KCET · Chemistry · Structure of Atom
Impossible orbital among the following is
- A \(3 f\)
- B \(2 p\)
- C \(4 d\)
- D \(2 s\)
Answer & Solution
Correct Answer
(A) \(3 f\)
Step-by-step Solution
Detailed explanation
According to Bohr - Burry's scheme, Maximum number of \(\mathrm{e}^{-}\)in 3 rd orbital \(=3 \mathrm{n}^{2}\)
\[
=3(3)^{2}=18
\]
and maximum number of \(\mathrm{e}^{-}\)in
s-subshell \(=2\)
p-subshell \(=6\)
d-subshell \(=10\)
f-subshell \(=14\)
Therefore, \(18 \mathrm{e}^{-}\)present in 3rd orbital enter in s, \(\mathrm{p}\) and \(\mathrm{d}\) subshell. Thereafter, there is no \(\mathrm{e}^{-}\) available for f-subshell. That's why \(3 \mathrm{f}\) subshell is not possible.
\[
=3(3)^{2}=18
\]
and maximum number of \(\mathrm{e}^{-}\)in
s-subshell \(=2\)
p-subshell \(=6\)
d-subshell \(=10\)
f-subshell \(=14\)
Therefore, \(18 \mathrm{e}^{-}\)present in 3rd orbital enter in s, \(\mathrm{p}\) and \(\mathrm{d}\) subshell. Thereafter, there is no \(\mathrm{e}^{-}\) available for f-subshell. That's why \(3 \mathrm{f}\) subshell is not possible.
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