A) \[{{N}_{2}}\]
B) \[{{C}_{2}}\]
C) \[{{C}_{2}}^{+}\]
D) \[{{O}_{2}}^{2-}\]
Correct Answer: C
Solution :
Key Idea: Write the electronic configuration according to molecular orbital theory to find the paramagnetic species. The paramagnetic species has at least one unpaired electron and in diamagnetic species all the electrons are paired. (i) \[{{C}_{2}}=(6+6=12)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}}\] \[\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{2},\pi p_{y}^{2}\] \[\because \] It has no unpaired electron. \[\therefore \] It is diamagnetic. (ii) \[{{N}_{2}}(7+7=14)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}}\], \[\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{2},\pi 2p_{y}^{2},2p_{z}^{2}\] \[\because \] All the electrons are paired. \[\therefore \] It is diamagnetic. (iii) \[O_{2}^{2-}=(8+8+2=18)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}}\], \[\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2},\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\] \[\because \] All the electrons are paired. \[\therefore \] It is diamagnetic. (iv)\[N_{2}^{+}(7+7-1=13)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\] \[\pi 2p_{x}^{2,}\pi p_{y}^{2},\sigma 2p_{z}^{1}\] \[\because \] It has one unpaired electron. \[\therefore \] It is paramagnetic.
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