The difference between bond orders of CO and $\ce{NO+}$ is $\frac{X}{2}$ where X = ______ (Round off to the Nearest integer)

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Chemical Bonding and Molecular Structure

The difference between bond orders of CO and

NO X^{+}

\frac{X}{2}

where X = ______
(Round off to the Nearest integer)

Answer: 0

CO has

6 + 8 = 14

electrons and

N O^{+}

has

7 + 8 - 1 = 14

, so the two species are isoelectronic and fill the same set of molecular orbitals:

σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}

Each has

10

bonding and

4

antibonding electrons, so for both

bond order = \frac{10 - 4}{2} = 3

The difference of the two bond orders is

0

, and since this difference equals

X /2

, the value of

X

0

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer