Bond Angle Doddle

Here we doddle as we think about dissolution.

This is the chloride member of the salt $NaCl$, lattice with the singular unpaired orbit drawn in.  Below this single orbit is three paired orbits with six electrons.  Each of this paired orbits forms two ionic bonds with two $Na$ members.  The six ionic bonds arrange themselves perpendicularly to the planes of a square and this unit forms a square lattice.

This is the $Na$ member of the salt $NaCl$ lattice.  Three paired orbits from the inner $Ne$ nucleus core forms six ionic bonds with six $Cl$ members.  The bonds also arrange themselves into a square.

Strictly speaking,

Bond $ON$ is free to rotate in the circle $C_n$ and $OP$ is free to rotate in the circle $C_p$.  In general, this makes,

$\angle S\ne\cfrac{180^o}{n}$

The bond angle is not solely determine by the total number of orbits, $n$, in the nucleus shell participating in bond formations.  If however, only one of the bond is free to rotate along its orbit then,

$\cfrac{180^o}{n}\le\angle S\le \left(180^ o-\cfrac{180^o}{n}\right)$

where $n$ is the principal quantum number of the nucleus shell participating in bond formations.  If both bonds are free to rotate, and we move one of the bond to one intersection of the two orbits, the bond angle,

$0\le\angle S\le 180^ o$

is not restricted at all.