A) \[{{F}_{2}}\]
B) \[{{N}_{2}}\]
C) \[{{O}_{2}}\]
D) \[H_{2}^{+}\]
Correct Answer: D
Solution :
\[{{F}_{2}}(9+9=18)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{2}\] \[\text{Bond}\,\text{order}=\frac{10-8}{2}=1\] \[{{N}_{2}}(7+7=14)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] \[\text{Bonder}\,\text{order = }\frac{10-4}{2}=3\] \[{{O}_{2}}(8+8=16)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] \[\text{Bond}\,\text{order}=\frac{10-6}{2}=2\] \[H_{2}^{+}(1+1-1=1)=\sigma 1{{s}^{1}}\] \[\text{Bond order =}\frac{1-0}{2}=0.5\] Hence, \[{{\text{O}}_{\text{2}}}\]and \[\text{H}_{\text{2}}^{\text{+}}\]is paramagnetic (contains unpaired electron) but the\[\text{H}_{\text{2}}^{\text{+}}\]has the bond order half (0.5).
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