In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields?

(a) n = 1, l = 0, m = 0 (b) n = 2, l = 0, m = 0

 (c) n = 2, l = 1, m = 1 (d) n = 3, l = 2, m = 1

 (e) n = 3, l = 2, m = 0

 (A) (b) and (c) (B) (a) and (b) (C) (d) and (e) (D) (c) and (d)

In the absence of magnetic and electric fields, the orbitals defined by magnetic quantum number are degenerate (of same energy). So, energy of the orbital, in the absence of magnetic and electric fields depends on the (n + l) value.

Higher the (n + l) value, larger the energy of the orbital.

Orbitals with same ‘n’ and ‘n + l” values are degenerate and have same energy.

In the given combinations,’d’ and ‘e’ have same ‘n’ and ‘n + l’ value and so, have same energy in the absence of magnetic and electric fields.

Correct choice is (C).

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