JEE Chemistry Classification Of Elements and Periodicity In Properties MCQs Set 18

Answer: (c) C < N < F < O

Explanation:
1. The second IE removes an electron from the monopositive ion.
2. O+ has the stable half-filled 2p3 configuration, so removing the next electron is very hard and O has the highest IE2.
3. F+ (2p4) comes next, then N+ (2s2 2p2) and C+ (2s2 2p1) is the lowest.
4. So C < N < F < O.

Teacher's Note:
1. Always write the configuration of the uni-positive ion before comparing second ionisation enthalpies.
2. Half-filled and fully filled subshells give unusually high values.

Answer: (a) Pb

Explanation:
1. In group 14 the inert pair effect becomes stronger down the group.
2. Pb prefers +2 over +4 because the 6s electrons are reluctant to take part in bonding.
3. So Pb2+ is the most stable among the choices.

Teacher's Note:
1. Inert pair effect increases down groups 13 to 15.
2. Pb(II) compounds are stable while Pb(IV) compounds are oxidising.

Answer: (a) Terbium (Z = 65)

Explanation:
1. Actinoids are the elements from Ac (89) to Lr (103).
2. Cm, Cf and U lie in this range.
3. Terbium (65) is a lanthanoid.

Teacher's Note:
1. Lanthanoids run from Z = 57 or 58 to 71.
2. Actinoids run from Z = 89 or 90 to 103.

Answer: (a) \([Xe]4f^7 5d^1 6s^2\)

Explanation:
1. Gd places one electron in 5d to keep the 4f subshell exactly half filled (4f7).
2. The configuration is therefore \([Xe]4f^7 5d^1 6s^2\).

Teacher's Note:
1. Gd and Cm show extra stability of half-filled f subshell.
2. Lu and La have a 5d electron too.

Answer: (c) electron repulsion outweighs the stability gained by achieving a noble gas configuration

Explanation:
1. The first electron gain enthalpy of oxygen is negative.
2. The second electron is added to an already negative ion, and the strong inter-electronic repulsion makes the second step endothermic.
3. This repulsion outweighs the stability gained from the neon configuration.

Teacher's Note:
1. Second electron gain enthalpies of all elements are positive.
2. The oxide ion is stable in solids only because lattice energy pays for the energy cost.

Answer: (c) Sets 1 and 2

Explanation:
1. Set 1: K+, Ca2+, Sc3+ and Cl- all have 18 electrons.
2. Set 2: P3-, S2-, Cl- and K+ all have 18 electrons.
3. Set 3 mixes 28, 18, 28 and 10 electron ions, and Set 4 has Cr3+ with 21 electrons.
4. So only Sets 1 and 2 qualify.

Teacher's Note:
1. Count electrons as Z minus the positive charge or plus the negative charge.
2. Species with 18 electrons form the argon isoelectronic series.

Answer: (d) Good conductor of electricity, solid and metallic

Explanation:
1. The element is aluminium.
2. Aluminium is a solid metal at room temperature and an excellent conductor of electricity.

Teacher's Note:
1. Group 13 in period 3 is Al.
2. Boron, in period 2, is the only non-metal in the group.

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

Explanation:
1. Across a period the nuclear charge increases while shielding by electrons of the same shell is small.
2. The effective nuclear charge rises, so electrons are held more tightly and ionisation enthalpy rises.
3. The reason therefore explains the assertion.

Teacher's Note:
1. Same-shell electrons shield each other poorly.
2. Effective nuclear charge increases across a period.

Answer: (d) \(\ce{Al2O3 < MgO < Na2O < K2O}\)

Explanation:
1. Basic character of oxides increases with increasing metallic and electropositive character.
2. Al2O3 is amphoteric, MgO is basic, Na2O is strongly basic and K2O is more so.
3. So Al2O3 < MgO < Na2O < K2O.

Teacher's Note:
1. Oxides become more basic down a group and less basic across a period.
2. Amphoteric Al2O3 sits between acidic and basic oxides.

Answer: (a) \(\ce{O^{2-} > F- > Na+ > Mg^{2+} > Al^{3+}}\)

Explanation:
1. These ions are isoelectronic (10 electrons).
2. Ionic radius decreases with increasing nuclear charge.
3. The nuclear charges increase in the order O < F < Na < Mg < Al, so the radii decrease in that order.

Teacher's Note:
1. In an isoelectronic series the anion with the lowest Z is the largest.
2. Arrange them by atomic number first.

Answer: (a) \(\ce{Na+ > Li+ > Mg^{2+} > Be^{2+}}\)

Explanation:
1. Na+ (102 pm) is bigger than Li+ (76 pm) since it has one more shell.
2. Mg2+ (72 pm) is bigger than Be2+ (45 pm).
3. Li+ and Mg2+ are close (diagonal pair), with Li+ slightly larger.
4. The order is Na+ > Li+ > Mg2+ > Be2+.

