Classification of Elements

p-block elements (groups 13 to 18):

1. Their last electron enters the p subshell (valence configuration $ns^2 np^{1-6}$).
2. They include metals, non-metals and metalloids (and the inert gases in group 18); most non-metals form covalent compounds.

d-block elements (groups 3 to 12, transition elements):

1. Their last electron enters the d subshell ($(n-1)d$).
2. They are metals, mostly hard with high melting points (mercury is a liquid exception), show variable valency, form coloured compounds and are often used as catalysts (e.g. Fe, Cu).

s-block elements (groups 1 and 2):

1. Their last electron enters the s subshell (valence configuration $ns^1$ or $ns^2$).
2. They are very reactive metals (except H and He) that readily lose electrons to form positive ions.

p-block elements (groups 13 to 18):

1. Their last electron enters the p subshell (valence configuration $ns^2 np^{1-6}$).
2. They include metals, non-metals and metalloids, and most non-metals form compounds by gaining or sharing electrons.

(iii) p-block. In group 15 (VA) the last electron enters the p subshell ($ns^2 np^3$), so these elements belong to the p-block.

Modern periodic law: The physical and chemical properties of elements are the periodic functions of their atomic numbers.

Groups:

• Metal group: Group 1 (IA), the alkali metals (Li, Na, K).
• Non-metal group: Group 17 (VIIA), the halogens (F, Cl, Br).

Change in atomic size: Atomic size decreases from left to right in a period. The number of shells stays the same, but the nuclear charge (number of protons) increases, so the nucleus pulls the electrons more strongly.

(c) Na$^+$. Sodium has 11 electrons; losing one gives 10. The others: O$^-$ has 8 + 1 = 9, Mg$^+$ has 12 - 1 = 11, N$^{--}$ has 7 + 2 = 9.

Chlorine has 7 valence electrons and 3 shells; magnesium has 2 valence electrons and 3 shells. The group number comes from the valence electrons and the period from the number of shells.

(d) Z. Z has three shells with 7 electrons in the outermost shell (chlorine), so it is a halogen that needs only one more electron to complete its octet and gains it very readily. W and X have too few valence electrons to be reactive non-metals, and Y (6 valence electrons) needs two electrons, so Z is comparatively the most reactive non-metal.

Atomic number 19 (potassium) has 19 electrons, filled in the order 1s, 2s, 2p, 3s, 3p, 4s:

(i) The element has 2 + 8 + 8 + 1 = 19 electrons (potassium):

(ii) A has only one valence electron, which is easily lost to form $A^+$ with the stable argon configuration (2, 8, 8). To form $A^{++}$ a second electron must be removed from this stable, complete inner shell, which needs a very large amount of energy. So the formation of $A^{++}$ is very difficult.

(d) Iron and copper. Both are d-block (transition) elements of groups 3 to 12. Lithium and sodium are s-block elements, aluminium is a p-block element and krypton is a noble gas, so none of them is a transition element.

Periodic table: A systematic arrangement of all known elements in horizontal rows (periods) and vertical columns (groups) in the order of increasing atomic number, so that elements with similar properties fall in the same group.

Modern periodic law: The physical and chemical properties of elements are the periodic functions of their atomic numbers.

Placement:

• Alkali metals are kept in group 1 (IA).
• Halogens are kept in group 17 (VIIA).

The first atom has 3 electrons (2, 1), so it is lithium (valency 1). The second atom has 9 electrons (2, 7), so it is fluorine (valency 1). Lithium loses one electron and fluorine gains it:

LiF (lithium fluoride).

The elements are Q = Na ($3s^1$), X = Mg ($3s^2$), Y = Al ($3s^2 3p^1$), Z = Cl ($3s^2 3p^5$).

(i) Q (sodium) forms the strong alkali NaOH, because it is the most metallic and most reactive metal.

(ii) Metallic character decreases from left to right in a period. In decreasing order of metallic character the elements are Q > X > Y > Z (Na > Mg > Al > Cl), so in order of lesser metallic character they run Z < Y < X < Q (Cl < Al < Mg < Na).

