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Impossible orbital among the following is
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The correct answer is:
$3 f$
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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