (a) 4 (b) 2 (c) 1 (d) 3

Answer

(d) 3

$[\mathrm{Co(en)_2Cl_2}]^+$ is octahedral of type $M(AA)_2b_2$ : the trans form is optically inactive (1) and the cis form exists as a pair of enantiomers (2), giving 3 stereoisomers in total.

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2015 · NEET-UG (India) · 4 marks

Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium? (a) $[\mathrm{Xe}]4f^6 5d^2 6s^2$ (b) $[\mathrm{Xe}]4f^8 6d^2$ (c) $[\mathrm{Xe}]4f^9 5s^1$ (d) $[\mathrm{Xe}]4f^7 5d^1 6s^2$

Answer

(d) $[\mathrm{Xe}]4f^7 5d^1 6s^2$

Gd ( $Z=64$ ) has the half-filled $4f^7$ subshell, which is extra stable, so one electron goes into $5d$ rather than a $4f^8$ arrangement: $[\mathrm{Xe}]4f^75d^16s^2$ .

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2015 · NEET-UG (India) · 4 marks

Decreasing order of stability of $\mathrm{O_2}$ , $\mathrm{O_2^-}$ , $\mathrm{O_2^+}$ and $\mathrm{O_2^{2-}}$ is: (a) $\mathrm{O_2^->O_2^{2-}>O_2^+>O_2}$ (b) $\mathrm{O_2^+>O_2>O_2^->O_2^{2-}}$ (c) $\mathrm{O_2^{2-}>O_2^->O_2>O_2^+}$ (d) $\mathrm{O_2>O_2^+>O_2^{2-}>O_2^-}$

Answer

(b) $\mathrm{O_2^+>O_2>O_2^->O_2^{2-}}$

Stability follows bond order (MO theory): $\mathrm{O_2^+}$ 2.5, $\mathrm{O_2}$ 2, $\mathrm{O_2^-}$ 1.5, $\mathrm{O_2^{2-}}$ 1. Higher bond order means greater stability.

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2015 · NEET-UG (India) · 4 marks

The hybridization involved in complex $[\mathrm{Ni(CN)_4}]^{2-}$ is: (At. No. Ni = 28) (a) $d^2sp^3$ (b) $dsp^2$ (c) $sp^3$ (d) $d^2sp^2$

Answer

(b) $dsp^2$

$\mathrm{Ni^{2+}}$ is $d^8$ . The strong-field ligand $\mathrm{CN^-}$ pairs the electrons, freeing one $3d$ orbital, which with one $4s$ and two $4p$ orbitals gives $dsp^2$ and a square planar, diamagnetic complex.

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2015 · NEET-UG (India) · 4 marks

The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence (a) $\mathrm{In<Tl<Ga<Al}$ (b) $\mathrm{Ga<In<Al<Tl}$ (c) $\mathrm{Al<Ga<In<Tl}$ (d) $\mathrm{Tl<In<Ga<Al}$

Answer

(c) $\mathrm{Al<Ga<In<Tl}$

Down group 13, the inert pair effect makes the $+1$ state increasingly stable, so $\mathrm{Tl^+}$ is the most stable and $\mathrm{Al^+}$ the least.

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2015 · NEET-UG (India) · 4 marks

The sum of coordination number and oxidation number of the metal M in the complex $[\mathrm{M(en)_2(C_2O_4)}]\mathrm{Cl}$ (where en is ethylenediamine) is: (a) 8 (b) 9 (c) 6 (d) 7

Answer

(b) 9

The ligands en (neutral) and $\mathrm{C_2O_4^{2-}}$ are all bidentate, so three ligands give coordination number $2+2+2=6$ . Charge balance: $x+0+(-2)+(-1)=0$ gives $x=+3$ for the metal (the ion $\mathrm{Cl^-}$ is outside the sphere). Sum $=6+3=9$ . Option (c) 6 gives only the coordination number.

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2015 · NEET-UG (India) · 4 marks

Which of the statements given below is incorrect? (a) $\mathrm{OF_2}$ is an oxide of fluorine (b) $\mathrm{Cl_2O_7}$ is an anhydride of perchloric acid (c) $\mathrm{O_3}$ molecule is bent (d) ONF is isoelectronic with $\mathrm{O_2N^-}$

Answer

(a) $\mathrm{OF_2}$ is an oxide of fluorine

$\mathrm{OF_2}$ is a fluoride of oxygen, not an oxide of fluorine: fluorine is more electronegative, so oxygen is in the $+2$ state. An oxide would need oxygen at $-2$ , so the statement is incorrect. $\mathrm{Cl_2O_7}+\mathrm{H_2O}\rightarrow 2\mathrm{HClO_4}$ , so (b) is true; $\mathrm{O_3}$ is bent; ONF and $\mathrm{NO_2^-}$ both have 24 electrons, so (d) is true.

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2015 · NEET-UG (India) · 4 marks

The formation of the oxide ion, $\mathrm{O^{2-}(g)}$ , from oxygen atom requires first an exothermic and then an endothermic step as shown below: $\mathrm{O(g)+e^-\rightarrow O^-(g)}$ ; $\Delta_fH^\ominus=-141\ \mathrm{kJ\ mol^{-1}}$ $\mathrm{O^-(g)+e^-\rightarrow O^{2-}(g)}$ ; $\Delta_fH^\ominus=+780\ \mathrm{kJ\ mol^{-1}}$ Thus process of formation of $\mathrm{O^{2-}}$ in gas phase is unfavourable even though $\mathrm{O^{2-}}$ is isoelectronic with neon. It is due to the fact that: (a) addition of electron in oxygen results in larger size of the ion. (b) electron repulsion outweighs the stability gained by achieving noble gas configuration. (c) $\mathrm{O^-}$ ion has comparatively smaller size than oxygen atom. (d) oxygen is more electronegative.

Answer

(b) electron repulsion outweighs the stability gained by achieving noble gas configuration.

Adding a second electron to the already negative $\mathrm{O^-}$ ion is strongly opposed by electrostatic repulsion, giving the large positive $+780$ kJ/mol. This repulsion outweighs the stability of the neon configuration, so gas-phase $\mathrm{O^{2-}}$ is unfavourable (it is stabilised only by lattice energy in solids).

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2015 · NEET-UG (India) · 4 marks

The variation of the boiling points of the hydrogen halides is in the order $\mathrm{HF>HI>HBr>HCl}$ . What explains the higher boiling point of hydrogen fluoride? (a) The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule. (b) The electronegativity of fluorine is much higher than for other elements in the group. (c) There is strong hydrogen bonding between HF molecules. (d) The bond energy of HF molecules is greater than in other hydrogen halides.

Answer

(c) There is strong hydrogen bonding between HF molecules.

The small, highly electronegative F atom makes the H-F bond very polar, so HF molecules associate through strong intermolecular H-bonds, raising the boiling point. Electronegativity (b) is the cause of H-bonding, not the direct explanation; boiling point depends on intermolecular forces, not on bond energy (d).

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2015 · NEET-UG (India) · 4 marks

The name of complex ion, $[\mathrm{Fe(CN)_6}]^{3-}$ is: (a) Hexacyanidoferrate (III) ion (b) Hexacyanoiron (III) ion (c) Hexacyanitoferrate (III) ion (d) Tricyanoferrate (III) ion

Answer

(a) Hexacyanidoferrate (III) ion

Six $\mathrm{CN^-}$ ligands give prefix hexa- and the ligand name cyanido (IUPAC 2005). The anionic complex takes the metal name ferrate, and the oxidation state is $x-6=-3$ , so $x=+3$ , written (III). "Hexacyanoiron" is wrong because an anionic complex uses -ate.

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2015 · NEET-UG (India) · 4 marks

The reducing behaviour of $\mathrm{H_3PO_2}$ is due to: (a) Presence of two $-$ OH groups and one P $-$ H bond (b) Presence of one $-$ OH group and two P $-$ H bonds (c) High electron gain enthalpy of phosphorus (d) High oxidation state of phosphorus

Answer

(b) Presence of one $-$ OH group and two P $-$ H bonds

In hypophosphorous acid $\mathrm{H_3PO_2}$ the structure is $\mathrm{H_2P(O)OH}$ : one ionisable $-$ OH and two P-H bonds. The P-H bonds are the reducing sites, since they are oxidised to P-OH. Option (a) describes $\mathrm{H_3PO_3}$ , which has two OH and one P-H; and the oxidation state of P in $\mathrm{H_3PO_2}$ is only $+1$ , not high.

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2014 · NEET-UG (India) · 4 marks

The reaction of aqueous $\mathrm{KMnO_4}$ with $\mathrm{H_2O_2}$ in acidic conditions gives: (a) $\mathrm{Mn^{2+}}$ and $\mathrm{O_2}$ (b) $\mathrm{Mn^{2+}}$ and $\mathrm{O_3}$ (c) $\mathrm{Mn^{4+}}$ and $\mathrm{MnO_2}$ (d) $\mathrm{Mn^{4+}}$ and $\mathrm{O_2}$

Answer

(a) $\mathrm{Mn^{2+}}$ and $\mathrm{O_2}$

In acid, $\mathrm{MnO_4^-}$ is reduced to $\mathrm{Mn^{2+}}$ while $\mathrm{H_2O_2}$ acts as a reductant and is oxidised to $\mathrm{O_2}$ : $2\mathrm{MnO_4^-}+6\mathrm{H^+}+5\mathrm{H_2O_2}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{O_2}$ . $\mathrm{MnO_2}$ forms in neutral or alkaline medium, and $\mathrm{O_3}$ is not produced.

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2014 · NEET-UG (India) · 4 marks

In acidic medium, $\mathrm{H_2O_2}$ changes $\mathrm{Cr_2O_7^{2-}}$ to $\mathrm{CrO_5}$ which has two ( $-$ O $-$ O $-$ ) bonds. Oxidation state of Cr in $\mathrm{CrO_5}$ is: (a) +3 (b) +6 (c) $-10$ (d) +5

Answer

(b) +6

$\mathrm{CrO_5}$ has two peroxide linkages ( $-$ O $-$ O $-$ ) and one oxo oxygen. Four peroxide oxygens are $-1$ each and the fifth is $-2$ . $x+4(-1)+(-2)=0$ , so $x=+6$ . A value of $-10$ results from wrongly taking all oxygens as $-2$ , and +6 matches the chromium in the starting dichromate.

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2014 · NEET-UG (India) · 4 marks

Which of the following orders of ionic radii is correctly represented? (a) $\mathrm{Na^+>F^->O^{2-}}$ (b) $\mathrm{F^->O^{2-}>Na^+}$ (c) $\mathrm{Al^{3+}>Mg^{2+}>N^{3-}}$ (d) $\mathrm{H^->H^+>H}$

Answer

(d) $\mathrm{H^->H>H^+}$

Anions are larger than the parent atom (extra electron, same nuclear charge) and cations are smaller, so the order is $\mathrm{H^-}>\mathrm{H}>\mathrm{H^+}$ ; this is the intended option, whose printed order (H $^+$ before H) is a typo of the official one. The isoelectronic series is $\mathrm{O^{2-}>F^->Na^+}$ and $\mathrm{N^{3-}>Mg^{2+}>Al^{3+}}$ , which makes (a), (b) and (c) wrong.

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2014 · NEET-UG (India) · 4 marks

$\mathrm{Be^{2+}}$ is isoelectronic with which of the following ions? (a) $\mathrm{Li^+}$ (b) $\mathrm{Na^+}$ (c) $\mathrm{Mg^{2+}}$ (d) $\mathrm{H^+}$

Answer

(a) $\mathrm{Li^+}$

$\mathrm{Be^{2+}}$ has $4-2=2$ electrons. $\mathrm{Li^+}$ also has $3-1=2$ electrons. $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$ have 10 electrons, and $\mathrm{H^+}$ has none.

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2014 · NEET-UG (India) · 4 marks

Magnetic moment 2.83 BM is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28) (a) $\mathrm{Ni^{2+}}$ (b) $\mathrm{Cr^{3+}}$ (c) $\mathrm{Mn^{2+}}$ (d) $\mathrm{Ti^{3+}}$

Answer

(a) $\mathrm{Ni^{2+}}$

$\mu=\sqrt{n(n+2)}$ BM, and $2.83=\sqrt8$ means $n=2$ unpaired electrons. $\mathrm{Ni^{2+}}$ is $d^8$ with 2 unpaired electrons. $\mathrm{Cr^{3+}}$ ( $d^3$ ) has 3, $\mathrm{Mn^{2+}}$ ( $d^5$ ) has 5 and $\mathrm{Ti^{3+}}$ ( $d^1$ ) has 1.

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2014 · NEET-UG (India) · 4 marks

The pair of compounds that can exist together is: (a) $\mathrm{HgCl_2,\ SnCl_2}$ (b) $\mathrm{FeCl_2,\ SnCl_2}$ (c) $\mathrm{FeCl_3,\ KI}$ (d) $\mathrm{FeCl_3,\ SnCl_2}$

Answer

(b) $\mathrm{FeCl_2,\ SnCl_2}$

Two compounds can coexist only if neither oxidises or reduces the other. $\mathrm{Fe^{2+}}$ and $\mathrm{Sn^{2+}}$ are both reducing and do not react with each other. In (a) $\mathrm{HgCl_2}$ is reduced by $\mathrm{SnCl_2}$ to $\mathrm{Hg_2Cl_2}$ and then Hg. In (c) $\mathrm{Fe^{3+}}$ oxidises $\mathrm{I^-}$ to $\mathrm{I_2}$ . In (d) $\mathrm{Sn^{2+}}$ reduces $\mathrm{Fe^{3+}}$ to $\mathrm{Fe^{2+}}$ .

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2014 · NEET-UG (India) · 4 marks

Which of the following complexes is used to be as an anticancer agent? (a) cis- $[\mathrm{PtCl_2(NH_3)_2}]$ (b) cis- $\mathrm{K_2[PtCl_2Br_2]}$ (c) $\mathrm{Na_2CoCl_4}$ (d) mer- $[\mathrm{Co(NH_3)_3Cl_3}]$

Answer

(a) cis- $[\mathrm{PtCl_2(NH_3)_2}]$

This is cisplatin, which binds to DNA and blocks cell division, so it is used in cancer chemotherapy. The trans isomer is not active, which is why the cis geometry matters. The other complexes have no anticancer use.

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2014 · NEET-UG (India) · 4 marks

Among the following complexes the one which shows Zero crystal field stabilization energy (CFSE) is: (a) $[\mathrm{Fe(H_2O)_6}]^{3+}$ (b) $[\mathrm{Co(H_2O)_6}]^{2+}$ (c) $[\mathrm{Co(H_2O)_6}]^{3+}$ (d) $[\mathrm{Mn(H_2O)_6}]^{3+}$

Answer

(a) $[\mathrm{Fe(H_2O)_6}]^{3+}$

$\mathrm{Fe^{3+}}$ is $d^5$ . $\mathrm{H_2O}$ is a weak-field ligand, so the configuration is high spin $t_{2g}^3e_g^2$ . $\mathrm{CFSE}=3(-0.4\Delta_o)+2(+0.6\Delta_o)=0$ . The others are not zero: $\mathrm{Co^{2+}}$ ( $d^7$ , $t_{2g}^5e_g^2$ ) gives $-0.8\Delta_o$ , $\mathrm{Co^{3+}}$ ( $d^6$ ) and $\mathrm{Mn^{3+}}$ ( $d^4$ ) are also non-zero.

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2014 · NEET-UG (India) · 4 marks

Which one of the following species has plane triangular shape? (a) $\mathrm{NO_3^-}$ (b) $\mathrm{NO_2^-}$ (c) $\mathrm{CO_2}$ (d) $\mathrm{N_3^-}$

Answer

(a) $\mathrm{NO_3^-}$

In $\mathrm{NO_3^-}$ nitrogen has three bond pairs and no lone pair ( $sp^2$ ), so it is trigonal planar. $\mathrm{NO_2^-}$ is bent (one lone pair), while $\mathrm{CO_2}$ and $\mathrm{N_3^-}$ are linear ( $sp$ ).

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2014 · NEET-UG (India) · 4 marks

Which of the following molecules has the maximum dipole moment? (a) $\mathrm{CH_4}$ (b) $\mathrm{NH_3}$ (c) $\mathrm{NF_3}$ (d) $\mathrm{CO_2}$

Answer

(b) $\mathrm{NH_3}$

$\mathrm{CH_4}$ and $\mathrm{CO_2}$ are symmetrical with zero net dipole moment. In $\mathrm{NH_3}$ the N-H bond dipoles and the lone-pair dipole point the same way, giving about 1.47 D. In $\mathrm{NF_3}$ the N-F bond dipoles oppose the lone-pair dipole, so the net value is small (about 0.24 D).

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2014 · NEET-UG (India) · 4 marks

Acidity of diprotic acids in aqueous solutions increases in the order: (a) $\mathrm{H_2Se<H_2S<H_2Te}$ (b) $\mathrm{H_2Te<H_2S<H_2Se}$ (c) $\mathrm{H_2Se<H_2Te<H_2S}$ (d) $\mathrm{H_2S<H_2Se<H_2Te}$

Answer

(d) $\mathrm{H_2S<H_2Se<H_2Te}$

Down group 16 the H-E bond length increases and bond strength falls, so the H-E bond breaks more easily and the hydride gives up $\mathrm{H^+}$ more readily. Acidity therefore increases as $\mathrm{H_2S<H_2Se<H_2Te}$ .

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2014 · NEET-UG (India) · 4 marks

Reason of lanthanoid contraction is: (a) Increasing nuclear charge (b) Decreasing nuclear charge (c) Decreasing screening effect (d) Negligible screening effect of ' $f$ ' orbitals

Answer

(d) Negligible screening effect of ' $f$ ' orbitals

Across the lanthanoids the 4f electrons are added. They shield each other very poorly because of the diffuse shape of f orbitals. The effective nuclear charge therefore rises steadily and pulls the outer electrons in, so the size contracts. Increasing nuclear charge alone (a) is not the specific reason, since it also rises in other series where shielding is better.

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2013 · NEET-UG (India) · 4 marks

A magnetic moment of 1.73 BM will be shown by one among the following (a) $[\mathrm{Cu(NH_3)_4}]^{2+}$ (b) $[\mathrm{Ni(CN)_4}]^{2-}$ (c) $\mathrm{TiCl_4}$ (d) $[\mathrm{CoCl_6}]^{4-}$

Answer

(a) $[\mathrm{Cu(NH_3)_4}]^{2+}$

$\mu=\sqrt{n(n+2)}=1.73$ BM means one unpaired electron. $\mathrm{Cu^{2+}}$ is $d^9$ with one unpaired electron. $[\mathrm{Ni(CN)_4}]^{2-}$ is $d^8$ square planar (0), $\mathrm{TiCl_4}$ is $d^0$ (0), and $[\mathrm{CoCl_6}]^{4-}$ is $d^7$ with three unpaired electrons (3.87 BM).

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2013 · NEET-UG (India) · 4 marks

Roasting of sulphides gives the gas X as a by-product. This is a colourless gas with choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic, acts as a reducing agent and its acid has never been isolated. The gas X is (a) $\mathrm{H_2S}$ (b) $\mathrm{SO_2}$ (c) $\mathrm{CO_2}$ (d) $\mathrm{SO_3}$

Answer

(b) $\mathrm{SO_2}$

Roasting of sulphide ores gives $\mathrm{SO_2}$ , a colourless gas with a choking smell that causes acid rain. It dissolves to give sulphurous acid ( $\mathrm{H_2SO_3}$ ), which cannot be isolated in pure form, and it acts as a reducing agent.

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2013 · NEET-UG (India) · 4 marks

Which is the strongest acid in the following? (a) $\mathrm{H_2SO_4}$ (b) $\mathrm{HClO_3}$ (c) $\mathrm{HClO_4}$ (d) $\mathrm{H_2SO_3}$

Answer

(c) $\mathrm{HClO_4}$

Among oxoacids of the same element, acid strength increases with oxidation state, so $\mathrm{HClO_4}$ (Cl +7) is stronger than $\mathrm{HClO_3}$ (+5). Its conjugate base $\mathrm{ClO_4^-}$ is the most stabilised by resonance, and $\mathrm{HClO_4}$ is the strongest known common acid, stronger than $\mathrm{H_2SO_4}$ .

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2013 · NEET-UG (India) · 4 marks

Which of the following is paramagnetic? (a) CO (b) $\mathrm{O_2^-}$ (c) $\mathrm{CN^-}$ (d) $\mathrm{NO^+}$

Answer

(b) $\mathrm{O_2^-}$

Superoxide $\mathrm{O_2^-}$ has 17 electrons, so its molecular orbital configuration ends with $\pi^*2p^3$ , leaving one unpaired electron. CO, $\mathrm{CN^-}$ and $\mathrm{NO^+}$ have 14 electrons each, with all electrons paired (diamagnetic).

