The correct option is C n=3, l=2, m=1, s=+12
(a) n=3, l=0 means 3s− orbital and n+1=3
(b) n=3, l=1 means 3p− orbital n+1=4
(c) n=3, l=2 means 3d− orbital n+1=5
(d) n=4, l=0 means 4s− orbital n+1=4
Increasing order of energy among these orbitals is
3s<3p<4s<3d
∴ 3d has highest energy.