(iii) Z (chlorine) has the smallest atomic radius. All four are in the same period (3 shells), but Z has the largest nuclear charge (17 protons), which pulls the electrons closest to the nucleus.

Modern table is more advanced than Mendeleev's (any two):

1. It is based on atomic number, not atomic mass, so there are no anomalies such as Ar and K, or Te and I.
2. Isotopes have the same atomic number and so find a place in the same position; the table is divided into s, p, d and f blocks based on electronic configuration.

Differences between s-block and p-block:

(a) s-block. Group IIA (2) elements have the valence configuration $ns^2$, so the last electron enters the s subshell.

Advantages of the modern periodic table over Mendeleev's (any four):

1. It is based on atomic number, which is a more fundamental property than atomic mass.
2. The anomalous positions (e.g. Ar before K, Te before I) of Mendeleev's table are removed.
3. Isotopes have the same atomic number, so they are placed in the same position without separate places.
4. It is based on electronic configuration, so it is divided into s, p, d and f blocks.
5. The noble gases have a proper place (group 18) and transition elements and inner transition elements are placed clearly.
6. Elements are easier to study because the periods and groups are clearly defined.

Here X = Na (11 electrons), Y = Cl (17 electrons), Z = Ar (18 electrons).

(i) X has the configuration ending in $3s^1$, so its last electron enters the s subshell: s-block.

(ii) Z has a complete outermost shell (8 electrons), so its valency is zero.

(iii) Sodium (valency 1) reacts with chlorine (valency 1):

(iii) KCl. A metal with one valence electron is in group 1 and has valency 1, which rules out MgCl$_2$ and CaCl$_2$ (group 2 metals). The halogen of the third period is chlorine, so NaBr (bromine is in period 4) is out. The compound is potassium chloride, KCl.

The elements are the third period: Na, Mg, Al, Si, P, S, Cl, Ar.

• Most reactive metal: Na (1s², 2s² 2p⁶, 3s¹). It has 1 valence electron, so its valency is 1.
• Most reactive non-metal: Cl (1s², 2s² 2p⁶, 3s² 3p⁵). It has 7 valence electrons and needs one more, so its valency is 1.

(i) Modern periodic law: The physical and chemical properties of elements are the periodic functions of their atomic numbers.

(ii) A = $1s^2, 2s^2 2p^6, 3s^2$ is magnesium (12); B = $1s^2, 2s^2 2p^4$ is oxygen (8).

(a) B has 6 valence electrons, so it lies in group 16 (VIA).

(b) Mg (valency 2) combines with O (valency 2):

(c) 3. In $2Y + 6HCl \rightarrow 2YCl_3 + 3H_2$, each Y atom combines with three Cl atoms (each valency 1), so the valency of Y is 3.

P = F (9 electrons), Q = Mg (12 electrons), R = Cl (17 electrons).

(i) Q has 2 valence electrons ($3s^2$), so its valency is 2.

(ii) P (fluorine) is more reactive. P and R are halogens (7 valence electrons), but P has a smaller atomic size (2 shells), so the nucleus attracts the incoming electron more strongly and it is gained more easily than in R.

(iii) Mg (valency 2) reacts with Cl (valency 1):

A = $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$ is potassium (19); B = $1s^2, 2s^2 2p^6, 3s^2 3p^5$ is chlorine (17).

(i) A (potassium) is the metal; it has one valence electron which it loses easily.

(ii) B has three shells (K, L, M), so it lies in period 3.

(iii) Potassium reacts with chlorine to form potassium chloride:

From the table: c = Li (group IA, period 2), b = Be (IIA, period 2), a = Cl (VIIA, period 3), d = Ar (group 0, period 3).

(i) Element a (chlorine, 17 electrons): $1s^2, 2s^2 2p^6, 3s^2 3p^5$.