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2013 · NEET-UG (India) · 4 marks

Which of the following structure is similar to graphite? (a) BN (b) B (c) $\mathrm{B_4C}$ (d) $\mathrm{B_2H_6}$

Answer

(a) BN

Boron nitride (hexagonal form) has a layered structure with six-membered rings, like graphite, and is isoelectronic with carbon. It is called 'white graphite'. B is a different structure, $\mathrm{B_4C}$ is an extremely hard covalent solid, and $\mathrm{B_2H_6}$ is a molecular hydride.

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2013 · NEET-UG (India) · 4 marks

The basic structural unit of silicates is (a) $\mathrm{SiO^-}$ (b) $\mathrm{SiO_4^{4-}}$ (c) $\mathrm{SiO_3^{2-}}$ (d) $\mathrm{SiO_4^{2-}}$

Answer

(b) $\mathrm{SiO_4^{4-}}$

Silicates are built from $\mathrm{SiO_4^{4-}}$ tetrahedra, with Si at the centre bonded to four oxygen atoms ( $sp^3$ ). The tetrahedra link through shared oxygen atoms to give chains, sheets and 3D networks.

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2013 · NEET-UG (India) · 4 marks

Which of the following does not give oxygen on heating? (a) $\mathrm{KClO_3}$ (b) $\mathrm{Zn(ClO_3)_2}$ (c) $\mathrm{K_2Cr_2O_7}$ (d) $\mathrm{(NH_4)_2Cr_2O_7}$

Answer

(d) $\mathrm{(NH_4)_2Cr_2O_7}$

Ammonium dichromate decomposes as $\mathrm{(NH_4)_2Cr_2O_7\rightarrow N_2+Cr_2O_3+4H_2O}$ , so it gives nitrogen, not oxygen. Here the oxygen goes into water. $\mathrm{KClO_3}$ and $\mathrm{Zn(ClO_3)_2}$ give $\mathrm{O_2}$ with chloride, and $\mathrm{K_2Cr_2O_7}$ gives $\mathrm{O_2}$ on strong heating.

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2013 · NEET-UG (India) · 4 marks

Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70) (a) $\mathrm{Ce^{2+}}$ (b) $\mathrm{Sm^{2+}}$ (c) $\mathrm{Eu^{2+}}$ (d) $\mathrm{Yb^{2+}}$

Answer

(d) $\mathrm{Yb^{2+}}$

Diamagnetic means no unpaired electrons. $\mathrm{Yb^{2+}}$ is $4f^{14}$ , a full subshell. $\mathrm{Ce^{2+}}$ is $4f^2$ , $\mathrm{Sm^{2+}}$ is $4f^6$ and $\mathrm{Eu^{2+}}$ is $4f^7$ , so all three are paramagnetic.

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2013 · NEET-UG (India) · 4 marks

$\mathrm{XeF_2}$ is isostructural with (a) $\mathrm{TeF_2}$ (b) $\mathrm{ICl_2^-}$ (c) $\mathrm{SbCl_3}$ (d) $\mathrm{BaCl_2}$

Answer

(b) $\mathrm{ICl_2^-}$

$\mathrm{XeF_2}$ has 2 bond pairs and 3 lone pairs ( $sp^3d$ ), giving a linear shape. $\mathrm{ICl_2^-}$ has the same count (2 bond pairs, 3 lone pairs) and is linear. $\mathrm{TeF_2}$ and $\mathrm{SbCl_3}$ have fewer lone pairs and are bent or pyramidal, and $\mathrm{BaCl_2}$ is ionic.

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2013 · NEET-UG (India) · 4 marks

An excess of $\mathrm{AgNO_3}$ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be (a) 0.001 (b) 0.002 (c) 0.003 (d) 0.01

Answer

(a) 0.001

The complex is $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}$ , so only the one outer-sphere chloride ionises and precipitates with $\mathrm{Ag^+}$ . Moles of complex $=0.01\times0.1=0.001$ , so AgCl formed $=0.001$ mol.

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2013 · NEET-UG (India) · 4 marks

Which of these is least likely to act as a Lewis base? (a) CO (b) $\mathrm{F^-}$ (c) $\mathrm{BF_3}$ (d) $\mathrm{PF_3}$

Answer

(c) $\mathrm{BF_3}$

A Lewis base donates an electron pair. $\mathrm{BF_3}$ has an incomplete octet and is an electron-pair acceptor, so it is a Lewis acid. CO, $\mathrm{F^-}$ and $\mathrm{PF_3}$ all have lone pairs to donate.

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2013 · NEET-UG (India) · 4 marks

$\mathrm{KMnO_4}$ can be prepared from $\mathrm{K_2MnO_4}$ as per the reaction: $\mathrm{3MnO_4^{2-}+2H_2O\rightleftharpoons 2MnO_4^-+MnO_2+4OH^-}$ The reaction can go to completion by removing $\mathrm{OH^-}$ ions by adding (a) HCl (b) KOH (c) $\mathrm{CO_2}$ (d) $\mathrm{SO_2}$

Answer

(c) $\mathrm{CO_2}$

The forward reaction produces $\mathrm{OH^-}$ , so removing $\mathrm{OH^-}$ shifts the equilibrium to the right. $\mathrm{CO_2}$ neutralises $\mathrm{OH^-}$ (forming carbonate or bicarbonate) without reacting with manganate or permanganate. HCl and $\mathrm{SO_2}$ would also reduce the manganese species, and KOH adds $\mathrm{OH^-}$ .

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2013 · NEET-UG (India) · 4 marks

Which of the following is electron-deficient? (a) $\mathrm{(CH_3)_2}$ (b) $\mathrm{(SiH_3)_2}$ (c) $\mathrm{(BH_3)_2}$ (d) $\mathrm{PH_3}$

Answer

(c) $\mathrm{(BH_3)_2}$

Diborane $\mathrm{B_2H_6}$ has 12 valence electrons, too few for all the bonds as ordinary 2-centre-2-electron bonds, so it uses two 3-centre-2-electron bridging bonds and is electron deficient. $\mathrm{(CH_3)_2}$ and $\mathrm{(SiH_3)_2}$ have enough electrons, and $\mathrm{PH_3}$ is electron rich.

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2013 · NEET-UG (India) · 4 marks

Which of the following statements about the interstitial compounds is incorrect? (a) They retain metallic conductivity (b) They are chemically reactive (c) They are much harder than the pure metal (d) They have higher melting points than the pure metal

Answer

(b) They are chemically reactive

Interstitial compounds (small atoms such as H, C, N in metal lattice voids) are chemically inert, so statement (b) is incorrect. They keep metallic conductivity and are harder and have higher melting points than the parent metal.

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2013 · NEET-UG (India) · 4 marks

Which one of the following molecules contains no $\pi$ bond? (a) $\mathrm{CO_2}$ (b) $\mathrm{H_2O}$ (c) $\mathrm{SO_2}$ (d) $\mathrm{NO_2}$

Answer

(b) $\mathrm{H_2O}$

$\pi$ bonds arise from multiple bonds. $\mathrm{H_2O}$ has only two O-H single ( $\sigma$ ) bonds. $\mathrm{CO_2}$ (two C=O), $\mathrm{SO_2}$ and $\mathrm{NO_2}$ all contain double bonds, so they have $\pi$ bonds.

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2013 · NEET-UG (India) · 4 marks

Which of the following is a polar molecule? (a) $\mathrm{BF_3}$ (b) $\mathrm{SF_4}$ (c) $\mathrm{SiF_4}$ (d) $\mathrm{XeF_4}$

Answer

(b) $\mathrm{SF_4}$

$\mathrm{SF_4}$ has 4 bond pairs and 1 lone pair (sp $^3$ d), giving a see-saw shape. The bond dipoles do not cancel, so it is polar. $\mathrm{BF_3}$ (trigonal planar), $\mathrm{SiF_4}$ (tetrahedral) and $\mathrm{XeF_4}$ (square planar, 2 lone pairs opposite each other) are symmetrical, so their dipoles cancel and they are non-polar.

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2017 · NEET-UG (India) · 4 marks

Which of the following pairs of compounds is isoelectronic and isostructural?
(a) $\mathrm{BeCl_2}$ , $\mathrm{XeF_2}$ (b) $\mathrm{TeI_2}$ , $\mathrm{XeF_2}$ (c) $\mathrm{IBr_2^-}$ , $\mathrm{XeF_2}$ (d) $\mathrm{IF_3}$ , $\mathrm{XeF_2}$

Answer

(c) $\mathrm{IBr_2^-}$ , $\mathrm{XeF_2}$

Count valence electrons: $\mathrm{XeF_2}$ = 8 + 14 = 22; $\mathrm{IBr_2^-}$ = 7 + 14 + 1 = 22, so they are isoelectronic. Both have $sp^3d$ hybridisation with 3 lone pairs on the central atom, giving a linear shape, so they are isostructural. $\mathrm{BeCl_2}$ is linear but has only 16 valence electrons, and $\mathrm{IF_3}$ is T-shaped.

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2017 · NEET-UG (India) · 4 marks

$\mathrm{HgCl_2}$ and $\mathrm{I_2}$ both when dissolved in water containing $\mathrm{I^-}$ ions the pair of species formed is
(a) $\mathrm{HgI_2}$ , $\mathrm{I_3^-}$ (b) $\mathrm{HgI_2}$ , $\mathrm{I^-}$ (c) $\mathrm{HgI_4^{2-}}$ , $\mathrm{I_3^-}$ (d) $\mathrm{Hg_2I_2}$ , $\mathrm{I^-}$

Answer

(c) $\mathrm{HgI_4^{2-}}$ , $\mathrm{I_3^-}$

$\mathrm{HgCl_2}$ reacts with excess iodide: $\mathrm{HgCl_2 + 4I^- \rightarrow HgI_4^{2-} + 2Cl^-}$ . The first-formed red $\mathrm{HgI_2}$ dissolves in excess $\mathrm{I^-}$ to give the soluble complex. Iodine dissolves in iodide as $\mathrm{I_2 + I^- \rightarrow I_3^-}$ . So the pair is $\mathrm{HgI_4^{2-}}$ and $\mathrm{I_3^-}$ , not simple $\mathrm{HgI_2}$ .

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2017 · NEET-UG (India) · 4 marks

The species, having bond angles of $120^\circ$ is
(a) $\mathrm{PH_3}$ (b) $\mathrm{ClF_3}$ (c) $\mathrm{NCl_3}$ (d) $\mathrm{BCl_3}$

Answer

(d) $\mathrm{BCl_3}$

$\mathrm{BCl_3}$ is $sp^2$ hybridised with no lone pair on boron, so it is trigonal planar with Cl-B-Cl = $120^\circ$ . $\mathrm{PH_3}$ and $\mathrm{NCl_3}$ are pyramidal (lone pair, angles below $109.5^\circ$ ), and $\mathrm{ClF_3}$ is T-shaped with angles near $87.5^\circ$ .

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2017 · NEET-UG (India) · 4 marks

It is because of inability of $ns^2$ electrons of the valence shell to participate in bonding that
(a) $\mathrm{Sn^{2+}}$ is reducing while $\mathrm{Pb^{4+}}$ is oxidising (b) $\mathrm{Sn^{2+}}$ is oxidising while $\mathrm{Pb^{4+}}$ is reducing (c) $\mathrm{Sn^{2+}}$ and $\mathrm{Pb^{2+}}$ are both oxidising and reducing (d) $\mathrm{Sn^{4+}}$ is reducing while $\mathrm{Pb^{4+}}$ is oxidising

Answer

(a) $\mathrm{Sn^{2+}}$ is reducing while $\mathrm{Pb^{4+}}$ is oxidising

This is the inert pair effect: down group 14 the $ns^2$ electrons resist bonding, so the +2 state becomes more stable than +4. $\mathrm{Sn^{2+}}$ readily goes to the more stable $\mathrm{Sn^{4+}}$ , so it is a reducing agent. $\mathrm{Pb^{4+}}$ readily goes to the stable $\mathrm{Pb^{2+}}$ , so it is an oxidising agent. Options (b) and (d) reverse this.

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2017 · NEET-UG (India) · 4 marks

Which one is the wrong statement?
(a) de-Broglie's wavelength is given by $\lambda = \dfrac{h}{mv}$ , where $m$ = mass of the particle, $v$ = group velocity of the particle (b) The uncertainty principle is $\Delta E \times \Delta t \geq \dfrac{h}{4\pi}$ (c) Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement (d) The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms

Answer

(d) The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms

This is the wrong statement. In hydrogen and hydrogen-like (one-electron) species the energy depends only on $n$ , so 2s and 2p are degenerate. The splitting of s and p subshells arises only in multi-electron atoms. Statements (a), (b) and (c) are all correct.

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2017 · NEET-UG (India) · 4 marks

In which pair of ions both the species contain S-S bond?
(a) $\mathrm{S_2O_7^{2-}}$ , $\mathrm{S_2O_3^{2-}}$ (b) $\mathrm{S_4O_6^{2-}}$ , $\mathrm{S_2O_3^{2-}}$ (c) $\mathrm{S_2O_7^{2-}}$ , $\mathrm{S_2O_8^{2-}}$ (d) $\mathrm{S_4O_6^{2-}}$ , $\mathrm{S_2O_7^{2-}}$

Answer

(b) $\mathrm{S_4O_6^{2-}}$ , $\mathrm{S_2O_3^{2-}}$

Tetrathionate $\mathrm{S_4O_6^{2-}}$ has an S-S-S-S chain (the two central S atoms are bonded to each other). Thiosulphate $\mathrm{S_2O_3^{2-}}$ has one S bonded to the central S (S-S bond). In $\mathrm{S_2O_7^{2-}}$ (disulphate) and $\mathrm{S_2O_8^{2-}}$ (peroxodisulphate) the two S atoms are linked through O or O-O, with no S-S bond.

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2017 · NEET-UG (India) · 4 marks

The correct order of the stoichiometries of AgCl formed when $\mathrm{AgNO_3}$ in excess is treated with the complexes: $\mathrm{CoCl_3\cdot6NH_3}$ , $\mathrm{CoCl_3\cdot5NH_3}$ , $\mathrm{CoCl_3\cdot4NH_3}$ respectively is
(a) 1 AgCl, 3 AgCl, 2 AgCl (b) 3 AgCl, 1 AgCl, 2 AgCl (c) 3 AgCl, 2 AgCl, 1 AgCl (d) 2 AgCl, 3 AgCl, 1 AgCl

Answer

(c) 3 AgCl, 2 AgCl, 1 AgCl

Only free (ionisable) chloride precipitates with $\mathrm{AgNO_3}$ ; coordinated chloride does not.

$\mathrm{[Co(NH_3)_6]Cl_3}$ gives 3 Cl $^-$ , so 3 AgCl.

$\mathrm{[Co(NH_3)_5Cl]Cl_2}$ gives 2 AgCl.

$\mathrm{[Co(NH_3)_4Cl_2]Cl}$ gives 1 AgCl.

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2017 · NEET-UG (India) · 4 marks

Match the interhalogen compounds of column I with the geometry in column II and assign the correct code.
Column I: (a) $\mathrm{XX'}$ ; (b) $\mathrm{XX'_3}$ ; (c) $\mathrm{XX'_5}$ ; (d) $\mathrm{XX'_7}$
Column II: (i) T-shape; (ii) Pentagonal bipyramidal; (iii) Linear; (iv) Square-pyramidal; (v) Tetrahedral
(a) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) (b) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) (c) (a)-(v), (b)-(iv), (c)-(iii), (d)-(ii) (d) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Answer

(b) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

By VSEPR: $\mathrm{XX^\prime}$ is linear; $\mathrm{XX^\prime_3}$ ( $sp^3d$ , 2 lone pairs) is T-shaped; $\mathrm{XX^\prime_5}$ ( $sp^3d^2$ , 1 lone pair) is square pyramidal; $\mathrm{XX^\prime_7}$ ( $sp^3d^3$ , no lone pair) is pentagonal bipyramidal. No interhalogen is tetrahedral, which rules out option (c).

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2017 · NEET-UG (India) · 4 marks

The reason for greater range of oxidation states in actinoids is attributed to
(a) The radioactive nature of actinoids (b) Actinoid contraction (c) 5f, 6d and 7s levels having comparable energies (d) 4f and 5d levels being close in energies

Answer

(c) 5f, 6d and 7s levels having comparable energies

In actinoids the 5f, 6d and 7s subshells lie close in energy, so electrons from all of them can take part in bonding. This gives the wide range of oxidation states (+3 to +7). Radioactivity and actinoid contraction do not explain it, and option (d) describes the lanthanoids.

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2017 · NEET-UG (India) · 4 marks

The element Z = 114 has been discovered recently. It will belong to which of the following family group and electronic configuration?
(a) Halogen family, $\mathrm{[Rn]\,5f^{14}6d^{10}7s^27p^5}$ (b) Carbon family, $\mathrm{[Rn]\,5f^{14}6d^{10}7s^27p^2}$ (c) Oxygen family, $\mathrm{[Rn]\,5f^{14}6d^{10}7s^27p^4}$ (d) Nitrogen family, $\mathrm{[Rn]\,5f^{14}6d^{10}7s^27p^6}$

Answer

(b) Carbon family, $\mathrm{[Rn]\,5f^{14}6d^{10}7s^27p^2}$

Period 7 ends with Z = 118 ( $7p^6$ ). Counting back from 118, Z = 114 has 4 fewer electrons, i.e. $7p^2$ , with the complete $5f^{14}6d^{10}7s^2$ below it. A $ns^2np^2$ outer configuration belongs to group 14, the carbon family.

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2017 · NEET-UG (India) · 4 marks

Correct increasing order for the wavelengths of absorption in the visible region for the complexes of $\mathrm{Co^{3+}}$ is
(a) $\mathrm{[Co(en)_3]^{3+}}$ , $\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Co(H_2O)_6]^{3+}}$ (b) $\mathrm{[Co(H_2O)_6]^{3+}}$ , $\mathrm{[Co(en)_3]^{3+}}$ , $\mathrm{[Co(NH_3)_6]^{3+}}$ (c) $\mathrm{[Co(H_2O)_6]^{3+}}$ , $\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Co(en)_3]^{3+}}$ (d) $\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Co(en)_3]^{3+}}$ , $\mathrm{[Co(H_2O)_6]^{3+}}$

Answer

(a) $\mathrm{[Co(en)_3]^{3+}}$ , $\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Co(H_2O)_6]^{3+}}$

A stronger ligand field gives a larger $\Delta_o$ , so the complex absorbs higher-energy (shorter-wavelength) light, since $\lambda \propto 1/\Delta_o$ . The spectrochemical order is $\mathrm{H_2O} < \mathrm{NH_3} < \mathrm{en}$ . So the wavelength absorbed increases in the order en < $\mathrm{NH_3}$ < $\mathrm{H_2O}$ .

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2017 · NEET-UG (India) · 4 marks

Which one of the following pairs of species have the same bond order?
(a) CO, NO (b) $\mathrm{O_2}$ , $\mathrm{NO^+}$ (c) $\mathrm{CN^-}$ , CO (d) $\mathrm{N_2}$ , $\mathrm{O_2^-}$

Answer

(c) $\mathrm{CN^-}$ , CO

$\mathrm{CN^-}$ and CO each have 10 valence electrons (isoelectronic with $\mathrm{N_2}$ ), so by MO theory both have bond order 3. For the others: NO = 2.5, $\mathrm{O_2}$ = 2, $\mathrm{NO^+}$ = 3, $\mathrm{N_2}$ = 3, $\mathrm{O_2^-}$ = 1.5, so none of those pairs match.

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2017 · NEET-UG (India) · 4 marks

Name the gas that can readily decolourises acidified $\mathrm{KMnO_4}$ solution:
(a) $\mathrm{CO_2}$ (b) $\mathrm{SO_2}$ (c) $\mathrm{NO_2}$ (d) $\mathrm{P_2O_5}$

Answer

(b) $\mathrm{SO_2}$

$\mathrm{SO_2}$ is a reducing agent and reduces purple $\mathrm{MnO_4^-}$ to colourless $\mathrm{Mn^{2+}}$ in acid, decolourising the solution. $\mathrm{CO_2}$ , $\mathrm{NO_2}$ and $\mathrm{P_2O_5}$ cannot reduce permanganate (the S in $\mathrm{SO_2}$ is +4 and can be oxidised to +6, while C and P are already at their highest oxidation states).

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2017 · NEET-UG (India) · 4 marks

Pick out the correct statement with respect to $\mathrm{[Mn(CN)_6]^{3-}}$ :
(a) It is $sp^3d^2$ hybridised and octahedral (b) It is $sp^3d^2$ hybridised and tetrahedral (c) It is $d^2sp^3$ hybridised and octahedral (d) It is $dsp^2$ hybridised and square planar

Answer

(c) It is $d^2sp^3$ hybridised and octahedral

In $\mathrm{[Mn(CN)_6]^{3-}}$ , Mn is +3 ( $3d^4$ ). $\mathrm{CN^-}$ is a strong-field ligand, so the four electrons pair up in two $3d$ orbitals, leaving two $3d$ orbitals empty. These with one $4s$ and three $4p$ orbitals give $d^2sp^3$ hybridisation (inner orbital complex), octahedral and low spin.

