Using molecular orbital theory for the…

Using molecular orbital theory for the species O2^+, O2, O2^- and O2^2-, which option correctly gives the order of increasing O–O bond length and, in…

Single correct option

Using molecular orbital theory for the species

\mathrm{O_2^{+}}

\mathrm{O_2}

\mathrm{O_2^{-}}

and

\mathrm{O_2^{2-}}

, which option correctly gives the order of increasing O–O bond length and, in the same order, the number of unpaired electrons in the four species?

A. $\mathrm{O_2^{+}}<\mathrm{O_2}<\mathrm{O_2^{-}}<\mathrm{O_2^{2-}}$ ; unpaired electrons $1,\,2,\,1,\,0$
B. $\mathrm{O_2^{+}}<\mathrm{O_2}<\mathrm{O_2^{-}}<\mathrm{O_2^{2-}}$ ; unpaired electrons $1,\,2,\,3,\,0$
C. $\mathrm{O_2^{2-}}<\mathrm{O_2^{-}}<\mathrm{O_2}<\mathrm{O_2^{+}}$ ; unpaired electrons $0,\,1,\,2,\,1$
D. $\mathrm{O_2^{+}}<\mathrm{O_2}<\mathrm{O_2^{2-}}<\mathrm{O_2^{-}}$ ; unpaired electrons $1,\,2,\,0,\,1$

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Answer: (A) $\mathrm{O_2^{+}}<\mathrm{O_2}<\mathrm{O_2^{-}}<\mathrm{O_2^{2-}}$ ; unpaired electrons $1,\,2,\,1,\,0$

Fill the valence MOs (for

\mathrm{O_2}

-type species

\sigma_{2p}

lies below

\pi_{2p}

\ldots\sigma_{2p}^2\,\pi_{2p}^4\,\pi^{*}_{2p}{}^{x}

with

x=1,2,3,4

for

\mathrm{O_2^{+}},\mathrm{O_2},\mathrm{O_2^{-}},\mathrm{O_2^{2-}}

. Bond length decreases with bond order, so the length order is

\mathrm{O_2^{+}}<\mathrm{O_2}<\mathrm{O_2^{-}}<\mathrm{O_2^{2-}}

. The two degenerate

\pi^*

orbitals are filled singly first (Hund):

x=1

gives 1 unpaired,

x=2

gives 2,

x=3

gives 1 (one orbital doubly filled, one singly), and

x=4

gives 0. Option B wrongly gives 3 unpaired for the superoxide; C reverses the length order; D swaps the last two species.

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Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake