A) Oxygen molecule
B) Boron molecule
C) \[{{N}_{2}}^{+}\]
D) None of the above
Correct Answer: D
Solution :
Key Idea: In diamagnetic species all the electrons are paired when we write configuration according to molecular orbital theory. Whereas a paramagnetic species has at least one unpaired electron. (i) \[{{O}_{2}}=(8+8=16)\] \[=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\], \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2},\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1},\overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] \[\because \] It has two unpaired electrons. \[\therefore \] It is paramagnetic. (ii) \[{{B}_{2}}=(5+5=10)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\], \[\pi 2p_{x}^{1},\pi 2p_{y}^{1}\] \[\because \] It has two unpaired electrons. \[\therefore \] It is paramagnetic. (iii) \[N_{2}^{+}=(7+7-1=13)=\] \[\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2s,\pi 2p_{x}^{2},\pi p_{y}^{2},\sigma p_{z}^{1}\] \[\because \] It has one unpaired electron. \[\therefore \] It is paramagnetic.
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