The total number of orbital for is — Atomic Structure Chemistry Question

💡 Solution & Explanation

The combinations for $(n+l)=4$ are: $n=4, l=0$ (4s, 1 orbital); $n=3, l=1$ (3p, 3 orbitals). Combinations like $n=2, l=2$ are not allowed since $l < n$. Total orbitals = $1 + 3 = 4$. Therefore, correct answer is A.