The outer electron configuration of Gd (Atomic No: 64) is:

The ground-state electronic configuration of any element is obtained by filling the subshells in the order of increasing energy following the $$n + \ell$$ rule and applying Hund’s rule and the Pauli exclusion principle.

For the lanthanides (atomic numbers 57-71), electrons are added mainly to the $$4f$$ subshell while $$5d$$ and $$6s$$ remain close in energy. Gadolinium has atomic number $$Z = 64$$, so its 64 electrons must be distributed starting from $$1s$$ upward.

Till xenon ($$Z = 54$$) the configuration is complete: $$[Xe]$$.

Electrons $$55 \text{ to } 64$$ are then filled as follows:

 • $$55^{\text{th}}$$ electron enters $$6s$$ ⇒ $$6s^1$$
 • $$56^{\text{th}}$$ electron also enters $$6s$$ ⇒ $$6s^2$$ (now $$6s$$ is filled).
 • Next electrons are expected to enter $$4f$$ because after $$6s$$ the $$4f$$ and $$5d$$ subshells are very close in energy. According to the usual order, we begin to fill $$4f$$: $$4f^1, 4f^2, \dots$$

This continues until we have introduced seven electrons into $$4f$$. The half-filled $$4f^7$$ configuration provides extra stability (Hund’s rule for maximum multiplicity).

After reaching the half-filled state, the next electron preferentially goes to the nearby higher subshell $$5d$$ rather than pairing inside $$4f$$; this again keeps energy minimum because both $$4f$$ (half-filled) and $$5d$$ (now singly occupied) achieve relatively stable arrangements.

Thus electrons $$55$$ to $$64$$ are distributed as:

$$6s^2$$ (2 electrons) + $$4f^7$$ (7 electrons) + $$5d^1$$ (1 electron)  → 10 electrons in total, matching $$64 - 54 = 10$$.

Hence the outer (valence-shell) electronic configuration of gadolinium is

$$[Xe]\,4f^7\,5d^1\,6s^2$$.

Therefore, the correct choice is Option C which is: $$4f^7 5d^1 6s^2$$.