JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 40

Answer: (d) A bond is formed when attractive forces overcome repulsive forces

Explanation:
1. A covalent bond forms by sharing of electrons, not transfer, so (a) is wrong.
2. Pure water is slightly self-ionised (\(\ce{H3O+}\) and \(\ce{OH-}\)), so (b) is wrong.
3. A bond forms only if the net attraction between nuclei and electrons exceeds the repulsions: the energy of the system is then lower than for the separate atoms. (c) is correct.
4. F is more electronegative than Br, so HF is more polar than HBr: (d) is wrong.

Teacher's Note:
1. Bond formation is always a lowering of energy: attraction wins over repulsion.

Answer: (c) \(\ce{NH4+}\)

Explanation:
1. \(109^\circ28'\) is the ideal tetrahedral angle, found when four bond pairs surround the central atom with no lone pair.
2. \(\ce{NH4+}\) has four N-H bond pairs and no lone pair: exactly tetrahedral.
3. \(\ce{NH3}\) (107 degrees) and \(\ce{H2O}\) (104.5 degrees) are reduced by lone pair repulsion; \(\ce{CH3+}\) is trigonal planar (120 degrees).

Teacher's Note:
1. Four bond pairs and no lone pairs give the tetrahedral angle.

Answer: (b) \(\ce{AlH3}\) changes to \(\ce{AlH4-}\) (Al)

Explanation:
1. \(\ce{AlH3}\): three bond pairs, no lone pair, \(sp^2\). \(\ce{AlH4-}\): four bond pairs, \(sp^3\). The hybridisation changes.
2. \(\ce{H2O}\) (two bond pairs + two lone pairs) and \(\ce{H3O+}\) (three + one) are both \(sp^3\) on oxygen.
3. \(\ce{NH3}\) and \(\ce{NH4+}\) are both \(sp^3\) on nitrogen. So only (a).

Teacher's Note:
1. Count bond pairs plus lone pairs before and after.

Answer: (b) \(\ce{FeCl3}\) is more covalent than \(\ce{FeCl2}\)

Explanation:
1. By Fajans' rules, a cation with higher charge and smaller size polarises the anion more, increasing covalent character.
2. \(\ce{Fe^3+}\) is smaller and more highly charged than \(\ce{Fe^2+}\), so \(\ce{FeCl3}\) is more covalent than \(\ce{FeCl2}\).

Teacher's Note:
1. Higher positive charge means greater polarising power and more covalent character.

Answer: (b) \(sp^3,\ sp^3d,\ sp^3d^2\)

Explanation:
1. \(\ce{SF2}\): 2 bond pairs + 2 lone pairs = 4 electron pairs: \(sp^3\).
2. \(\ce{SF4}\): 4 bond pairs + 1 lone pair = 5 electron pairs: \(sp^3d\).
3. \(\ce{SF6}\): 6 bond pairs = 6 electron pairs: \(sp^3d^2\).

Teacher's Note:
1. Hybridisation is decided by the number of electron pairs: 4 gives sp3, 5 gives sp3d, 6 gives sp3d2.

Answer: (a) it has two three-centre two-electron bonds

Explanation:
1. Diborane has two bridging hydrogens, each forming a three-centre two-electron (banana) B-H-B bond.
2. The terminal B-H bonds (about 119 pm) are shorter than the bridging ones (about 134 pm), and the HBH angles are about 120 degrees (terminal) and 97 degrees (bridge), so (a) and (b) are false.
3. The two borons and four terminal H atoms lie in one plane; the two bridging H atoms lie above and below it, so (d) is false.

Teacher's Note:
1. Remember the banana bonds: two 3c-2e bonds, with bridging H above and below the plane.

Answer: (b) \(\ce{He2+} < \ce{O2-} < \ce{NO} < \ce{C2^2-}\)

Explanation:
1. Bond orders from MO theory: \(\ce{He2+}\): \(\frac{2-1}{2}=0.5\); \(\ce{O2-}\) (13 valence electrons): \(\frac{8-5}{2}=1.5\); NO (11 valence electrons): \(\frac{8-3}{2}=2.5\); \(\ce{C2^2-}\) has 10 valence electrons like \(\ce{N2}\): bond order 3.
2. Increasing order: \(\ce{He2+}\) (0.5) < \(\ce{O2-}\) (1.5) < NO (2.5) < \(\ce{C2^2-}\) (3).

