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JEE MainChemistryStructure of Atom

Consider the following six electrons characterized by their quantum numbers: (I) n=4, l=2, m_l=-2 (II) n=5, l=1, m_l=0 (III) n=4, l=2, m_l=+1 (IV) n=5, l=1, m_l=+1 (V) n=4, l=1, m_l=0 (VI) n=4, l=1, m_l=+1 Which of the following sets of electrons occupy degenerate orbitals with the highest energy among the given set?

Options

  1. AOnly (I) and (III)
  2. BOnly (V) and (VI)
  3. C(I), (II), (III) and (IV)
  4. DOnly (II) and (IV)

Correct answer

D. Only (II) and (IV)

Step-by-step solution

For a multi-electron atom, the energy of an orbital is determined by the (n+l) rule. Let us find the subshell and (n+l) value for each electron: (I) 4d orbital, n+l = 4+2 = 6 (II) 5p orbital, n+l = 5+1 = 6 (III) 4d orbital, n+l = 4+2 = 6 (IV) 5p orbital, n+l = 5+1 = 6 (V) 4p orbital, n+l = 4+1 = 5 (VI) 4p orbital, n+l = 4+1 = 5 Degenerate orbitals must belong to the same subshell (same n and same l ). The degenerate pairs are: (I) and (III) in the 4d subshell. (II) and (IV) in the 5p subshell. (V) and (VI) in the 4

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