(ii) Element c has one valence electron and b has two. c (Li) has a larger atomic size and a smaller nuclear charge than b (Be), so its valence electron is farther from the nucleus and held less strongly. It is therefore lost more easily, so c is more reactive than b.

(iii) Both a and d are in period 3 with the same number of shells, but d has a larger nuclear charge (18 against 17), so d pulls its electrons closer and has the smaller size. Hence the size of a is greater than d.

Potassium is more reactive than sodium. Both have one valence electron, but potassium has four shells and sodium has three. The valence electron of K is farther from the nucleus and is held more weakly (greater shielding), so it is lost more easily.

Chlorine (17 electrons): $1s^2, 2s^2 2p^6, 3s^2 3p^5$

It has 7 valence electrons, so it lies in group 17 (VIIA).

(ii) $1s^2, 2s^2 2p^6, 3s^2$. This is magnesium, with 2 valence electrons that it loses, so its valency is 2. The others: (i) carbon (valency 4), (iii) silicon (valency 4), (iv) argon (valency 0).

Group 1 elements are called alkali metals because they react with water to form strongly basic solutions called alkalis (e.g. $2Na + 2H_2O \rightarrow 2NaOH + H_2$).

Metallic property decreases from left to right in a period, because the nuclear charge increases and atomic size decreases, so atoms lose electrons less easily.

From the table: X = Li, A = Be, M = F (period 2); Y = Na, B = Mg, J = Cl (period 3).

(a) Very active metal: Y (Na). Very active non-metal: M (F).

(b) B (Mg) is more active than A (Be). Both have two valence electrons, but B has an extra shell, so its valence electrons are farther from the nucleus and are lost more easily.

(c) Sodium (valency 1) reacts with fluorine (valency 1):

(a) Differences between s-block and p-block elements (any two):

(b) The third-period element of group 1 is sodium (Na); the second-period element of group 17 is fluorine (F). Both have valency 1:

(d) F < O < N. N, O and F are in the same period (period 2). Atomic radius decreases from left to right because the nuclear charge increases while the number of shells stays the same. Hence N (group 15) > O (group 16) > F (group 17).

Elements in the same group have the same number of valence electrons (e.g. all group 1 elements have one), and valency depends on the number of valence electrons, so they have the same valency.

X = $1s^2, 2s^2 2p^4$ is oxygen (8); Y = $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$ is potassium (19).

(a) X has 6 valence electrons, so it lies in group 16 (VIA).

(b) Y is a highly reactive metal (electropositive): it has one valence electron which it readily loses to form $Y^+$, and its oxide and hydroxide are basic.

From the table: A = Li, B = K (group IA); X = F, Y = Cl (group VIIA); C = Mg (IIA); Z = Ar (18).

(a) Most reactive metal: B (K, group IA with the largest size in the table). Most reactive non-metal: X (F, smallest halogen).

(b) Z is a noble gas with a complete outermost shell, so its valency is zero.

(c) C (valency 2) reacts with Y (valency 1):

The molecular formula is $CY_2$ (that is, $MgCl_2$).

From the table: X = Li, Y = Na (group 1); Z = Mg (group 2); M = Cl (group 17); W = Ar (group 18).

(a) The very active non-metal is M (chlorine).

(b) W (Ar, 18 electrons): $1s^2, 2s^2 2p^6, 3s^2 3p^6$.

(c) Y (Na) has one more shell than X (Li), so its valence electron is farther from the nucleus and is held more weakly. It is lost more easily, so Y is more reactive.

(d) Mg (valency 2) reacts with Cl (valency 1):

(ii) 2p, 3s, 3p, 4s. The Aufbau principle says electrons fill the subshells in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d. Only option (ii) is in this order.

Group 2 (IIA). The element is magnesium; it has 2 valence electrons ($3s^2$).

Lithium, sodium and potassium belong to group 1 (IA), the alkali metals.

Their chemical reactivity differs because the atomic size increases down the group (Li has 2 shells, Na 3, K 4). The valence electron is farther from the nucleus and is held more weakly, so it is lost more easily. The reactivity increases in the order Li < Na < K.

