# How many subshells are possible with (n+1) equal to 6?

Jul 4, 2018

Well, there are five subshells POSSIBLE, but only the first two are ACCESSIBLE.

Well... apparently, you want the principal quantum number PLUS ONE to be:

$n + 1 = 6$

so the energy level specified, is $n = 5$. For a given $n$, a certain maximum angular momentum $l$ exists:

$l = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , . . . , {\overbrace{n - 1}}^{{l}_{\max}}$

and each of these values of $l$ correspond to a unique subshell $s , p , d , f , g , h , i , k , . . .$.

Since $n = 5$ and $n - 1 \equiv {l}_{\max} = 4$,

$\textcolor{b l u e}{l = 0 , 1 , 2 , 3 , 4}$

are allowed. Clearly, it means there are FIVE subshells possible (five values of $l$ are allowed), and they are:

$s , p , d , f , g$

i.e. there are one $5 s$, three $5 p$, five $5 d$, and seven $5 g$ orbitals theoretically possible for any given atom among $Z = 37 - 54$.

However, only the $5 s$ and $5 p$ are accessible for these atoms.