CBSE Class 11 Chemistry Classification Of Elements And Periodicity In Properties Of Elements Notes Set 08

This set of Class 11 CBSE Chemistry notes for Chapter 3 goes deeper into the periodic properties themselves. It covers successive ionisation enthalpies and the irregularities in the trend, electron gain enthalpy and its sign convention, the Pauling scale of electronegativity, valence and oxidation states, how the reactivity of metals and non-metals and the nature of oxides change across the table, and the anomalous behaviour of the second-period elements. Seven solved examples include a data-based problem that identifies elements from their ionisation and electron gain enthalpies.

Important Notes and Formulas

Ionisation Enthalpy in Detail

1. The ionisation enthalpy ΔiH is expressed in kJ mol⁻¹ or in electron volts per atom. 1 eV per atom = 96.49 kJ mol⁻¹. Ionisation is always endothermic.
2. The successive steps are M(g) → M⁺(g) + e⁻ (first, ΔiH₁), M⁺(g) → M²⁺(g) + e⁻ (second, ΔiH₂) and so on. Always ΔiH₁ < ΔiH₂ < ΔiH₃, because each electron is removed from a more positive, smaller ion in which the same nuclear charge holds fewer electrons.
3. The factors are the size of the atom, the nuclear charge, the shielding by inner electrons, the penetration of the orbital (s > p > d > f within a shell) and the stability of filled and half-filled sub-shells.

The ionisation enthalpies are in kJ mol⁻¹. Each element shows one very large jump immediately after its last valence electron has been removed (after the 1st for Na, the 2nd for Mg and the 3rd for Al), because the next electron has to come from a stable noble-gas core. This is how the number of valence electrons can be read from a list of successive values.

Trends and Irregularities

1. Across a period ΔiH₁ generally increases, because the nuclear charge rises and the atomic size falls. Down a group it decreases, because the size and the shielding increase.
2. In group 1 the values fall from 520 kJ mol⁻¹ (Li) to 376 kJ mol⁻¹ (Cs). Helium has the highest value of all the elements (2372 kJ mol⁻¹).
3. Be > B: the 2s electron of Be is more penetrating and more strongly held than the 2p electron of B.
4. N > O: N (2p³) has a stable half-filled sub-shell, while O (2p⁴) has a pair of electrons in one 2p orbital and loses one of them more easily. In the same way, Mg > Al and P > S.
5. Low ΔiH₁ means a strong tendency to form cations (metallic character). High ΔiH₁ goes with non-metallic character.

The graph shows the first ionisation enthalpies of the elements of periods 2 and 3. Notice the dips at B (after Be) and O (after N) in period 2, and at Al (after Mg) and S (after P) in period 3.

Electron Gain Enthalpy in Detail

1. For X(g) + e⁻ → X⁻(g), the enthalpy change is the electron gain enthalpy ΔegH. It is negative when energy is released (most elements) and positive when energy is absorbed. The older term electron affinity has the opposite sign: electron affinity = −ΔegH.
2. Halogens have the most negative values because one added electron gives them a noble-gas configuration. Noble gases have large positive values (He +48, Ne +116, Ar +96 kJ mol⁻¹) because the electron must enter the next higher shell. Be, Mg (filled s sub-shell) and N (half-filled p sub-shell) have values that are positive or close to zero.
3. Trend: in a period ΔegH becomes more negative from left to right; in a group it generally becomes less negative from top to bottom.
4. Exceptions: the second-period elements O and F have less negative values than S and Cl. The 2p sub-shell is small and compact, so electron–electron repulsion reduces the energy released. The order for the halogens is Cl > F > Br > I in magnitude.
5. Second electron gain enthalpy is always positive, because the electron is added to a negative ion and is repelled. For oxygen, O(g) + e⁻ → O⁻(g) is −141 kJ mol⁻¹ and O⁻(g) + e⁻ → O²⁻(g) is +780 kJ mol⁻¹.

Electronegativity and the Pauling Scale

1. Electronegativity has no unit. It is a relative number, and the Pauling scale gives fluorine, the most electronegative element, the value 4.0. Caesium and francium are the least electronegative (0.7).
2. Electronegativity rises across a period (smaller size, larger effective nuclear charge) and falls down a group (larger size).
3. Metals have low electronegativities; non-metals have high ones. The greater the difference in electronegativity between two bonded atoms, the more polar the bond.
4. Electronegativity belongs to an atom in a molecule, whereas electron gain enthalpy belongs to an isolated gaseous atom. Chlorine has a more negative ΔegH than fluorine, but fluorine is more electronegative than chlorine.

Periodicity of Valence and Oxidation State

1. The valence of a main group element is generally equal to the number of its valence electrons (groups 1 to 4) or to eight minus that number (groups 5 to 8, i.e. 15 to 18). So it rises from 1 to 4 and then falls to 0 across a period.
2. Valence can also be taken as the number of hydrogen atoms, or twice the number of oxygen atoms, that combine with one atom of the element. With oxygen, the valence rises from 1 to 7 across period 3.
3. The oxidation state is the charge an atom would have if the bonding electrons were assigned to the more electronegative atom. In Na₂O, sodium is +1 and oxygen is −2.
4. All elements of a group have the same valence because they have the same number of valence electrons. Noble gases are zerovalent. Transition and inner transition elements show variable valence, because d (or f) electrons also take part in bonding.

