Which of the following ions possess a spin-only magnetic moment of...

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$, where $n$ is the number of unpaired electrons.

Given $\mu \approx 2.83 \text{ BM}$, we have $\sqrt{n(n+2)} = 2.83 \Rightarrow n(n+2) \approx 8 \Rightarrow n = 2$.

Electronic configuration of $\mathrm{V}^{3+}$ is $[\mathrm{Ar}] 3d^2$. Number of unpaired electrons $n = 2$.

Electronic configuration of $\mathrm{Co}^{2+}$ is $[\mathrm{Ar}] 3d^7$. Number of unpaired electrons $n = 3$.

Electronic configuration of $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}] 3d^8$. Number of unpaired electrons $n = 2$.

Electronic configuration of $\mathrm{Cu}^{+}$ is $[\mathrm{Ar}] 3d^{10}$. Number of unpaired electrons $n = 0$.

Electronic configuration of $\mathrm{Mn}^{3+}$ is $[\mathrm{Ar}] 3d^4$. Number of unpaired electrons $n = 4$.

Thus, $\mathrm{V}^{3+}$ and $\mathrm{Ni}^{2+}$ have $2$ unpaired electrons and a magnetic moment of $2.83 \text{ BM}$.

Answer: A and C only