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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