The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion M$$^{Z+}$$ are $$-0.8\Delta_0$$ and 3.87 BM, respectively. Identify M$$^{Z+}$$:

We are told that the complex is octahedral and the ligand is water. Because $$\mathrm{H_2O}$$ is a weak-field ligand, we first assume a high-spin distribution of the d electrons. In an octahedral field the energy separation of the two sets of orbitals is $$\Delta_0$$, and the crystal-field stabilisation energy (CFSE) is obtained from

$$\text{CFSE}=(-0.4\,n_{t_{2g}}+0.6\,n_{e_g})\Delta_0$$

where $$n_{t_{2g}}$$ and $$n_{e_g}$$ are the numbers of electrons present in the $$t_{2g}$$ and $$e_g$$ sets, respectively.

The given CFSE is $$-0.8\Delta_0$$. We now examine which high-spin d configurations yield this value:

For $$d^2$$ (high spin) the distribution is $$t_{2g}^2e_g^0$$, so

$$\text{CFSE}=(-0.4\times2+0.6\times0)\Delta_0=-0.8\Delta_0$$

For $$d^7$$ (high spin) the distribution is $$t_{2g}^5e_g^2$$ (three electrons remain unpaired in the manner explained later). Hence

$$\text{CFSE}=(-0.4\times5+0.6\times2)\Delta_0=(-2.0+1.2)\Delta_0=-0.8\Delta_0$$

Thus both $$d^2$$ and $$d^7$$ configurations give the required CFSE, so we must use the magnetic moment to decide which one is correct.

The spin-only magnetic moment is related to the number of unpaired electrons $$n$$ by the formula

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

The observed value is $$3.87\;\text{BM}$$. We now compute $$\mu$$ for each of the two candidate configurations.

Case 1: $$d^2$$ (high spin)
All two d electrons remain unpaired, so $$n=2$$.

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

This is much smaller than the observed $$3.87\;\text{BM}$$, hence $$d^2$$ is ruled out.

Case 2: $$d^7$$ (high spin)
To minimise pairing, the first five electrons occupy the five different d orbitals singly (three in $$t_{2g}$$ and two in $$e_g$$). The sixth and seventh electrons must pair with two of the original $$t_{2g}$$ electrons, leaving one $$t_{2g}$$ electron and both $$e_g$$ electrons unpaired. Therefore $$n=3$$.

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

This matches the given magnetic moment exactly, so the metal ion must possess a high-spin $$d^7$$ configuration.

Now we identify which of the options carries a $$d^7$$ configuration:

• $$\mathrm{V^{3+}}$$ : 23 − 3 = 20 electrons ⇒ $$d^2$$ (not suitable)
• $$\mathrm{Co^{2+}}$$ : 27 − 2 = 25 electrons ⇒ $$d^7$$ (suitable)
• $$\mathrm{Cr^{3+}}$$ : 24 − 3 = 21 electrons ⇒ $$d^3$$ (not suitable)
• $$\mathrm{Mn^{4+}}$$ : 25 − 4 = 21 electrons ⇒ $$d^3$$ (not suitable)

Only $$\mathrm{Co^{2+}}$$ has the required $$d^7$$ electron count.

Hence, the correct answer is Option B.