JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 26

Answer: (a) \( \ce{BCl3} \)

Explanation:
1. In \( \ce{BCl3} \) boron has only six electrons around it
2. In \( \ce{ICl} \), \( \ce{NH3} \) and \( \ce{PCl3} \) the central atoms have octets.

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) both \( \ce{BCl3} \) and \( \ce{PCl5} \)

Explanation:
1. \( \ce{BCl3} \) has a sextet on B and \( \ce{PCl5} \) has ten electrons on P (expanded octet)
2. \( \ce{CO2} \) obeys the octet rule.

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) \( \ce{N3-} \), \( \ce{CNO-} \) and \( \ce{NCN^2-} \) are all isoelectronic with \( \ce{CO2} \)

Explanation:
1. \( \ce{CO2} \) has 22 electrons
2. \( \ce{N3-} \): 21+1 = 22; \( \ce{CNO-} \): 6+7+8+1 = 22; \( \ce{NCN^2-} \): 7+6+7+2 = 22
3. All have 22 electrons.

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) \( \ce{CO2} \)

Explanation:
1. In \( \ce{CO2} \) each atom has eight electrons
2. In \( \ce{BrF5} \), \( \ce{SF6} \) and \( \ce{IF7} \) the central atom has 12, 12 and 14 electrons.

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) 1.5

Explanation:
1. Two Kekule structures give one double bond and one single bond for each pair: (2+1)/2 = 1.5.
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: (c) \( \ce{CO3^2-} \)

Explanation:
1. Bond orders are about 1.33 in \( \ce{CO3^2-} \), 2 in \( \ce{CO2} \), 3 in \( \ce{CO} \) and 1.5 in \( \ce{CH3COO-} \)
2. The lowest bond order gives the weakest bond.

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) CO < \( \ce{CO2} \) < \( \ce{CO3^2-} \)

Explanation:
1. Bond order: CO 3, \( \ce{CO2} \) 2, \( \ce{CO3^2-} \) 1.33
2. A higher bond order means a shorter bond.

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) p orbitals can only overlap sidewise

Explanation:
1. p orbitals can overlap head-on and give sigma bonds (as in \( \ce{F2} \)), so the first statement is false
2. The others are correct.

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 sigma bond is weaker than a pi bond

Explanation:
1. End-on overlap (sigma) is better than sidewise overlap (pi), so a sigma bond is stronger than a pi bond
2. The first statement is wrong.

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) 5

Explanation:
1. \( \ce{CH3OH} \) has three C-H bonds, one C-O bond and one O-H bond
2. All five are sigma 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: (d) 4

Explanation:
1. The bonds are H-O, O-N, N-O and N=O. Each has one sigma bond, so 4
2. The pi bond in N=O adds one pi bond.

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) 7 and 3

Explanation:
1. The carbon chain has 3 C-C links (3 sigma) and 4 C-H bonds (4 sigma): 7 sigma
2. The three double bonds give 3 pi 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: (c) 4 and 0

Explanation:
1. N has no lone pair in \( \ce{NO3-} \); it forms three sigma bonds and takes part in one delocalised pi bond, so 4 bond pairs and 0 lone pairs.
2. Only option (c) 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: (a) \( \ce{BeCl2} \) (gas phase)

Explanation:
1. Be in monomeric \( \ce{BeCl2} \) has only four electrons in its valence shell
2. The others have complete octets.

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) \( \ce{SnCl4} \)

Explanation:
1. \( \ce{SnCl4} \) follows the octet rule (4 bonds, 8 electrons)
2. \( \ce{BF3} \) is electron-deficient, \( \ce{XeF6} \) is expanded and \( \ce{ClO3} \) has an odd number of electrons.

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) \( \ce{ClO4-} \) > \( \ce{ClO3-} \) > \( \ce{ClO2-} \) > \( \ce{ClO-} \)

Explanation:
1. Chlorine uses all its valence bonds spread over the oxygens: +7/4 = 1.75, +5/3 = 1.67, +3/2 = 1.5 and +1/1 = 1.
2. Only option (b) 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) The sigma bond energy is about 264 kJ/mol and the pi bond energy about 347 kJ/mol

Explanation:
1. The sigma bond energy (about 347 kJ/mol) is greater than that of a pi bond (about 264 kJ/mol), so the first statement has the values reversed.
2. Only option (b) 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: (a) 3 and 4

Explanation:
1. \( \ce{N#C-C#N} \) has three sigma bonds (two C-N and one C-C) and two triple bonds that give 4 pi bonds.
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: (c) 2

Explanation:
1. Oxygen forms two single bonds to fluorine, so there are 2 bond pairs and 2 lone pairs.
2. Only option (c) 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: (c) 3, 2, 1

Explanation:
1. Xe has 8 valence electrons: 2, 4 and 6 are used in bonds, leaving 6, 4 and 2 electrons, i.e. 3, 2 and 1 lone pairs.
2. Only option (c) 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: (a) different, with 1, 0 and 2 lone pairs on the central atom

Explanation:
1. \( \ce{SF4} \) is see-saw (1 lp), \( \ce{CF4} \) tetrahedral (0 lp) and \( \ce{XeF4} \) square planar (2 lp).
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: (d) 1

Explanation:
1. Xe has 8 valence electrons; four electrons go to the four Xe-F bonds and two to the Xe-O bond, leaving 2 electrons, i.e. 1 lone pair.
2. Only option (d) 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.