Teacher's Note:
1. Down a group radius increases; across a period cation radius decreases.
2. Diagonal pairs have similar sizes.

Answer: (c) \(\ce{K+ < Ca^{2+} < Mg^{2+} < Be^{2+}}\)

Explanation:
1. Polarising power increases with charge/size ratio.
2. K+ has a low charge and large size; Ca2+ has a higher charge but is bigger than Mg2+; Be2+ is the smallest.
3. The order is K+ < Ca2+ < Mg2+ < Be2+.

Teacher's Note:
1. Fajans' rules: small cation with high charge polarises strongly.
2. Be compounds are therefore often covalent.

Answer: (c) NaCl

Explanation:
1. Lattice energy decreases with increasing cation size, which would favour LiCl.
2. LiCl, however, has appreciable covalent character due to the high polarising power of Li+.
3. Experimentally NaCl melts at 801 degrees C, higher than LiCl (605), KCl (770) and RbCl (718).

Teacher's Note:
1. Covalent character lowers the melting point of LiCl.
2. Use data: NaCl is the highest among alkali chlorides.

Answer: (d) Cr

Explanation:
1. The second IE removes an electron from \(\ce{M+}\).
2. Cr+ is \(3d^5\), a stable half-filled configuration, so removing an electron from it is difficult.
3. Cr has the highest second ionisation enthalpy.

Teacher's Note:
1. Cr and Cu have very high IE2 because the uni-positive ions are d5 and d10.
2. Mn+ is 3d5 4s1, with an easily removable 4s electron.

Answer: (c) Yb < Pm < Ce < La

Explanation:
1. These are lanthanoid ions of increasing atomic number: La (57), Ce (58), Pm (61), Yb (70).
2. Because of lanthanoid contraction the ionic radius decreases with increasing Z.
3. So the increasing order is Yb < Pm < Ce < La.

Teacher's Note:
1. Radius falls along the 4f series.
2. Arrange by atomic number and reverse it.

Answer: (b) F > O > N : first ionisation enthalpy

Explanation:
1. First ionisation enthalpies are F (1681) > N (1402) > O (1314).
2. So the order F > O > N is not correct.
3. The other three orders are correct.

Teacher's Note:
1. N has higher IE than O because of its half-filled 2p3.
2. Chlorine has the largest negative electron gain enthalpy.

Answer: (a) Among isoelectronic species, the smaller the positive charge on the cation, the smaller its ionic radius

Explanation:
1. For isoelectronic cations a larger positive charge pulls the electrons in more strongly, so the radius is smaller.
2. The statement (a) says the opposite, so it is wrong.
3. (b), (c) and (d) are correct.

Teacher's Note:
1. More protons holding the same number of electrons gives a smaller ion.
2. Check each statement for the direction of the trend.

Answer: (b) \(\ce{F2 > Cl2 > Br2 > I2}\) : bond dissociation energy

Explanation:
1. Bond dissociation energy of Cl2 (242 kJ/mol) is higher than that of F2 (158 kJ/mol) due to repulsion between lone pairs in the small F2 molecule.
2. The real order is Cl2 > Br2 > F2 > I2.
3. The other three orders are correct.

Teacher's Note:
1. F2 is an exception to the trend of bond energies.
2. Learn the order Cl2 > Br2 > F2 > I2.

Answer: (b) a significant decrease from O2- to Al3+

Explanation:
1. With the same number of electrons, an increase in nuclear charge from 8 to 13 contracts the electron cloud.
2. So the radius decreases from O2- (140 pm) to Al3+ (54 pm).

Teacher's Note:
1. Isoelectronic species: radius falls steadily as Z rises.
2. The trend is monotonic.

Answer: (c) 3

Explanation:
1. Se2- (below O in group 16) and P3- (a third period anion) are larger than O2-.
2. N3- has the same electrons as O2- but fewer protons, so it is larger.
3. F- has more protons and is smaller. Hence the count is 3.

Teacher's Note:
1. Compare isoelectronic ions by Z and different-period ions by number of shells.
2. More shells means a bigger ion.

Answer: (c) 7

Explanation:
1. Cs is a 6th period alkali metal (about 265 pm).
2. Li, Na, Mg, Rb, Ba, Sr and Ca are all smaller.
3. Fr, being below Cs in group 1, is larger. So 7 are smaller.

Teacher's Note:
1. Atomic radius increases down a group and decreases across a period.
2. Ba is in the same period as Cs but to its right, hence smaller.