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Impossible orbital among the following is

Options

  1. A3 f
  2. B2 p
  3. C4 d
  4. D2 s

Correct answer

A. 3 f

Step-by-step solution

According to Bohr - Burry's scheme, Maximum number of e ⁻ in 3 rd orbital =3 n ² =3(3)²=18 and maximum number of e ⁻ in s-subshell =2 p-subshell =6 d-subshell =10 f-subshell =14 Therefore, 18 e ⁻ present in 3rd orbital enter in s, p and d subshell. Thereafter, there is no e ⁻ available for f-subshell. That's why 3 f subshell is not possible.

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