JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 38

Answer: (a) p_x + p_x

Explanation:
1. Overlap along the z-axis (head-on) gives sigma MOs: s with s, s with p_z and p_z with p_z. Sideways overlap of p_x with p_x (or p_y with p_y) gives pi MOs.
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) \( \ce{N2^+} \) and \( \ce{O2^+} \)

Explanation:
1. \( \ce{N2^+} \) has 9 bonding and 4 antibonding electrons: bond order 2.5
2. \( \ce{O2^+} \) has 10 and 5: bond order 2.5
3. The other pairs have orders 1 and 0, 2 and 1, and 2 and 3.

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) both \( \ce{B2} \) and \( \ce{C2^2-} \)

Explanation:
1. Mixing of sigma 2s and sigma 2pz is significant when the 2s-2p energy gap is small, up to \( \ce{N2} \)
2. It is seen in \( \ce{B2} \) and \( \ce{C2^2-} \) (isoelectronic with \( \ce{N2} \)) but not in \( \ce{O2^+} \), which has the normal order.

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) bond order two and isoelectronic

Explanation:
1. Each has 16 valence electrons (so they are isoelectronic)
2. They fill the MOs up to two electrons in pi* orbitals: Nb = 10, Na = 6, bond order = 2.

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) \( \ce{C2} \)

Explanation:
1. Bond orders: \( \ce{B2} \) 1, \( \ce{C2} \) 2, \( \ce{F2} \) 1 and \( \ce{O2^-} \) 1.5
2. Shorter bond means higher order, so \( \ce{C2} \) is the shortest.

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) both \( \ce{O2^-} \) and \( \ce{NO} \)

Explanation:
1. \( \ce{O2^-} \) has one unpaired electron in pi* (17 electrons) and \( \ce{NO} \) has one unpaired electron in pi* (11 valence electrons)
2. \( \ce{CN^-} \) has none.

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) F, F, T, T

Explanation:
1. \( \ce{F2^-} \) has 19 electrons and the odd one goes into sigma* 2pz, so S1 is false
2. The bond orders are 1.5 for \( \ce{O2^-} \) and 2.5 for \( \ce{O2^+} \), so S2 is false
3. \( \ce{NO^+} \) has order 3 against 2.5 for \( \ce{N2^+} \) (S3 true), and \( \ce{C2} \) has order 2 against 1.5 for \( \ce{C2^+} \) (S4 true).

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) metallic bonds are stronger in iron

Explanation:
1. Iron has more electrons (s and d) contributing to the metallic bond, so its metallic bonding is much stronger than in sodium, which supplies only one electron per atom.
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) the delocalisation of valence electrons between metallic kernels

Explanation:
1. In a metal the valence electrons are delocalised over the whole crystal and hold the positive kernels together
2. This gives strong, non-directional cohesion.

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) mercury

Explanation:
1. Mercury has a filled 5d subshell and a very poor contribution of d electrons to bonding
2. Its metallic bonding is the weakest, which is why it is a liquid at room temperature.

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) 3

Explanation:
1. Bonding electrons: 2 + 2 + 4 + 2 = 10
2. Antibonding electrons: 2 + 2 = 4
3. Bond order = (10 - 4)/2 = 3.

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) 1/2

Explanation:
1. \( \ce{He2^+} \) has 3 electrons: 2 in sigma 1s and 1 in sigma* 1s
2. Bond order = (2 - 1)/2 = 1/2.

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) \( \ce{B2} \)

Explanation:
1. \( \ce{B2} \) has bond order 1
2. \( \ce{Be2} \), \( \ce{Ne2} \) and \( \ce{He2} \) have equal numbers of bonding and antibonding electrons (bond order zero) and do not exist as stable molecules.

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) \( \ce{K2O2} \)

Explanation:
1. \( \ce{K2O2} \) contains peroxide \( \ce{O2^2-} \) (bond order 1, 149 pm)
2. \( \ce{KO2} \) contains superoxide (1.5, 128 pm), \( \ce{O2} \) is 2 (121 pm) and \( \ce{O2^+} \) is 2.5 (112 pm).

