CBSE Class 11 Chemistry Classification of Elements and Periodicity in Properties MCQs Set 06

Answer: (c) alkali metals

Explanation:
1. Alkali metals have a single ns electron outside a noble-gas core.
2. They have the largest atoms and the lowest effective nuclear charge in their period.
3. So the ns¹ electron is removed most easily.

Teacher's Note:
1. Ionisation enthalpy is lowest at the left end and highest at the right end of a period.
2. Group 18 has the highest values.

Answer: (d) s sub-shell

Explanation:
1. Within one shell the penetration towards the nucleus is in the order s > p > d > f.
2. An s electron is therefore attracted most strongly and is shielded least by the other electrons.
3. So the most energy is needed to remove an s electron.

Teacher's Note:
1. More penetration means a stronger attraction and a higher ionisation enthalpy.
2. This is also why 2s electrons of Be are harder to remove than the 2p electron of B.

Answer: (b) 1s² 2s² 2p⁶ 3s¹

Explanation:
1. The configuration 1s² 2s² 2p⁶ 3s¹ is that of sodium (Z = 11).
2. Loss of one 3s electron gives Na⁺ with the stable neon core.
3. It has the lowest first ionisation enthalpy of the four atoms.

Teacher's Note:
1. A single outer s electron over a noble-gas core is lost most easily.
2. Neon (third option) is a noble gas and does not lose electrons readily.

Answer: (a) Hydrogen

Explanation:
1. First ionisation enthalpies (kJ mol⁻¹): H 1312, B 801, Li 520, Na 496.
2. H has a single electron close to the nucleus with no shielding, so it is held strongly.

Teacher's Note:
1. Li to Ne is a different period from H, so compare the values rather than the positions.
2. Na is the lowest of these four.

Answer: (d) S²⁻

Explanation:
1. All four ions have 18 electrons.
2. S²⁻ has the smallest nuclear charge (+16), so its outer electron is held least strongly and its ion is the largest.
3. Hence S²⁻ has the lowest ionisation enthalpy.

Teacher's Note:
1. For the same number of electrons, the higher the nuclear charge, the higher the ionisation enthalpy.
2. Compare K (+19), Ca (+20), Cl (+17) and S (+16).

Answer: (d) +2

Explanation:
1. The jump from the second (12.5 eV) to the third (42.5 eV) ionisation energy is very large.
2. So two electrons can be removed fairly easily, but the third is removed from a stable core.
3. The element therefore forms a +2 ion.

Teacher's Note:
1. A very large jump after the nth ionisation energy means n valence electrons.
2. Here n = 2.

Answer: (c) three

Explanation:
1. The first three values rise gradually (284, 412, 656).
2. The fourth value (3210) is about five times the third.
3. So the first three electrons are valence electrons and the fourth comes from an inner shell.

Teacher's Note:
1. Look for the largest jump between successive values.
2. The jump after the nth electron tells you the number of valence electrons.

Answer: (d) O > F > N > C

Explanation:
1. The second ionisation removes an electron from the unipositive ion.
2. O⁺ is 2s² 2p³ (half-filled, very stable), so removing an electron from it is the hardest.
3. F⁺ (2p⁴) comes next, then N⁺ (2p²) and C⁺ (2p¹). So O > F > N > C.

Teacher's Note:
1. Write the configuration of each M⁺ ion.
2. The ion with a half-filled sub-shell has the highest next ionisation enthalpy.

Answer: (c) Be, Mg (filled s sub-shells) and N, P (half-filled p sub-shells) all show unusually high values

Explanation:
1. Filled and half-filled sub-shells are specially stable, so removing an electron from them needs extra energy.
2. Be (2s²) and Mg (3s²) have filled s sub-shells, and N (2p³) and P (3p³) have half-filled p sub-shells.
3. These atoms have higher values than their neighbours B, Al, O and S, which causes the dips in the graph.

Teacher's Note:
1. Look at the graph in the notes: the dips come at B, O, Al and S.
2. Size and nuclear charge give the general rise; sub-shell stability gives the exceptions.

Answer: (b) 1s² 2s² 2p⁶ 3s¹

Explanation:
1. The four configurations are Mg, Na, Ne and F.
2. After the first ionisation, Na⁺ has the stable neon core, so removing a second electron is extremely difficult (4562 kJ mol⁻¹).
3. Mg⁺ still has a 3s¹ electron, and for Ne and F the second values are 3952 and 3374 kJ mol⁻¹, which are lower than that of Na.

