The correct order of spin-only magnetic moments among the following is: (Atomic number: Mn = 25, Co = 27, Ni = 28, Zn = 30)

We have to compare the spin-only magnetic moments of the tetra-hedral complexes $$[\text{MnCl}_4]^{2-},\;[\text{CoCl}_4]^{2-},\;[\text{NiCl}_4]^{2-},\;[\text{ZnCl}_4]^{2-}.$$

First, every chloride ion carries charge $$-1.$$ Because four chloride ions are present, the total anionic charge supplied by the ligands is

$$4 \times (-1)= -4.$$

Each complex as a whole has charge $$-2.$$ If we denote the oxidation state of the metal by $$x,$$ we can write

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

So in every complex the central metal is in the $$+2$$ oxidation state, that is, $$\text{Mn}^{2+},\;\text{Co}^{2+},\;\text{Ni}^{2+},\;\text{Zn}^{2+}.$$

Now we determine the number of $$d$$-electrons for each $$M^{2+}$$ ion. The ground-state electronic configurations for the neutral atoms are

Mn : $$[ \text{Ar} ]\,3d^5 4s^2,$$   Co : $$[ \text{Ar} ]\,3d^7 4s^2,$$   Ni : $$[ \text{Ar} ]\,3d^8 4s^2,$$   Zn : $$[ \text{Ar} ]\,3d^{10} 4s^2.$$

In forming the $$+2$$ ions, the two $$4s$$ electrons are lost first, so we get

$$\text{Mn}^{2+}: 3d^5,$$

$$\text{Co}^{2+}: 3d^7,$$

$$\text{Ni}^{2+}: 3d^8,$$

$$\text{Zn}^{2+}: 3d^{10}.$$

Chloride is a weak-field ligand and the complexes are tetra-hedral, so all four complexes are high-spin. Hence the number of unpaired electrons $$n$$ equals that of the free high-spin ions:

$$n_{\text{Mn}^{2+}} = 5,$$

$$n_{\text{Co}^{2+}} = 3,$$

$$n_{\text{Ni}^{2+}} = 2,$$

$$n_{\text{Zn}^{2+}} = 0.$$

The spin-only magnetic moment is given by the well-known formula

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

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

Substituting each value of $$n$$ one by one, we obtain

For $$[\text{MnCl}_4]^{2-}:$$

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

For $$[\text{CoCl}_4]^{2-}:$$

$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\;\text{BM}.$$

For $$[\text{NiCl}_4]^{2-}:$$

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

For $$[\text{ZnCl}_4]^{2-}:$$

$$\mu = \sqrt{0(0+2)} = 0\;\text{BM}.$$

The magnetic moments decrease in the sequence

$$[\text{MnCl}_4]^{2-} > [\text{CoCl}_4]^{2-} > [\text{NiCl}_4]^{2-} > [\text{ZnCl}_4]^{2-}.$$

Comparing with the given options, this order matches Option D.

Hence, the correct answer is Option D.