Beryllium and calcium lie in group 2 (group IIA, the alkaline earth metals).

Potassium (K) is more reactive, because it is lower in group 1, has a larger atom, and loses its single outer electron more easily than Na.

Group 17 (the halogens: F, Cl, Br, I).

Mendeleev's periodic law is based on atomic mass: the properties of elements are a periodic function of their atomic masses.

Fluorine has atomic number $9$ and electronic configuration $1s^2, 2s^2 2p^5$ (2, 7). It has 7 electrons in its valence shell. In the modern periodic table, elements with 7 valence electrons are placed in group $10+7=17$, so fluorine is kept in group 17 (the halogen family) and, since it has two shells, it is in period 2.

(a) Element X has the configuration $1s^2$ (helium). Its only shell, which holds a maximum of two electrons, is completely filled (a duplet). Such an atom is very stable and neither gains, loses nor shares electrons, so X is inert.

(b) Y is fluorine (Z $=9$, $2s^2 2p^5$) and Z is sodium ($Z=11$, $3s^1$).

$2\text{Na}+\text{F}_2\longrightarrow 2\text{NaF}$

OR The compound formed is sodium fluoride (NaF), an ionic compound.

A is in group 1 and period 4, so it is potassium ($Z=19$). B has $Z=17$, so it is chlorine.

(a) Electronic configuration of A:

$1s^2,\;2s^2 2p^6,\;3s^2 3p^6,\;4s^1$

(b) $2\text{K}+\text{Cl}_2\longrightarrow 2\text{KCl}$

OR The reaction is a combination (synthesis) reaction; it is also a redox reaction, since K loses an electron and Cl gains one to form the ionic compound KCl.

A metal that forms $A^{3+}$ and has $14$ neutrons is aluminium: its atomic number is $13$ and mass number is $13+14=27$.

(a) Electronic configuration of the atom: $1s^2,\;2s^2 2p^6,\;3s^2 3p^1$ (that is, $2,8,3$).

(b) The atom has $13$ protons and $13$ electrons. It loses its $3$ outer electrons to form $A^{3+}$.

Atom A (Z = 13)              Ion A3+
  nucleus: 13 p, 14 n          nucleus: 13 p, 14 n
  shells:  2, 8, 3             shells:  2, 8
  electrons: 13                electrons: 10  (charge = 13 - 10 = +3)

OR $A^{3+}$ has $8$ electrons in its last (second) shell.

(c) The atom has three valence electrons. If it formed $A^{2+}$, one electron would remain in the third shell ($2,8,1$), which is not stable. By losing all three, it gets a stable octet ($2,8$) like neon. So it forms $A^{3+}$ and not $A^{2+}$.

From the configurations: X is sodium ($Z=11$), Y is chlorine ($Z=17$) and Z is iron ($Z=26$).

(a) Y and Z form iron(II) chloride, FeCl$_2$ (or iron(III) chloride, FeCl$_3$, since iron shows valencies 2 and 3).

(b) The element just right of Na is magnesium (Mg). With chlorine it forms MgCl$_2$.

(c) Iron is a component of haemoglobin in red blood cells, which carries oxygen from the lungs to the body cells.

(a) Sodium is more reactive than magnesium. Its single valence electron is lost very easily, so it reacts vigorously even with cold water to form the soluble alkali sodium hydroxide: $2Na + 2H_2O \to 2NaOH + H_2$. Magnesium holds its two valence electrons more strongly and does not react with cold water. It reacts only with steam, and the product is the oxide, not an alkali.

(b) Magnesium and calcium react similarly because:

1. Both are in the same group (group II) and have the same number of valence electrons (2).
2. Both lose these two electrons to form $M^{2+}$ ions, and their atomic sizes are close in the same group, so their ionisation tendencies are similar.

(c) The main reason is its atomic number (12). It lies between Be (4) and Ca (20), so its properties (size, reactivity) change gradually and fall between those of Be and Ca. Like them, it has two valence electrons.

