- A$[Ni(NH_3)_6]Cl_2$
- B$Na_3[FeF_6 ]$
- C$[Cr(H_2O)_6]SO_4$
- ✓$K_4[Fe(CN)_6 ]$
$N{a_3}[Fe{F_6}]$ $\to $ $s{p^3}{d^2}$ hybridisation $\to $ $3$ unpaired electrons
$[Cr{({H_2}O)_6}]S{O_4}$ $\to $ ${d^2}s{p^3}$ hybridisation $\to $ $3$ unpaired electrons
${K_4}[Fe{(CN)_6}]$ $\to $ ${d^2}s{p^3}$ hybridisation $\to $ No unpaired electrons
Zero magnetic moment means all the electrons paired.
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$(A)$ $SnCl _2 \cdot 2 H _2 O$ is a reducing agent.
$(B)$ $SnO _2$ reacts with $KOH$ to form $K _2\left[ Sn ( OH )_6\right]$.
$(C)$ A solution of $PbCl _2$ in $HCl$ contains $Pb ^{2+}$ and $Cl ^{-}$ions.
$(D)$ The reaction of $Pb _3 O _4$ with hot dilute nitric acid to give $PbO _2$ is a redoxreaction.
Reason : $sp$ hybridised carbon atoms of acetylene are considerably electronegative.