The species that has a spin-only magnetic moment of 5.9 BM, is: ($$T_d$$ = tetrahedral)

For a transition-metal complex, the spin-only magnetic moment is given by the formula

$$\mu_{\text{spin}} = \sqrt{n(n+2)}\ \text{BM},$$

where $$n$$ is the number of unpaired electrons.

The question gives $$\mu_{\text{spin}} = 5.9\ \text{BM}$$. We equate and solve for $$n$$.

$$5.9 \approx \sqrt{n(n+2)}$$

Squaring both sides, we obtain

$$5.9^{2} \approx n(n+2).$$

Since $$5.9^{2} = 34.81 \approx 35,$$ we write

$$n(n+2) \approx 35.$$

Trying integral values, $$n=5$$ gives $$5(5+2)=35,$$ exactly the required value. Hence the complex we are looking for must possess five unpaired electrons.

Now we examine each option one by one, always finding

• the oxidation state of the metal,
• its $$d$$-electron count,
• whether the ligand field is strong (low-spin) or weak (high-spin), and
• the number of unpaired electrons $$n$$, followed by its calculated moment.

Option A : $$[\text{Ni(CN)}_4]^{2-}$$, square planar

The total charge is $$-2,$$ the four $$\text{CN}^-$$ ligands contribute $$-4,$$ so the oxidation state of Ni is $$+2.$$ Thus $$\text{Ni}^{2+}$$ has the configuration $$3d^8.$$ $$\text{CN}^-$$ is a strong-field ligand; in a square-planar environment all eight $$d$$ electrons pair up, giving $$n=0.$$ Hence $$\mu = 0\ \text{BM},$$ far below $$5.9\ \text{BM}.$$

Option B : $$[\text{NiCl}_4]^{2-}$$, tetrahedral

Again the complex charge is $$-2,$$ the four $$\text{Cl}^-$$ ligands supply $$-4,$$ so Ni is $$+2,$$ giving $$d^8.$$ Chloride is a weak-field ligand and the tetrahedral splitting is small, so the complex is high-spin. Filling eight electrons in the five $$d$$ orbitals yields two unpaired electrons, so $$n=2.$$ Therefore

$$\mu = \sqrt{2(2+2)} = \sqrt8 = 2.83\ \text{BM},$$

which is much smaller than $$5.9\ \text{BM}.$$

Option C : $$\text{Ni(CO)}_4$$, tetrahedral

The molecule is neutral; each CO is neutral, so Ni is in oxidation state $$0,$$ hence configuration $$3d^{10}4s^{0}.$$ All ten $$d$$ electrons are paired, giving $$n=0$$ and $$\mu = 0\ \text{BM}.$$

Option D : $$[\text{MnBr}_4]^{2-}$$, tetrahedral

The complex has overall charge $$-2.$$ Four $$\text{Br}^-$$ ligands contribute $$-4,$$ thus Mn must be in the $$+2$$ oxidation state:

$$x + (-4) = -2 \;\;\Longrightarrow\;\; x = +2.$$

Therefore $$\text{Mn}^{2+}$$ possesses a $$3d^5$$ configuration. Bromide is a weak-field ligand and the geometry is tetrahedral, so the complex is high-spin. The five $$d$$ electrons occupy all five orbitals singly, hence $$n = 5.$$ The spin-only moment is then

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

in excellent agreement with the required $$5.9\ \text{BM}.$$

Only Option D furnishes a magnetic moment of the correct magnitude.

Hence, the correct answer is Option D.