Teacher's Note:
1. C2 2- is isoelectronic with N2 and has bond order 3.

Answer: (a) 4

Explanation:
1. Without 2s-2p mixing, the order is \(\sigma2p<\pi2p\) for every molecule (as in \(\ce{O2}\)).
2. \(\ce{Be2}\) (4 valence e) and \(\ce{B2}\) (6 e, the last pair goes into \(\sigma2p\)) are diamagnetic; \(\ce{N2}\) (10 e) is diamagnetic.
3. \(\ce{C2}\) (8 e) puts 2 electrons in the degenerate \(\pi2p\) pair: 2 unpaired. \(\ce{O2}\), \(\ce{O2+}\) and NO each have unpaired electrons in \(\pi^*2p\).
4. Paramagnetic: \(\ce{C2}\), \(\ce{O2}\), \(\ce{O2+}\), NO = 4.

Teacher's Note:
1. Fill the MO order with sigma 2p below pi 2p and look for unpaired electrons.

Answer: (d) 2

Explanation:
1. (a) Allene: the two \(\ce{CH2}\) planes are perpendicular, so it is not planar. (c) \(\ce{SiF4}\) is tetrahedral, not planar.
2. (b) Borazine is a planar hexagon (\(D_{3h}\)) with zero dipole moment: planar and non-polar.
3. (d) Carbon suboxide \(\ce{O=C=C=C=O}\) is linear (all atoms in a line, hence in a plane) with zero dipole moment.
4. (e) \(\ce{NO2}\) is bent and polar.
5. So 2 molecules: (b) and (d).

Teacher's Note:
1. Check geometry first (planar?), then symmetry for the dipole.

Answer: (a) 4

Explanation:
1. Gerade HOMOs: \(\sigma2p\) and \(\pi^*2p\) are g; \(\pi2p\) is u.
2. \(\ce{B2}\) (6 valence e), \(\ce{C2}\) (8 e) and \(\ce{N2^2+}\) (8 e) end in \(\pi2p\): u.
3. \(\ce{N2}\) (10 e) ends in \(\sigma2p\): g. \(\ce{O2}\) and \(\ce{N2^2-}\) (12 e) and \(\ce{O2^2-}\) (14 e) end in \(\pi^*2p\): g.
4. Count of g: \(\ce{N2}\), \(\ce{O2}\), \(\ce{O2^2-}\), \(\ce{N2^2-}\) = 4.

Teacher's Note:
1. Count valence electrons and see whether the last electron goes into a sigma 2p or pi* 2p (g) or a pi 2p (u) orbital.

Answer: (a) 4

Explanation:
1. \(\ce{O2^2-}\) (14 valence e, all paired), \(\ce{C2}\) (8 e, both pairs in \(\pi2p\)) and \(\ce{N2}\) are diamagnetic.
2. NO (11 valence e), \(\ce{ClO2}\) (19 e) and OF (13 e) have an odd number of electrons: paramagnetic. \(\ce{B2}\) has two unpaired electrons in \(\pi2p\): paramagnetic.
3. Paramagnetic: NO, \(\ce{ClO2}\), OF, \(\ce{B2}\) = 4.

Teacher's Note:
1. Odd-electron species are always paramagnetic; check even-electron ones with the MO diagram.

Answer: (c) 3

Explanation:
1. \(\ce{B2}\): paramagnetic, bond order 1 (integer). \(\ce{C2^2-}\) (bond order 3) and \(\ce{NO+}\) (bond order 3) are diamagnetic.
2. \(\ce{N2+}\) (9 valence e): one unpaired electron, bond order \(\frac{7-2}{2}=2.5\). \(\ce{O2-}\): one unpaired electron, bond order 1.5. OF (13 valence e, like \(\ce{O2-}\)): one unpaired electron, bond order 1.5.
3. Species with both properties: \(\ce{N2+}\), \(\ce{O2-}\), OF = 3.

Teacher's Note:
1. Combine the two tests: unpaired electrons present and a bond order that is not a whole number.