Atomic Structure, Periodicity, and Chemical Bonding

(d) Hund's rule.

Hund's rule states that orbitals of equal energy (the three 2p orbitals here) are first filled singly with parallel spins before pairing begins. Nitrogen's 2p³ configuration therefore has three unpaired electrons, one in each 2p orbital.

(b) AB₂

Element A has configuration 2,8,2 → it loses 2 electrons to form A²⁺. Element B has configuration 2,8,7 → it gains 1 electron to form B⁻.

To balance charges: one A²⁺ needs two B⁻, giving the formula AB₂.

(d) 1/25

By Graham's law, rate of diffusion is inversely proportional to the square root of density:

$\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}$

Given $\frac{r_A}{r_B} = 5$:

$5 = \sqrt{\frac{d_B}{d_A}} \Rightarrow \frac{d_B}{d_A} = 25 \Rightarrow \frac{d_A}{d_B} = \frac{1}{25}$

(d) $6\times9650$

$\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$, so 1 mole of Cu²⁺ needs 2 faradays, i.e. $2\times96500$ C. For 0.3 mole: $Q=0.3\times2\times96500=57900\ \text{C}=6\times9650\ \text{C}$

(d) law of conservation of mass

A balanced chemical equation has an equal number of atoms of each element on both sides, which means the total mass of the reactants equals the total mass of the products — this is exactly what the law of conservation of mass states.

(b) loss of electrons

Oxidation is defined as the loss of electrons (or an increase in oxidation number); the species that loses electrons is oxidised.

(b) 4.0 m mole

Reason:
For Na₂CO₃, the neutralization reaction with HCl is:

$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$

Moles of Na₂CO₃ in 10 ml of 0.2 M solution:

$= 0.2 \times \frac{10}{1000} = 0.002 \text{ mol} = 2.0 \text{ m mol}$

Since 1 mole of Na₂CO₃ requires 2 moles of HCl,

$\text{HCl required} = 2 \times 2.0 = 4.0 \text{ m mol}$

(c) free energy decreases

A reaction is spontaneous when the change in Gibbs free energy is negative (ΔG < 0), i.e. free energy decreases. Low activation energy only affects the rate, not the spontaneity, and a decrease in entropy alone does not guarantee spontaneity (ΔG = ΔH − TΔS must be negative overall).

(c) decreasing the activation energy

A catalyst provides an alternative reaction pathway with a lower activation energy, so a larger fraction of molecules can cross the energy barrier and the reaction proceeds faster. It does not change the free energy, heat of reaction, or the average speed of the molecules.

(b) 208 gm

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

One mole of butane needs $\dfrac{13}{2}=6.5$ mol of $O_2$. Mass $=6.5\times32=208$ gm.

(b) increases with dilution

The degree of dissociation (α) of a weak electrolyte increases as the solution is diluted. This is because dilution lowers the concentration of ions, so by Le Chatelier's principle the equilibrium shifts towards more ionisation to restore balance. This relationship is expressed by Ostwald's dilution law: α ∝ 1/√C.

For the reaction N₂ + O₂ ⇋ 2 NO, the equilibrium constant expression is:

$K_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]}$

(c) $H_2O$ and $OH^-$

A conjugate acid-base pair differs by exactly one proton. $H_2O$ loses $H^+$ to give $OH^-$, so $H_2O$/$OH^-$ is a pair. The other pair in this reaction is $NH_3$ (base) and $NH_4^+$ (conjugate acid).

(d) 1 N

$H_2SO_4$: $0.5\,M \times 2 = 1\,N$, so 100 mL contains $1 \times 100 = 100$ meq. HCl: $1\,M = 1\,N$, so 100 mL contains 100 meq.

$N = \frac{100 + 100}{100 + 100} = 1\ N$

(c) 7.0

At $10^{-12}$ M the acid is far less than the $10^{-7}$ M of $H^+$ from water, so water dominates. Solving $x(x + 10^{-12}) = 10^{-14}$ gives $x \approx 1.0\times10^{-7}$ M, so pH $\approx 7$ (slightly below 7). An acid solution cannot have pH 12 or a negative pH here.

