# Question 2e9c4

Nov 11, 2017

Number of Period = Row number in table, so it concerns Li to Ne...

4 Quantum numbers:
n = Principal number (basically row number);
$l$  = Azimuthal (Orbital) number;
m = Magnetic number;
s = Spin number.

Maximum values:
n = 1,2,3,4,5 etc. (remember, this is the row number)
$l$ ranges from 0 to n-1.
m ranges from -l"" to $+ l$

If n =1 (row 1) then $l$ can only be 0. (n-1 = 1-1 = 0)
If $l$=0 then m can only be 0. (-l"" to $+ l$ )
s is always either -1/2 or +1/2
This describes the $1 S$-orbital, with max. 2 ${e}^{-}$ in it.

If n=2, then $l$ ranges from 0 to n-1, which is 1.

Once again:
If $l$=0 then m can only be 0. (-l"" to $+ l$ )
s is always either -1/2 or +1/2
since n=2, this describes the $2 S$-orbital, with max. 2 ${e}^{-}$ in it.

If however $l$=1 then m can be $\textcolor{red}{\text{-1, 0 or 1}}$ (-l""# to $+ l$ )

$l$ = 1 describes the $P$-orbital, or rather set of:
m=-1: $\rightarrow {P}_{x}$
m=0: $\rightarrow {P}_{y}$
m=+1: $\rightarrow {P}_{z}$

Therefore, in period 2 there are 4 orbitals: $2 S , 2 {P}_{x} , 2 {P}_{y}$ and $2 {P}_{z}$

2 electrons in each orbital: $\rightarrow$ 4 x 2 = 8 ${e}^{-}$...

The $2 S$-orbital (S for Spherical) has a larger diameter than the $1 S$, and therefore completely envelopes it.

Hence the two ${e}^{-}$ of the $1 S$ are sheltered and play no part in normal reactions. They are therefore not Valence Eectrons.