# Why can elements in the 3rd period exceed 8 valence electrons?

## Sulfur can have 12 valence electrons in $S {O}_{4}^{2 -}$ and Chlorine has 10 in ${\left[C l {O}_{4}\right]}^{-}$ Why?

Jul 25, 2018

What's new in $n = 3$?

Recall that the angular momentum quantum number $l$ tells you what orbital subshell you have, $s , p , d , f , \ldots$ Well, you should take note that

$\text{ } \textcolor{w h i t e}{/} s , p , d , f , . . .$
$l = 0 , 1 , 2 , 3 , . . . , n - 1$,

i.e. that the maximum $l$ is one less than $n$, the principal quantum number (which indicates the energy level), where:

$n = 1 , 2 , 3 , . . .$

Hence, if we are on the third period, we introduce $n = 3$, and so, $n - 1 = 2$ and orbitals with UP TO $l = 2$, $d$ orbitals, are possible. That is, $3 s$, $3 p$, AND $3 d$ orbitals are usable.

This is especially notable in silicon, phosphorus, sulfur, and chlorine if we consider the third period.

Usage of those $3 d$ orbitals allows for extra space to hold electrons, and as a result, hypervalency is possible.

This expansion of "orbital space" is known in, for example:

• ${\text{PF}}_{5}$, where phosphorus has $10$ valence electrons around it arranged in a trigonal bipyramidal geometry.

• ${\text{SF}}_{6}$, where sulfur has $12$ valence electrons around it arranged in an octahedral geometry.

• ${\text{ClF}}_{5}$, where chlorine has $12$ valence electrons around it arranged in a square pyramidal geometry (two of which are in one lone pair).