Question

Impossible orbital among the following is

• A. $2s$
• B. $3f$
• C. $2p$
• D. $4d$

Answer 1

Answer: (B)

$3f$

Answer 2

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 $3\, rd$ 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.