Question: 51. Which of the following are paramagnetic?
A. \left[\mathrm{NiCl}_{4}\right]^{2-}
B. \mathrm{Ni}(\mathrm{CO})_{4}
C. \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}
D. \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}
E. \mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{4}
Choose the correct answer from the options given below:
(1) A and C only
(2) B and E only
(3) A and D only
(4) A, D and E only
Answer: Option (3)
Explanation:
Nickel has atomic number 28, so its ground-state electronic configuration is
[Ar]\,3d^{8}4s^{2}.
In complexes where nickel is in the oxidation state +2,
the configuration becomes 3d^{8}.
Paramagnetism depends on the presence of unpaired electrons.
Statement A: [NiCl_{4}]^{2-}. Here nickel is Ni^{2+} (3d^{8}).
The complex is tetrahedral with weak-field ligand Cl^{-}.
In tetrahedral d^{8} with weak ligands,
electrons remain unpaired, giving two unpaired electrons.
Hence it is paramagnetic. So A is true.
Statement B: Ni(CO)_{4}. Oxidation state of nickel is 0,
so configuration is 3d^{10}.
All electrons are paired; therefore the complex is diamagnetic.
So B is not paramagnetic.
Statement C: [Ni(CN)_{4}]^{2-}.
Nickel is Ni^{2+} (3d^{8})
with strong-field ligand CN^{-}.
The complex is square planar and low spin;
all d-electrons get paired, giving 0 unpaired electrons.
Hence it is diamagnetic. So C is not paramagnetic.
Statement D: [Ni(H_{2}O)_{6}]^{2+}.
Nickel is Ni^{2+} (3d^{8}) in an octahedral field with weak-field ligand H_{2}O.
The complex is high spin with two unpaired electrons, so it is paramagnetic.
Thus D is true.
Statement E: Ni(PPh_{3})_{4}.
Nickel is in oxidation state 0 with configuration 3d^{10}.
All electrons are paired, so the complex is diamagnetic.
Hence E is not paramagnetic.
Therefore, only complexes A and D are paramagnetic,
so the correct option is (3) A and D only.