JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 39

Answer: (c) the bond order decreases when \( \ce{N2} \) ionises to \( \ce{N2^+} \) and increases when \( \ce{O2} \) ionises to \( \ce{O2^+} \)

Explanation:
1. Ionisation removes a bonding electron from \( \ce{N2} \) (order 3 to 2.5) but an antibonding electron from \( \ce{O2} \) (order 2 to 2.5).
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) Sc > Ca > K

Explanation:
1. Boiling is a better guide to metallic bond strength than melting
2. More valence electrons (d electrons for Sc) and smaller atoms give stronger bonding, so the boiling points are about 2836 K for Sc, 1757 K for Ca and 1032 K for K.

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) Zn > Cd > Hg for both

Explanation:
1. Metallic bonding weakens from Zn to Hg because the filled d subshell contributes less and the atoms are larger
2. Boiling points are 1180, 1040 and 630 K, and melting points are 693, 594 and 234 K. Mercury is the lowest in both.

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) gold

Explanation:
1. Caesium (302 K), gallium (303 K) and mercury (234 K) all melt near or below room temperature
2. Gold, a d-block metal with strong bonding, melts at 1337 K.

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) all of these: electrical conduction, thermal conduction and malleability

Explanation:
1. The mobile electrons carry charge and heat
2. When layers of ions slide over one another the electron sea readjusts, so the metal deforms without breaking.

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) (IV), (i), (S)

Explanation:
1. \( \ce{C2} \) has bond order 2, all electrons paired, and its HOMO is the bonding pi 2p set, so (IV), (i), (S) is correct
2. \( \ce{F2} \) has order 1, not 2, and no s-p mixing.

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) (I), (iii), (R)

Explanation:
1. \( \ce{B2} \) has bond order 1 and two unpaired electrons, so it is paramagnetic, which matches (I), (iii), (R)
2. \( \ce{F2} \) has order 1 (not 2.5), no s-p mixing, and \( \ce{O2} \) has order 2 and is paramagnetic.

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) \( \ce{O2^-} \)

Explanation:
1. \( \ce{O2^-} \) has 17 electrons and one unpaired electron in a pi* orbital
2. \( \ce{N2} \), \( \ce{F2} \) and \( \ce{O2^2-} \) have 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: (c) It is paramagnetic and has a higher bond order than \( \ce{O2} \)

Explanation:
1. \( \ce{O2^+} \) has 15 valence electrons with one unpaired electron (paramagnetic)
2. Its bond order is 2.5, higher than the 2 of \( \ce{O2} \).

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) \( \ce{NO^-} \)

Explanation:
1. \( \ce{CO} \), \( \ce{NO^+} \), \( \ce{CN^-} \) and \( \ce{N2} \) are isoelectronic with 10 valence electrons and bond order 3
2. \( \ce{NO^-} \) has 16 valence electrons and 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) \( \ce{C2} \)

Explanation:
1. Without mixing the order is sigma 2pz below pi 2p
2. \( \ce{B2} \) has 6 valence electrons: sigma 2s2, sigma* 2s2 and sigma 2pz2, all paired, so it is diamagnetic
3. \( \ce{C2} \) has 8: after sigma 2pz2 the last two electrons enter separate pi orbitals, so it has two unpaired electrons
4. \( \ce{Be2} \) and \( \ce{N2} \) are diamagnetic.

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 and C only

Explanation:
1. \( \ce{C2^2-} \) has all electrons paired
2. \( \ce{O2^2+} \) has order 3, a shorter bond than \( \ce{O2} \)
3. \( \ce{N2^+} \) and \( \ce{N2^-} \) both have order 2.5
4. \( \ce{He2^+} \) (order 0.5) is lower in energy than \( \ce{He} \) + \( \ce{He^+} \).

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

Explanation:
1. The bond orders are 1 for \( \ce{O2^2-} \), 1.5 for \( \ce{O2^-} \), 2 for \( \ce{O2} \) and 2.5 for \( \ce{O2^+} \), and bond strength increases with bond order.
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.