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2017 · NEET-UG (India) · 4 marks

Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
(a) $\mathrm{Na^+}$ (b) $\mathrm{K^+}$ (c) $\mathrm{Rb^+}$ (d) $\mathrm{Li^+}$

Answer

(d) $\mathrm{Li^+}$

$\mathrm{Li^+}$ is the smallest ion and has the highest charge density, so it is the most heavily hydrated. The large hydrated radius makes it move slowly in an electric field, so its ionic mobility is the lowest in the group. The hydrated radius order is $\mathrm{Li^+} > \mathrm{Na^+} > \mathrm{K^+} > \mathrm{Rb^+}$ .

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2018 · NEET-UG (India) · 4 marks

The correct order of atomic radii in group 13 elements is
(a) B < Al < In < Ga < Tl (b) B < Ga < Al < Tl < In (c) B < Al < Ga < In < Tl (d) B < Ga < Al < In < Tl

Answer

(d) B < Ga < Al < In < Tl

Atomic radii (pm): B 85, Ga 135, Al 143, In 167, Tl 170. Ga is smaller than Al because the poorly shielding $3d^{10}$ electrons increase the effective nuclear charge. Radius then increases normally down the group to In and Tl.

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2018 · NEET-UG (India) · 4 marks

In the structure of $\mathrm{ClF_3}$ , the number of lone pairs of electrons on central atom 'Cl' is
(a) One (b) Four (c) Two (d) Three

Answer

(c) Two

Cl in $\mathrm{ClF_3}$ has 7 valence electrons; three are used in bonds to F, leaving 4 electrons, i.e. 2 lone pairs. With 3 bond pairs and 2 lone pairs ( $sp^3d$ ) the shape is T-shaped.

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2018 · NEET-UG (India) · 4 marks

The correct order of N-compounds in its decreasing order of oxidation states is
(a) $\mathrm{HNO_3}$ , NO, $\mathrm{N_2}$ , $\mathrm{NH_4Cl}$ (b) $\mathrm{HNO_3}$ , $\mathrm{NH_4Cl}$ , NO, $\mathrm{N_2}$ (c) $\mathrm{HNO_3}$ , NO, $\mathrm{NH_4Cl}$ , $\mathrm{N_2}$ (d) $\mathrm{NH_4Cl}$ , $\mathrm{N_2}$ , NO, $\mathrm{HNO_3}$

Answer

(a) $\mathrm{HNO_3}$ , NO, $\mathrm{N_2}$ , $\mathrm{NH_4Cl}$

Oxidation states of N: $\mathrm{HNO_3}$ = +5, NO = +2, $\mathrm{N_2}$ = 0, $\mathrm{NH_4Cl}$ = -3. This is already the decreasing order. The other options misplace $\mathrm{NH_4Cl}$ (-3) or $\mathrm{N_2}$ (0).

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2018 · NEET-UG (India) · 4 marks

Which of the following statements is not true for halogens?
(a) All form monobasic oxyacids (b) All but fluorine show positive oxidation states (c) All are oxidizing agents (d) Chlorine has the highest electron-gain enthalpy

Answer

(a) All form monobasic oxyacids

The question asks for the false statement. Fluorine forms only HOF, but the heavier halogens form oxyacids that are not all monobasic, for example $\mathrm{H_5IO_6}$ (paraperiodic acid) is pentabasic, so "all form monobasic oxyacids" is not true. The other statements hold: only F lacks positive oxidation states, all halogens are oxidising agents, and Cl has the highest electron-gain enthalpy (more negative than F because of F's small size and inter-electronic repulsion).

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2018 · NEET-UG (India) · 4 marks

Which one of the following elements is unable to form $\mathrm{MF_6^{3-}}$ ion?
(a) Ga (b) B (c) Al (d) In

Answer

(b) B

$\mathrm{MF_6^{3-}}$ needs six fluoride ligands, so the central atom must use $d$ orbitals ( $sp^3d^2$ ). Boron is in period 2 and has no $d$ orbitals, so it can only form $\mathrm{BF_4^-}$ . Al, Ga and In have vacant $d$ orbitals and form $\mathrm{AlF_6^{3-}}$ , $\mathrm{GaF_6^{3-}}$ and $\mathrm{InF_6^{3-}}$ .

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2018 · NEET-UG (India) · 4 marks

Which of the following oxides is most acidic in nature?
(a) MgO (b) BaO (c) BeO (d) CaO

Answer

(c) BeO

Acidic character of oxides increases as metallic character decreases. Be is the smallest and least electropositive in group 2, so BeO is amphoteric, while MgO, CaO and BaO are progressively more basic. BeO is therefore the most acidic of the four.

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2018 · NEET-UG (India) · 4 marks

The type of isomerism shown by the complex $\mathrm{[CoCl_2(en)_2]}$ is
(a) Geometrical isomerism (b) Ionization isomerism (c) Coordination isomerism (d) Linkage isomerism

Answer

(a) Geometrical isomerism

$\mathrm{[CoCl_2(en)_2]^+}$ is an octahedral $\mathrm{[M(AA)_2X_2]}$ complex, so the two $\mathrm{Cl^-}$ ligands can sit cis or trans to each other. The cis form is also optically active. There is no ionisable counter-ion exchange, no second complex ion for coordination isomerism, and no ambidentate ligand for linkage isomerism.

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2018 · NEET-UG (India) · 4 marks

Which one of the following ions exhibits d-d transition and paramagnetism as well?
(a) $\mathrm{CrO_4^{2-}}$ (b) $\mathrm{MnO_4^{-}}$ (c) $\mathrm{Cr_2O_7^{2-}}$ (d) $\mathrm{MnO_4^{2-}}$

Answer

(d) $\mathrm{MnO_4^{2-}}$

d-d transitions need a partly filled $d$ subshell, which also gives unpaired electrons and paramagnetism. In $\mathrm{MnO_4^{2-}}$ , Mn is $+6$ ( $d^1$ ), so it has one unpaired electron and can show a d-d transition. In $\mathrm{CrO_4^{2-}}$ (Cr $^{6+}$ ), $\mathrm{Cr_2O_7^{2-}}$ and $\mathrm{MnO_4^-}$ (Mn $^{7+}$ ) the metal is $d^0$ , so they are diamagnetic with colour only from charge transfer.

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2018 · NEET-UG (India) · 4 marks

Iron carbonyl, $\mathrm{Fe(CO)_5}$ is
(a) Tetranuclear (b) Trinuclear (c) Mononuclear (d) Dinuclear

Answer

(c) Mononuclear

$\mathrm{Fe(CO)_5}$ contains a single metal atom surrounded by five CO ligands (trigonal bipyramidal), so it is mononuclear. Dinuclear and polynuclear carbonyls such as $\mathrm{Fe_2(CO)_9}$ and $\mathrm{Fe_3(CO)_{12}}$ contain metal-metal bonds, which is not the case here.

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2018 · NEET-UG (India) · 4 marks

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code:
Column I: a. $\mathrm{Co^{3+}}$ ; b. $\mathrm{Cr^{3+}}$ ; c. $\mathrm{Fe^{3+}}$ ; d. $\mathrm{Ni^{2+}}$
Column II: i. $\sqrt{8}$ BM; ii. $\sqrt{35}$ BM; iii. $\sqrt{3}$ BM; iv. $\sqrt{24}$ BM; v. $\sqrt{15}$ BM
(a) a-iv, b-v, c-ii, d-i (b) a-iv, b-i, c-ii, d-iii (c) a-i, b-ii, c-iii, d-iv (d) a-iii, b-v, c-i, d-ii

Answer

(a) a-iv, b-v, c-ii, d-i

Use $\mu=\sqrt{n(n+2)}$ . $\mathrm{Co^{3+}}$ ( $d^6$ , 4 unpaired) gives $\sqrt{24}$ ; $\mathrm{Cr^{3+}}$ ( $d^3$ , 3 unpaired) gives $\sqrt{15}$ ; $\mathrm{Fe^{3+}}$ ( $d^5$ , 5 unpaired) gives $\sqrt{35}$ ; $\mathrm{Ni^{2+}}$ ( $d^8$ , 2 unpaired) gives $\sqrt{8}$ . That is a-iv, b-v, c-ii, d-i.

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2018 · NEET-UG (India) · 4 marks

The geometry and magnetic behaviour of the complex $\mathrm{[Ni(CO)_4]}$ are
(a) Square planar geometry and diamagnetic (b) Square planar geometry and paramagnetic (c) Tetrahedral geometry and diamagnetic (d) Tetrahedral geometry and paramagnetic

Answer

(c) Tetrahedral geometry and diamagnetic

In $\mathrm{Ni(CO)_4}$ nickel is in the zero oxidation state ( $3d^8 4s^2$ ). CO is a strong-field ligand, so the 4s electrons shift into 3d giving $3d^{10}$ with no unpaired electrons. The 4s and three 4p orbitals hybridise ( $sp^3$ ), giving a tetrahedral, diamagnetic complex. Square planar $dsp^2$ geometry would need a $d^8$ ion such as $\mathrm{Ni^{2+}}$ .

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2018 · NEET-UG (India) · 4 marks

Among $\mathrm{CaH_2}$ , $\mathrm{BeH_2}$ , $\mathrm{BaH_2}$ , the order of ionic character is
(a) $\mathrm{BeH_2} < \mathrm{CaH_2} < \mathrm{BaH_2}$ (b) $\mathrm{BeH_2} < \mathrm{BaH_2} < \mathrm{CaH_2}$ (c) $\mathrm{CaH_2} < \mathrm{BeH_2} < \mathrm{BaH_2}$ (d) $\mathrm{BaH_2} < \mathrm{BeH_2} < \mathrm{CaH_2}$

Answer

(a) $\mathrm{BeH_2} < \mathrm{CaH_2} < \mathrm{BaH_2}$

Ionic character increases as the metal becomes more electropositive, i.e. down the group: Be < Ca < Ba. $\mathrm{BeH_2}$ is largely covalent (polymeric), whereas $\mathrm{BaH_2}$ is strongly ionic. Larger cation size and lower polarising power give greater ionic character.

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2018 · NEET-UG (India) · 4 marks

Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is $1s^2 2s^2 2p^3$ , the simplest formula for this compound is
(a) $\mathrm{Mg_2X_3}$ (b) $\mathrm{Mg_2X}$ (c) $\mathrm{MgX_2}$ (d) $\mathrm{Mg_3X_2}$

Answer

(d) $\mathrm{Mg_3X_2}$

The configuration $1s^2 2s^2 2p^3$ is nitrogen, which gains three electrons to form $\mathrm{X^{3-}}$ . Magnesium loses two electrons to form $\mathrm{Mg^{2+}}$ . Balancing charges needs 3 Mg ions and 2 N ions, so the formula is $\mathrm{Mg_3N_2}$ .

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2018 · NEET-UG (India) · 4 marks

Consider the following species: $\mathrm{CN^+}$ , $\mathrm{CN^-}$ , NO and CN. Which one of these will have the highest bond order?
(a) NO (b) $\mathrm{CN^+}$ (c) $\mathrm{CN^-}$ (d) CN

Answer

(c) $\mathrm{CN^-}$

Bond orders from MO theory: $\mathrm{CN^+}$ (12 electrons) is 2, $\mathrm{CN}$ (13 electrons) is 2.5, NO (15 electrons) is 2.5, and $\mathrm{CN^-}$ (14 electrons, isoelectronic with $\mathrm{N_2}$ and CO) is 3. So $\mathrm{CN^-}$ has the highest bond order.

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2019 · NEET-UG (India) · 4 marks

Which is the correct thermal stability order for $\mathrm{H_2E}$ (E = O, S, Se, Te and Po)?
(a) $\mathrm{H_2O} < \mathrm{H_2S} < \mathrm{H_2Se} < \mathrm{H_2Te} < \mathrm{H_2Po}$ (b) $\mathrm{H_2Po} < \mathrm{H_2Te} < \mathrm{H_2Se} < \mathrm{H_2S} < \mathrm{H_2O}$ (c) $\mathrm{H_2Se} < \mathrm{H_2Te} < \mathrm{H_2Po} < \mathrm{H_2O} < \mathrm{H_2S}$ (d) $\mathrm{H_2S} < \mathrm{H_2O} < \mathrm{H_2Se} < \mathrm{H_2Te} < \mathrm{H_2Po}$

Answer

(b) $\mathrm{H_2Po} < \mathrm{H_2Te} < \mathrm{H_2Se} < \mathrm{H_2S} < \mathrm{H_2O}$

Thermal stability of group 16 hydrides falls down the group because the E-H bond lengthens and weakens as the size of E increases. Water has the strongest bond and $\mathrm{H_2Po}$ the weakest.

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2019 · NEET-UG (India) · 4 marks

4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is
(a) 6p > 5f > 5p > 4d (b) 6p > 5f > 4d > 5p (c) 5f > 6p > 4d > 5p (d) 5f > 6p > 5p > 4d

Answer

(d) 5f > 6p > 5p > 4d

Use the $(n+l)$ rule: 5f = 8, 6p = 7, 5p = 6, 4d = 6. For equal $(n+l)$ the orbital with the higher $n$ has higher energy, so 5p > 4d. The order is 5f > 6p > 5p > 4d.

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2019 · NEET-UG (India) · 4 marks

For the second period elements the correct increasing order of first ionisation enthalpy is:
(a) Li < B < Be < C < O < N < F < Ne (b) Li < B < Be < C < N < O < F < Ne (c) Li < Be < B < C < O < N < F < Ne (d) Li < Be < B < C < N < O < F < Ne

Answer

(a) Li < B < Be < C < O < N < F < Ne

First ionisation enthalpy generally rises across the period, with two dips. Be (filled $2s$ ) is higher than B, and N (half-filled $2p^3$ ) is higher than O. This gives Li < B < Be < C < O < N < F < Ne.

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2019 · NEET-UG (India) · 4 marks

Identify the incorrect statement related to $\mathrm{PCl_5}$ from the following:
(a) Two axial P-Cl bonds make an angle of $180^\circ$ with each other (b) Axial P-Cl bonds are longer than equatorial P-Cl bonds (c) $\mathrm{PCl_5}$ molecule is non-reactive (d) Three equatorial P-Cl bonds make an angle of $120^\circ$ with each other

Answer

(c) $\mathrm{PCl_5}$ molecule is non-reactive

This is the incorrect statement. $\mathrm{PCl_5}$ is trigonal bipyramidal and is quite reactive, with its longer axial bonds easily broken (for example it hydrolyses readily). Statements (a), (b) and (d) are correct: axial-axial angle $180^\circ$ , axial bonds longer than equatorial, and equatorial angles $120^\circ$ .

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2019 · NEET-UG (India) · 4 marks

Which of the following diatomic molecular species has only $\pi$ bonds according to Molecular Orbital Theory?
(a) $\mathrm{N_2}$ (b) $\mathrm{C_2}$ (c) $\mathrm{Be_2}$ (d) $\mathrm{O_2}$

Answer

(b) $\mathrm{C_2}$

$\mathrm{C_2}$ has 12 electrons with configuration $\sigma_{1s}^2\sigma^{*2}_{1s}\sigma_{2s}^2\sigma^{*2}_{2s}\pi_{2p_x}^2\pi_{2p_y}^2$ . The two bonding pairs are both in $\pi$ orbitals, so it has two $\pi$ bonds and no $\sigma$ bond (bond order 2). $\mathrm{N_2}$ has $\pi$ bonds plus a $\sigma$ bond from $\sigma_{2p_z}$ , so it is the tempting wrong option.

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2019 · NEET-UG (India) · 4 marks

Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region?
(a) Balmer series (b) Paschen series (c) Brackett series (d) Lyman series

Answer

(a) Balmer series

The Balmer series is the set of transitions to $n=2$ from $n\ge3$ . It falls in the visible region (about 656 to 365 nm). Lyman is ultraviolet, while Paschen and Brackett are infrared.

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2019 · NEET-UG (India) · 4 marks

Which of the following is an amphoteric hydroxide?
(a) $\mathrm{Ca(OH)_2}$ (b) $\mathrm{Mg(OH)_2}$ (c) $\mathrm{Be(OH)_2}$ (d) $\mathrm{Sr(OH)_2}$

Answer

(c) $\mathrm{Be(OH)_2}$

$\mathrm{Be(OH)_2}$ reacts with both acids and bases, forming $\mathrm{BeCl_2}$ with HCl and $[\mathrm{Be(OH)_4}]^{2-}$ with NaOH, so it is amphoteric. The hydroxides of Mg, Ca and Sr are basic, and basicity increases down the group.

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2019 · NEET-UG (India) · 4 marks

What is the correct electronic configuration of the central atom in $\mathrm{K_4[Fe(CN)_6]}$ based on crystal field theory?
(a) $t_{2g}^{6}e_g^{0}$ (b) $e^{3}t_2^{3}$ (c) $e^{4}t_2^{2}$ (d) $t_{2g}^{4}e_g^{2}$

Answer

(a) $t_{2g}^{6}e_g^{0}$

In $\mathrm{K_4[Fe(CN)_6]}$ , iron is $\mathrm{Fe^{2+}}$ ( $d^6$ ). $\mathrm{CN^-}$ is a strong field ligand, so pairing occurs in the octahedral field. All six electrons go into $t_{2g}$ , giving $t_{2g}^6e_g^0$ (low spin, diamagnetic). Options (b) and (c) use $e$ / $t_2$ labels, which belong to tetrahedral fields.

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2019 · NEET-UG (India) · 4 marks

The manganate and permanganate ions are tetrahedral, due to:
(a) There is no $\pi$ -bonding (b) The $\pi$ -bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese (c) The $\pi$ -bonding involves overlap of d-orbitals of oxygen with d-orbitals of manganese (d) The $\pi$ -bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese

Answer

(d) The $\pi$ -bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese

In $\mathrm{MnO_4^-}$ and $\mathrm{MnO_4^{2-}}$ , Mn is $sp^3$ hybridised, giving a tetrahedral shape. The extra $\pi$ bonding is $p\pi-d\pi$ , from filled p orbitals of O overlapping with empty d orbitals of Mn. Oxygen has no d orbitals in its valence shell, so option (c) is impossible.

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2019 · NEET-UG (India) · 4 marks

Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code:
Column-I: (a) $\mathrm{XeF_4}$ ; (b) $\mathrm{XeF_6}$ ; (c) $\mathrm{XeOF_4}$ ; (d) $\mathrm{XeO_3}$
Column-II: (i) Pyramidal; (ii) Square planar; (iii) Distorted octahedral; (iv) Square pyramidal
(a) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (b) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv) (c) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) (d) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Answer

(a) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

$\mathrm{XeF_4}$ ( $sp^3d^2$ , 2 lone pairs) is square planar. $\mathrm{XeF_6}$ (1 lone pair) is distorted octahedral. $\mathrm{XeOF_4}$ (1 lone pair) is square pyramidal. $\mathrm{XeO_3}$ (1 lone pair, $sp^3$ ) is pyramidal.

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2019 · NEET-UG (India) · 4 marks

Which of the following is incorrect statement?
(a) $\mathrm{SiCl_4}$ is easily hydrolysed (b) $\mathrm{GeX_4}$ (X = F, Cl, Br, I) is more stable than $\mathrm{GeX_2}$ (c) $\mathrm{SnF_4}$ is ionic in nature (d) $\mathrm{PbF_4}$ is covalent in nature

Answer

(d) $\mathrm{PbF_4}$ is covalent in nature

$\mathrm{PbF_4}$ is ionic. $\mathrm{Pb^{4+}}$ is large and fluoride is small and hard, so it is not appreciably polarised. $\mathrm{SnF_4}$ is also ionic, $\mathrm{SiCl_4}$ hydrolyses easily because Si has vacant d orbitals, and Ge is more stable in the +4 state than +2 (the inert pair effect only becomes strong for Sn and Pb).

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2019 · NEET-UG (India) · 4 marks

Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is:
(a) Mg (b) Ca (c) Sr (d) Be

Answer

(a) Mg

Enzymes that use ATP for phosphate transfer need $\mathrm{Mg^{2+}}$ , which binds to the phosphate groups of ATP. Ca is involved in muscle contraction and nerve impulses, while Be is toxic.

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2019 · NEET-UG (India) · 4 marks

Which of the following species is not stable?
(a) $\mathrm{[GeCl_6]^{2-}}$ (b) $\mathrm{[Sn(OH)_6]^{2-}}$ (c) $\mathrm{[SiCl_6]^{2-}}$ (d) $\mathrm{[SiF_6]^{2-}}$

Answer

(c) $\mathrm{[SiCl_6]^{2-}}$

Si can expand its coordination to six with small fluoride, as in $\mathrm{[SiF_6]^{2-}}$ . Six large $\mathrm{Cl^-}$ ions cannot fit around the small Si atom, so $\mathrm{[SiCl_6]^{2-}}$ is not stable. Ge and Sn are larger and form stable hexa-coordinate species.

