JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 21

Answer: (c) B, C, E only

Explanation:
1. By Fajans' rules, covalent character rises with a smaller cation, a larger anion and a higher cation charge.
2. KF < KI and LiF > KF are both true, so B is correct. A is false because KF > KI is wrong.
3. SnCl4 > SnCl2 (higher charge) and CuCl > NaCl (pseudo-inert-gas Cu+ polarises more) are true, so C is correct.
4. D is false because CuCl is more covalent than NaCl. E is correct.
5. Correct: B, C and E.

Teacher's Note:
1. Apply Fajans' rules to each pair separately.
2. Cu+ (18-electron core) polarises anions more than Na+.

Answer: (a) 0, 1, 2, 1

Explanation:
1. In N2 the HOMO is sigma 2pz with 2 electrons, 0 unpaired.
2. In N2+ one electron is removed from sigma 2pz, 1 unpaired.
3. In O2 the HOMO is pi* 2p with 2 electrons in separate orbitals, 2 unpaired.
4. In O2+ one pi* electron is removed, 1 unpaired.
5. Sequence: 0, 1, 2, 1.

Teacher's Note:
1. Fill MOs in the correct order: for N2 the sigma 2pz lies above pi 2p.
2. For O2 the HOMO is the degenerate pi* pair.

Answer: (c) 6

Explanation:
1. Ozone has a central O with one lone pair, forming one double and one single bond.
2. The terminal O with the double bond has 2 lone pairs.
3. The terminal O with the single bond carries a negative charge and has 3 lone pairs.
4. Total = 1 + 2 + 3 = 6.

Teacher's Note:
1. Draw the Lewis structure with formal charges.
2. Resonance does not change the total number of lone pairs.

Answer: (b) Statement I is incorrect but Statement II is correct

Explanation:
1. By the chemical convention the arrow has its tail on the positive centre and its head towards the negative centre.
2. Statement I states the reverse, so it is incorrect.
3. The crossed tail sits on the positive end and the arrow shows the shift of charge towards the negative end, so Statement II is correct.

Teacher's Note:
1. In chemistry the dipole arrow points from + to -.
2. Physics uses the opposite direction.

Answer: (d) diamagnetic, diamagnetic and paramagnetic

Explanation:
1. Li2O contains O2- (oxide) with all electrons paired, so it is diamagnetic.
2. Na2O2 contains the peroxide ion O2 2-, which has bond order 1 and no unpaired electrons, so it is diamagnetic.
3. KO2 contains the superoxide ion O2-, which has one unpaired electron in pi*, so it is paramagnetic.

Teacher's Note:
1. Peroxides are diamagnetic and superoxides are paramagnetic.
2. Check the MO configuration of the anion.

Answer: (b) 3

Explanation:
1. Total electrons: NO2 = 23 (odd), ICl4- = 53 + 68 + 1 = 122 (even), BrF3 = 35 + 27 = 62 (even), ClO2 = 33 (odd), NO2+ = 22 (even), NO = 15 (odd).
2. Even-electron species: ICl4-, BrF3 and NO2+, which are 3.

Teacher's Note:
1. Odd-electron species are paramagnetic radicals.
2. Count total electrons including charge.

Answer: (a) 1, 3 and 3

Explanation:
1. O2 2- has 18 electrons, bond order = (10 - 8)/2 = 1.
2. CO has 14 electrons, bond order = (10 - 4)/2 = 3.
3. NO+ has 14 electrons, bond order = (10 - 4)/2 = 3.

Teacher's Note:
1. CO, NO+, CN- and N2 are isoelectronic with 14 electrons and bond order 3.
2. Bond order = half of (bonding minus antibonding electrons).

Answer: (b) A - IV, B - III, C - II, D - I

Explanation:
1. IF7 has 7 bond pairs and 0 lone pairs.
2. ICl4- has 4 bond pairs and 2 lone pairs.
3. XeF6 has 6 bond pairs and 1 lone pair.
4. XeF2 has 2 bond pairs and 3 lone pairs.
5. So A - IV, B - III, C - II, D - I.

