Question

Bond energies in $NO,N{O}^{+}$ and $N{O}^{-}$ are such as

• A. $N{O}^{-}>NO>N{O}^{+}$
• B. $NO>N{O}^{-}>N{O}^{+}$
• C. $N{O}^{+}>NO>N{O}^{-}$
• D. $N{O}^{+}>N{O}^{-}>NO$

Answer 1

The correct option is C

$N{O}^{+}>NO>N{O}^{-}$

We know that,

$\to$ Its harder to break a triple bond than a double and single bond.

$\Rightarrow$ Bond order ∝ Bond energy


Lets take a look at the bond order of these species.

You will learn to calculate the following in the section "Molecular Orbital Theory". For now, let us just make a note of these bond orders shall we?

$\left\{\begin{array}{c}N{O}^{-}\Rightarrow 2\\ N{O}^{+}\Rightarrow 3\\ NO\Rightarrow 2.5\end{array}\right\}$

Therefore, bond energies will be in the order of

$N{O}^{+}>NO>N{O}^{-}$