| Four diatomic species are listed below. |
| Identify the correct order in which the bond order is increasing in them. [AIPMT (M) 2012] |
A) \[NO<O_{2}^{-}<C_{2}^{2-}<He_{2}^{+}\]
B) \[O_{2}^{-}<NO<C_{2}^{2-}<He_{2}^{+}\]
C) \[C_{2}^{2-}<He_{2}^{+}<O_{2}^{-}<NO\]
D) \[He_{2}^{+}<O_{2}^{-}<NO<C_{2}^{2-}\]
Correct Answer: D
Solution :
| \[Bond\,order=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] |
| In No, total electrons \[=7+8=15\] |
| \[\therefore \]Configuration of NO |
| \[=KK,\sigma (2{{s}^{2}}),\overset{*}{\mathop{\sigma }}\,(2{{s}^{2}}),\sigma (2p_{z}^{2}),\pi (2p_{x}^{2})\] |
| \[\approx \pi (2p_{y}^{2})\overset{*}{\mathop{\pi }}\,(2p_{x}^{1})\] |
| \[\therefore \] \[Bond\,order=\frac{8-3}{2}=\frac{5}{2}=2.5\] |
| In \[O_{2}^{-},\]total electrons = 16 + 1 = 17 |
| \[\therefore \]Configuration of \[O_{2}^{-}\] |
| \[=KK,\sigma (2{{s}^{2}}),\sigma (2p_{z}^{2}),\pi (2p_{x}^{2})\] |
| \[\approx \pi (2p_{y}^{2}),\overset{*}{\mathop{\pi }}\,(2p_{x}^{2})\approx \overset{*}{\mathop{\pi }}\,(2p_{y}^{1})\] |
| \[\therefore \] \[Bond\,order=\frac{8-5}{2}=\frac{3}{2}=1.5\] |
| In \[C_{2}^{2-},\]total electrons \[=12+2=14\] |
| \[\therefore \]Configuration of \[C_{2}^{2-}\] |
| \[=KK,\sigma (2{{s}^{2}}),\overset{*}{\mathop{\sigma }}\,(2{{s}^{2}}),\sigma (2p_{z}^{2}),\approx \pi (2p_{x}^{2}),\pi (2p_{y}^{2})\] |
| \[\therefore \] \[Bond\,order=\frac{8-2}{2}\] |
| \[=\frac{6}{2}=3\] |
| In \[He_{2}^{+},\]total electrons \[=4-1=3\] |
| \[\therefore \]Configuration of \[He_{2}^{+}=\sigma (1{{s}^{2}}),\overset{*}{\mathop{\sigma }}\,(1{{s}^{1}})\] |
| \[\therefore \]\[Bond\,order=\frac{2-1}{2}\] |
| \[=\frac{1}{2}=0.5\] |
| Hence, correct order of bond order is |
| \[He_{2}^{+}<O_{2}^{-}<NO<C_{2}^{2-}\] |
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