JEE Main · 2004 · 24 Apr 2004 · Offline · Atomic Structure

In a 4f orbital the principal quantum number is fixed at $n=4$, and because the f subshell corresponds to the azimuthal quantum number $l=3$, that value is forced too. With $l=3$ the magnetic quantum number may be any integer from $-3$ through $+3$, and the spin quantum number is either $+\frac{1}{2}$ or $-\frac{1}{2}$. The only set that honours every one of these limits is $n=4,l=3,m=+1,s=+\frac{1}{2}$, since $m=+1$ sits inside the permitted range. A set carrying $l=4$ or $m=+4$ is simply impossible for an f electron.