Question

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

Sulfur can have 12 valence electrons in `SO_4^(2-)` and Chlorine has 10 in `[ClO_4]^-` Why?

Answer 1

What's new in `n = 3`?

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

`" "color(white)(/)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, `3s`, `3p`, AND `3d` orbitals are usable.

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

Usage of those `3d` 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:

• `"PF"_5`, where phosphorus has `10` valence electrons around it arranged in a trigonal bipyramidal geometry.

• `"SF"_6`, where sulfur has `12` valence electrons around it arranged in an octahedral geometry.

• `"ClF"_5`, where chlorine has `12` valence electrons around it arranged in a square pyramidal geometry (two of which are in one lone pair).