A) \[C{{l}^{-}}\]
B) Be
C) \[N{{e}^{2+}}\]
D) \[A{{s}^{+}}\]
Correct Answer: A
Solution :
| Paramagnetic character is based upon presence of unpaired electron |
| \[_{17}C{{l}^{-}}=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},\,3{{s}^{2}}3p_{x}^{2}3p_{y}^{2}3p_{z}^{2}\] |
| Unpaired electron about Diamagnetic |
| \[_{4}Be=1{{s}^{2}},2{{s}^{1}}\,2p_{x}^{1}\] |
| \[1{{s}^{2}}\,2{{s}^{2}}\] |
| \[_{10}N{{e}^{2+}}=1{{s}^{2}},2{{s}^{2}}2p_{x}^{2}2p_{y}^{1}2p_{z}^{1}\] |
| \[_{33}A{{s}^{+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{2}}\,4p_{x}^{1}4p_{y}^{1}\,4p_{z}^{0}\] |
| All have unpaired electrons. So, these are paramagnetic in nature. |
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