A) \[[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}\]
B) \[[Ag{{(N{{H}_{3}})}_{2}}]Cl\]
C) \[NO\]
D) \[N{{O}_{2}}\]
Correct Answer: B
Solution :
\[[Ag{{(N{{H}_{3}})}_{2}}]Cl\] shows diamagnetism due to the absence of unpaired electron. \[{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}\] \[Ag=[Ar]\]
\[A{{g}^{+}}=[Ar]\] 
\[{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}=[Ar]\] 
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