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Which of the following are paramagnetic?
A. $[\text{NiCl}_4]^{2-}$
B. $\text{Ni(CO)}_4$
C. $[\text{Ni(CN)}_4]^{2-}$
D. $[\text{Ni(H}_2\text{O)}_6]^{2+}$
E. $\text{Ni(PPh}_3)_4$
Choose the correct answer from the options given below:
A. $[\text{NiCl}_4]^{2-}$
B. $\text{Ni(CO)}_4$
C. $[\text{Ni(CN)}_4]^{2-}$
D. $[\text{Ni(H}_2\text{O)}_6]^{2+}$
E. $\text{Ni(PPh}_3)_4$
Choose the correct answer from the options given below:
A
A and C only
B
B and E only
C
A and D only
D
A, D and E only
Explanation
$[\text{NiCl}_4]^{2-}$ (tetrahedral, $sp^3$) and $[\text{Ni(H}_2\text{O)}_6]^{2+}$ (octahedral, $sp^3d^2$) both have two unpaired electrons ($d^8$ high-spin) and are paramagnetic.
Detailed Solution
$\text{Ni}^{2+}$ has $3d^8$ electronic configuration. With weak field ligands: $[\text{NiCl}_4]^{2-}$ is tetrahedral ($sp^3$) with 2 unpaired electrons (paramagnetic); $[\text{Ni(H}_2\text{O)}_6]^{2+}$ is octahedral ($sp^3d^2$) with 2 unpaired electrons (paramagnetic). With strong field ligand: $[\text{Ni(CN)}_4]^{2-}$ undergoes pairing to form square planar ($dsp^2$) low-spin with 0 unpaired electrons (diamagnetic). For $\text{Ni}(0)$ complexes $\text{Ni(CO)}_4$ and $\text{Ni(PPh}_3)_4$, configuration is $3d^{10}$ (all paired, diamagnetic). Thus only A and D are paramagnetic.