JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 25

Answer: (a) \(\ce{BF4-}\)

Explanation:
1. BF4- has four bond pairs and no lone pair on boron, so it is a regular tetrahedron (sp3).
2. XeF4 is square planar, SF4 is see-saw and [Ni(CN)4]2- is square planar (dsp2).

Teacher's Note:
1. Ni(II) with strong field CN- gives square planar complexes.
2. Regular tetrahedral means identical bonds and no lone pair.

Answer: (c) H2S < NH3 < SiH4 < BF3

Explanation:
1. Bond angles: H2S about 92 degrees, NH3 107 degrees, SiH4 109.5 degrees, BF3 120 degrees.
2. So the order is H2S < NH3 < SiH4 < BF3.

Teacher's Note:
1. More lone pairs decrease the angle.
2. Planar sp2 gives 120 degrees, which is the largest here.

Answer: (b) sp2 and sp3

Explanation:
1. Boron has three sigma bonds and no lone pair, so it is sp2.
2. Each oxygen in B-O-H has two sigma bonds and two lone pairs, so it is sp3.

Teacher's Note:
1. Count sigma bonds plus lone pairs.
2. Boric acid is planar around B with hydrogen bonding between the layers.

Answer: (a) \(\ce{SH2}\)

Explanation:
1. SO2 is about 119 degrees, H2O 104.5 degrees, NH3 107 degrees and H2S about 92 degrees.
2. The large, less electronegative S pulls bond pairs far away from the centre, so H2S has the smallest angle.

Teacher's Note:
1. Hydrides of group 16: H2O 104.5, H2S 92, H2Se 91, H2Te 90 degrees.
2. Lower electronegativity of the central atom gives a smaller angle.

Answer: (c) \(\ce{NO2}\) and \(\ce{O3}\)

Explanation:
1. NO2 is bent and polar, O3 is bent with a small dipole moment.
2. SiF4 (tetrahedral) and CO2 (linear) have zero dipole moments because of symmetry.

Teacher's Note:
1. Symmetric molecules have a zero net dipole.
2. Bent molecules with unequal vectors have a net dipole.

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

Explanation:
1. A chemical bond forms when the net force between atoms is attractive, i.e. attraction overcomes repulsion.
2. Covalent bonds share electrons rather than transfer them.
3. Pure water has a small ionisation (H3O+ and OH-), and HF is more polar than HBr.

Teacher's Note:
1. Bond formation leads to a lower energy of the system.
2. The potential energy curve has its minimum at the bond length.

Answer: (b) FeCl3 is more covalent than FeCl2

Explanation:
1. Fe3+ has a higher charge and smaller size than Fe2+, so its polarising power is higher.
2. FeCl3 therefore has more covalent character than FeCl2.

Teacher's Note:
1. Higher oxidation state of the cation means greater covalency.
2. This is Fajans' rule.

Answer: (c) HCl

Explanation:
1. HF has strong hydrogen bonding and a high boiling point.
2. The others rise with size and van der Waals forces in the order HCl < HBr < HI.
3. So HCl has the lowest boiling point.

Teacher's Note:
1. The trend in boiling points is HCl < HBr < HI < HF.
2. HF is the exception because of hydrogen bonding.

Answer: (a) equally shared between them

Explanation:
1. Two identical atoms have equal electronegativity.
2. The shared pair is therefore attracted equally, forming a non-polar covalent bond.

Teacher's Note:
1. A non-polar covalent bond forms between identical atoms.
2. The shared electrons have opposite spins.

Answer: (c) It has two three-centre two-electron bonds

Explanation:
1. Diborane has four terminal B-H bonds and two bridging B-H-B three-centre two-electron (banana) bonds.
2. Terminal and bridge bonds differ in length and angle, and the bridge hydrogens lie above and below the plane.
3. Only (c) is correct.

Teacher's Note:
1. Diborane is electron deficient.
2. Its 12 valence electrons cannot form eight normal two-centre bonds.