Answer: (a) the bond length in \( \ce{NO} \) is greater than in \( \ce{NO^+} \)

Explanation:
1. \( \ce{NO^+} \) has a higher bond order, since the antibonding electron of \( \ce{NO} \) is lost on ionisation, so its bond is shorter (106 pm against 115 pm).
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) \( \ce{H2} \)

Explanation:
1. \( \ce{H2} \) has two electrons paired in sigma 1s
2. \( \ce{He2^+} \) (3 electrons), \( \ce{H2^+} \) (1 electron) and \( \ce{H2^-} \) (3 electrons) each have an unpaired electron.

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) \( \ce{NO -> NO^+} \)

Explanation:
1. \( \ce{NO} \) (order 2.5, paramagnetic) becomes \( \ce{NO^+} \) (order 3, diamagnetic)
2. \( \ce{O2} \) to \( \ce{O2^+} \) increases the order but stays paramagnetic
3. \( \ce{N2} \) and \( \ce{C2} \) lose bonding electrons, so their orders fall.

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) \( \ce{CN^-} \) and \( \ce{NO^+} \)

Explanation:
1. \( \ce{CN^-} \) and \( \ce{NO^+} \) each have 10 valence electrons with bond order 3
2. \( \ce{CN^+} \) has 8 (order 2), and \( \ce{O2^-} \) has 1.5.

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) \( \ce{O2^2+} \)

Explanation:
1. \( \ce{O2^2+} \) has 14 valence electrons like \( \ce{N2} \) and so bond order 3, the highest of the four, so its bond is the shortest.
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) \( \ce{C2} \) and \( \ce{N2} \) only

Explanation:
1. \( \ce{C2} \) (all electrons paired in pi orbitals) and \( \ce{N2} \) are diamagnetic
2. \( \ce{O2} \) and its heavier analogue \( \ce{S2} \) have two unpaired electrons in pi* orbitals and are paramagnetic.

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) \( \ce{H2^2+} \) and \( \ce{He2} \)

Explanation:
1. \( \ce{H2^2+} \) has no electrons and \( \ce{He2} \) has equal bonding and antibonding electrons (order 0), so neither exists
2. The other pairs each contain a species with a positive bond order such as \( \ce{H2^+} \) (0.5) or \( \ce{He2^2+} \) (1).

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) \( \ce{CO} \)

Explanation:
1. \( \ce{CO} \) has 10 valence electrons, all paired
2. \( \ce{O2} \) and \( \ce{B2} \) each have two unpaired electrons and \( \ce{NO} \) has one.

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) \( \ce{N2} \) > \( \ce{O2} \) > \( \ce{O2^-} \)

Explanation:
1. The bond orders are 3, 2 and 1.5 respectively, so the dissociation energies fall in the order \( \ce{N2} \) (942 kJ per mole), \( \ce{O2} \) (498) and \( \ce{O2^-} \) (about 395).
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) \( \ce{NO} \)

Explanation:
1. \( \ce{NO} \) has 11 valence electrons and so one unpaired electron
2. \( \ce{N2} \), \( \ce{CO} \) and \( \ce{O3} \) have 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: (d) Assertion is correct, reason is incorrect

Explanation:
1. The bonding MO (psi A + psi B) arises from constructive interference of the two waves, which piles up electron density between the nuclei
2. So the assertion is right but the reason (destructive interference) 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) both are stable

Explanation:
1. \( \ce{Li2^+} \) has 5 electrons (sigma 2s1) and \( \ce{Li2^-} \) has 7 (sigma 2s2 sigma* 2s1), so each has bond order (2-1)/2 = 0.5 or (1-0)/2 = 0.5
2. Both orders are positive, so both can exist as weakly bound species.

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) \( \ce{N2^+} \)

Explanation:
1. \( \ce{N2^+} \) has 4 electrons in two pi orbitals (two pi bonds) and one electron in the sigma 2p orbital (half a sigma bond)
2. \( \ce{N2} \) has a full sigma bond
3. In \( \ce{O2} \) and \( \ce{O2^+} \) the pi* electrons reduce the pi bonding.

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.