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1 questions with worked solutions.
Answer: Number of possible orbitals satisfying these conditions is 4
The valid (n, l) pairs are (4,1) and (3,2), each contributing 2 orbitals (mₗ = +1 and -1), giving 4 orbitals total. Each can hold 1 electron with mₛ = +1/2, so max electrons with mₛ = +1/2 is 4 (not 6). For Fe ([Ar] 3d⁶ 4s²), the 3d orbitals with mₗ = ±1 each hold one electron from the first 5 fills, giving 2 electrons satisfying the condition (not 1). Hence only statement A is correct.