Question

Which one of the following pairs has only paramagnetic species?

• A. $[Cu(NH_{3})_{4}]Cl_{2}$ and $O_{2}$
• B. $[Cu(NH_{3})_{4}]Cl_{2}$ and $N_{2}$
• C. $[Zn(H_{2}O)_{6}]Cl_{2}$ and $O_{2}$
• D. $[Cu(NH_{3})_{4}]Cl_{2}$ and $K_{4}[Fe(CN)_{6}]$

Hint:

Unpaired electrons = Paramagnetic; All paired = Diamagnetic.

Answer 1

The Correct Option is A

Step 1: Concept
Paramagnetic species possess at least one unpaired electron.
Step 2: Analysis
$[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ has $3d^9$ configuration (one unpaired electron). Paramagnetic. $O_2$: According to Molecular Orbital Theory, $O_2$ has two unpaired electrons in antibonding orbitals. Paramagnetic. $N_2$ and $Zn^{2+}$ ($3d^{10}$) are diamagnetic (all electrons paired). $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ in strong field ($CN^-$) has all electrons paired ($t_{2g}^6$). Diamagnetic.
Step 3: Conclusion
Pair (A) contains only paramagnetic species.
Final Answer:(A)