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Chemical Bonding and Molecular Structure question

2022 · 25 Jul · Shift 1 · Q12
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Chemical Bonding and Molecular Structure question

2022 · 25 Jul · Shift 1 · Q12

JEE MainChemistryChemical Bonding and Molecular StructureNumerical+4 / −1
Among the following species N2, N2+,N2−,N22−,O2,O2+,O2−,O22−\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{N}_{2}^{-}, \mathrm{N}_{2}^{2-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}N2​, N2+​,N2−​,N22−​,O2​,O2+​,O2−​,O22−​ the number of species showing diamagnesim is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Criterion for diamagnetism

A species is diamagnetic if all electrons are paired in its molecular orbitals.

So we examine the MO electronic configurations of the given species.

  1. For nitrogen species

For molecules up to nitrogen, the MO order is:

σ(1s), σ∗(1s), σ(2s), σ∗(2s), π(2px)=π(2py), σ(2pz)\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \pi(2p_x)=\pi(2p_y),\ \sigma(2p_z)σ(1s), σ∗(1s), σ(2s), σ∗(2s), π(2px​)=π(2py​), σ(2pz​)

We only need the valence part.

(a) N2\mathrm{N_2}N2​

Each N has 7 electrons, so total electrons =14=14=14. Valence MO filling:

(σ2s)2(σ∗2s)2(π2p)4(σ2p)2(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p)^4(\sigma 2p)^2(σ2s)2(σ∗2s)2(π2p)4(σ2p)2

All electrons are paired. So, N2\mathrm{N_2}N2​ is diamagnetic.

(b) N2+\mathrm{N_2^+}N2+​

One electron removed from the HOMO of N2\mathrm{N_2}N2​, i.e. from σ(2p)\sigma(2p)σ(2p):

(σ2s)2(σ∗2s)2(π2p)4(σ2p)1(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p)^4(\sigma 2p)^1(σ2s)2(σ∗2s)2(π2p)4(σ2p)1

One unpaired electron is present. So, N2+\mathrm{N_2^+}N2+​ is paramagnetic.

(c) N2−\mathrm{N_2^-}N2−​

One electron added to the next MO after σ(2p)\sigma(2p)σ(2p), i.e. π∗(2p)\pi^*(2p)π∗(2p):

(σ2s)2(σ∗2s)2(π2p)4(σ2p)2(π∗2p)1(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p)^4(\sigma 2p)^2(\pi^*2p)^1(σ2s)2(σ∗2s)2(π2p)4(σ2p)2(π∗2p)1

One unpaired electron is present. So, N2−\mathrm{N_2^-}N2−​ is paramagnetic.

(d) N22−\mathrm{N_2^{2-}}N22−​

Two electrons added to degenerate π∗(2p)\pi^*(2p)π∗(2p) orbitals:

(σ2s)2(σ∗2s)2(π2p)4(σ2p)2(π∗2p)2(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p)^4(\sigma 2p)^2(\pi^*2p)^2(σ2s)2(σ∗2s)2(π2p)4(σ2p)2(π∗2p)2

By Hund’s rule, the two electrons occupy separate degenerate orbitals with parallel spins. Hence there are two unpaired electrons. So, N22−\mathrm{N_2^{2-}}N22−​ is paramagnetic.

  1. For oxygen species

For oxygen and fluorine, the MO order is:

σ(1s), σ∗(1s), σ(2s), σ∗(2s), σ(2pz), π(2px)=π(2py), π∗(2px)=π∗(2py)\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \sigma(2p_z),\ \pi(2p_x)=\pi(2p_y),\ \pi^*(2p_x)=\pi^*(2p_y)σ(1s), σ∗(1s), σ(2s), σ∗(2s), σ(2pz​), π(2px​)=π(2py​), π∗(2px​)=π∗(2py​)

Again, valence part is enough.

(e) O2\mathrm{O_2}O2​

Each O has 8 electrons, so total electrons =16=16=16. Valence MO filling:

(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2(\sigma 2s)^2(\sigma^*2s)^2(\sigma 2p)^2(\pi 2p)^4(\pi^*2p)^2(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2

The two electrons in π∗(2p)\pi^*(2p)π∗(2p) occupy separate degenerate orbitals. So O2\mathrm{O_2}O2​ has two unpaired electrons and is paramagnetic.

(f) O2+\mathrm{O_2^+}O2+​

Remove one electron from π∗(2p)\pi^*(2p)π∗(2p):

(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1(\sigma 2s)^2(\sigma^*2s)^2(\sigma 2p)^2(\pi 2p)^4(\pi^*2p)^1(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1

One unpaired electron is present. So, O2+\mathrm{O_2^+}O2+​ is paramagnetic.

(g) O2−\mathrm{O_2^-}O2−​

Add one electron to π∗(2p)\pi^*(2p)π∗(2p):

(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3(\sigma 2s)^2(\sigma^*2s)^2(\sigma 2p)^2(\pi 2p)^4(\pi^*2p)^3(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3

One orbital in π∗\pi^*π∗ has a pair and the other has one electron, so there is one unpaired electron. Thus O2−\mathrm{O_2^-}O2−​ is paramagnetic.

(h) O22−\mathrm{O_2^{2-}}O22−​

Add two electrons to π∗(2p)\pi^*(2p)π∗(2p):

(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)4(\sigma 2s)^2(\sigma^*2s)^2(\sigma 2p)^2(\pi 2p)^4(\pi^*2p)^4(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)4

All electrons are paired. So, O22−\mathrm{O_2^{2-}}O22−​ is diamagnetic.

  1. Count diamagnetic species

Diamagnetic species are:

N2, O22−\mathrm{N_2},\ \mathrm{O_2^{2-}}N2​, O22−​

So the total number is:

2\boxed{2}2​
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