Consider the following pairs of electrons
(A) (a) $$n = 3, l = 1, m_l = 1, m_s = +\frac{1}{2}$$
(b) $$n = 3, l = 2, m_l = 1, m_s = +\frac{1}{2}$$
(B) (a) $$n = 3, l = 2, m_l = -2, m_s = -\frac{1}{2}$$
(b) $$n = 3, l = 2, m_l = -1, m_s = -\frac{1}{2}$$
(C) (a) $$n = 4, l = 2, m_l = 2, m_s = +\frac{1}{2}$$
(b) $$n = 3, l = 2, m_l = 2, m_s = +\frac{1}{2}$$
The pairs of electrons present in degenerate orbitals is/are

We need to identify which pairs of electrons are in degenerate orbitals. Degenerate orbitals are orbitals that have the same energy. In a hydrogen-like atom, orbitals with the same $$n$$ are degenerate, but in multi-electron atoms, orbitals with the same $$n$$ and $$l$$ are degenerate (due to shielding effects).

Pair (A):

(a) $$n = 3, l = 1$$ → 3p orbital

(b) $$n = 3, l = 2$$ → 3d orbital

These are in different subshells (3p vs 3d). In a multi-electron atom, 3p and 3d have different energies. Not degenerate.

Pair (B):

(a) $$n = 3, l = 2, m_l = -2$$ → one of the 3d orbitals

(b) $$n = 3, l = 2, m_l = -1$$ → another of the 3d orbitals

Both electrons are in 3d orbitals (same $$n$$ and same $$l$$). Orbitals with the same $$n$$ and $$l$$ but different $$m_l$$ are degenerate. Degenerate.

Pair (C):

(a) $$n = 4, l = 2$$ → 4d orbital

(b) $$n = 3, l = 2$$ → 3d orbital

These have different principal quantum numbers (4d vs 3d). They have different energies. Not degenerate.

Only pair (B) has electrons in degenerate orbitals.

The correct answer is Option B: Only (B).