Periodic Trends and Chemical Reactivity

1. Reactivity of metals depends on how easily they lose electrons. It decreases across a period (Na > Mg > Al) and increases down a group (Li < Na < K < Rb < Cs).
2. Reactivity of non-metals depends on how readily they gain electrons. It increases across a period (C < N < O < F) and decreases down a group (F > Cl > Br > I).
3. Nature of oxides: the oxides on the left of a period are strongly basic and those on the right are strongly acidic. Oxides in the middle are amphoteric or neutral.
4. Amphoteric oxides (Al₂O₃, As₂O₃) behave as acidic with bases and as basic with acids. Neutral oxides (CO, NO, N₂O) have no acidic or basic properties.

1. Na₂O + H₂O → 2NaOH (a strong base).
2. Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (basic behaviour) and Al₂O₃ + 6NaOH + 3H₂O → 2Na₃[Al(OH)₆] (acidic behaviour).
3. P₄O₁₀ + 6H₂O → 4H₃PO₄ and SO₃ + H₂O → H₂SO₄ (acidic oxides give acids).
4. Cl₂O₇ + H₂O → 2HClO₄ (a very strong acid).

Anomalous Properties of Second-Period Elements

1. The first member of each group (Li, Be, B, C, N, O, F) differs in many properties from the rest of its group. The reasons are the small size, the large charge to radius ratio and the high electronegativity.
2. These elements have only four valence orbitals (2s and 2p), so their maximum covalency is 4 (for example boron forms [BF₄]⁻), whereas the later members can expand their octets using d orbitals (for example aluminium forms [AlF₆]³⁻).
3. The first member of a p-block group forms pπ–pπ multiple bonds with itself and with other second-period elements much more readily than the later members. Examples: C=C, C≡C, N≡N, C=O, C≡N and N=O.

Solved Examples

Example 1. The ionisation energy of a hydrogen atom is 13.6 eV per atom. Express it in kJ mol⁻¹. Compare the ionisation energy of He⁺ (54.4 eV per atom) with the second ionisation enthalpy of helium.
Answer:
1 eV per atom = 96.49 kJ mol⁻¹.
H: 13.6 × 96.49 = 1312 kJ mol⁻¹, which is the first ionisation enthalpy of hydrogen.
He⁺: 54.4 × 96.49 = 5249 kJ mol⁻¹, which agrees with the second ionisation enthalpy of helium (about 5251 kJ mol⁻¹).
He⁺ has the larger value because its single electron is held by a nuclear charge of +2.

Example 2. Among the second-period elements the actual first ionisation enthalpies follow the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has a higher ΔiH₁ than B, and (ii) O has a lower ΔiH₁ than N and F.
Answer:
(i) The 2s electron of Be (1s² 2s²) is more strongly attracted by the nucleus than the 2p electron of B (1s² 2s² 2p¹). So the 2p electron of B is removed with less energy.
(ii) N has the half-filled 2p³ configuration, which is extra stable, so its ΔiH₁ is higher than that of O (2p⁴). F (2p⁵) has a higher nuclear charge (+9), which outweighs the half-filled stability of N, so F is higher than both.

Example 3. The first ionisation enthalpy of magnesium is higher than that of sodium, but the second ionisation enthalpy of sodium is much higher than that of magnesium. Explain.
Answer:
Na is 1s² 2s² 2p⁶ 3s¹ and Mg is 1s² 2s² 2p⁶ 3s². In both cases the first electron comes from 3s, but Mg has the higher nuclear charge (+12 against +11), so ΔiH₁ of Mg (738) is higher than that of Na (496 kJ mol⁻¹).
Na⁺ is 1s² 2s² 2p⁶, the stable neon configuration, so removing a second electron is extremely difficult (4562 kJ mol⁻¹).
Mg⁺ is 1s² 2s² 2p⁶ 3s¹ and still has a 3s electron, which is removed fairly easily (1451 kJ mol⁻¹). So the second ionisation enthalpy of Na is much higher.

Example 4. Among the elements B, Al, C and Si, (i) which has the highest first ionisation enthalpy and (ii) which has the most metallic character? Justify your answers.
Answer:
Placing them in the table: B (group 13, period 2), C (group 14, period 2), Al (group 13, period 3) and Si (group 14, period 3).
(i) The ionisation enthalpy increases along a period and decreases down a group. So carbon (top right of the four) has the highest ΔiH₁.
(ii) Metallic character decreases along a period and increases down a group. So aluminium (bottom left of the four) is the most metallic.

Example 5. Among the elements of the third period, Na to Ar, pick out the element (i) with the highest first ionisation enthalpy, (ii) with the largest atomic radius, (iii) that is the most reactive non-metal, (iv) that is the most reactive metal.
Answer:
(i) Argon (the noble gas at the right end of the period).
(ii) Sodium (the radius falls across the period, so the first element is the largest).
(iii) Chlorine (the halogen has the most negative electron gain enthalpy of the period).
(iv) Sodium (it has the lowest ΔiH₁ and loses its 3s electron most readily).

Example 7. Among B, C, N, F and Si, the correct order of decreasing non-metallic character is:
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer:
Non-metallic character increases from left to right in a period, so among the second-period elements F > N > C > B.
It decreases down a group, so C is more non-metallic than Si (which lies below it in group 14). On the Pauling scale B is 2.0 and Si is 1.8, so Si is also less non-metallic than B and comes last.
The order is F > N > C > B > Si, which is option (c).