Which of the following are paramagnetic?

A. [NiCl₄]²⁻
B. Ni(CO)₄
C. [Ni(CN)₄]²⁻
D. [Ni(H₂O)₆]²⁺
E. Ni(PPh₃)₄

Choose the correct answer from the options given below:

The correct option is A and D only.

To determine which of the given complexes are paramagnetic, we must find the oxidation state of nickel in each complex, write its electronic configuration, and check for the presence of unpaired electrons using Crystal Field Theory (CFT) or Valence Bond Theory (VBT).

The atomic number of Nickel (Ni) is 28, and its ground-state electronic configuration is:
$\left[\mathrm{Ar}\right]3d^{8}4s^2$
Let us analyze each complex individually:

A. $[\mathrm{Ni}{\mathrm{Cl}}_4{]}^{2-}$
In this complex, the oxidation state of Ni is +2.
The electronic configuration of ${\mathrm{Ni}}^{2+}$ is $\left[\mathrm{Ar}\right]3d^8$.
Since chloride ion (${\mathrm{Cl}}^{-}$) is a weak field ligand, it cannot pair up the unpaired electrons in the 3d orbitals. Thus, the configuration remains as a high-spin tetrahedral arrangement with 2 unpaired electrons.
Therefore, $[\mathrm{Ni}{\mathrm{Cl}}_4{]}^{2-}$ is paramagnetic.

B. $\mathrm{Ni}(\mathrm{CO}{)}_4$
Here, Nickel is in the 0 oxidation state with configuration $\left[\mathrm{Ar}\right]3d^{8}4s^2$.
Carbon monoxide ($\mathrm{CO}$) is a very strong field ligand. It causes the pairing of the 4s electrons into the 3d orbitals, resulting in a fully filled $3d^{10}$ electronic configuration.
Since there are no unpaired electrons, this complex is diamagnetic.

C. $\left[\mathrm{Ni}\right(\mathrm{CN}{)}_4{]}^{2-}$
Here, the oxidation state of Ni is +2, giving the configuration ${\mathrm{Ni}}^{2+}$: $\left[\mathrm{Ar}\right]3d^8$.
Cyanide (${\mathrm{CN}}^{-}$) is a strong field ligand. It forces the pairing of the 3d electrons, leading to a square planar geometry with $ds{p}^2$ hybridization.
Since all electrons are paired, the complex is diamagnetic.

D. $\left[\mathrm{Ni}\right(H_{2}O{)}_6{]}^{2+}$
Here, the oxidation state of Ni is +2, and the configuration is ${\mathrm{Ni}}^{2+}$: $\left[\mathrm{Ar}\right]3d^8$.
Water ($H_{2}O$) is a weak field ligand.
In this octahedral complex, the 3d electrons are arranged in the crystal field split orbitals as $t_{2g}^6{e}_g^2$.
Since there are 2 unpaired electrons in the $e_g$ level, the complex is paramagnetic.

E. $\mathrm{Ni}({\mathrm{PPh}}_3{)}_4$
In this complex, Nickel is in the 0 oxidation state.
Triphenylphosphine (${\mathrm{PPh}}_3$) is a strong ligand that causes the 4s electrons to pair up inside the 3d orbitals, resulting in a stable tetrahedral geometry with a fully filled $3d^{10}$ configuration.
Therefore, this complex is diamagnetic.

In conclusion, only the complexes A and D are paramagnetic.