MCQ
According to $MO$ theory which of the lists ranks the nitrogen species in terms of increasing bond order?
- ✓$N_2^{2-} < N_2^- < N_2$
- B$N_2 < N_2^{2-} < N_2^-$
- C$N_2^- < N_2^{2-} < N_2$
- D$N_2^- < N_2 < N_2^{2-}$
$\therefore \quad \mathrm{B} . \mathrm{O}.=\frac{10-4}{2}=3$
$\mathrm{N}_{2}^{-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}$
$\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1}$
$\therefore \quad \mathrm{B}. \mathrm{O}.=\frac{10-5}{2}=2.5$
$\mathrm{N}_{2}^{2-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}$
$\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1}\left(\pi^{*} 2 p_{y}\right)^{1}$
$\therefore \quad \mathrm{B} . \mathrm{O} .=\frac{10-6}{2}=2$
Hence the order: $\mathrm{N}_{2}^{2-}<\mathrm{N}_{2}^{-}<\mathrm{N}_{2}$
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$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,;$ $K_1$
$N_2 + O_2 \rightleftharpoons 2NO\,;$ $K_2$
$H_2 + 2 O_2 \rightleftharpoons H_2O\,;$ $K_3$
The equilibrium constant $(K)$ of the reaction :
$2NH_3 + \frac{5}{2} \overset K \leftrightarrows 2NO + 3H_2O$
where $pK_1 , pK_2, pK_3$ are first ionization constants. Incorrect order is