The spin magnetic moment ($\mu$) of a transition metal ion is calculated using the 'spin-only' formula: $\mu = \sqrt{n(n+2)}\text{ B.M.}$ where $n$ is the number of unpaired electrons in the central metal ion.

Let's find the number of unpaired electrons for each ion:

• $\text{Co}^{3+}$: The electronic configuration of Co is $[\text{Ar}] 3d^7 4s^2$. For $\text{Co}^{3+}$, it is $[\text{Ar}] 3d^6$. Number of unpaired electrons, $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24}\text{ B.M.}$ $\rightarrow$ (A) matches (iv).
• $\text{Cr}^{3+}$: The electronic configuration of Cr is $[\text{Ar}] 3d^5 4s^1$. For $\text{Cr}^{3+}$, it is $[\text{Ar}] 3d^3$. Number of unpaired electrons, $n = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15}\text{ B.M.}$ $\rightarrow$ (B) matches (iii).
• $\text{Fe}^{3+}$: The electronic configuration of Fe is $[\text{Ar}] 3d^6 4s^2$. For $\text{Fe}^{3+}$, it is $[\text{Ar}] 3d^5$. Number of unpaired electrons, $n = 5$. $\mu = \sqrt{5(5+2)} = \sqrt{35}\text{ B.M.}$ $\rightarrow$ (C) matches (ii).
• $\text{Ni}^{2+}$: The electronic configuration of Ni is $[\text{Ar}] 3d^8 4s^2$. For $\text{Ni}^{2+}$, it is $[\text{Ar}] 3d^8$. Number of unpaired electrons, $n = 2$. $\mu = \sqrt{2(2+2)} = \sqrt{8}\text{ B.M.}$ $\rightarrow$ (D) matches (i).

Therefore, the correct matching is A-(iv), B-(iii), C-(ii), D-(i).