Element A is Na (11), B is Al (13), C is Cl (17) and D is Ar (18).

(a) Element B has three shells, so it belongs to period 3.

(b) Element D has a complete octet ($3s^2 3p^6$), so it is a noble (inert) gas. It is chemically unreactive and monoatomic.

(c) Aluminium reacts with chlorine to form aluminium chloride: $2Al + 3Cl_2 \to 2AlCl_3$

OR The compound formed is aluminium chloride, $AlCl_3$. Aluminium (3 valence electrons) shares electrons with three chlorine atoms (7 valence electrons each), so the bond is mainly covalent.

A is sodium (Z = 11) and B is oxygen (Z = 8).

(a) $A^{+}$: $1s^2, 2s^2 2p^6$

(b) Yes. A has only one electron in its outer shell and a low ionisation energy, so it loses that electron easily. The resulting $A^{+}$ has the stable octet of neon.

(c) Sodium reacts with oxygen to form sodium oxide, $Na_2O$.

(d) A (the atom) is bigger. On losing its outer electron, $A^{+}$ loses the whole third shell, and the nucleus (11 protons) now pulls the remaining 10 electrons more strongly.

Given: A = $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$ (19 electrons, potassium, K); B = $1s^2, 2s^2 2p^4$ (8 electrons, oxygen, O).

i. Element A has 1 valence electron, so it belongs to group IA (group 1), the alkali metals.

ii. Element B has electrons in two shells (K and L), so it belongs to period 2.

iii. A loses 1 electron (valency 1) and B gains 2 electrons (valency 2), so two K atoms combine with one O atom:

The molecular formula of the compound is $K_2O$ (potassium oxide).

iv. The last electron of B enters the p subshell ($2p^4$). Elements whose last electron enters a p subshell are placed in the p-block, so B is kept in the p-block.

Group VII (17) elements (F, Cl, Br, I) are called halogens because the word means "salt producers": they react directly with metals to form salts such as NaCl and KBr.

A group IIA metal (e.g. calcium) has valency 2 and a group VIIA non-metal (e.g. chlorine) has valency 1, so the metal reacts with two chlorine atoms:

(Similarly, $Mg + Cl_2 \rightarrow MgCl_2$.) The equation is balanced: 1 Ca and 2 Cl on each side.

Given: X = $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^2$ (20 electrons, calcium); Y = $1s^2, 2s^2 2p^6, 3s^2 3p^5$ (17 electrons, chlorine).

(a) X is the metal. It has 2 valence electrons which it easily loses to form $X^{2+}$.

(b) Y has electrons in three shells (K, L, M), so it lies in period 3.

(c) X (valency 2) combines with Y (valency 1):

Here X = Ca and Y = Cl, so the compound is $XY_2$ (calcium chloride).

The eight elements have the same number of orbits (shells) but valence electrons increasing from 1 to 8, so they are the elements of one period: A = Na, B = Mg, C = Al, D = Si, E = P, F = S, G = Cl, H = Ar. A and G form NaCl, common salt, which is added to vegetables while cooking.

(i) The elements belong to period 3.

(ii) Highly reactive metal: Na (A). Highly reactive non-metal: Cl (G).

(iii) The element that does not take part in chemical reactions is Ar (H), because its outermost shell is complete (8 electrons), so it is inert.

(iv) B (Mg) is a metal; it loses electrons and forms a basic oxide (MgO). F (S) is a non-metal; it gains electrons and forms an acidic oxide ($SO_2$).

(b) 2. In $X + 2HCl \rightarrow XCl_2 + H_2$, one atom of X combines with two Cl atoms (each of valency 1), so the valency of X is 2.

(i) The configuration $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$ has 19 electrons: Potassium (K).

(ii) An alkali metal other than sodium: Lithium (or potassium).

(iii) A metalloid other than silicon: Germanium (or boron, arsenic).

(iv) An alkaline earth metal other than magnesium: Calcium (or beryllium, barium).