Which of the following sets of quantum numbers is correct for an electron in $$4f$$ orbital?

For an electron in a $$4f$$ orbital, the quantum numbers must satisfy the following conditions.

The principal quantum number, $$n$$, is given by the shell number. Since the electron is present in the fourth shell,

$$n=4$$

The azimuthal quantum number, $$l$$, depends on the type of orbital:

$$s \rightarrow l=0$$

$$p \rightarrow l=1$$

$$d \rightarrow l=2$$

$$f \rightarrow l=3$$

Since the electron is in an $$f$$-orbital,

$$l=3$$

The magnetic quantum number, $$m_l$$, can have values from $$-l$$ to $$+l$$.

For $$l=3$$,

$$m_l=-3,-2,-1,0,+1,+2,+3$$

Hence, any value outside this range is not possible.

The spin quantum number, $$m_s$$, can have only two values:

$$m_s=+\frac{1}{2}\ \text{or}\ -\frac{1}{2}$$

Checking option (C):

$$n=4,\ l=3,\ m_l=+1,\ m_s=+\frac{1}{2}$$

All these values satisfy the allowed conditions for a $$4f$$ electron.

Therefore, the correct set of quantum numbers is

$$n=4,\ l=3,\ m_l=+1,\ m_s=+\frac{1}{2}$$

Hence, option (C) is the correct answer.