Which of the following compound is paramagnetic in nature?

The correct answer is [NiCl4]^2-.

To determine which compound is paramagnetic, we need to find the number of unpaired electrons in the central metal ion of each complex. A compound with one or more unpaired electrons is paramagnetic, while a compound with all paired electrons is diamagnetic.

Let's analyze the given complexes step-by-step:

1. Analysis of ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$:
- Oxidation state of Nickel (Ni): Let the oxidation state of Ni be x.
$x+4(-1)=-2\Rightarrow x=+2$
Thus, nickel is in the +2 oxidation state (${\mathrm{Ni}}^{2+}$).
- Electronic configuration:
Atomic number of Ni = 28
$\mathrm{Ni}=\left[\mathrm{Ar}\right]{\mathrm{3d}}^8{\mathrm{4s}}^2$
${\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right]{\mathrm{3d}}^8{\mathrm{4s}}^0$
- Nature of Ligand: ${\mathrm{Cl}}^{-}$ is a weak field ligand. According to the spectrochemical series, it cannot cause pairing of electrons in the 3d orbitals.
- Electron Distribution: In the 3d subshell (5 orbitals), the 8 electrons are distributed as:
$(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow )(\uparrow )$
This leaves 2 unpaired electrons. Therefore, ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$ is paramagnetic (with a tetrahedral geometry via ${\mathrm{sp}}^3$ hybridization).

2. Analysis of other options (why they are diamagnetic):
- $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$: Ni is in 0 oxidation state (${\mathrm{3d}}^8{\mathrm{4s}}^2$). CO is a very strong field ligand that forces the 4s electrons to shift and pair up in the 3d orbitals, resulting in a completely filled ${\mathrm{3d}}^{10}$ configuration. There are 0 unpaired electrons (diamagnetic).
- ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$: Ni is in +2 oxidation state (${\mathrm{3d}}^8$). ${\mathrm{CN}}^{-}$ is a strong field ligand that forces the pairing of the 3d electrons, resulting in 0 unpaired electrons and a square planar geometry via ${\mathrm{dsp}}^2$ hybridization (diamagnetic).
- ${\left[\mathrm{Co}{\left(H_{2}O\right)}_6\right]}^{3+}$: Co is in +3 oxidation state (${\mathrm{3d}}^6$). Due to the high charge on ${\mathrm{Co}}^{3+}$, $H_{2}O$ acts as a strong field ligand and causes all 6 d-electrons to pair up in the low-lying $t_{2g}$ orbitals, leaving 0 unpaired electrons (diamagnetic).