Answer: (a) longer, since the bond order in NO is 2.5 while that in NO+ is 3

Explanation:
1. A higher bond order means a shorter bond.
2. NO has bond order 2.5 and NO+ has 3, so NO+ has the shorter bond and NO the longer.

Teacher's Note:
1. Bond length is inversely related to bond order.
2. NO+ has bond order 3 like N2 and CO.

Answer: (a) \(\ce{ClO3-}\), \(\ce{SO3^{2-}}\)

Explanation:
1. ClO3- and SO3 2- both have 42 electrons and a pyramidal structure with one lone pair on the central atom.
2. The other pairs have different electron counts or shapes (CO3 2- and NO3- are planar).

Teacher's Note:
1. Isoelectronic species have the same total electrons.
2. They are iso-structural only if the lone pair count is also equal.

Answer: (c) \(\ce{PCl3}\)

Explanation:
1. PCl3 has three P-Cl bonds and one lone pair on P.
2. H2O and NH2- have two bond pairs and two lone pairs, and BF3 has three bond pairs and no lone pair.

Teacher's Note:
1. Pyramidal molecules have 3 bond pairs and 1 lone pair.
2. NH3, PCl3 and H3O+ are examples.

Answer: (d) Every AB5 molecule does in fact have a square pyramidal structure

Explanation:
1. AB5 molecules such as PCl5 are trigonal bipyramidal, not square pyramidal.
2. Only AB5E molecules like BrF5 are square pyramidal.
3. The other statements are correct.

Teacher's Note:
1. Shape depends on lone pairs as well as bond pairs.
2. AB5 with no lone pair is trigonal bipyramidal.

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

Explanation:
1. NH4+ is regular tetrahedral with 109.5 degrees.
2. NH3 is 107, PCl3 is about 100 and SCl2 is about 103 degrees because of lone pairs.

Teacher's Note:
1. Lone pairs reduce angles from the ideal 109.5 degrees.
2. No lone pair gives the largest angle in this set.

Answer: (c) 5

Explanation:
1. Cyclic metaphosphoric acid forms a ring of alternating P and O atoms.
2. A ring of 5 phosphorus atoms contains 5 bridging oxygens, i.e. 5 P-O-P linkages.

Teacher's Note:
1. In a cyclic n-mer the number of bridging linkages equals n.
2. Each P also carries one =O and one OH.

Answer: (c) 6, 5, 3, 3

Explanation:
1. P4O10 has 6 bridging oxygen atoms, i.e. 6 P-O-P bonds.
2. The borate anion \(\ce{[B4O5(OH)4]^{2-}}\) has 5 B-O-B bridges.
3. The cyclic trisilicic acid ring has 3 Si-O-Si linkages and cyclic S3O9 has 3 S-O-S linkages.

Teacher's Note:
1. Count bridging oxygens carefully; terminal O atoms do not count.
2. The cyclic trimers have as many bridges as ring members.

Answer: (a) 4

Explanation:
1. Regular tetrahedra need four identical bonds and no lone pair: CCl4, SiF4, BF4- and SiO4 4-.
2. CHCl3 is distorted since H and Cl differ.
3. SO3 is planar and NH3 and H2O have lone pairs. So the count is 4.

Teacher's Note:
1. Regular means all bond lengths and bond angles are equal.
2. CHCl3 is tetrahedral but irregular.

Answer: (c) \(\ce{O2+}\)

Explanation:
1. Bond orders: O2 2- is 1, O2- is 1.5, O2 is 2 and O2+ is 2.5.
2. The highest bond order gives the shortest bond, i.e. O2+.

Teacher's Note:
1. Bond length is inversely proportional to bond order.
2. O2+ is formed by removing an antibonding electron from O2.

Answer: (a) \(\ce{O2^{2-} < O2- < O2 < O2+}\)

Explanation:
1. Bond orders are O2 2- = 1, O2- = 1.5, O2 = 2 and O2+ = 2.5.
2. Bond strength increases with bond order, so O2 2- < O2- < O2 < O2+.

Teacher's Note:
1. Add or remove electrons from the antibonding pi* orbitals to change the bond order.
2. Fewer antibonding electrons means a stronger bond.