Teacher's Note:
1. Lone pairs on the central atom = (valence electrons - electrons used in bonds + charge) / 2.
2. Use this for every species in the list.

Answer: (d) A, B, D only

Explanation:
1. Oxygen in OF2 has 2 bond pairs and 2 lone pairs, so (A) is correct and the shape is bent, so (D) is correct and (E) is wrong.
2. F is more electronegative than O, so the oxidation state of O is +2 and (C) is wrong.
3. The bond pairs are pulled towards F, so the bond-pair repulsion is less than in H2O. The angle is about 103 degrees, less than 104.5 degrees, so (B) is correct.
4. Correct: A, B and D.

Teacher's Note:
1. OF2 is the only common compound where oxygen has a positive oxidation state.
2. Compare the bond angle with H2O using electronegativity.

Answer: (c) strong hydrogen bond in boric acid

Explanation:
1. Boric acid molecules are linked into layers by hydrogen bonds between the B-OH groups.
2. This extensive hydrogen bonding makes it a solid, whereas BF3 has only weak van der Waals forces between molecules and is a gas.

Teacher's Note:
1. Physical state depends on intermolecular forces.
2. Boric acid has a layered structure held by hydrogen bonds.

Answer: (d) A - II, B - I, C - IV, D - III

Explanation:
1. XeF4 has 4 bond pairs and 2 lone pairs (octahedral arrangement), so it is square planar.
2. SF4 has 4 bond pairs and 1 lone pair (trigonal bipyramidal arrangement), so it is see-saw.
3. NH4+ has 4 bond pairs, so it is tetrahedral.
4. BrF3 has 3 bond pairs and 2 lone pairs, so it is bent T-shaped.
5. So A - II, B - I, C - IV, D - III.

Teacher's Note:
1. Find bond pairs and lone pairs, then use VSEPR.
2. Lone pairs sit in equatorial positions of a trigonal bipyramid.

Answer: (d) BI3 > BBr3 > BCl3 > BF3

Explanation:
1. In BX3 the halogen lone pair back-donates into the empty 2p orbital of boron.
2. The p-pi to p-pi overlap is best for F (2p-2p) and poorer for the larger halogens.
3. Less back-bonding means a more electron-deficient boron and a stronger Lewis acid, so acidity rises BF3 < BCl3 < BBr3 < BI3.

Teacher's Note:
1. The Lewis acidity of boron halides increases down the group.
2. This is opposite to the order based on electronegativity.

Answer: (c) 5

Explanation:
1. Linear: XeF2 (sp3d, three lone pairs), C3O2 (O=C=C=C=O), I3- (three lone pairs on central I), CO2 and BeCl2.
2. Not linear: I3+ (bent), SO2 (bent) and BCl3 (trigonal planar).
3. Total linear species = 5.

Teacher's Note:
1. Three lone pairs on the central atom of an AX2 species give a linear shape.
2. Remember that C3O2 is linear.

Answer: (d) CO3 2- has a single structure, the resonance hybrid of the above three structures

Explanation:
1. Canonical forms are only imaginary and cannot be isolated or identified.
2. They are not in equilibrium and do not exist for any time.
3. The real ion has one structure, the resonance hybrid, with three equal C-O bonds of bond order 4/3.

Teacher's Note:
1. Resonance is not an equilibrium between structures.
2. The hybrid is more stable than any contributing form.

Answer: (d) X - longer, Y - shorter

Explanation:
1. In F2O2 the O-O bond is very short (about 122 pm) with partial double bond character, while in H2O2 it is about 148 pm. So X is longer.
2. O-H (about 97 pm) is shorter than O-F (about 158 pm), as H is smaller than F. So Y is shorter.

Teacher's Note:
1. Electronegative F pulls electron density and gives O-O multiple bond character.
2. Compare O-H and O-F with atomic sizes.