JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 34

Answer: (a) The electron pairs are at the corners of a trigonal bipyramid with one equatorial pair unshared, and its structure is analogous to \( \ce{SF4} \)

Explanation:
1. Iodine has 8 valence electrons here (7 + 1 charge); four sigma bonds use 4 pairs of the 5 sigma-framework positions and one lone pair remains, giving sp³d with the lone pair equatorial
2. The shape is see-saw, like \( \ce{SF4} \), not T-shaped.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) \( \ce{BF3 + F^- -> BF4^-} \) (boron)

Explanation:
1. Boron changes from 3 pairs (trigonal planar, sp²) to 4 pairs (tetrahedral, sp³)
2. In the other cases the electron-pair geometry stays tetrahedral (N, O) or trigonal planar (S in \( \ce{SO2} \) and \( \ce{SO3} \)).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (b) \( \ce{I3^-} \)

Explanation:
1. The central I in \( \ce{I3^-} \) has 2 bond pairs and 3 lone pairs
2. In \( \ce{H2O} \), \( \ce{SCl2} \) and each O of \( \ce{O2F2} \), the central atom has 2 bond pairs and 2 lone pairs.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (b) sp, sp², sp

Explanation:
1. \( \ce{N3^-} \) (linear, 2 sigma bonds, no lone pair on central N) is sp
2. In \( \ce{NOCl} \) the central N has 2 sigma bonds and 1 lone pair, so sp² (bent)
3. In \( \ce{N2O} \) (N=N=O) the central N has 2 sigma bonds and no lone pair, so sp.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) sp and sp³

Explanation:
1. The first carbon is in a triple bond, so sp
2. The last carbon is the saturated methyl group with four single bonds, so sp³.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) sp², sp³

Explanation:
1. \( \ce{BF3} \) has 3 sigma bonds and no lone pair, so sp². \( \ce{BF4^-} \) has 4 sigma bonds, so sp³.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

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

Explanation:
1. \( \ce{NH4^+} \) has no lone pair and is tetrahedral (109.5 degrees)
2. \( \ce{NH3} \) is about 107, \( \ce{SCl2} \) about 103 and \( \ce{PCl3} \) about 100 degrees.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) 3

Explanation:
1. The cyclic trimer \( \ce{(SO3)3} \) has a six-membered ring of alternating S and O atoms with each S carrying two terminal oxygens
2. The ring contains three S-O-S bridges.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) \( \ce{H2S2O7} \)

Explanation:
1. \( \ce{H2S2O7} \) (pyrosulphuric acid) has the two S atoms bridged through oxygen (S-O-S)
2. \( \ce{H2S2O5} \), \( \ce{H2S2O3} \) and the polythionic acid \( \ce{H2S4O6} \) each have S-S bonds.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) There is one Si-Si bond

Explanation:
1. Two \( \ce{SiO4} \) tetrahedra share one bridging oxygen
2. Each Si has four Si-O bonds, giving 8 in all, and each Si is sp³. There is no Si-Si bond; the units are linked through oxygen.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) the value of n is 12

Explanation:
1. Each Si is bonded to four oxygens (two bridging and two terminal), so statement B is false
2. Of the 18 oxygens, 6 are bridging and 12 are terminal, so C is false
3. The charge is 6(+4) + 18(-2) = -12, so n = 12.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) b and c

Explanation:
1. In \( \ce{NH3} \) and \( \ce{PCl3} \) the central atoms already have lone pairs and three unpaired p electrons, so they hybridise from the ground state
2. For \( \ce{BCl3} \) and \( \ce{BeF2} \) one s electron must first be promoted to a p orbital to give the required unpaired electrons.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (c) p_z and d_z2

Explanation:
1. The sp³d set can be viewed as sp² (s, p_x, p_y) forming the equatorial plane, plus p_z and d_z2 forming the two axial bonds.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (c) 2

Explanation:
1. A T-shape comes from a trigonal bipyramid with 3 bond pairs and 2 lone pairs in the equatorial plane, as in \( \ce{ClF3} \).
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) \( \ce{H2S} \) < \( \ce{NH3} \) < \( \ce{CH4} \) < \( \ce{BF3} \)

Explanation:
1. The angles are \( \ce{H2S} \) about 92, \( \ce{NH3} \) 107, \( \ce{CH4} \) 109.5 and \( \ce{BF3} \) 120 degrees.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) \( \ce{ClF3} \)

Explanation:
1. \( \ce{ClF3} \) is T-shaped, with two axial Cl-F bonds that are longer than the single equatorial bond
2. In \( \ce{NF3} \) and \( \ce{BF3} \) all bonds are equivalent.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (b) \( \ce{NO2^+} \) < \( \ce{NO2^-} \)

Explanation:
1. \( \ce{NO2^+} \) has N-O bond order 2 (115 pm) and \( \ce{NO2^-} \) has bond order 1.5 (124 pm), so the first is shorter
2. \( \ce{BF3} \) has partial pi bonding, so its B-F is shorter than that in \( \ce{BF4^-} \)
3. C-F is shorter than C-Cl, and \( \ce{CH3^+} \) (sp²) has shorter C-H bonds than \( \ce{CH4} \).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) A single N-N bond is weaker than a single P-P bond

Explanation:
1. The N-N single bond is weak (about 160 kJ per mole) because of lone pair-lone pair repulsion between the small N atoms, while P-P is about 200 kJ per mole
2. Triple bonds show the reverse, \( \ce{N2} \) (945 kJ per mole) being much stronger than \( \ce{P2} \).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) \( \ce{S3O9} \) contains no S-S linkage

Explanation:
1. The \( \ce{S3O9} \) trimer is a ring of S-O-S links
2. In dithionate \( \ce{S2O6^2-} \) the S atoms are directly bonded (S-S) with no O-O. \( \ce{(HPO3)3} \) has P-O-P ring links
3. \( \ce{S2O8^2-} \) has an O-O peroxide link between two sulphate units.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (c) The 2p_x and 2p_y orbitals of carbon can hybridise to give two new, more stable orbitals

Explanation:
1. Mixing two p orbitals gives no gain in stability, because hybrids are stabilised by including s-character
2. The other three statements are standard facts about hybridisation.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.