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2020 · NEET-UG (India) · 4 marks

Match the following.
Oxide: (a) CO; (b) BaO; (c) $\mathrm{Al_2O_3}$ ; (d) $\mathrm{Cl_2O_7}$
Nature: (i) Basic; (ii) Neutral; (iii) Acidic; (iv) Amphoteric
Which one of the following is correct?
(a) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) (b) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) (c) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) (d) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Answer

(d) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

CO is a neutral oxide. BaO is basic (an alkaline earth metal oxide). $\mathrm{Al_2O_3}$ is amphoteric, and $\mathrm{Cl_2O_7}$ is the anhydride of $\mathrm{HClO_4}$ , so it is acidic.

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2020 · NEET-UG (India) · 4 marks

The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.
(a) Calcium (b) Potassium (c) Iron (d) Copper

Answer

(b) Potassium

$\mathrm{K^+}$ activates many enzymes and takes part in glucose oxidation to give ATP. Together with $\mathrm{Na^+}$ it carries nerve signals through the sodium-potassium pump. Calcium is mainly linked to bones, muscle and blood clotting.

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2020 · NEET-UG (India) · 4 marks

Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(a) $\mathrm{F^-} < \mathrm{SCN^-} < \mathrm{C_2O_4^{2-}} < \mathrm{CN^-}$ (b) $\mathrm{CN^-} < \mathrm{C_2O_4^{2-}} < \mathrm{SCN^-} < \mathrm{F^-}$ (c) $\mathrm{SCN^-} < \mathrm{F^-} < \mathrm{C_2O_4^{2-}} < \mathrm{CN^-}$ (d) $\mathrm{SCN^-} < \mathrm{F^-} < \mathrm{CN^-} < \mathrm{C_2O_4^{2-}}$

Answer

(c) $\mathrm{SCN^-} < \mathrm{F^-} < \mathrm{C_2O_4^{2-}} < \mathrm{CN^-}$

This follows the spectrochemical series: $\mathrm{I^-}<\mathrm{Br^-}<\mathrm{SCN^-}<\mathrm{Cl^-}<\mathrm{F^-}<\mathrm{OH^-}<\mathrm{C_2O_4^{2-}}<\mathrm{H_2O}<\mathrm{NH_3}<\mathrm{en}<\mathrm{CN^-}<\mathrm{CO}$ . $\mathrm{CN^-}$ is a strong $\pi$ -acceptor and sits near the top, so options with it at the low end are wrong.

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2020 · NEET-UG (India) · 4 marks

The number of protons, neutrons and electrons in $\mathrm{{}^{175}_{71}Lu}$ , respectively, are:
(a) 71, 71 and 104 (b) 175, 104 and 71 (c) 71, 104 and 71 (d) 104, 71 and 71

Answer

(c) 71, 104 and 71

For $^{175}_{71}\mathrm{Lu}$ , the atomic number is the number of protons, 71. Neutrons = mass number minus protons = $175-71=104$ . A neutral atom has electrons equal to protons, 71.

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2020 · NEET-UG (India) · 4 marks

Identify a molecule which does not exist.
(a) $\mathrm{C_2}$ (b) $\mathrm{O_2}$ (c) $\mathrm{He_2}$ (d) $\mathrm{Li_2}$

Answer

(c) $\mathrm{He_2}$

$\mathrm{He_2}$ has the configuration $\sigma_{1s}^2\sigma^{*2}_{1s}$ . Bond order = $(2-2)/2=0$ , so the molecule does not exist. $\mathrm{C_2}$ has bond order 2, $\mathrm{O_2}$ has 2 and $\mathrm{Li_2}$ has 1.

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2020 · NEET-UG (India) · 4 marks

Identify the incorrect statement.
(a) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. (b) The oxidation states of chromium in $\mathrm{CrO_4^{2-}}$ and $\mathrm{Cr_2O_7^{2-}}$ are not the same. (c) $\mathrm{Cr^{2+}}$ ( $d^4$ ) is a stronger reducing agent than $\mathrm{Fe^{2+}}$ ( $d^6$ ) in water. (d) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.

Answer

(b) The oxidation states of chromium in $\mathrm{CrO_4^{2-}}$ and $\mathrm{Cr_2O_7^{2-}}$ are not the same.

This is the incorrect statement. In $\mathrm{CrO_4^{2-}}$ : $x + 4(-2) = -2$ gives $x = +6$ . In $\mathrm{Cr_2O_7^{2-}}$ : $2x + 7(-2) = -2$ gives $x = +6$ . They are the same. The others are true: $\mathrm{Cr^{2+}}$ is a strong reducing agent because it becomes the stable $d^3$ ( $\mathrm{Cr^{3+}}$ ), more easily than $\mathrm{Fe^{2+}}$ goes to $\mathrm{Fe^{3+}}$ .

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2020 · NEET-UG (India) · 4 marks

The calculated spin only magnetic moment of $\mathrm{Cr^{2+}}$ ion is:
(a) 5.92 BM (b) 2.84 BM (c) 3.87 BM (d) 4.90 BM

Answer

(d) 4.90 BM

$\mathrm{Cr}$ is $3d^5 4s^1$ , so $\mathrm{Cr^{2+}}$ is $3d^4$ with $n = 4$ unpaired electrons. $\mu = \sqrt{n(n+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM.

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2020 · NEET-UG (India) · 4 marks

Which of the following oxoacids of sulphur has $-\mathrm{O-O}-$ linkage?
(a) $\mathrm{H_2S_2O_8}$ , peroxodisulphuric acid (b) $\mathrm{H_2S_2O_7}$ , pyrosulphuric acid (c) $\mathrm{H_2SO_3}$ , sulphurous acid (d) $\mathrm{H_2SO_4}$ , sulphuric acid

Answer

(a) $\mathrm{H_2S_2O_8}$ , peroxodisulphuric acid

Its structure is $\mathrm{HO_3S-O-O-SO_3H}$ , containing the peroxo $-\mathrm{O-O}-$ linkage. Pyrosulphuric acid has an $\mathrm{S-O-S}$ bridge, while sulphurous and sulphuric acids have no oxygen-oxygen bond.

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2020 · NEET-UG (India) · 4 marks

Which of the following sets of molecules will have zero dipole moment?
(a) Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene (b) Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene (c) Ammonia, beryllium difluoride, water, 1,4-dichlorobenzene (d) Boron trifluoride, hydrogen fluoride, carbon dioxide, 1,3-dichlorobenzene

Answer

(b) Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene

$\mathrm{BF_3}$ (trigonal planar), $\mathrm{BeF_2}$ and $\mathrm{CO_2}$ (linear) and para-dichlorobenzene are symmetric, so bond dipoles cancel. $\mathrm{NF_3}$ , $\mathrm{H_2O}$ , $\mathrm{NH_3}$ , HF and 1,3-dichlorobenzene have a net dipole, which rules out the other options.

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2021 · NEET-UG (India) · 4 marks

$\mathrm{BF_3}$ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are
(a) $sp^2$ and 6 (b) $sp^2$ and 8 (c) $sp^3$ and 4 (d) $sp^3$ and 6

Answer

(a) $sp^2$ and 6

Boron in $\mathrm{BF_3}$ has three bond pairs and no lone pair, so it is $sp^2$ hybridised and the molecule is trigonal planar. Boron shares only three pairs, so it has 6 electrons in its valence shell, an incomplete octet, which makes it electron deficient. The "8 electrons" options would need a coordinate bond, as in $\mathrm{BF_4^-}$ .

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2021 · NEET-UG (India) · 4 marks

The structure of beryllium chloride in solid state and vapour phase, are:
(a) Dimer and Linear, respectively (b) Chain in both (c) Chain and dimer, respectively (d) Linear in both

Answer

(c) Chain and dimer, respectively

In the solid state $\mathrm{BeCl_2}$ is a polymeric chain with chlorine bridges. In the vapour phase it exists as a chlorine-bridged dimer, which dissociates into the linear monomer only at about 1200 K. So "linear in both" and "dimer in solid" are wrong.

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2021 · NEET-UG (India) · 4 marks

Zr ( $Z = 40$ ) and Hf ( $Z = 72$ ) have similar atomic and ionic radii because of:
(a) lanthanoid contraction (b) having similar chemical properties (c) belonging to same group (d) diagonal relationship

Answer

(a) lanthanoid contraction

Between Zr (4d) and Hf (5d) the 4f subshell is filled. The 4f electrons shield the nuclear charge poorly, so the steady fall in size across the lanthanoids (lanthanoid contraction) cancels the normal increase in radius down the group. Hence Zr and Hf have almost identical radii, and very similar properties follow from that, not the other way round.

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2021 · NEET-UG (India) · 4 marks

A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is: [Speed of light, $c = 3\times10^{8}\ \mathrm{m\,s^{-1}}$ ]
(a) 2192 m (b) 21.92 cm (c) 219.3 m (d) 219.2 m

Answer

(c) 219.3 m

$\lambda=\dfrac{c}{\nu}=\dfrac{3\times10^{8}\ \mathrm{m\,s^{-1}}}{1368\times10^{3}\ \mathrm{s^{-1}}}=219.3\ \mathrm{m}$ . The frequency must be converted from kHz to Hz ( $10^3$ ); forgetting this gives the wrong power of ten, as in 2192 m.

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2021 · NEET-UG (India) · 4 marks

Ethylenediaminetetraacetate (EDTA) ion is
(a) Bidentate ligand with two "N" donor atoms (b) Tridentate ligand with three "N" donor atoms (c) Hexadentate ligand with four "O" and two "N" donor atoms (d) Unidentate ligand

Answer

(c) Hexadentate ligand with four "O" and two "N" donor atoms

EDTA $^{4-}$ , $(\mathrm{^-OOCCH_2)_2NCH_2CH_2N(CH_2COO^-)_2}$ , binds through two nitrogen atoms and four carboxylate oxygen atoms. Six donor atoms make it hexadentate, and it forms very stable chelates with metal ions.

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2021 · NEET-UG (India) · 4 marks

Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is:
(a) Magnesium chloride (b) Beryllium chloride (c) Calcium chloride (d) Strontium chloride

Answer

(b) Beryllium chloride

Be $^{2+}$ is very small with a high charge density, so it strongly polarises the chloride ion (Fajans rules). This makes $\mathrm{BeCl_2}$ largely covalent and soluble in organic solvents. The chlorides of Mg, Ca and Sr are progressively more ionic and are not soluble in organic solvents.

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2021 · NEET-UG (India) · 4 marks

Match List-I with List-II. List-I A. $\mathrm{PCl_5}$ B. $\mathrm{SF_6}$ C. $\mathrm{BrF_5}$ D. $\mathrm{BF_3}$ List-II I. Square pyramidal II. Trigonal planar III. Octahedral IV. Trigonal bipyramidal
(a) A-III, B-I, C-IV, D-II (b) A-IV, B-III, C-II, D-I (c) A-IV, B-III, C-I, D-II (d) A-II, B-III, C-IV, D-I

Answer

(c) A-IV, B-III, C-I, D-II

$\mathrm{PCl_5}$ : $sp^3d$ , trigonal bipyramidal. $\mathrm{SF_6}$ : $sp^3d^2$ , octahedral. $\mathrm{BrF_5}$ : 5 bond pairs and 1 lone pair, so square pyramidal. $\mathrm{BF_3}$ : $sp^2$ , trigonal planar. Matching these gives A-IV, B-III, C-I, D-II.

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2021 · NEET-UG (India) · 4 marks

The incorrect statement among the following is:
(a) Lanthanoids are good conductors of heat and electricity (b) Actinoids are highly reactive metals, especially when finely divided (c) Actinoid contraction is greater for element to element than Lanthanoid contraction (d) Most of the trivalent Lanthanoid ions are colorless in the solid state

Answer

(d) Most of the trivalent Lanthanoid ions are colorless in the solid state

Many trivalent lanthanoid ions (for example $\mathrm{Pr^{3+}, Nd^{3+}, Sm^{3+}, Eu^{3+}}$ ) are coloured, both in the solid state and in solution, because of f-f transitions, so this statement is incorrect. The other three are true: lanthanoids conduct well, actinoids are reactive, and actinoid contraction is larger because 5f electrons shield even more poorly.

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2021 · NEET-UG (India) · 4 marks

Statement I: Acid strength increases in the order given as HF $\ll$ HCl $\ll$ HBr $\ll$ HI. Statement II: As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases.
(a) Statement I is correct but Statement II is false. (b) Statement I is incorrect but Statement II is true. (c) Both Statement I and Statement II are true. (d) Both Statement I and Statement II are false.

Answer

(c) Both Statement I and Statement II are true.

Going down the group the H-X bond length increases and bond enthalpy falls (H-F is about 570 kJ/mol, H-I about 298 kJ/mol), so the H-X bond breaks more easily in water. Hence acid strength rises as $\mathrm{HF<HCl<HBr<HI}$ , and Statement II correctly explains Statement I.

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2021 · NEET-UG (India) · 4 marks

Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.
(a) Noble gases have weak dispersion forces (b) Noble gases have large positive values of electron gain enthalpy (c) Noble gases are sparingly soluble in water (d) Noble gases have very high melting and boiling points

Answer

(d) Noble gases have very high melting and boiling points

Noble gases have only weak dispersion forces, so their melting and boiling points are very low, and they are gases at room temperature. The other statements are true: they have large positive electron gain enthalpy (filled shells) and are sparingly soluble in water.

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2021 · NEET-UG (India) · 4 marks

Which of the following molecules is non-polar in nature?
(a) $\mathrm{SbCl_5}$ (b) $\mathrm{NO_2}$ (c) $\mathrm{POCl_3}$ (d) $\mathrm{CH_2O}$

Answer

(a) $\mathrm{SbCl_5}$

$\mathrm{SbCl_5}$ is symmetric trigonal bipyramidal, so the bond dipoles cancel and it is non-polar. $\mathrm{NO_2}$ is bent, $\mathrm{POCl_3}$ is distorted tetrahedral and $\mathrm{CH_2O}$ is planar with an unsymmetrical C=O dipole, so all three are polar.

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2021 · NEET-UG (India) · 4 marks

Match List-I with List-II. List-I A. $\mathrm{[Fe(CN)_6]^{3-}}$ B. $\mathrm{[Fe(H_2O)_6]^{3+}}$ C. $\mathrm{[Fe(CN)_6]^{4-}}$ D. $\mathrm{[Fe(H_2O)_6]^{2+}}$ List-II I. 5.92 BM II. 0 BM III. 4.90 BM IV. 1.73 BM
(a) A-I, B-III, C-IV, D-II (b) A-IV, B-I, C-II, D-III (c) A-IV, B-II, C-I, D-III (d) A-II, B-IV, C-III, D-I

Answer

(b) A-IV, B-I, C-II, D-III

$[\mathrm{Fe(CN)_6}]^{3-}$ : $\mathrm{Fe^{3+}}$ ( $d^5$ ) low spin, 1 unpaired electron, 1.73 BM. $[\mathrm{Fe(H_2O)_6}]^{3+}$ : $d^5$ high spin, 5 unpaired electrons, 5.92 BM. $[\mathrm{Fe(CN)_6}]^{4-}$ : $\mathrm{Fe^{2+}}$ ( $d^6$ ) low spin, 0 unpaired, 0 BM. $[\mathrm{Fe(H_2O)_6}]^{2+}$ : $d^6$ high spin, 4 unpaired, 4.90 BM.

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2021 · NEET-UG (India) · 4 marks

From the following pairs of ions which one is not an iso-electronic pair?
(a) $\mathrm{Mn^{2+}, Fe^{3+}}$ (b) $\mathrm{Fe^{2+}, Mn^{2+}}$ (c) $\mathrm{O^{2-}, F^{-}}$ (d) $\mathrm{Na^{+}, Mg^{2+}}$

Answer

(b) $\mathrm{Fe^{2+}, Mn^{2+}}$

$\mathrm{Fe^{2+}}$ has 24 electrons and $\mathrm{Mn^{2+}}$ has 23, so they are not iso-electronic. $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ both have 23, $\mathrm{O^{2-}}$ and $\mathrm{F^-}$ both have 10, and $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$ both have 10.

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2021 · NEET-UG (India) · 4 marks

In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?
(a) $\mathrm{NH_3 < PH_3 < AsH_3 < SbH_3}$ : Increasing acidic character (b) $\mathrm{CO_2 < SiO_2 < SnO_2 < PbO_2}$ : Increasing oxidizing power (c) HF < HCl < HBr < HI : Increasing acidic strength (d) $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$ : Increasing $pK_a$ values

Answer

(d) $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$ : Increasing $pK_a$ values

Going down the group the H-E bond weakens, so acidity rises and $pK_a$ falls, so the $pK_a$ order is $\mathrm{H_2O>H_2S>H_2Se>H_2Te}$ . The sequence given is therefore wrong. The others are right: hydride acidity increases down group 15, oxidising power increases down group 14 (inert pair effect), and acid strength of HX increases from HF to HI.

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2022 · NEET-UG (India) · 4 marks

Gadolinium has a low value of third ionisation enthalpy because of
(a) small size (b) high exchange enthalpy (c) high electronegativity (d) high basic character

Answer

(b) high exchange enthalpy

Removing the third electron from $\mathrm{Gd^{2+}}$ ( $4f^75d^1$ ) leaves $\mathrm{Gd^{3+}}$ with the stable half-filled $4f^7$ configuration, which has large exchange energy. That stability makes the third ionisation enthalpy low.

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2022 · NEET-UG (India) · 4 marks

Identify the incorrect statement from the following
(a) Alkali metals react with water to form their hydroxides (b) The oxidation number of K in $\mathrm{KO_2}$ is +4. (c) Ionisation enthalpy of alkali metals decreases from top to bottom in the group. (d) Lithium is the strongest reducing agent among the alkali metals.

Answer

(b) The oxidation number of K in $\mathrm{KO_2}$ is +4.

$\mathrm{KO_2}$ is a superoxide containing $\mathrm{K^+}$ and $\mathrm{O_2^-}$ , so the oxidation number of K is +1, not +4. The other statements are true: alkali metals form hydroxides with water, ionisation enthalpy falls down the group, and Li is the strongest reducing agent in aqueous solution because of its very high hydration enthalpy.

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2022 · NEET-UG (India) · 4 marks

The IUPAC name of an element with atomic number 119 is
(a) ununennium (b) unnilennium (c) unununnium (d) ununoctium

Answer

(a) ununennium

IUPAC systematic names use the roots for digits: 1 = un, 9 = enn. So 119 gives un-un-enn plus "ium", i.e. ununennium (symbol Uue). Unnilennium is the older name system, and ununoctium is for Z = 118.

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2022 · NEET-UG (India) · 4 marks

Choose the correct statement:
(a) Diamond and graphite have two dimensional network. (b) Diamond is covalent and graphite is ionic. (c) Diamond is $sp^3$ hybridised and graphite is $sp^2$ hybridized. (d) Both diamond and graphite are used as dry lubricants.

Answer

(c) Diamond is $sp^3$ hybridised and graphite is $sp^2$ hybridized.

In diamond each carbon is bonded to four others ( $sp^3$ , 3D network). In graphite each carbon is bonded to three others ( $sp^2$ , layers). Both are covalent. Only graphite is a dry lubricant, since its layers slip over each other.

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2022 · NEET-UG (India) · 4 marks

Which amongst the following is incorrect statement?
(a) The bond orders of $\mathrm{O_2^+}$ , $\mathrm{O_2}$ , $\mathrm{O_2^-}$ and $\mathrm{O_2^{2-}}$ are 2.5, 2, 1.5 and 1, respectively. (b) $\mathrm{C_2}$ molecule has four electrons in its two degenerate $\pi$ molecular orbitals. (c) $\mathrm{H_2^+}$ ion has one electron. (d) $\mathrm{O_2^+}$ ion is diamagnetic.

Answer

(d) $\mathrm{O_2^+}$ ion is diamagnetic.

$\mathrm{O_2^+}$ has 15 electrons, giving the configuration $\ldots(\pi^*2p_x)^1$ with one unpaired electron, so it is paramagnetic. The other statements hold: bond orders are $2.5, 2, 1.5, 1$ for $\mathrm{O_2^+, O_2, O_2^-, O_2^{2-}}$ ; $\mathrm{C_2}$ has $\pi 2p_x^2\,\pi 2p_y^2$ (four electrons in two degenerate $\pi$ MOs); $\mathrm{H_2^+}$ has one electron.