(a) CO

CO is a neutral oxide: it shows neither acidic nor basic properties and does not react with water, acids or alkalis under ordinary conditions. $SnO_2$ and ZnO are amphoteric, and $SiO_2$ is acidic.

(a) $HNO_3$ and HCl

Aqua regia is a mixture of concentrated HCl and concentrated $HNO_3$ in a 3:1 ratio by volume. It forms nascent chlorine and $NOCl$, which dissolve noble metals such as gold and platinum.

(d) $N_2$

Hot CuO oxidises ammonia to nitrogen, being itself reduced to copper:

$3CuO + 2NH_3 \xrightarrow{\Delta} 3Cu + N_2 + 3H_2O$

(b) Mg and Ca

Alkaline earth metals are the Group 2 elements: Be, Mg, Ca, Sr, Ba and Ra. Na and K are alkali metals, Cu and Ag are coinage (transition) metals, and Al and Fe are not Group 2.

(c) flux

Limestone decomposes to CaO, which combines with the acidic gangue silica to form fusible calcium silicate:

$CaCO_3 \rightarrow CaO + CO_2,\qquad CaO + SiO_2 \rightarrow CaSiO_3$

The calcium silicate is the slag; limestone itself is the flux.

(b) 750

Pure gold is 24 carat. Fineness is parts of gold per 1000 parts of alloy:

$\frac{18}{24} \times 1000 = 750$

(b) phenyl hydroxylamine

Reduction in a neutral medium ($Zn/NH_4Cl$) stops at the hydroxylamine stage:

$C_6H_5NO_2 + 4[H] \rightarrow C_6H_5NHOH + H_2O$

Aniline needs acidic medium (Sn/HCl), while azoxybenzene and hydrazobenzene form in alkaline medium.

(a) $FCH_2COOH$

The halogen withdraws electrons by the $-I$ effect, stabilising the carboxylate ion and increasing acidity. Fluorine is the most electronegative, so its $-I$ effect is strongest, giving the order $FCH_2COOH > ClCH_2COOH > BrCH_2COOH > ICH_2COOH$.

(d) carbon readily forms covalent bonds with other carbon atoms

This property is catenation: carbon atoms bond strongly to each other in long chains, branched chains and rings, with single, double and triple bonds. Together with its tetravalence and isomerism, this gives the very large number of carbon compounds.

(d) 6

Place two Cl atoms on n-butane, $C-C-C-C$ (carbons 1 to 4):

• 1,1-dichlorobutane
• 1,2-dichlorobutane
• 1,3-dichlorobutane
• 1,4-dichlorobutane
• 2,2-dichlorobutane
• 2,3-dichlorobutane

Other positions repeat these by symmetry, so there are 6 structural isomers.

(c) HCHO

The Cannizzaro reaction occurs for aldehydes with no $\alpha$-hydrogen, in concentrated alkali, giving an alcohol and a carboxylate:

$2HCHO + NaOH \rightarrow CH_3OH + HCOONa$

$C_2H_5CHO$ has $\alpha$-hydrogens and gives aldol condensation instead. $CH_3COOH$ and $CH_3OCH_3$ are not aldehydes.

(d) $CH_3NH_2$

Only primary amines react with $CHCl_3$ and alcoholic KOH on heating to give foul-smelling isocyanide:

$CH_3NH_2 + CHCl_3 + 3KOH \rightarrow CH_3NC + 3KCl + 3H_2O$

Secondary, tertiary and quaternary ions lack the two N-H hydrogens needed.

(b) addition polymer

PVC (polyvinyl chloride) is formed by addition polymerisation of vinyl chloride from a single monomer, with no loss of small molecules:

$n\,CH_2{=}CHCl \rightarrow [-CH_2-CHCl-]_n$

It is a homopolymer and is thermoplastic.

(b) glucose and fructose

Table sugar is sucrose. Hydrolysis with dilute acid or invertase gives an equimolar mixture of D-glucose and D-fructose (invert sugar):

$C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6$

(c) methane

Aluminium carbide is a methanide and reacts with water to give methane:

$Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$

Ethyne comes from calcium carbide.