Quantum number n = 3, l = 2, m = +2 represent an orbital with

s = ±¹⁄₂ (3d_xy or 3d_{x² - y²})

which is possible only for one electron.

We know that atomic number of gadolinium is 64. Therefore the electronic configuration of gadolinium is [Xe] 4f⁷ 5d¹ 6s². Because the half filled and fully filled orbitals are more stable.

According to Hund’s rule electron pairing in P, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied.

λ = ^h⁄_mv;

∴ λ ∝ ¹⁄_v hence answer (c).

n + l = 7

7 + 0 = 7s ; 6 + 1 = 6p ; 5 + 2 = 5d ; 4 + 3 = 4f

The sub-shell are 3d, 4d, 4p and 4s, 4d has highest energy as n + l value is maximum for this.

If m = – 3; l = 3,

[m ranges from -l to +l]

So n = 4 as nature of l ranges from

0 to (n – 1).

So option (c) is the answer.

l = 3 means f-subshell. Maximum no. of electrons = 4l + 2 = 4 × 3 + 2 = 14

n = 2, l = 1 means 2p–orbital. Electrons that can be accommodated = 6 as p sub-shell has 3 orbital and each orbital contains 2 electrons.