Question

Arrange the following species in the correct order of their stability.
$C_2,L{i}_2,O_2^{+},H{e}_2^{+}$.

• A. $L{i}_2<H{e}_2^{+}<O_2^{+}<C_2$
• B. $C_2<O_2^{+}<L{i}_2<H{e}_2^{+}$
• C. $H{e}_2^{+}<L{i}_2<C_2<O_2^{+}$
• D. $O_2^{+}<C_2<L{i}_2<H{e}_2^{+}$
• E. $C_2<L{i}_2<H{e}_2^{+}<O_2^{+}$

Answer 1

Answer: (C)

$H{e}_2^{+}<L{i}_2<C_2<O_2^{+}$

Higher the bond order, more is the stability.
The bond order of the given species can be calculated as
$MO$ configuration of $C_2=(6+6=12)$
$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2$
$BO=\frac{N_b-N_a}{2}=\frac{8-4}{2}=\frac{4}{2}=2$
$MO$ configuration of $L{i}_2=(3+3=6)$
$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2$
$BO=\frac{4-2}{2}=1$MO$configurationof$O_{2} (8 + 8 - 1 = 15)= \sigma 1s^{2}, \sigma^{\ast} 1s^{2}, \sigma 2s^{2}, \sigma^{\ast} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2}\approx \pi 2p_{y}^{2}, \overset {\ast}{\pi} 2p_{x}^{1}BO = \dfrac {10 - 5}{2} = 2.5MO$configurationof$He_{2} = (2 + 2 - 1 = 3)= \sigma 1s^{2}, \sigma^{\ast} 1s^{1}BO = \dfrac {2 - 1}{2} = \dfrac {1}{2} = 0.5$Thus,theorderofbondordersofthegivenspeciesis$He_{2}^{+} < Li_{2} < C_{2} < O_{2}^{+}$Sincebondorder$\alpha$ stability.
Thus, order of stability will also be the same.