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JEE MainChemistryStructure of Atom

Consider five orbitals A, B, C, D, and E in a multi-electron atom, described by their number of radial and angular nodes: A: 2 radial nodes, 0 angular nodes B: 0 radial nodes, 2 angular nodes C: 1 radial node, 1 angular node D: 1 radial node, 2 angular nodes E: 3 radial nodes, 0 angular nodes The correct increasing order of energy of these orbitals is:

Options

  1. AA < C < B < E < D
  2. BA < C < E < B < D
  3. CA = B = C < E = D
  4. DA < E < C < B < D

Correct answer

B. A < C < E < B < D

Step-by-step solution

The number of angular nodes is equal to the azimuthal quantum number ( l ). The number of radial nodes is given by n - l - 1 , where n is the principal quantum number. Let us determine n and l for each orbital: A: l=0 , radial nodes = n - 0 - 1 = 2 n=3 . This is the 3s orbital. B: l=2 , radial nodes = n - 2 - 1 = 0 n=3 . This is the 3d orbital. C: l=1 , radial nodes = n - 1 - 1 = 1 n=3 . This is the 3p orbital. D: l=2 , radial nodes = n - 2 - 1 = 1 n=4 . This is the 4d orbital. E: l=0 , radial nodes = n - 0 - 1 = 3

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