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Correct answer: It can be zero for a 3p orbital
Probability density psi² is always >= 0, so it can never be negative (ruling out option A). For 1s, there are no nodes (psi² = 0 only at r = infinity), so option C is wrong. For 2s, there is one radial node where psi² = 0, so option D is wrong. For 2p, there is one angular node (psi = 0 on a nodal plane), so psi² can be zero. For 3p, there is one radial node and one angular node, so psi² can also be zero. Option B (3p can be zero) is correct.