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2022 · NEET-UG (India) · 4 marks

The IUPAC name of the complex $\mathrm{[Ag(H_2O)_2][Ag(CN)_2]}$ is:
(a) dicyanidosilver(II) diaquaargentate(II) (b) diaquasilver(II) dicyanidoargentate(II) (c) dicyanidosilver(I) diaquaargentate(I) (d) diaquasilver(I) dicyanidoargentate(I)

Answer

(d) diaquasilver(I) dicyanidoargentate(I)

In the cation $\mathrm{[Ag(H_2O)_2]^+}$ , silver is $+1$ (water is neutral), and in the anion $\mathrm{[Ag(CN)_2]^-}$ it is $+1$ again ( $x - 2 = -1$ ). The cation is named first, and the anion takes the "-ate" form: argentate. Oxidation state (II) options are therefore wrong.

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2022 · NEET-UG (India) · 4 marks

Amongst the following which one will have maximum 'lone pair - lone pair' electron repulsions?
(a) $\mathrm{ClF_3}$ (b) $\mathrm{IF_5}$ (c) $\mathrm{SF_4}$ (d) $\mathrm{XeF_2}$

Answer

(d) $\mathrm{XeF_2}$

Count lone pairs on the central atom: $\mathrm{ClF_3}$ has 2, $\mathrm{IF_5}$ has 1, $\mathrm{SF_4}$ has 1 and $\mathrm{XeF_2}$ has 3 (8 valence electrons on Xe, 2 used in bonds, 3 lone pairs). The three lone pairs sit in the equatorial plane of the trigonal bipyramid, giving the most lone pair-lone pair repulsion, with a linear shape.

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2022 · NEET-UG (India) · 4 marks

Identify the incorrect statement from the following.
(a) All the five 5d orbitals are different in size when compared to the respective 4d orbitals. (b) All the five 4d orbitals have shapes similar to the respective 3d orbitals. (c) In an atom, all the five 3d orbitals are equal in energy in free state. (d) The shapes of $d_{xy}$ , $d_{yz}$ and $d_{zx}$ orbitals are similar to each other; and $d_{x^2-y^2}$ and $d_{z^2}$ are similar to each other.

Answer

(d) The shapes of $d_{xy}$ , $d_{yz}$ , $d_{zx}$ are similar to each other; and $d_{x^2-y^2}$ and $d_{z^2}$ are similar to each other.

This is the incorrect statement: $d_{z^2}$ has a distinct shape (a dumbbell with a doughnut-shaped collar in the xy-plane), different from the four-lobed $d_{x^2-y^2}$ . The other statements are true: higher shells give larger orbitals of similar shapes, and the five 3d orbitals are degenerate in a free atom.

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2022 · NEET-UG (India) · 4 marks

Assertion (A): ICl is more reactive than $\mathrm{I_2}$ Reason (R): I-Cl bond is weaker than I-I bond Choose the most appropriate answer from the options given below:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A) (b) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (c) (A) is correct but (R) is not correct (d) (A) is not correct but (R) is correct

Answer

(a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

The I-Cl bond is polar and weaker than the nonpolar I-I bond, so ICl breaks more easily and is the more reactive interhalogen. The weaker bond is the reason for its higher reactivity.

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2022 · NEET-UG (India) · 4 marks

Statement I: The boiling points of the following hydrides of group 16 elements increases in the order - $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$ Statement II: The boiling points of these hydrides increases with increase in molar mass. In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are correct. (b) Both Statement I and Statement II are incorrect. (c) Statement I is correct but Statement II is incorrect. (d) Statement I is incorrect but Statement II is correct.

Answer

(b) Both Statement I and Statement II are incorrect.

$\mathrm{H_2O}$ has by far the highest boiling point because of extensive hydrogen bonding. The actual order is $\mathrm{H_2S < H_2Se < H_2Te < H_2O}$ , so Statement I is wrong. Statement II fails as a general rule: boiling point rises with molar mass only from $\mathrm{H_2S}$ to $\mathrm{H_2Te}$ , and water, the lightest, is the exception.

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2022 · NEET-UG (India) · 4 marks

Match List-I with List-II. List-I A. Li B. Na C. KOH D. Cs List-II I. absorbent for carbon dioxide II. electrochemical cells III. coolant in fast breeder reactors IV. photoelectric cell
(a) A-IV, B-I, C-III, D-II (b) A-III, B-IV, C-II, D-I (c) A-I, B-III, C-IV, D-II (d) A-II, B-III, C-I, D-IV

Answer

(d) A-II, B-III, C-I, D-IV

Li is used in electrochemical cells (lithium batteries, strongly negative $E^\circ$ ), liquid Na as coolant in fast breeder reactors, KOH as absorbent for $\mathrm{CO_2}$ , and Cs in photoelectric cells because of its very low ionisation enthalpy.

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2022 · NEET-UG (India) · 4 marks

Which of the following statement is not correct about diborane?
(a) There are two 3-centre-2-electron bonds. (b) The four terminal B-H bonds are two centre two electron bonds. (c) The four terminal hydrogen atoms and the two boron atoms lie in one plane. (d) Both the boron atoms are $sp^2$ hybridised.

Answer

(d) Both the boron atoms are $sp^2$ hybridised.

In $\mathrm{B_2H_6}$ each boron is $sp^3$ hybridised, using two $sp^3$ orbitals for terminal 2c-2e B-H bonds and two for the two bridging 3c-2e (banana) bonds. The other statements are correct: two 3c-2e bonds, four terminal 2c-2e bonds, and the four terminal H and two B atoms lie in one plane.

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2022 · NEET-UG (India) · 4 marks

The order of energy absorbed which is responsible for the color of complexes (A) $\mathrm{[Ni(H_2O)_2(en)_2]^{2+}}$ (B) $\mathrm{[Ni(H_2O)_4(en)]^{2+}}$ (C) $\mathrm{[Ni(en)_3]^{2+}}$
(a) A > B > C (b) C > B > A (c) C > A > B (d) B > A > C

Answer

(c) C > A > B

The colour depends on the energy absorbed, which is proportional to the crystal field splitting $\Delta_o$ . Ethylenediamine (en) is a stronger field ligand than $\mathrm{H_2O}$ , so more en ligands means larger $\Delta_o$ and greater energy absorbed: $\mathrm{[Ni(en)_3]^{2+}}$ (3 en) > $\mathrm{[Ni(H_2O)_2(en)_2]^{2+}}$ (2 en) > $\mathrm{[Ni(H_2O)_4(en)]^{2+}}$ (1 en).

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2022 · NEET-UG (India) · 4 marks

If radius of second Bohr orbit of the $\mathrm{He^+}$ ion is 105.8 pm, what is the radius of third Bohr orbit of $\mathrm{Li^{2+}}$ ion?
(a) 158.7 pm (b) 15.87 pm (c) 1.587 pm (d) 158.7 Å

Answer

(a) 158.7 pm

The Bohr radius varies as $r_n \propto n^2/Z$ . For $\mathrm{He^+}$ , $n=2$ : $4/2 = 2$ units, so one unit is $105.8/2 = 52.9$ pm. For $\mathrm{Li^{2+}}$ , $n=3$ : $9/3 = 3$ units, so $r = 3\times52.9 = 158.7$ pm.

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2022 · NEET-UG (India) · 4 marks

In the neutral or faintly alkaline medium, $\mathrm{KMnO_4}$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from
(a) +7 to +4 (b) +6 to +4 (c) +7 to +3 (d) +6 to +5

Answer

(a) +7 to +4

In neutral or faintly alkaline medium, $\mathrm{MnO_4^-}$ (Mn $+7$ ) is reduced to $\mathrm{MnO_2}$ (Mn $+4$ ), e.g. $2\mathrm{MnO_4^-} + \mathrm{H_2O} + \mathrm{I^-} \rightarrow 2\mathrm{MnO_2} + 2\mathrm{OH^-} + \mathrm{IO_3^-}$ . Reduction to $\mathrm{Mn^{2+}}$ happens only in acid.

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2023 · NEET-UG (India) · 4 marks

Select the correct statements from the following: A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is $9.10939\times10^{-31}$ kg. C. All the isotopes of a given element show same chemical properties. D. Protons and electrons are, collectively known as nucleons. E. Dalton's atomic theory, regarded the atom as an ultimate particle of matter. Choose the correct answer from the options given below.
(a) A, B and C only (b) C, D and E only (c) A and E only (d) B, C and E only

Answer

(d) B, C and E only

B is correct ( $m_e = 9.10939\times10^{-31}$ kg), C is correct (isotopes have the same electronic configuration, hence same chemical properties) and E is correct (Dalton regarded the atom as indivisible). A is false (atoms have three fundamental particles) and D is false (nucleons are protons and neutrons).

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2023 · NEET-UG (India) · 4 marks

Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic. Reason R: The deep blue solution is due to the formation of amide. In the light of the above statements, choose the correct answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true and R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Answer

(c) A is true but R is false

Na dissolves in liquid $\mathrm{NH_3}$ to give a deep blue, paramagnetic solution. The colour is due to ammoniated electrons $\mathrm{e^-(NH_3)_x}$ absorbing visible light, not amide formation (amide forms slowly on standing with a catalyst, giving hydrogen).

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2023 · NEET-UG (India) · 4 marks

Amongst the following, the total number of species NOT having eight electrons around central atom in its outer most shell, is $\mathrm{NH_3,\ AlCl_3,\ BeCl_2,\ CCl_4,\ PCl_5}$
(a) 3 (b) 2 (c) 4 (d) 1

Answer

(a) 3

$\mathrm{NH_3}$ (8 electrons) and $\mathrm{CCl_4}$ (8) obey the octet. $\mathrm{AlCl_3}$ has 6, $\mathrm{BeCl_2}$ has 4 and $\mathrm{PCl_5}$ has 10 electrons around the central atom, so three species do not have eight.

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2023 · NEET-UG (India) · 4 marks

Homoleptic complex from the following complexes is
(a) Potassium trioxalatoaluminate(III) (b) Diamminechloridonitrito-N-platinum(II) (c) Pentaamminecarbonatocobalt(III) chloride (d) Triamminetriaquachromium(III) chloride

Answer

(a) Potassium trioxalatoaluminate(III)

A homoleptic complex has only one kind of ligand. $\mathrm{K_3[Al(C_2O_4)_3]}$ has only oxalate. The others have mixed ligands: NH $_3$ , Cl, NO $_2$ in (b); NH $_3$ and CO $_3$ in (c); NH $_3$ and H $_2$ O in (d).

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2023 · NEET-UG (India) · 4 marks

The relation between $n_m$ ( $n_m$ = the number of permissible values of magnetic quantum number $m$ ) for a given value of azimuthal quantum number $l$ , is
(a) $l = \frac{n_m - 1}{2}$ (b) $l = 2n_m + 1$ (c) $n_m = 2l^2 + 1$ (d) $n_m = l + 2$

Answer

(a) $l = \dfrac{n_m - 1}{2}$

For a given $l$ , $m$ takes values from $-l$ to $+l$ , so $n_m = 2l+1$ . Rearranging gives $l = \dfrac{n_m-1}{2}$ .

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2023 · NEET-UG (India) · 4 marks

The stability of $\mathrm{Cu^{2+}}$ is more than $\mathrm{Cu^{+}}$ salts in aqueous solution due to
(a) First ionisation enthalpy (b) Enthalpy of atomization (c) Hydration energy (d) Second ionisation enthalpy

Answer

(c) Hydration energy

Cu $^+$ ( $3d^{10}$ ) would seem more stable, but the very high hydration enthalpy of the small, doubly charged $\mathrm{Cu^{2+}}$ more than compensates for its high second ionisation enthalpy. So in aqueous solution Cu(II) is more stable.

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2023 · NEET-UG (India) · 4 marks

Taking stability as the factor, which one of the following represents correct relationship?
(a) $\mathrm{TlCl_3 > TlCl}$ (b) $\mathrm{InI_3 > InI}$ (c) $\mathrm{AlCl > AlCl_3}$ (d) $\mathrm{TlI > TlI_3}$

Answer

(d) $\mathrm{TlI > TlI_3}$

Due to the inert pair effect, the $+1$ state is more stable than $+3$ for Tl, so TlI is more stable than $\mathrm{TlI_3}$ (which is really $\mathrm{Tl^+I_3^-}$ ). $\mathrm{TlCl_3 > TlCl}$ is wrong for the same reason, and for Al the $+3$ state is far more stable, so $\mathrm{AlCl > AlCl_3}$ is wrong. $\mathrm{InI_3 > InI}$ is not accepted by the key.

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2023 · NEET-UG (India) · 4 marks

The correct order of energies of molecular orbitals of $\mathrm{N_2}$ molecule, is
(a) $\sigma_{1s} < \sigma^_{1s} < \sigma_{2s} < \sigma^_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z} < (\pi^_{2p_x} = \pi^_{2p_y}) < \sigma^_{2p_z}$ (b) $\sigma_{1s} < \sigma^_{1s} < \sigma_{2s} < \sigma^_{2s} < \sigma_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^_{2p_x} = \pi^_{2p_y}) < \pi^_{2p_z}$ (c) $\sigma_{1s} < \sigma^_{1s} < \sigma_{2s} < \sigma^_{2s} < \sigma_{2p_z} < \sigma^_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^_{2p_x} = \pi^_{2p_y})$ (d) $\sigma_{1s} < \sigma^_{1s} < \sigma_{2s} < \sigma^_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^_{2p_x} = \pi^_{2p_y}) < \sigma_{2p_z} < \sigma^_{2p_z}$

Answer

(a) $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$

For $\mathrm{N_2}$ (and other molecules up to 14 electrons) $s$ - $p$ mixing pushes $\sigma_{2p_z}$ above the two degenerate $\pi_{2p}$ orbitals. The order with $\sigma_{2p_z}$ below $\pi_{2p}$ (options b, c) holds only for $\mathrm{O_2}$ and heavier, and option (d) wrongly places $\pi^*$ below $\sigma_{2p_z}$ .

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2023 · NEET-UG (India) · 4 marks

Assertion A: Helium is used to dilute oxygen in diving apparatus. Reason R: Helium has high solubility in $\mathrm{O_2}$ . In the light of the above statements, choose the correct answer from the options given below
(a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true and R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Answer

(c) A is true but R is false

Helium is used to dilute oxygen in diving gas because it is very sparingly soluble in blood, so it avoids the bends (nitrogen narcosis and decompression problems). The reason is false: helium has very low solubility, not high solubility.

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2023 · NEET-UG (India) · 4 marks

The element expected to form largest ion to achieve the nearest noble gas configuration is
(a) O (b) F (c) N (d) Na

Answer

(c) N

The ions formed are $\mathrm{N^{3-}}$ , $\mathrm{O^{2-}}$ , $\mathrm{F^-}$ and $\mathrm{Na^+}$ , all isoelectronic with Ne. For isoelectronic ions, size decreases as nuclear charge increases, so $\mathrm{N^{3-}}$ (Z = 7) is the largest. $\mathrm{Na^+}$ is the smallest, as it has the highest nuclear charge.

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2023 · NEET-UG (India) · 4 marks

Which one of the following statements is correct?
(a) The bone in human body is an inert and unchanging substance. (b) Mg plays roles in neuromuscular function and interneuronal transmission (c) The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g. (d) All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor

Answer

(b) Mg plays roles in neuromuscular function and interneuronal transmission

Mg ions take part in neuromuscular function and interneuronal transmission, so (b) is correct. Bone is not inert; it is continuously remodelled, so (a) is wrong. The daily requirement of Mg and Ca is about 1 g or more, not 0.2-0.3 g, so (c) is wrong. ATP-utilising phosphate-transfer enzymes need Mg, not Ca, as cofactor, so (d) is wrong.

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2023 · NEET-UG (India) · 4 marks

Match List-I with List-II. List-I A. Coke B. Diamond C. Fullerene D. Graphite List-II I. Carbon atoms are $sp^3$ hybridised II. Used as a dry lubricant III. Used as a reducing agent IV. Cage like molecules
(a) A-II, B-IV, C-I, D-III (b) A-IV, B-I, C-II, D-III (c) A-III, B-I, C-IV, D-II (d) A-III, B-IV, C-I, D-II

Answer

(c) A-III, B-I, C-IV, D-II

Coke is used as a reducing agent in metallurgy (III). Diamond has $sp^3$ carbon atoms (I). Fullerenes are cage-like molecules (IV). Graphite is soft and slippery, used as a dry lubricant (II).

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2023 · NEET-UG (India) · 4 marks

Which of the following statements are INCORRECT? A. All the transition metals except scandium form MO oxides which are ionic. B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in $\mathrm{Sc_2O_3}$ to $\mathrm{Mn_2O_7}$ . C. Basic character increases from $\mathrm{V_2O_3}$ to $\mathrm{V_2O_4}$ to $\mathrm{V_2O_5}$ . D. $\mathrm{V_2O_4}$ dissolves in acids to give $\mathrm{VO_4^{3-}}$ salts. E. CrO is basic but $\mathrm{Cr_2O_3}$ is amphoteric. Choose the correct answer from the options given below:
(a) A and E only (b) B and D only (c) C and D only (d) B and C only

Answer

(c) C and D only

C is incorrect: acidic character increases from $\mathrm{V_2O_3}$ to $\mathrm{V_2O_4}$ to $\mathrm{V_2O_5}$ as oxidation state rises (basic character decreases). D is incorrect: $\mathrm{V_2O_4}$ dissolves in acids to give $\mathrm{VO^{2+}}$ salts, not $\mathrm{VO_4^{3-}}$ ( $\mathrm{V_2O_5}$ gives $\mathrm{VO_4^{3-}}$ with alkali). A, B and E are correct.

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2023 · NEET-UG (India) · 4 marks

Match List-I (Oxoacids of Sulphur) with List-II (Bonds). List-I A. Peroxodisulphuric acid B. Sulphuric acid C. Pyrosulphuric acid D. Sulphurous acid List-II I. Two S-OH, Four S=O, One S-O-S II. Two S-OH, One S=O III. Two S-OH, Four S=O, One S-O-O-S IV. Two S-OH, Two S=O
(a) A-I, B-III, C-II, D-IV (b) A-III, B-IV, C-I, D-II (c) A-I, B-III, C-IV, D-II (d) A-III, B-IV, C-II, D-I

Answer

(b) A-III, B-IV, C-I, D-II

Peroxodisulphuric acid $\mathrm{H_2S_2O_8}$ has S-O-O-S, two S-OH and four S=O (III). Sulphuric acid has two S-OH and two S=O (IV). Pyrosulphuric acid $\mathrm{H_2S_2O_7}$ has S-O-S, two S-OH and four S=O (I). Sulphurous acid has two S-OH and one S=O (II).

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2024 · NEET-UG (India) · 4 marks

'Spin only' magnetic moment is same for which of the following ions? A. $\mathrm{Ti^{3+}}$ , B. $\mathrm{Cr^{2+}}$ C. $\mathrm{Mn^{2+}}$ , D. $\mathrm{Fe^{2+}}$ E. $\mathrm{Sc^{3+}}$ Choose the most appropriate answer from the options given below:
(a) A and E only (b) B and C only (c) A and D only (d) B and D only

Answer

(d) B and D only

Spin-only moment depends on unpaired electrons. $\mathrm{Ti^{3+}}$ ( $d^1$ ) has 1, $\mathrm{Cr^{2+}}$ ( $d^4$ ) has 4, $\mathrm{Mn^{2+}}$ ( $d^5$ ) has 5, $\mathrm{Fe^{2+}}$ ( $d^6$ ) has 4, $\mathrm{Sc^{3+}}$ ( $d^0$ ) has 0. So $\mathrm{Cr^{2+}}$ and $\mathrm{Fe^{2+}}$ share the same value, $\sqrt{24}$ BM.

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2024 · NEET-UG (India) · 4 marks

The energy of an electron in the ground state ( $n = 1$ ) for $\mathrm{He^+}$ ion is $-x$ J, then that for an electron in $n = 2$ state for $\mathrm{Be^{3+}}$ ion in J is
(a) $-\frac{x}{9}$ (b) $-4x$ (c) $-\frac{4}{9}x$ (d) $-x$

Answer

(d) $-x$

For a hydrogen-like ion $E_n\propto -Z^2/n^2$ . $\mathrm{He^+}$ , $n=1$ : $Z^2/n^2=4$ , giving $-x$ . $\mathrm{Be^{3+}}$ , $n=2$ : $Z^2/n^2=16/4=4$ . The ratio is the same, so the energy is $-x$ .

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2024 · NEET-UG (India) · 4 marks

Match List-I (Molecule) with List-II (Number and types of bond/s between two carbon atoms). List-I A. ethane B. ethene C. carbon molecule, $\mathrm{C_2}$ D. ethyne List-II I. one $\sigma$ -bond and two $\pi$ -bonds II. two $\pi$ -bonds III. one $\sigma$ -bond IV. one $\sigma$ -bond and one $\pi$ -bond
(a) A-IV, B-III, C-II, D-I (b) A-III, B-IV, C-II, D-I (c) A-III, B-IV, C-I, D-II (d) A-I, B-IV, C-II, D-III

Answer

(b) A-III, B-IV, C-II, D-I

Ethane has a single C-C $\sigma$ -bond. Ethene has one $\sigma$ and one $\pi$ . $\mathrm{C_2}$ molecule has two $\pi$ -bonds (no $\sigma$ ) as per MO theory. Ethyne has one $\sigma$ and two $\pi$ -bonds.

