A) \[N{{a}_{2}}CuC{{l}_{4}}\]
B) \[N{{a}_{2}}CdC{{l}_{4}}\]
C) \[{{K}_{4}}Fe{{(CN)}_{6}}\]
D) \[{{K}_{3}}Fe{{(CN)}_{6}}\]
Correct Answer: B
Solution :
\[N{{a}_{2}}CdC{{l}_{4}}\]- no unpaired electrons.
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