JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 35

Answer: (c) borax

Explanation:
1. In the borax anion \( \ce{[B4O5(OH)4]^2-} \) two boron atoms are 3-coordinate (sp²) and two are 4-coordinate (sp³)
2. Diborane has only sp³ boron and borazole only sp² boron.

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) 5

Explanation:
1. \( \ce{BF3} \), \( \ce{BCl3} \), \( \ce{CO3^2-} \), \( \ce{SO3} \) and \( \ce{XeF4} \) are planar
2. \( \ce{NH3} \), \( \ce{NCl3} \) and \( \ce{PCl3} \) are pyramidal
3. So there are 5.

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) 6

Explanation:
1. \( \ce{H2S} \), \( \ce{NH3} \), \( \ce{PH3} \), \( \ce{PF3} \), \( \ce{NH2^-} \) and \( \ce{H2O} \) have lone pairs on the central atom with angles under 109.5 degrees
2. \( \ce{SO4^2-} \), \( \ce{CCl4} \), \( \ce{SiH4} \) and \( \ce{NH4^+} \) are tetrahedral at 109.5, and \( \ce{SO3} \) is planar at 120
3. So there are 6.

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) 16

Explanation:
1. Each Xe=O bond has one pi bond
2. The O counts are 1, 2, 3, 4, 3, 1 and 2, giving 16.

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) 6

Explanation:
1. \( \ce{P4O6} \) has a \( \ce{P4} \) tetrahedron with six edges, and each edge is bridged by one oxygen
2. So there are six P-O-P links.

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) A and C only

Explanation:
1. Hybridisation is the mixing of orbitals, not of electrons
2. Filled, half-filled and empty orbitals can all hybridise (for example the lone-pair orbital of N in \( \ce{NH3} \))
3. Hybrid orbitals may hold lone pairs, so (D) is false.

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) \( \ce{H2C=CH-CN} \) and \( \ce{H2C=CH-C#CH} \) both

Explanation:
1. In \( \ce{H2C=CH-C#N} \) the sequence of C, C, C, N is sp², sp², sp, sp
2. In \( \ce{H2C=CH-C#CH} \) it is sp², sp², sp, sp
3. The cumulene \( \ce{H2C=C=C=CH2} \) is sp², sp, sp, sp², and the diyne is all 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: (c) A and C only

Explanation:
1. All three have sp³ nitrogen, with 2, 1 and 0 lone pairs respectively
2. Their shapes are bent, pyramidal and tetrahedral and their angles are about 104, 107 and 109.5 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) B, C and D

Explanation:
1. \( \ce{NH3} \) is 107 degrees
2. \( \ce{H2S} \) (92), \( \ce{PH3} \) (93.5) and the T-shaped \( \ce{ICl3} \) (about 88) all have angles near 90 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: (d) B, C and D

Explanation:
1. All boron atoms in borax are +3, so (A) is false
2. In \( \ce{B2H6} \) boron is also +3 on the usual count
3. Two borons are 4-coordinate (sp³) and two are 3-coordinate (sp²), so (C) and (D) are true.

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) A, B and C

Explanation:
1. Sulphur reaches the +6 state with strongly oxidising ligands such as F or O, but H and I cannot stabilise S(VI), so \( \ce{SH6} \) and \( \ce{SI6} \) are not known
2. Fluorine has no d orbitals and cannot reach the +7 state, so \( \ce{HFO4} \) cannot exist
3. \( \ce{HClO3} \) is chloric acid.

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) (I), (ii), (S)

Explanation:
1. \( \ce{XeF4} \) has 4 bond pairs and 2 lone pairs, so 6 pairs, sp³d² and square planar.
2. Only option (d) 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) (IV), (iii), (P)

Explanation:
1. In \( \ce{CH3OCH3} \) the C-O-C angle is about 111 degrees, because the two methyl groups repel each other more than lone pairs would allow in water
2. It is sp³ and bent
3. In \( \ce{SiF4} \) the angle is exactly 109.5.

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) (II), (iv), (R)

Explanation:
1. \( \ce{ClF3} \) is T-shaped, planar and sp³d
2. \( \ce{SiF4} \) is a tetrahedron
3. \( \ce{XeF4} \) is planar but is square planar sp³d², not T-shaped sp³, so the third option mixes entries.

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) sp, sp², sp³

Explanation:
1. \( \ce{NO2^+} \) has 2 sigma bonds and no lone pair (sp)
2. \( \ce{NO3^-} \) has 3 sigma bonds (sp²)
3. \( \ce{NH4^+} \) has 4 sigma bonds (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) 3

Explanation:
1. Trimetaphosphoric acid is a six-membered ring of alternating P and O atoms, each P carrying one =O and one OH. The ring has three P-O-P bridges.
2. Only option (a) agrees with this working.

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) zero

Explanation:
1. \( \ce{S3O9} \) is a cyclic trimer in which the sulphur atoms are linked through bridging oxygens (S-O-S)
2. There is no S-S bond, so the answer is zero.

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) \( \ce{NO3^-} \) and \( \ce{CO3^2-} \)

Explanation:
1. \( \ce{NO3^-} \) has 32 electrons and \( \ce{CO3^2-} \) also has 32; both are planar triangular
2. \( \ce{SO3} \) has 40 electrons and \( \ce{ClO3^-} \) has 42, and \( \ce{ClO3^-} \) is pyramidal.

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) 75

Explanation:
1. The P-P-P angle in \( \ce{P4} \) is only 60 degrees, so the bonds are formed from almost pure p orbitals with the lone pair in an s-rich orbital
2. The accepted value for the p-character of the P-P bonding orbitals is 75 per cent.

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) \( \ce{OSF2} \)

Explanation:
1. \( \ce{OSF2} \) has 3 sigma bonds and 1 lone pair on S, so it is pyramidal
2. \( \ce{SO3} \) and \( \ce{SiO3^2-} \) are planar triangular and \( \ce{BrF3} \) is 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.