Magnetic moment of $$\text{Gd}^{3+}$$ ion ($$Z = 64$$) is

Atomic number of gadolinium, $$Z = 64$$. Its ground-state electronic configuration is
$$\text{Gd}: [Xe]\;4f^{7}\,5d^{1}\,6s^{2}$$

To obtain $$\text{Gd}^{3+}$$ we remove the two $$6s$$ electrons and the one $$5d$$ electron:
$$\text{Gd}^{3+}: [Xe]\;4f^{7}$$

The 4f subshell can accommodate 14 electrons. A configuration of $$4f^{7}$$ is exactly half-filled, so every one of the seven 4f orbitals contains a single unpaired electron. Therefore,

Number of unpaired electrons, $$n = 7$$.

For lanthanide ions the theoretical magnetic moment is given by
$$\mu_{\text{eff}} = \sqrt{n\,(n+2)}\ \text{BM}$$ when the orbital contribution is either negligible or vanishes (as in a half-filled shell where the total orbital angular momentum $$L=0$$).

Substituting $$n = 7$$:
$$\mu_{\text{eff}} = \sqrt{7\,(7+2)} = \sqrt{63} \approx 7.94\ \text{BM}$$

Rounded to one decimal place, $$\mu_{\text{eff}} \approx 7.9\ \text{BM}$$.

Option C which is: $$7.9$$ BM