Teacher's Note:
1. The second ionisation enthalpy is the first one of M⁺.
2. A noble-gas-like M⁺ ion has a very high next ionisation enthalpy.

Answer: (b) +526 kcal

Explanation:
1. The two steps are Mg → Mg⁺ + e⁻ (178 kcal) and Mg⁺ → Mg²⁺ + e⁻ (348 kcal).
2. Total energy = 178 + 348 = 526 kcal mol⁻¹.
3. Ionisation is endothermic, so the sign is positive.

Teacher's Note:
1. Add the successive ionisation energies to remove several electrons.
2. Do not subtract them.

Answer: (a) Na < Mg > Al < Si

Explanation:
1. The values are Na 496, Mg 738, Al 577 and Si 786 kJ mol⁻¹.
2. Mg is above Al because 3s² is more stable and more penetrating than 3p¹.
3. So the pattern is up, down, up.

Teacher's Note:
1. Learn the exception 'Mg > Al'.
2. Compare the actual values to avoid mistakes in the order.

Answer: (c) the outermost electron in boron occupies a 2p orbital

Explanation:
1. Be is 1s² 2s² and B is 1s² 2s² 2p¹.
2. A 2s electron is more penetrating and more strongly attracted by the nucleus than a 2p electron.
3. So less energy is needed to remove the 2p electron of B than the 2s electron of Be.

Teacher's Note:
1. Penetration order is s > p > d > f in the same shell.
2. Nuclear charge is actually higher in B, so it cannot be the reason.

Answer: (b) III > II > I

Explanation:
1. Each successive electron is removed from a more positive ion of the same nucleus.
2. The remaining electrons are drawn closer and held more strongly, so ΔiH rises at every step.
3. For nitrogen the values are 1402, 2856 and 4578 kJ mol⁻¹, so III > II > I.

Teacher's Note:
1. For any element, ΔiH₁ < ΔiH₂ < ΔiH₃.
2. Positive ions are smaller, so the next electron is harder to remove.

Answer: (b) S, Cl, O, F

Explanation:
1. On the Pauling scale: S = 2.5, Cl = 3.0, O = 3.5 and F = 4.0.
2. So the order of increasing electronegativity is S < Cl < O < F.

Teacher's Note:
1. F is the most electronegative element, then O, then Cl.
2. Keep the Pauling values of period 2 in mind: Li 1.0 to F 4.0.

Answer: (a) 1s² 2s² 2p⁶ 3s² 3p⁵

Explanation:
1. The configurations are Cl, N, O and F.
2. Chlorine has the most negative electron gain enthalpy (−349 kJ mol⁻¹) of all elements; F is −328 kJ mol⁻¹.
3. N has about zero and O is −141 kJ mol⁻¹.

Teacher's Note:
1. F is smaller than Cl and its added electron is repelled more in the compact 2p sub-shell.
2. So the highest value is for Cl and not F.

Answer: (a) O < S < F < Cl

Explanation:
1. Electron gain enthalpies (kJ mol⁻¹): O −141, S −200, F −328, Cl −349.
2. Within each of groups 16 and 17, the second-period element has a lower value than the third-period one.
3. So O < S < F < Cl.

Teacher's Note:
1. Magnitudes rise from group 16 to group 17.
2. Second-period elements O and F lie below S and Cl in magnitude.

Answer: (a) F > Br > I

Explanation:
1. Electron gain enthalpies: F −328, Br −325 and I −295 kJ mol⁻¹.
2. Electron affinity generally falls down the group, and F stays above Br in spite of its small size.
3. So F > Br > I.

Teacher's Note:
1. F is out of order only with Cl.
2. Do not compare with Cl here; just use the three values.

Answer: (d) F(g) + e⁻ → F⁻(g)

Explanation:
1. Removing an electron from an atom (first two options) always needs energy.
2. Adding a second electron to O⁻ needs energy because of repulsion by the negative ion.
3. Adding an electron to F releases energy (−328 kJ mol⁻¹).

Teacher's Note:
1. Ionisation is endothermic; first electron gain by halogens is exothermic.
2. Second electron gain is endothermic.

Answer: (c) Li < Be < B < C: first ionisation enthalpy

Explanation:
1. The correct order of ionisation enthalpy is Li < B < Be < C, since Be (899) is higher than B (801).
2. Si 1.8 < P 2.1 < Cl 3.0 < F 4.0 is the correct electronegativity order.
3. Basic character of oxides rises as the metal becomes more electropositive, and ionic radii increase down a group.

Teacher's Note:
1. Be and B are the usual trap in ionisation enthalpy orders.
2. Check each statement separately.