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) \( \ce{O2} \) < \( \ce{O3} \) < \( \ce{H2O2} \)

Explanation:
1. \( \ce{O2} \) has a double bond (121 pm)
2. \( \ce{O3} \) has a bond order of 1.5 (128 pm)
3. \( \ce{H2O2} \) has a single O-O bond (147 pm).

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) \( \ce{He2^+} \) < \( \ce{O2^-} \) < \( \ce{C2} \) < \( \ce{NO} \)

Explanation:
1. The bond orders are \( \ce{He2^+} \) 0.5, \( \ce{O2^-} \) 1.5, \( \ce{C2} \) 2 and \( \ce{NO} \) 2.5, and the bond energies follow roughly this order (about 240, 395, 600 and 630 kJ per mole).
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 bond length in \( \ce{N2^+} \) is less than in \( \ce{N2} \)

Explanation:
1. Losing a bonding electron from \( \ce{N2} \) gives \( \ce{N2^+} \) with bond order 2.5, so its bond is longer and weaker
2. For \( \ce{O2^+} \) an antibonding electron is lost, so its bond is shorter and stronger than in \( \ce{O2} \)
3. \( \ce{NO^+} \) (order 3) is shorter than \( \ce{NO} \) (2.5)
4. So the statement about \( \ce{N2^+} \) is incorrect.

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) \( \ce{N2O4} \)

Explanation:
1. \( \ce{N2O4} \) has an N-N bond with all electrons paired
2. \( \ce{O2^-} \), \( \ce{NO2} \) and \( \ce{ClO2} \) have odd numbers of electrons and one unpaired electron each.

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) mobile valence electrons

Explanation:
1. Metallic bonding is non-directional and arises from the attraction between positive ions and mobile, delocalised valence electrons.
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) 4

Explanation:
1. \( \ce{N2^+} \) 2.5, \( \ce{N2^-} \) 2.5, \( \ce{O2^+} \) 2.5 and \( \ce{C2^+} \) 1.5 are fractional
2. \( \ce{O2} \) 2, \( \ce{F2} \) 1, \( \ce{B2} \) 1, \( \ce{CN^-} \) 3 and \( \ce{NO^+} \) 3 are whole numbers.

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) 9

Explanation:
1. The orders are \( \ce{NO3^-} \) 1.33, \( \ce{CO3^2-} \) 1.33, \( \ce{F2} \) 1, \( \ce{Cl2} \) 1, \( \ce{Br2} \) 1, \( \ce{O2^2-} \) 1, \( \ce{O2^-} \) 1.5, \( \ce{Li2^+} \) 0.5 and \( \ce{He2^+} \) 0.5, which are all below 2
2. \( \ce{N2^-} \) is 2.5 and \( \ce{O2^2+} \) is 3
3. So there are 9.

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) A, B and C only

Explanation:
1. \( \ce{O2^2+} \), \( \ce{NO^+} \) and \( \ce{CN^-} \) each have 10 valence electrons like \( \ce{N2} \), so order 3
2. \( \ce{CN^+} \) has 8 and has bond order 2.

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, B and C only

Explanation:
1. \( \ce{NO} \), \( \ce{NO2} \) and \( \ce{ClO2} \) all have an odd number of electrons and an unpaired electron
2. \( \ce{N2O4} \) has an even number with all electrons paired.

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, B and C only

Explanation:
1. \( \ce{O2^+} \) (one in pi*), \( \ce{NO} \) (one in pi*) and \( \ce{O2^-} \) (one in pi*, three electrons in the pi* set) each have one
2. \( \ce{B2} \) has two unpaired electrons in its pi orbitals.

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) A and C only

Explanation:
1. Alkali metals get weaker down the group as the atoms grow larger and the single electron is further from the nucleus
2. In transition metals the bonding (the 4d to 5d metals use more diffuse d orbitals) generally gets stronger down the group, so (A) and (C) 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.