1. Use the spin-only magnetic moment formula

The spin-only magnetic moment is

$\mu = \sqrt{n(n+2)}\ \text{BM}$

where $n$ is the number of unpaired electrons.

We are given:

$\mu \approx 5.9\ \text{BM}$

Now check values:

• For $n=5$, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}$

So the required species must have 5 unpaired electrons.

2. Examine each option

Option A: $[MnBr_4]^{2-}$ (tetrahedral)

• Oxidation state of Mn: $x + 4(-1) = -2 \Rightarrow x=+2$
• So Mn is $Mn^{2+}$.
• Electronic configuration of Mn: $Mn: [Ar]3d^54s^2$ $Mn^{2+}: [Ar]3d^5$

In a tetrahedral field, $Br^-$ is a weak field ligand, so the complex is high spin.

For $d^5$ high spin tetrahedral, all five $d$ electrons remain unpaired.

Thus,

$n=5 \Rightarrow \mu = \sqrt{35} \approx 5.92\ \text{BM}$

So Option A matches.

Option B: $[NiCl_4]^{2-}$ (tetrahedral)

• Oxidation state of Ni: $x + 4(-1) = -2 \Rightarrow x=+2$
• So Ni is $Ni^{2+}$.
• Electronic configuration: $Ni: [Ar]3d^84s^2$ $Ni^{2+}: [Ar]3d^8$

Tetrahedral $d^8$ is usually high spin with 2 unpaired electrons.

Thus,

$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\ \text{BM}$

So Option B does not match.

Option C: $Ni(CO)_4$ (tetrahedral)

• CO is neutral, so Ni oxidation state is $0$.
• $Ni^0$ configuration: $[Ar]3d^84s^2$

In $Ni(CO)_4$, due to strong-field CO and metal-ligand bonding, the complex is effectively diamagnetic with all electrons paired.

So,

$n=0 \Rightarrow \mu = 0$

So Option C does not match.

Option D: $[Ni(CN)_4]^{2-}$ (square planar)

• Oxidation state of Ni: $x + 4(-1) = -2 \Rightarrow x=+2$
• So Ni is $Ni^{2+}$, i.e. $d^8$.

$CN^-$ is a strong field ligand, and square planar $d^8$ complexes are typically low spin and diamagnetic.

Thus,

$n=0 \Rightarrow \mu = 0$

So Option D does not match.

3. Conclusion

The only species having spin-only magnetic moment close to $5.9$ BM is

$[MnBr_4]^{2-}$

So the correct option is A.