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2024 · NEET-UG (India) · 4 marks

Match List I (Complex) with List II (types of isomerism). List I A. $\mathrm{[Co(NH_3)_5(NO_2)]Cl_2}$ B. $\mathrm{[Co(NH_3)_5(SO_4)]Br}$ C. $\mathrm{[Co(NH_3)_6][Cr(CN)_6]}$ D. $\mathrm{[Co(H_2O)_6]Cl_3}$ List II I. Solvate isomerism II. Linkage isomerism III. Ionization isomerism IV. Coordination isomerism
(a) A-I, B-III, C-IV, D-II (b) A-I, B-IV, C-III, D-II (c) A-II, B-IV, C-III, D-I (d) A-II, B-III, C-IV, D-I

Answer

(d) A-II, B-III, C-IV, D-I

A has $\mathrm{NO_2}$ that can bind through N or O: linkage isomerism. B can swap $\mathrm{SO_4^{2-}}$ and $\mathrm{Br^-}$ between sphere and counter ion: ionization isomerism. C can exchange ligands between the two complex ions: coordination isomerism. D is a hydrate with water inside or outside the sphere: solvate isomerism.

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2024 · NEET-UG (India) · 4 marks

The $E^\circ$ value for the $\mathrm{Mn^{3+}/Mn^{2+}}$ couple is more positive than that of $\mathrm{Cr^{3+}/Cr^{2+}}$ or $\mathrm{Fe^{3+}/Fe^{2+}}$ due to change of
(a) $d^5$ to $d^2$ configuration (b) $d^4$ to $d^5$ configuration (c) $d^3$ to $d^5$ configuration (d) $d^5$ to $d^4$ configuration

Answer

(b) $d^4$ to $d^5$ configuration

$\mathrm{Mn^{3+}}$ is $d^4$ and reduces to $\mathrm{Mn^{2+}}$ , which is the stable half-filled $d^5$ . This strong tendency to gain an electron makes $E^\circ(\mathrm{Mn^{3+}/Mn^{2+}})$ highly positive; $\mathrm{Cr^{3+}}$ ( $d^3$ ) is stable, so its couple is negative.

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2024 · NEET-UG (India) · 4 marks

Match List-I (Quantum Number) with List-II (Information provided). List-I A. $m_l$ B. $m_s$ C. $l$ D. $n$ List-II I. shape of orbital II. size of orbital III. orientation of orbital IV. orientation of spin of electrons
(a) A-III, B-IV, C-I, D-II (b) A-III, B-IV, C-II, D-I (c) A-II, B-I, C-IV, D-III (d) A-I, B-III, C-II, D-IV

Answer

(a) A-III, B-IV, C-I, D-II

$m_l$ gives the orientation of the orbital, $m_s$ gives the spin orientation of the electron. $l$ gives the shape of the orbital and $n$ gives its size (and energy).

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2024 · NEET-UG (India) · 4 marks

Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N. Choose the answer from the options given below:
(a) Li < B < Be < C < N (b) Li < Be < C < B < N (c) Li < Be < N < B < C (d) Li < Be < B < C < N

Answer

(a) Li < B < Be < C < N

Ionization enthalpy generally increases across a period, but Be ( $2s^2$ , full subshell) is higher than B ( $2p^1$ ). Approximate values (kJ/mol): Li 520, B 801, Be 900, C 1086, N 1402.

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2024 · NEET-UG (India) · 4 marks

Match List-I (Compound) with List-II (Shape/geometry). List-I A. $\mathrm{NH_3}$ B. $\mathrm{BrF_5}$ C. $\mathrm{XeF_4}$ D. $\mathrm{SF_6}$ List-II I. Trigonal pyramidal II. Square planar III. Octahedral IV. Square pyramidal
(a) A-II, B-IV, C-III, D-I (b) A-III, B-IV, C-I, D-II (c) A-II, B-III, C-IV, D-I (d) A-I, B-IV, C-II, D-III

Answer

(d) A-I, B-IV, C-II, D-III

$\mathrm{NH_3}$ is trigonal pyramidal ( $sp^3$ , one lone pair). $\mathrm{BrF_5}$ is square pyramidal ( $sp^3d^2$ , one lone pair). $\mathrm{XeF_4}$ is square planar (two lone pairs). $\mathrm{SF_6}$ is octahedral.

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2024 · NEET-UG (India) · 4 marks

Among Group 16 elements, which one does NOT show $-2$ oxidation state?
(a) Se (b) Te (c) Po (d) O

Answer

(c) Po

O, S, Se and Te all form $-2$ compounds, but Po is metallic and mainly shows $+2$ and $+4$ ; it does not show the $-2$ state. Oxygen, being highly electronegative, shows $-2$ readily.

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2024 · NEET-UG (India) · 4 marks

Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si. Choose the correct answer from the options given below:
(a) Si < C < O < N < F (b) O < F < N < C < Si (c) F < O < N < C < Si (d) Si < C < N < O < F

Answer

(d) Si < C < N < O < F

Electronegativity rises across a period (C < N < O < F) and falls down a group (Si < C). Pauling values: Si 1.8, C 2.5, N 3.0, O 3.5, F 4.0.

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2024 · NEET-UG (India) · 4 marks

Statement I: The boiling point of hydrides of Group 16 elements follow the order $\mathrm{H_2O > H_2Te > H_2Se > H_2S}$ . Statement II: On the basis of molecular mass, $\mathrm{H_2O}$ is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in $\mathrm{H_2O}$ it has higher boiling point. In the light of the above statements, choose the correct answer from the options given below:
(a) Both statement I and statement II are false (b) Statement I is true but statement II is false (c) Statement I is false but statement II is true (d) Both statement I and statement II are true

Answer

(d) Both statement I and statement II are true

Boiling points: $\mathrm{H_2O}$ (373 K) > $\mathrm{H_2Te}$ > $\mathrm{H_2Se}$ > $\mathrm{H_2S}$ , so Statement I is true. By mass alone water would be lowest, but extensive hydrogen bonding raises it, so Statement II is true too.

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2024 · NEET-UG (India) · 4 marks

Statement I: Both $\mathrm{[Co(NH_3)_6]^{3+}}$ and $\mathrm{[CoF_6]^{3-}}$ complexes are octahedral but differ in their magnetic behaviour. Statement II: $\mathrm{[Co(NH_3)_6]^{3+}}$ is diamagnetic whereas $\mathrm{[CoF_6]^{3-}}$ is paramagnetic. In the light of above statements, choose the correct answer from the options given below:
(a) Both statement I and statement II are false (b) Statement I is true but statement II is false (c) Statement I is false but statement II is true (d) Both statement I and statement II are true

Answer

(d) Both statement I and statement II are true

Both complexes are octahedral $\mathrm{Co^{3+}}$ ( $d^6$ ). $\mathrm{NH_3}$ is a strong-field ligand, giving low spin $t_{2g}^6$ with no unpaired electrons (diamagnetic). $\mathrm{F^-}$ is weak-field, giving high spin $t_{2g}^4e_g^2$ with four unpaired electrons (paramagnetic).

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2024 · NEET-UG (India) · 4 marks

Statement I: $\mathrm{[Co(NH_3)_6]^{3+}}$ is a homoleptic complex whereas $\mathrm{[Co(NH_3)_4Cl_2]^+}$ is heteroleptic complex. Statement II: Complex $\mathrm{[Co(NH_3)_6]^{3+}}$ has one kind of ligands but $\mathrm{[Co(NH_3)_4Cl_2]^+}$ has more than one kind of ligands. In the light of the above statements, choose the correct answer from the options given below:
(a) Both statement I and statement II are false (b) Statement I is true but statement II is false (c) Statement I is false but statement II is true (d) Both statement I and statement II are true

Answer

(d) Both statement I and statement II are true

A homoleptic complex has only one kind of ligand, and a heteroleptic complex has more than one kind. $[\mathrm{Co(NH_3)_6}]^{3+}$ has only $\mathrm{NH_3}$ (homoleptic). $[\mathrm{Co(NH_3)_4Cl_2}]^+$ has $\mathrm{NH_3}$ and $\mathrm{Cl^-}$ (heteroleptic). Statement II is the correct explanation of Statement I.

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2024 · NEET-UG (India) · 4 marks

The pair of lanthanoid ions which are diamagnetic is
(a) $\mathrm{Ce^{3+}}$ and $\mathrm{Eu^{2+}}$ (b) $\mathrm{Gd^{3+}}$ and $\mathrm{Eu^{3+}}$ (c) $\mathrm{Pm^{3+}}$ and $\mathrm{Sm^{3+}}$ (d) $\mathrm{Ce^{4+}}$ and $\mathrm{Yb^{2+}}$

Answer

(d) $\mathrm{Ce^{4+}}$ and $\mathrm{Yb^{2+}}$

Diamagnetic means no unpaired electrons. $\mathrm{Ce^{4+}}$ is $4f^0$ and $\mathrm{Yb^{2+}}$ is $4f^{14}$ , so both are fully empty or filled. In the other pairs, $\mathrm{Ce^{3+}}$ ( $4f^1$ ), $\mathrm{Eu^{2+}}$ ( $4f^7$ ), $\mathrm{Gd^{3+}}$ ( $4f^7$ ), $\mathrm{Pm^{3+}}$ ( $4f^4$ ) and $\mathrm{Sm^{3+}}$ ( $4f^5$ ) all have unpaired electrons. $\mathrm{Eu^{3+}}$ ( $4f^6$ ) is also paramagnetic.

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2024 · NEET-UG (India) · 4 marks

Identify the correct answer.
(a) $\mathrm{BF_3}$ has non-zero dipole moment (b) Dipole moment of $\mathrm{NF_3}$ is greater than that of $\mathrm{NH_3}$ (c) Three canonical forms can be drawn for $\mathrm{CO_3^{2-}}$ ion. (d) Three resonance structures can be drawn for ozone.

Answer

(c) Three canonical forms can be drawn for $\mathrm{CO_3^{2-}}$ ion.

The double bond in carbonate can be placed on any one of three oxygens, giving three equivalent resonance forms. $\mathrm{BF_3}$ is planar, so its dipole moment is zero. In $\mathrm{NF_3}$ the N-F bond dipoles oppose the lone-pair dipole, so it is smaller than in $\mathrm{NH_3}$ . Ozone has only two canonical forms.

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2025 · NEET-UG (India) · 4 marks

Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Statement I is true but Statement II is false (b) Statement I is false but Statement II is true (c) Both Statement I and Statement II are true (d) Both Statement I and Statement II are false

Answer

(d) Both Statement I and Statement II are false

Bond order zero means no net bonding, so such a molecule is unstable and does not exist (for example $\mathrm{He_2}$ ). Statement I is false. Bond length decreases as bond order increases, because a stronger multiple bond pulls the atoms closer. Statement II is false.

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2025 · NEET-UG (India) · 4 marks

Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
(a) $\mathrm{[Co(NH_3)_6]Cl_3}$ (b) $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ (c) $\mathrm{[Co(NH_3)_3Cl_3]}$ (d) $\mathrm{[Co(NH_3)_4Cl_2]}$

Answer

(c) $\mathrm{[Co(NH_3)_3Cl_3]}$

Conductance depends on the number of ions produced in solution. $[\mathrm{Co(NH_3)_3Cl_3}]$ is a neutral complex with all three chlorides inside the coordination sphere, so it gives essentially no ions and has the minimum conductance. $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$ gives 4 ions and $[\mathrm{Co(NH_3)_5Cl}]\mathrm{Cl_2}$ gives 3 ions, so both conduct more.

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2025 · NEET-UG (India) · 4 marks

Statement I: Like nitrogen that can form ammonia, arsenic can form arsine. Statement II: Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Statement I is correct but Statement II is incorrect (b) Statement I is incorrect but Statement II is correct (c) Both Statement I and Statement II are correct (d) Both Statement I and Statement II are incorrect

Answer

(a) Statement I is correct but Statement II is incorrect

Arsenic forms arsine, $\mathrm{AsH_3}$ , just as nitrogen forms $\mathrm{NH_3}$ , so Statement I is correct. Antimony forms the pentoxide $\mathrm{Sb_2O_5}$ (oxidation state +5), as well as $\mathrm{Sb_2O_3}$ . Statement II is therefore incorrect.

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2025 · NEET-UG (India) · 4 marks

Statement I: Ferromagnetism is considered as an extreme form of paramagnetism. Statement II: The number of unpaired electrons in a $\mathrm{Cr^{2+}}$ ion ( $Z = 24$ ) is the same as that of a $\mathrm{Nd^{3+}}$ ion ( $Z = 60$ ). In the light of the above statements, choose the correct answer from the options given below:
(a) Statement I is true but Statement II is false (b) Statement I is false but Statement II is true (c) Both Statement I and Statement II are true (d) Both Statement I and Statement II are false

Answer

(a) Statement I is true but Statement II is false

Ferromagnetism is an extreme, cooperative form of paramagnetism, with domains of aligned spins. Statement I is true. $\mathrm{Cr^{2+}}$ is $3d^4$ with 4 unpaired electrons. $\mathrm{Nd^{3+}}$ is $4f^3$ with 3 unpaired electrons. These are not equal, so Statement II is false.

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2025 · NEET-UG (India) · 4 marks

Match List-I with List-II. List-I A. $\mathrm{XeO_3}$ B. $\mathrm{XeF_2}$ C. $\mathrm{XeOF_4}$ D. $\mathrm{XeF_6}$ List-II I. $sp^3d$ ; linear II. $sp^3$ ; pyramidal III. $sp^3d^3$ ; distorted octahedral IV. $sp^3d^2$ ; square pyramidal
(a) A-IV, B-II, C-III, D-I (b) A-IV, B-II, C-I, D-III (c) A-II, B-I, C-IV, D-III (d) A-II, B-I, C-III, D-IV

Answer

(c) A-II, B-I, C-IV, D-III

$\mathrm{XeO_3}$ has 3 bond pairs and 1 lone pair, so it is $sp^3$ and pyramidal. $\mathrm{XeF_2}$ has 2 bond pairs and 3 lone pairs, so it is $sp^3d$ and linear. $\mathrm{XeOF_4}$ is $sp^3d^2$ and square pyramidal. $\mathrm{XeF_6}$ is $sp^3d^3$ and distorted octahedral.

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2025 · NEET-UG (India) · 4 marks

Which of the following statements are true? A. Unlike Ga that has a very high melting point, Cs has a very low melting point. B. On Pauling scale, the electronegativity values of N and Cl are not the same. C. Ar, $\mathrm{K^+}$ , $\mathrm{Cl^-}$ , $\mathrm{Ca^{2+}}$ , and $\mathrm{S^{2-}}$ are all isoelectronic species. D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na. E. The atomic radius of Cs is greater than that of Li and Rb. Choose the correct answer from the options given below
(a) C and D only (b) A, C, and E only (c) A, B, and E only (d) C and E only

Answer

(d) C and E only

C is true: Ar, $\mathrm{K^+}$ , $\mathrm{Cl^-}$ , $\mathrm{Ca^{2+}}$ and $\mathrm{S^{2-}}$ all have 18 electrons. E is true: atomic radius is $\mathrm{Li<Rb<Cs}$ . A is false because Ga melts at about 30 C and Cs at about 28.5 C, so both are low. B is false because N and Cl both have electronegativity 3.0. D is false because the order is Si > Mg > Al > Na.

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2025 · NEET-UG (India) · 4 marks

The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes $n = 2 \rightarrow n = 3$ and $n = 4 \rightarrow n = 6$ transitions, respectively, is
(a) $\frac{1}{9}$ (b) $\frac{1}{4}$ (c) $\frac{1}{36}$ (d) $\frac{1}{16}$

Answer

(b) $\dfrac{1}{4}$

$\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ . For $2\to3$ : $\dfrac{1}{4}-\dfrac{1}{9}=\dfrac{5}{36}$ . For $4\to6$ : $\dfrac{1}{16}-\dfrac{1}{36}=\dfrac{5}{144}$ . $\dfrac{\lambda_1}{\lambda_2}=\dfrac{5/144}{5/36}=\dfrac{1}{4}$ .

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2025 · NEET-UG (India) · 4 marks

Which of the following are paramagnetic? A. $\mathrm{[NiCl_4]^{2-}}$ B. $\mathrm{Ni(CO)_4}$ C. $\mathrm{[Ni(CN)_4]^{2-}}$ D. $\mathrm{[Ni(H_2O)_6]^{2+}}$ E. $\mathrm{Ni(PPh_3)_4}$ Choose the correct answer from the options given below:
(a) A and D only (b) A, D and E only (c) A and C only (d) B and E only

Answer

(a) A and D only

$[\mathrm{NiCl_4}]^{2-}$ is tetrahedral (weak ligand) $d^8$ with 2 unpaired electrons. $[\mathrm{Ni(H_2O)_6}]^{2+}$ is octahedral $d^8$ with 2 unpaired electrons. $\mathrm{Ni(CO)_4}$ and $\mathrm{Ni(PPh_3)_4}$ are Ni(0) $d^{10}$ , so they are diamagnetic. $[\mathrm{Ni(CN)_4}]^{2-}$ is square planar and diamagnetic.

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2025 · NEET-UG (India) · 4 marks

Energy and radius of Bohr orbit of $\mathrm{He^+}$ and $\mathrm{Li^{2+}}$ are [Given $R_H = 2.18\times10^{-18}$ J, $a_0 = 52.9$ pm]
(a) $E_n(\mathrm{Li^{2+}}) = -19.62\times10^{-16}$ J; $r_n(\mathrm{Li^{2+}}) = 17.6$ pm, $E_n(\mathrm{He^+}) = 8.72\times10^{-16}$ J, $r_n(\mathrm{He^+}) = 26.4$ pm (b) $E_n(\mathrm{Li^{2+}}) = -8.72\times10^{-16}$ J; $r_n(\mathrm{Li^{2+}}) = 17.6$ pm, $E_n(\mathrm{He^+}) = -19.62\times10^{-16}$ J, $r_n(\mathrm{He^+}) = 17.6$ pm (c) $E_n(\mathrm{Li^{2+}}) = -19.62\times10^{-18}$ J; $r_n(\mathrm{Li^{2+}}) = 17.6$ pm, $E_n(\mathrm{He^+}) = -8.72\times10^{-18}$ J, $r_n(\mathrm{He^+}) = 26.4$ pm (d) $E_n(\mathrm{Li^{2+}}) = -8.72\times10^{-18}$ J; $r_n(\mathrm{Li^{2+}}) = 26.4$ pm, $E_n(\mathrm{He^+}) = -19.62\times10^{-18}$ J, $r_n(\mathrm{He^+}) = 17.6$ pm

Answer

(c) $E_n(\mathrm{Li^{2+}})=-19.62\times10^{-18}$ J, $r_n=17.6$ pm; $E_n(\mathrm{He^+})=-8.72\times10^{-18}$ J, $r_n=26.4$ pm

$E_n=-R_H\dfrac{Z^2}{n^2}$ and $r_n=a_0\dfrac{n^2}{Z}$ (for $n=1$ ). $\mathrm{Li^{2+}}$ ( $Z=3$ ): $E=-2.18\times10^{-18}\times9=-19.62\times10^{-18}$ J and $r=52.9/3=17.6$ pm. $\mathrm{He^+}$ ( $Z=2$ ): $E=-8.72\times10^{-18}$ J and $r=52.9/2=26.4$ pm.

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2025 · NEET-UG (India) · 4 marks

Identify the correct orders against the property mentioned A. $\mathrm{H_2O > NH_3 > CHCl_3}$ - dipole moment B. $\mathrm{XeF_4 > XeO_3 > XeF_2}$ - number of lone pairs on central atom C. O-H > C-H > N-O - bond length D. $\mathrm{N_2 > O_2 > H_2}$ - bond enthalpy Choose the answer from the options given below:
(a) A, C only (b) B, C only (c) A, D only (d) B, D only

Answer

(c) A, D only

A is correct: dipole moments are $\mathrm{H_2O}$ (1.85 D) > $\mathrm{NH_3}$ (1.47 D) > $\mathrm{CHCl_3}$ (1.04 D). D is correct: bond enthalpy is $\mathrm{N_2}$ (triple bond) > $\mathrm{O_2}$ > $\mathrm{H_2}$ . B is wrong because lone pairs on the central atom are $\mathrm{XeF_2}$ (3) > $\mathrm{XeF_4}$ (2) > $\mathrm{XeO_3}$ (1). C is wrong because N-O is the longest bond.

