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Chemical Bonding and Molecular Structure question

2021 · 27 Jul · Shift 1 · Q22

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Chemical Bonding and Molecular Structure
2021 · 27 Jul · Shift 1 · Q22

Chemical Bonding and Molecular Structure question

2021 · 27 Jul · Shift 1 · Q22

JEE MainChemistryChemical Bonding and Molecular StructureNumerical+4 / −1

The difference between bond orders of CO and NO

^ \oplus

is

{x \over 2}

where x =

\underline{\hspace{2cm}}

. (Round off to the Nearest Integer)

Numerical answer

View written solutionFree

Correct answer: 0

Find the bond order of CO

Carbon monoxide, CO, is isoelectronic with $N_2$ .

Total electrons in CO: $6 + 8 = 14$

Using MO theory for 14-electron diatomic molecules, the bond order is: $\text{Bond order of CO} = 3$
Find the bond order of $NO^+$

First count total electrons in $NO^+$ : $7 + 8 - 1 = 14$

So $NO^+$ is also isoelectronic with $N_2$ and CO.

Hence, $\text{Bond order of } NO^+ = 3$
Find the difference

$\text{Difference} = 3 - 3 = 0$

Given that the difference is $\dfrac{x}{2}$ , $\frac{x}{2} = 0$

Therefore, $x = 0$
Nearest integer

Already an integer, so the nearest integer is: $0$

Final Answer: $x = 0$

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