The correct option is D
9
n=3(third energy shell)l=n−1=3−1=2i.e.,l=0,1,2
l = 0, s-subshell-3s = 1 orbital(3s)
l =1, p-subshell-3p = 3 orbitals(3px,3py,3pz)
l = 2, d-subshell-3d = 5 orbitals(3dxy,3dyz,3dxz,3dx2−y2,3dz2)
The total number of orbitals is equal to 1+3+5=9.