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2025 · NEET-UG (India) · 4 marks

The correct order of the wavelength of light absorbed by the following complexes is, A. $\mathrm{[Co(NH_3)_6]^{3+}}$ B. $\mathrm{[Co(CN)_6]^{3-}}$ C. $\mathrm{[Cu(H_2O)_4]^{2+}}$ D. $\mathrm{[Ti(H_2O)_6]^{3+}}$ Choose the correct answer from the options given below:
(a) C < D < A < B (b) C < A < D < B (c) B < D < A < C (d) B < A < D < C

Answer

(d) B < A < D < C

The wavelength absorbed is inversely proportional to the crystal-field splitting $\Delta_o$ . Higher $\Delta$ means shorter wavelength. The field strength order is $\mathrm{CN^-}>\mathrm{NH_3}>\mathrm{H_2O}$ , and $\mathrm{Co^{3+}}$ gives a larger $\Delta$ than $\mathrm{Ti^{3+}}$ or $\mathrm{Cu^{2+}}$ with water. So $\Delta$ is $\mathrm{[Co(CN)_6]^{3-}}>\mathrm{[Co(NH_3)_6]^{3+}}>\mathrm{[Ti(H_2O)_6]^{3+}}>\mathrm{[Cu(H_2O)_4]^{2+}}$ , and the wavelengths run in the reverse order.

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2025 · NEET-UG (India) · 4 marks

Which among the following electronic configurations belong to main group elements? A. $\mathrm{[Ne]3s^1}$ B. $\mathrm{[Ar]3d^34s^2}$ C. $\mathrm{[Kr]4d^{10}5s^25p^5}$ D. $\mathrm{[Ar]3d^{10}4s^1}$ E. $\mathrm{[Rn]5f^06d^27s^2}$ Choose the correct answer from the option given below
(a) D and E only (b) A, C and D only (c) B and E only (d) A and C only

Answer

(d) A and C only

Main group (s- and p-block) elements are A, $[\mathrm{Ne}]3s^1$ (Na) and C, $[\mathrm{Kr}]4d^{10}5s^25p^5$ (I). B ( $[\mathrm{Ar}]3d^34s^2$ , V) is $d$ -block, and D, $[\mathrm{Ar}]3d^{10}4s^1$ (Cu), is $d$ -block. E, $[\mathrm{Rn}]6d^27s^2$ (Th), is an inner-transition (actinoid) element.

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Past questions and answers

10.The product obtained as a result of a reaction of nitrogen with CaC2\mathrm{CaC_2}CaC2​ is: (a) Ca(CN)2\mathrm{Ca(CN)_2}Ca(CN)2​ (b) CaCN\mathrm{CaCN}CaCN (c) CaCN3\mathrm{CaCN_3}CaCN3​ (d) Ca2CN\mathrm{Ca_2CN}Ca2​CN

13.Which is the correct statement for the given acids? (a) Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid. (b) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid. (c) Both are triprotic acids. (d) Both are diprotic acids.

14.Which one of the following compounds shows the presence of intramolecular hydrogen bond? (a) H2O2\mathrm{H_2O_2}H2​O2​ (b) HCN (c) Cellulose (d) Concentrated acetic acid

15.Boric acid is an acid because its molecule (a) contains replaceable H+\mathrm{H^+}H+ ion (b) gives up a proton (c) accepts OH−\mathrm{OH^-}OH− from water releasing proton (d) combines with proton from water molecule

16.AlF3\mathrm{AlF_3}AlF3​ is soluble in HF only in presence of KF. It is due to the formation of (a) K3[AlF3H3]\mathrm{K_3[AlF_3H_3]}K3​[AlF3​H3​] (b) K3[AlF6]\mathrm{K_3[AlF_6]}K3​[AlF6​] (c) AlH3\mathrm{AlH_3}AlH3​ (d) K[AlF3H]\mathrm{K[AlF_3H]}K[AlF3​H]

17.The suspension of slaked lime in water is known as (a) limewater (b) quicklime (c) milk of lime (d) aqueous solution of slaked lime

19.Which of the following fluoro-compounds is most likely to behave as a Lewis base? (a) BF3\mathrm{BF_3}BF3​ (b) PF3\mathrm{PF_3}PF3​ (c) CF4\mathrm{CF_4}CF4​ (d) SiF4\mathrm{SiF_4}SiF4​

21.In context with beryllium, which one of the following statements is incorrect? (a) It is rendered passive by nitric acid. (b) It forms Be2C\mathrm{Be_2C}Be2​C. (c) Its salts rarely hydrolyze. (d) Its hydride is electron-deficient and polymeric.

23.Which of the following pairs of d-orbitals will have electron density along the axes? (a) dz2, dxzd_{z^2},\ d_{xz}dz2​, dxz​ (b) dxz, dyzd_{xz},\ d_{yz}dxz​, dyz​ (c) dz2, dx2−y2d_{z^2},\ d_{x^2-y^2}dz2​, dx2−y2​ (d) dxy, dx2−y2d_{xy},\ d_{x^2-y^2}dxy​, dx2−y2​

24.The correct geometry and hybridization for XeF4\mathrm{XeF_4}XeF4​ are (a) octahedral, sp3d2sp^3d^2sp3d2 (b) trigonal bipyramidal, sp3dsp^3dsp3d (c) planar triangle, sp3d3sp^3d^3sp3d3 (d) square planar, sp3d2sp^3d^2sp3d2

28.Jahn-Teller effect is not observed in high spin complexes of (a) d7d^7d7 (b) d8d^8d8 (c) d4d^4d4 (d) d9d^9d9

29.In which of the following pairs, both the species are not isostructural? (a) XeF4, XeO4\mathrm{XeF_4,\ XeO_4}XeF4​, XeO4​ (b) SiCl4, PCl4+\mathrm{SiCl_4,\ PCl_4^+}SiCl4​, PCl4+​ (c) diamond, silicon carbide (d) NH3, PH3\mathrm{NH_3,\ PH_3}NH3​, PH3​

30.On heating which of the following releases CO2\mathrm{CO_2}CO2​ most easily? (a) CaCO3\mathrm{CaCO_3}CaCO3​ (b) K2CO3\mathrm{K_2CO_3}K2​CO3​ (c) Na2CO3\mathrm{Na_2CO_3}Na2​CO3​ (d) MgCO3\mathrm{MgCO_3}MgCO3​

31.Number of possible isomers for the complex [Co(en)2Cl2]Cl[\mathrm{Co(en)_2Cl_2}]\mathrm{Cl}[Co(en)2​Cl2​]Cl will be: (en = ethylenediamine) (a) 4 (b) 2 (c) 1 (d) 3

34.The hybridization involved in complex [Ni(CN)4]2−[\mathrm{Ni(CN)_4}]^{2-}[Ni(CN)4​]2− is: (At. No. Ni = 28) (a) d2sp3d^2sp^3d2sp3 (b) dsp2dsp^2dsp2 (c) sp3sp^3sp3 (d) d2sp2d^2sp^2d2sp2

35.The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence (a) In<Tl<Ga<Al\mathrm{In<Tl<Ga<Al}In<Tl<Ga<Al (b) Ga<In<Al<Tl\mathrm{Ga<In<Al<Tl}Ga<In<Al<Tl (c) Al<Ga<In<Tl\mathrm{Al<Ga<In<Tl}Al<Ga<In<Tl (d) Tl<In<Ga<Al\mathrm{Tl<In<Ga<Al}Tl<In<Ga<Al

36.The sum of coordination number and oxidation number of the metal M in the complex [M(en)2(C2O4)]Cl[\mathrm{M(en)_2(C_2O_4)}]\mathrm{Cl}[M(en)2​(C2​O4​)]Cl (where en is ethylenediamine) is: (a) 8 (b) 9 (c) 6 (d) 7

37.Which of the statements given below is incorrect? (a) OF2\mathrm{OF_2}OF2​ is an oxide of fluorine (b) Cl2O7\mathrm{Cl_2O_7}Cl2​O7​ is an anhydride of perchloric acid (c) O3\mathrm{O_3}O3​ molecule is bent (d) ONF is isoelectronic with O2N−\mathrm{O_2N^-}O2​N−

40.The name of complex ion, [Fe(CN)6]3−[\mathrm{Fe(CN)_6}]^{3-}[Fe(CN)6​]3− is: (a) Hexacyanidoferrate (III) ion (b) Hexacyanoiron (III) ion (c) Hexacyanitoferrate (III) ion (d) Tricyanoferrate (III) ion

41.The reducing behaviour of H3PO2\mathrm{H_3PO_2}H3​PO2​ is due to: (a) Presence of two −-−OH groups and one P−-−H bond (b) Presence of one −-−OH group and two P−-−H bonds (c) High electron gain enthalpy of phosphorus (d) High oxidation state of phosphorus

43.In acidic medium, H2O2\mathrm{H_2O_2}H2​O2​ changes Cr2O72−\mathrm{Cr_2O_7^{2-}}Cr2​O72−​ to CrO5\mathrm{CrO_5}CrO5​ which has two (−-−O−-−O−-−) bonds. Oxidation state of Cr in CrO5\mathrm{CrO_5}CrO5​ is: (a) +3 (b) +6 (c) −10-10−10 (d) +5

44.Which of the following orders of ionic radii is correctly represented? (a) Na+>F−>O2−\mathrm{Na^+>F^->O^{2-}}Na+>F−>O2− (b) F−>O2−>Na+\mathrm{F^->O^{2-}>Na^+}F−>O2−>Na+ (c) Al3+>Mg2+>N3−\mathrm{Al^{3+}>Mg^{2+}>N^{3-}}Al3+>Mg2+>N3− (d) H−>H+>H\mathrm{H^->H^+>H}H−>H+>H

45.Be2+\mathrm{Be^{2+}}Be2+ is isoelectronic with which of the following ions? (a) Li+\mathrm{Li^+}Li+ (b) Na+\mathrm{Na^+}Na+ (c) Mg2+\mathrm{Mg^{2+}}Mg2+ (d) H+\mathrm{H^+}H+

46.Magnetic moment 2.83 BM is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28) (a) Ni2+\mathrm{Ni^{2+}}Ni2+ (b) Cr3+\mathrm{Cr^{3+}}Cr3+ (c) Mn2+\mathrm{Mn^{2+}}Mn2+ (d) Ti3+\mathrm{Ti^{3+}}Ti3+

47.The pair of compounds that can exist together is: (a) HgCl2, SnCl2\mathrm{HgCl_2,\ SnCl_2}HgCl2​, SnCl2​ (b) FeCl2, SnCl2\mathrm{FeCl_2,\ SnCl_2}FeCl2​, SnCl2​ (c) FeCl3, KI\mathrm{FeCl_3,\ KI}FeCl3​, KI (d) FeCl3, SnCl2\mathrm{FeCl_3,\ SnCl_2}FeCl3​, SnCl2​

50.Which one of the following species has plane triangular shape? (a) NO3−\mathrm{NO_3^-}NO3−​ (b) NO2−\mathrm{NO_2^-}NO2−​ (c) CO2\mathrm{CO_2}CO2​ (d) N3−\mathrm{N_3^-}N3−​

51.Which of the following molecules has the maximum dipole moment? (a) CH4\mathrm{CH_4}CH4​ (b) NH3\mathrm{NH_3}NH3​ (c) NF3\mathrm{NF_3}NF3​ (d) CO2\mathrm{CO_2}CO2​

53.Reason of lanthanoid contraction is: (a) Increasing nuclear charge (b) Decreasing nuclear charge (c) Decreasing screening effect (d) Negligible screening effect of 'fff' orbitals

54.A magnetic moment of 1.73 BM will be shown by one among the following (a) [Cu(NH3)4]2+[\mathrm{Cu(NH_3)_4}]^{2+}[Cu(NH3​)4​]2+ (b) [Ni(CN)4]2−[\mathrm{Ni(CN)_4}]^{2-}[Ni(CN)4​]2− (c) TiCl4\mathrm{TiCl_4}TiCl4​ (d) [CoCl6]4−[\mathrm{CoCl_6}]^{4-}[CoCl6​]4−

56.Which is the strongest acid in the following? (a) H2SO4\mathrm{H_2SO_4}H2​SO4​ (b) HClO3\mathrm{HClO_3}HClO3​ (c) HClO4\mathrm{HClO_4}HClO4​ (d) H2SO3\mathrm{H_2SO_3}H2​SO3​

57.Which of the following is paramagnetic? (a) CO (b) O2−\mathrm{O_2^-}O2−​ (c) CN−\mathrm{CN^-}CN− (d) NO+\mathrm{NO^+}NO+

58.Which of the following structure is similar to graphite? (a) BN (b) B (c) B4C\mathrm{B_4C}B4​C (d) B2H6\mathrm{B_2H_6}B2​H6​

59.The basic structural unit of silicates is (a) SiO−\mathrm{SiO^-}SiO− (b) SiO44−\mathrm{SiO_4^{4-}}SiO44−​ (c) SiO32−\mathrm{SiO_3^{2-}}SiO32−​ (d) SiO42−\mathrm{SiO_4^{2-}}SiO42−​

60.Which of the following does not give oxygen on heating? (a) KClO3\mathrm{KClO_3}KClO3​ (b) Zn(ClO3)2\mathrm{Zn(ClO_3)_2}Zn(ClO3​)2​ (c) K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​ (d) (NH4)2Cr2O7\mathrm{(NH_4)_2Cr_2O_7}(NH4​)2​Cr2​O7​

61.Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70) (a) Ce2+\mathrm{Ce^{2+}}Ce2+ (b) Sm2+\mathrm{Sm^{2+}}Sm2+ (c) Eu2+\mathrm{Eu^{2+}}Eu2+ (d) Yb2+\mathrm{Yb^{2+}}Yb2+

62.XeF2\mathrm{XeF_2}XeF2​ is isostructural with (a) TeF2\mathrm{TeF_2}TeF2​ (b) ICl2−\mathrm{ICl_2^-}ICl2−​ (c) SbCl3\mathrm{SbCl_3}SbCl3​ (d) BaCl2\mathrm{BaCl_2}BaCl2​

63.An excess of AgNO3\mathrm{AgNO_3}AgNO3​ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be (a) 0.001 (b) 0.002 (c) 0.003 (d) 0.01

64.Which of these is least likely to act as a Lewis base? (a) CO (b) F−\mathrm{F^-}F− (c) BF3\mathrm{BF_3}BF3​ (d) PF3\mathrm{PF_3}PF3​

66.Which of the following is electron-deficient? (a) (CH3)2\mathrm{(CH_3)_2}(CH3​)2​ (b) (SiH3)2\mathrm{(SiH_3)_2}(SiH3​)2​ (c) (BH3)2\mathrm{(BH_3)_2}(BH3​)2​ (d) PH3\mathrm{PH_3}PH3​

67.Which of the following statements about the interstitial compounds is incorrect? (a) They retain metallic conductivity (b) They are chemically reactive (c) They are much harder than the pure metal (d) They have higher melting points than the pure metal

68.Which one of the following molecules contains no π\piπ bond? (a) CO2\mathrm{CO_2}CO2​ (b) H2O\mathrm{H_2O}H2​O (c) SO2\mathrm{SO_2}SO2​ (d) NO2\mathrm{NO_2}NO2​

69.Which of the following is a polar molecule? (a) BF3\mathrm{BF_3}BF3​ (b) SF4\mathrm{SF_4}SF4​ (c) SiF4\mathrm{SiF_4}SiF4​ (d) XeF4\mathrm{XeF_4}XeF4​

72.The species, having bond angles of 120∘120^\circ120∘ is (a) PH3\mathrm{PH_3}PH3​ (b) ClF3\mathrm{ClF_3}ClF3​ (c) NCl3\mathrm{NCl_3}NCl3​ (d) BCl3\mathrm{BCl_3}BCl3​

78.The reason for greater range of oxidation states in actinoids is attributed to (a) The radioactive nature of actinoids (b) Actinoid contraction (c) 5f, 6d and 7s levels having comparable energies (d) 4f and 5d levels being close in energies

81.Which one of the following pairs of species have the same bond order? (a) CO, NO (b) O2\mathrm{O_2}O2​, NO+\mathrm{NO^+}NO+ (c) CN−\mathrm{CN^-}CN−, CO (d) N2\mathrm{N_2}N2​, O2−\mathrm{O_2^-}O2−​

82.Name the gas that can readily decolourises acidified KMnO4\mathrm{KMnO_4}KMnO4​ solution: (a) CO2\mathrm{CO_2}CO2​ (b) SO2\mathrm{SO_2}SO2​ (c) NO2\mathrm{NO_2}NO2​ (d) P2O5\mathrm{P_2O_5}P2​O5​

84.Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field? (a) Na+\mathrm{Na^+}Na+ (b) K+\mathrm{K^+}K+ (c) Rb+\mathrm{Rb^+}Rb+ (d) Li+\mathrm{Li^+}Li+

85.The correct order of atomic radii in group 13 elements is (a) B < Al < In < Ga < Tl (b) B < Ga < Al < Tl < In (c) B < Al < Ga < In < Tl (d) B < Ga < Al < In < Tl

86.In the structure of ClF3\mathrm{ClF_3}ClF3​, the number of lone pairs of electrons on central atom 'Cl' is (a) One (b) Four (c) Two (d) Three

88.Which of the following statements is not true for halogens? (a) All form monobasic oxyacids (b) All but fluorine show positive oxidation states (c) All are oxidizing agents (d) Chlorine has the highest electron-gain enthalpy

89.Which one of the following elements is unable to form MF63−\mathrm{MF_6^{3-}}MF63−​ ion? (a) Ga (b) B (c) Al (d) In

90.Which of the following oxides is most acidic in nature? (a) MgO (b) BaO (c) BeO (d) CaO

91.The type of isomerism shown by the complex [CoCl2(en)2]\mathrm{[CoCl_2(en)_2]}[CoCl2​(en)2​] is (a) Geometrical isomerism (b) Ionization isomerism (c) Coordination isomerism (d) Linkage isomerism

92.Which one of the following ions exhibits d-d transition and paramagnetism as well? (a) CrO42−\mathrm{CrO_4^{2-}}CrO42−​ (b) MnO4−\mathrm{MnO_4^{-}}MnO4−​ (c) Cr2O72−\mathrm{Cr_2O_7^{2-}}Cr2​O72−​ (d) MnO42−\mathrm{MnO_4^{2-}}MnO42−​

93.Iron carbonyl, Fe(CO)5\mathrm{Fe(CO)_5}Fe(CO)5​ is (a) Tetranuclear (b) Trinuclear (c) Mononuclear (d) Dinuclear

95.The geometry and magnetic behaviour of the complex [Ni(CO)4]\mathrm{[Ni(CO)_4]}[Ni(CO)4​] are (a) Square planar geometry and diamagnetic (b) Square planar geometry and paramagnetic (c) Tetrahedral geometry and diamagnetic (d) Tetrahedral geometry and paramagnetic

98.Consider the following species: CN+\mathrm{CN^+}CN+, CN−\mathrm{CN^-}CN−, NO and CN. Which one of these will have the highest bond order? (a) NO (b) CN+\mathrm{CN^+}CN+ (c) CN−\mathrm{CN^-}CN− (d) CN

100.4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is (a) 6p > 5f > 5p > 4d (b) 6p > 5f > 4d > 5p (c) 5f > 6p > 4d > 5p (d) 5f > 6p > 5p > 4d

101.For the second period elements the correct increasing order of first ionisation enthalpy is: (a) Li < B < Be < C < O < N < F < Ne (b) Li < B < Be < C < N < O < F < Ne (c) Li < Be < B < C < O < N < F < Ne (d) Li < Be < B < C < N < O < F < Ne

103.Which of the following diatomic molecular species has only π\piπ bonds according to Molecular Orbital Theory? (a) N2\mathrm{N_2}N2​ (b) C2\mathrm{C_2}C2​ (c) Be2\mathrm{Be_2}Be2​ (d) O2\mathrm{O_2}O2​

104.Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region? (a) Balmer series (b) Paschen series (c) Brackett series (d) Lyman series

105.Which of the following is an amphoteric hydroxide? (a) Ca(OH)2\mathrm{Ca(OH)_2}Ca(OH)2​ (b) Mg(OH)2\mathrm{Mg(OH)_2}Mg(OH)2​ (c) Be(OH)2\mathrm{Be(OH)_2}Be(OH)2​ (d) Sr(OH)2\mathrm{Sr(OH)_2}Sr(OH)2​

106.What is the correct electronic configuration of the central atom in K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]}K4​[Fe(CN)6​] based on crystal field theory? (a) t2g6eg0t_{2g}^{6}e_g^{0}t2g6​eg0​ (b) e3t23e^{3}t_2^{3}e3t23​ (c) e4t22e^{4}t_2^{2}e4t22​ (d) t2g4eg2t_{2g}^{4}e_g^{2}t2g4​eg2​

110.Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is: (a) Mg (b) Ca (c) Sr (d) Be

111.Which of the following species is not stable? (a) [GeCl6]2−\mathrm{[GeCl_6]^{2-}}[GeCl6​]2− (b) [Sn(OH)6]2−\mathrm{[Sn(OH)_6]^{2-}}[Sn(OH)6​]2− (c) [SiCl6]2−\mathrm{[SiCl_6]^{2-}}[SiCl6​]2− (d) [SiF6]2−\mathrm{[SiF_6]^{2-}}[SiF6​]2−

