The correct electronic configuration and spin-only magnetic moment (BM) of $$\text{Gd}^{3+}$$ ($$Z = 64$$), respectively, are:

For gadolinium the atomic number is given as $$Z = 64$$, so the ground-state electronic configuration of a neutral atom is

$$\text{Gd} : [\text{Xe}]\,4f^{\,7}\,5d^{\,1}\,6s^{\,2}.$$

We are interested in the triply charged ion $$\text{Gd}^{3+}$$. Whenever electrons are removed to form a positive ion, we first remove them from the subshell having the highest principal quantum number $$n$$, and within that same $$n$$ value we follow the order $$s \rightarrow p \rightarrow d \rightarrow f$$. Hence we take away the two $$6s$$ electrons (because $$n = 6$$ is the outermost shell) and then one of the $$5d$$ electrons. No electrons are taken from the $$4f$$ subshell because it has a lower principal quantum number.

So, after removing three electrons we obtain

$$\text{Gd}^{3+} : [\text{Xe}]\,4f^{\,7}.$$

Next, we calculate the spin-only magnetic moment. The formula for the spin-only magnetic moment (in Bohr magnetons, BM) is

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

where $$n$$ is the number of unpaired electrons present in the ion.

In the half-filled $$4f^{\,7}$$ subshell each of the seven $$f$$ orbitals contains exactly one electron, so all seven electrons are unpaired. Therefore,

$$n = 7.$$

Substituting this value into the formula, we have

$$\mu_\text{spin} = \sqrt{7(7+2)}\;\text{BM} = \sqrt{7 \times 9}\;\text{BM} = \sqrt{63}\;\text{BM}.$$

Evaluating the square root,

$$\sqrt{63} \approx 7.94\;\text{BM},$$

which we round off to $$7.9\;\text{BM}$$ for practical purposes.

Putting the electronic configuration and the calculated magnetic moment together, the correct pair is

$$[\text{Xe}]\,4f^{\,7} \quad\text{and}\quad 7.9\;\text{BM}.$$

This matches Option B.

Hence, the correct answer is Option B.