113.The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals. (a) Calcium (b) Potassium (c) Iron (d) Copper

115.The number of protons, neutrons and electrons in 71175Lu\mathrm{{}^{175}_{71}Lu}71175​Lu, respectively, are: (a) 71, 71 and 104 (b) 175, 104 and 71 (c) 71, 104 and 71 (d) 104, 71 and 71

116.Identify a molecule which does not exist. (a) C2\mathrm{C_2}C2​ (b) O2\mathrm{O_2}O2​ (c) He2\mathrm{He_2}He2​ (d) Li2\mathrm{Li_2}Li2​

118.The calculated spin only magnetic moment of Cr2+\mathrm{Cr^{2+}}Cr2+ ion is: (a) 5.92 BM (b) 2.84 BM (c) 3.87 BM (d) 4.90 BM

121.BF3\mathrm{BF_3}BF3​ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are (a) sp2sp^2sp2 and 6 (b) sp2sp^2sp2 and 8 (c) sp3sp^3sp3 and 4 (d) sp3sp^3sp3 and 6

122.The structure of beryllium chloride in solid state and vapour phase, are: (a) Dimer and Linear, respectively (b) Chain in both (c) Chain and dimer, respectively (d) Linear in both

123.Zr (Z=40Z = 40Z=40) and Hf (Z=72Z = 72Z=72) have similar atomic and ionic radii because of: (a) lanthanoid contraction (b) having similar chemical properties (c) belonging to same group (d) diagonal relationship

125.Ethylenediaminetetraacetate (EDTA) ion is (a) Bidentate ligand with two "N" donor atoms (b) Tridentate ligand with three "N" donor atoms (c) Hexadentate ligand with four "O" and two "N" donor atoms (d) Unidentate ligand

126.Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is: (a) Magnesium chloride (b) Beryllium chloride (c) Calcium chloride (d) Strontium chloride

131.Which of the following molecules is non-polar in nature? (a) SbCl5\mathrm{SbCl_5}SbCl5​ (b) NO2\mathrm{NO_2}NO2​ (c) POCl3\mathrm{POCl_3}POCl3​ (d) CH2O\mathrm{CH_2O}CH2​O

133.From the following pairs of ions which one is not an iso-electronic pair? (a) Mn2+,Fe3+\mathrm{Mn^{2+}, Fe^{3+}}Mn2+,Fe3+ (b) Fe2+,Mn2+\mathrm{Fe^{2+}, Mn^{2+}}Fe2+,Mn2+ (c) O2−,F−\mathrm{O^{2-}, F^{-}}O2−,F− (d) Na+,Mg2+\mathrm{Na^{+}, Mg^{2+}}Na+,Mg2+

135.Gadolinium has a low value of third ionisation enthalpy because of (a) small size (b) high exchange enthalpy (c) high electronegativity (d) high basic character

137.The IUPAC name of an element with atomic number 119 is (a) ununennium (b) unnilennium (c) unununnium (d) ununoctium

138.Choose the correct statement: (a) Diamond and graphite have two dimensional network. (b) Diamond is covalent and graphite is ionic. (c) Diamond is sp3sp^3sp3 hybridised and graphite is sp2sp^2sp2 hybridized. (d) Both diamond and graphite are used as dry lubricants.

The product obtained as a result of a reaction of nitrogen with $\mathrm{CaC_2}$ is: (a) $\mathrm{Ca(CN)_2}$ (b) $\mathrm{CaCN}$ (c) $\mathrm{CaCN_3}$ (d) $\mathrm{Ca_2CN}$

Which is the correct statement for the given acids? (a) Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid. (b) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid. (c) Both are triprotic acids. (d) Both are diprotic acids.

Which one of the following compounds shows the presence of intramolecular hydrogen bond? (a) $\mathrm{H_2O_2}$ (b) HCN (c) Cellulose (d) Concentrated acetic acid

Boric acid is an acid because its molecule (a) contains replaceable $\mathrm{H^+}$ ion (b) gives up a proton (c) accepts $\mathrm{OH^-}$ from water releasing proton (d) combines with proton from water molecule

$\mathrm{AlF_3}$ is soluble in HF only in presence of KF. It is due to the formation of (a) $\mathrm{K_3[AlF_3H_3]}$ (b) $\mathrm{K_3[AlF_6]}$ (c) $\mathrm{AlH_3}$ (d) $\mathrm{K[AlF_3H]}$

The suspension of slaked lime in water is known as (a) limewater (b) quicklime (c) milk of lime (d) aqueous solution of slaked lime

Which of the following fluoro-compounds is most likely to behave as a Lewis base? (a) $\mathrm{BF_3}$ (b) $\mathrm{PF_3}$ (c) $\mathrm{CF_4}$ (d) $\mathrm{SiF_4}$

In context with beryllium, which one of the following statements is incorrect? (a) It is rendered passive by nitric acid. (b) It forms $\mathrm{Be_2C}$ . (c) Its salts rarely hydrolyze. (d) Its hydride is electron-deficient and polymeric.

Which of the following pairs of d-orbitals will have electron density along the axes? (a) $d_{z^2},\ d_{xz}$ (b) $d_{xz},\ d_{yz}$ (c) $d_{z^2},\ d_{x^2-y^2}$ (d) $d_{xy},\ d_{x^2-y^2}$

The correct geometry and hybridization for $\mathrm{XeF_4}$ are (a) octahedral, $sp^3d^2$ (b) trigonal bipyramidal, $sp^3d$ (c) planar triangle, $sp^3d^3$ (d) square planar, $sp^3d^2$

Jahn-Teller effect is not observed in high spin complexes of (a) $d^7$ (b) $d^8$ (c) $d^4$ (d) $d^9$

In which of the following pairs, both the species are not isostructural? (a) $\mathrm{XeF_4,\ XeO_4}$ (b) $\mathrm{SiCl_4,\ PCl_4^+}$ (c) diamond, silicon carbide (d) $\mathrm{NH_3,\ PH_3}$

On heating which of the following releases $\mathrm{CO_2}$ most easily? (a) $\mathrm{CaCO_3}$ (b) $\mathrm{K_2CO_3}$ (c) $\mathrm{Na_2CO_3}$ (d) $\mathrm{MgCO_3}$

Number of possible isomers for the complex $[\mathrm{Co(en)_2Cl_2}]\mathrm{Cl}$ will be: (en = ethylenediamine) (a) 4 (b) 2 (c) 1 (d) 3

The hybridization involved in complex $[\mathrm{Ni(CN)_4}]^{2-}$ is: (At. No. Ni = 28) (a) $d^2sp^3$ (b) $dsp^2$ (c) $sp^3$ (d) $d^2sp^2$

The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence (a) $\mathrm{In<Tl<Ga<Al}$ (b) $\mathrm{Ga<In<Al<Tl}$ (c) $\mathrm{Al<Ga<In<Tl}$ (d) $\mathrm{Tl<In<Ga<Al}$

The sum of coordination number and oxidation number of the metal M in the complex $[\mathrm{M(en)_2(C_2O_4)}]\mathrm{Cl}$ (where en is ethylenediamine) is: (a) 8 (b) 9 (c) 6 (d) 7

Which of the statements given below is incorrect? (a) $\mathrm{OF_2}$ is an oxide of fluorine (b) $\mathrm{Cl_2O_7}$ is an anhydride of perchloric acid (c) $\mathrm{O_3}$ molecule is bent (d) ONF is isoelectronic with $\mathrm{O_2N^-}$

The name of complex ion, $[\mathrm{Fe(CN)_6}]^{3-}$ is: (a) Hexacyanidoferrate (III) ion (b) Hexacyanoiron (III) ion (c) Hexacyanitoferrate (III) ion (d) Tricyanoferrate (III) ion

The reducing behaviour of $\mathrm{H_3PO_2}$ is due to: (a) Presence of two $-$ OH groups and one P $-$ H bond (b) Presence of one $-$ OH group and two P $-$ H bonds (c) High electron gain enthalpy of phosphorus (d) High oxidation state of phosphorus

In acidic medium, $\mathrm{H_2O_2}$ changes $\mathrm{Cr_2O_7^{2-}}$ to $\mathrm{CrO_5}$ which has two ( $-$ O $-$ O $-$ ) bonds. Oxidation state of Cr in $\mathrm{CrO_5}$ is: (a) +3 (b) +6 (c) $-10$ (d) +5

Which of the following orders of ionic radii is correctly represented? (a) $\mathrm{Na^+>F^->O^{2-}}$ (b) $\mathrm{F^->O^{2-}>Na^+}$ (c) $\mathrm{Al^{3+}>Mg^{2+}>N^{3-}}$ (d) $\mathrm{H^->H^+>H}$

$\mathrm{Be^{2+}}$ is isoelectronic with which of the following ions? (a) $\mathrm{Li^+}$ (b) $\mathrm{Na^+}$ (c) $\mathrm{Mg^{2+}}$ (d) $\mathrm{H^+}$

Magnetic moment 2.83 BM is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28) (a) $\mathrm{Ni^{2+}}$ (b) $\mathrm{Cr^{3+}}$ (c) $\mathrm{Mn^{2+}}$ (d) $\mathrm{Ti^{3+}}$

The pair of compounds that can exist together is: (a) $\mathrm{HgCl_2,\ SnCl_2}$ (b) $\mathrm{FeCl_2,\ SnCl_2}$ (c) $\mathrm{FeCl_3,\ KI}$ (d) $\mathrm{FeCl_3,\ SnCl_2}$

Which one of the following species has plane triangular shape? (a) $\mathrm{NO_3^-}$ (b) $\mathrm{NO_2^-}$ (c) $\mathrm{CO_2}$ (d) $\mathrm{N_3^-}$

Which of the following molecules has the maximum dipole moment? (a) $\mathrm{CH_4}$ (b) $\mathrm{NH_3}$ (c) $\mathrm{NF_3}$ (d) $\mathrm{CO_2}$

Reason of lanthanoid contraction is: (a) Increasing nuclear charge (b) Decreasing nuclear charge (c) Decreasing screening effect (d) Negligible screening effect of ' $f$ ' orbitals

A magnetic moment of 1.73 BM will be shown by one among the following (a) $[\mathrm{Cu(NH_3)_4}]^{2+}$ (b) $[\mathrm{Ni(CN)_4}]^{2-}$ (c) $\mathrm{TiCl_4}$ (d) $[\mathrm{CoCl_6}]^{4-}$

Which is the strongest acid in the following? (a) $\mathrm{H_2SO_4}$ (b) $\mathrm{HClO_3}$ (c) $\mathrm{HClO_4}$ (d) $\mathrm{H_2SO_3}$

Which of the following is paramagnetic? (a) CO (b) $\mathrm{O_2^-}$ (c) $\mathrm{CN^-}$ (d) $\mathrm{NO^+}$

Which of the following structure is similar to graphite? (a) BN (b) B (c) $\mathrm{B_4C}$ (d) $\mathrm{B_2H_6}$

The basic structural unit of silicates is (a) $\mathrm{SiO^-}$ (b) $\mathrm{SiO_4^{4-}}$ (c) $\mathrm{SiO_3^{2-}}$ (d) $\mathrm{SiO_4^{2-}}$

Which of the following does not give oxygen on heating? (a) $\mathrm{KClO_3}$ (b) $\mathrm{Zn(ClO_3)_2}$ (c) $\mathrm{K_2Cr_2O_7}$ (d) $\mathrm{(NH_4)_2Cr_2O_7}$

Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70) (a) $\mathrm{Ce^{2+}}$ (b) $\mathrm{Sm^{2+}}$ (c) $\mathrm{Eu^{2+}}$ (d) $\mathrm{Yb^{2+}}$

$\mathrm{XeF_2}$ is isostructural with (a) $\mathrm{TeF_2}$ (b) $\mathrm{ICl_2^-}$ (c) $\mathrm{SbCl_3}$ (d) $\mathrm{BaCl_2}$

An excess of $\mathrm{AgNO_3}$ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be (a) 0.001 (b) 0.002 (c) 0.003 (d) 0.01

Which of these is least likely to act as a Lewis base? (a) CO (b) $\mathrm{F^-}$ (c) $\mathrm{BF_3}$ (d) $\mathrm{PF_3}$

Which of the following is electron-deficient? (a) $\mathrm{(CH_3)_2}$ (b) $\mathrm{(SiH_3)_2}$ (c) $\mathrm{(BH_3)_2}$ (d) $\mathrm{PH_3}$

Which of the following statements about the interstitial compounds is incorrect? (a) They retain metallic conductivity (b) They are chemically reactive (c) They are much harder than the pure metal (d) They have higher melting points than the pure metal

Which one of the following molecules contains no $\pi$ bond? (a) $\mathrm{CO_2}$ (b) $\mathrm{H_2O}$ (c) $\mathrm{SO_2}$ (d) $\mathrm{NO_2}$

Which of the following is a polar molecule? (a) $\mathrm{BF_3}$ (b) $\mathrm{SF_4}$ (c) $\mathrm{SiF_4}$ (d) $\mathrm{XeF_4}$

The species, having bond angles of $120^\circ$ is
(a) $\mathrm{PH_3}$ (b) $\mathrm{ClF_3}$ (c) $\mathrm{NCl_3}$ (d) $\mathrm{BCl_3}$

The reason for greater range of oxidation states in actinoids is attributed to
(a) The radioactive nature of actinoids (b) Actinoid contraction (c) 5f, 6d and 7s levels having comparable energies (d) 4f and 5d levels being close in energies

Which one of the following pairs of species have the same bond order?
(a) CO, NO (b) $\mathrm{O_2}$ , $\mathrm{NO^+}$ (c) $\mathrm{CN^-}$ , CO (d) $\mathrm{N_2}$ , $\mathrm{O_2^-}$

Name the gas that can readily decolourises acidified $\mathrm{KMnO_4}$ solution:
(a) $\mathrm{CO_2}$ (b) $\mathrm{SO_2}$ (c) $\mathrm{NO_2}$ (d) $\mathrm{P_2O_5}$

Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
(a) $\mathrm{Na^+}$ (b) $\mathrm{K^+}$ (c) $\mathrm{Rb^+}$ (d) $\mathrm{Li^+}$

The correct order of atomic radii in group 13 elements is
(a) B < Al < In < Ga < Tl (b) B < Ga < Al < Tl < In (c) B < Al < Ga < In < Tl (d) B < Ga < Al < In < Tl

In the structure of $\mathrm{ClF_3}$ , the number of lone pairs of electrons on central atom 'Cl' is
(a) One (b) Four (c) Two (d) Three

Which of the following statements is not true for halogens?
(a) All form monobasic oxyacids (b) All but fluorine show positive oxidation states (c) All are oxidizing agents (d) Chlorine has the highest electron-gain enthalpy

Which one of the following elements is unable to form $\mathrm{MF_6^{3-}}$ ion?
(a) Ga (b) B (c) Al (d) In

Which of the following oxides is most acidic in nature?
(a) MgO (b) BaO (c) BeO (d) CaO

The type of isomerism shown by the complex $\mathrm{[CoCl_2(en)_2]}$ is
(a) Geometrical isomerism (b) Ionization isomerism (c) Coordination isomerism (d) Linkage isomerism

Which one of the following ions exhibits d-d transition and paramagnetism as well?
(a) $\mathrm{CrO_4^{2-}}$ (b) $\mathrm{MnO_4^{-}}$ (c) $\mathrm{Cr_2O_7^{2-}}$ (d) $\mathrm{MnO_4^{2-}}$

Iron carbonyl, $\mathrm{Fe(CO)_5}$ is
(a) Tetranuclear (b) Trinuclear (c) Mononuclear (d) Dinuclear

The geometry and magnetic behaviour of the complex $\mathrm{[Ni(CO)_4]}$ are
(a) Square planar geometry and diamagnetic (b) Square planar geometry and paramagnetic (c) Tetrahedral geometry and diamagnetic (d) Tetrahedral geometry and paramagnetic

Consider the following species: $\mathrm{CN^+}$ , $\mathrm{CN^-}$ , NO and CN. Which one of these will have the highest bond order?
(a) NO (b) $\mathrm{CN^+}$ (c) $\mathrm{CN^-}$ (d) CN

4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is
(a) 6p > 5f > 5p > 4d (b) 6p > 5f > 4d > 5p (c) 5f > 6p > 4d > 5p (d) 5f > 6p > 5p > 4d

For the second period elements the correct increasing order of first ionisation enthalpy is:
(a) Li < B < Be < C < O < N < F < Ne (b) Li < B < Be < C < N < O < F < Ne (c) Li < Be < B < C < O < N < F < Ne (d) Li < Be < B < C < N < O < F < Ne

Which of the following diatomic molecular species has only $\pi$ bonds according to Molecular Orbital Theory?
(a) $\mathrm{N_2}$ (b) $\mathrm{C_2}$ (c) $\mathrm{Be_2}$ (d) $\mathrm{O_2}$

Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region?
(a) Balmer series (b) Paschen series (c) Brackett series (d) Lyman series

Which of the following is an amphoteric hydroxide?
(a) $\mathrm{Ca(OH)_2}$ (b) $\mathrm{Mg(OH)_2}$ (c) $\mathrm{Be(OH)_2}$ (d) $\mathrm{Sr(OH)_2}$

What is the correct electronic configuration of the central atom in $\mathrm{K_4[Fe(CN)_6]}$ based on crystal field theory?
(a) $t_{2g}^{6}e_g^{0}$ (b) $e^{3}t_2^{3}$ (c) $e^{4}t_2^{2}$ (d) $t_{2g}^{4}e_g^{2}$

Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is:
(a) Mg (b) Ca (c) Sr (d) Be

Which of the following species is not stable?
(a) $\mathrm{[GeCl_6]^{2-}}$ (b) $\mathrm{[Sn(OH)_6]^{2-}}$ (c) $\mathrm{[SiCl_6]^{2-}}$ (d) $\mathrm{[SiF_6]^{2-}}$

The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.
(a) Calcium (b) Potassium (c) Iron (d) Copper

The number of protons, neutrons and electrons in $\mathrm{{}^{175}_{71}Lu}$ , respectively, are:
(a) 71, 71 and 104 (b) 175, 104 and 71 (c) 71, 104 and 71 (d) 104, 71 and 71

Identify a molecule which does not exist.
(a) $\mathrm{C_2}$ (b) $\mathrm{O_2}$ (c) $\mathrm{He_2}$ (d) $\mathrm{Li_2}$

The calculated spin only magnetic moment of $\mathrm{Cr^{2+}}$ ion is:
(a) 5.92 BM (b) 2.84 BM (c) 3.87 BM (d) 4.90 BM

$\mathrm{BF_3}$ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are
(a) $sp^2$ and 6 (b) $sp^2$ and 8 (c) $sp^3$ and 4 (d) $sp^3$ and 6

The structure of beryllium chloride in solid state and vapour phase, are:
(a) Dimer and Linear, respectively (b) Chain in both (c) Chain and dimer, respectively (d) Linear in both

Zr ( $Z = 40$ ) and Hf ( $Z = 72$ ) have similar atomic and ionic radii because of:
(a) lanthanoid contraction (b) having similar chemical properties (c) belonging to same group (d) diagonal relationship

Ethylenediaminetetraacetate (EDTA) ion is
(a) Bidentate ligand with two "N" donor atoms (b) Tridentate ligand with three "N" donor atoms (c) Hexadentate ligand with four "O" and two "N" donor atoms (d) Unidentate ligand

Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is:
(a) Magnesium chloride (b) Beryllium chloride (c) Calcium chloride (d) Strontium chloride

Which of the following molecules is non-polar in nature?
(a) $\mathrm{SbCl_5}$ (b) $\mathrm{NO_2}$ (c) $\mathrm{POCl_3}$ (d) $\mathrm{CH_2O}$

From the following pairs of ions which one is not an iso-electronic pair?
(a) $\mathrm{Mn^{2+}, Fe^{3+}}$ (b) $\mathrm{Fe^{2+}, Mn^{2+}}$ (c) $\mathrm{O^{2-}, F^{-}}$ (d) $\mathrm{Na^{+}, Mg^{2+}}$

Gadolinium has a low value of third ionisation enthalpy because of
(a) small size (b) high exchange enthalpy (c) high electronegativity (d) high basic character

The IUPAC name of an element with atomic number 119 is
(a) ununennium (b) unnilennium (c) unununnium (d) ununoctium

Choose the correct statement:
(a) Diamond and graphite have two dimensional network. (b) Diamond is covalent and graphite is ionic. (c) Diamond is $sp^3$ hybridised and graphite is $sp^2$ hybridized. (d) Both diamond and graphite are used as dry lubricants.

The IUPAC name of the complex $\mathrm{[Ag(H_2O)_2][Ag(CN)_2]}$ is:
(a) dicyanidosilver(II) diaquaargentate(II) (b) diaquasilver(II) dicyanidoargentate(II) (c) dicyanidosilver(I) diaquaargentate(I) (d) diaquasilver(I) dicyanidoargentate(I)