# Question 1e0d8

Aug 12, 2017

Consider the structure of the ammonia molecule versus that of boron trifluoride....

#### Explanation:

$\left(i\right)$ For stackrel(ddot)NH_3# there are 4 regions of electron density, 4 electron pairs, around the central nitrogen atom. The most stable geometric arrangement for the electron pairs is as a tetrahedron. And thus, TO A FIRST APPROX., the electronic geometry of the ammonia is tetrahedral.

But we decide molecular geometry on the basis of the disposition of atoms, not on the geometry of electron pairs. The geometry of ammonia descends from tetrahedral to trigonal pyramidal with $\angle H - N - H \cong 104 - {5}^{\circ}$.

On the other hand, $B {F}_{3}$, has 3 electron pairs around the central boron atoms, and a trigonal planar geometry is predicted with $\angle F - B - F = {120}^{\circ}$, and this prediction coincides with the observed geometry.

$\left(i i\right)$ For the physical state of ${H}_{2} O$ versus ${H}_{2} S$ at room temperature we must interrogate the intermolecular forces that act between molecules. For water, the most potent intermolecular force is hydrogen bonding, which occurs when hydrogen is bound to a strongly electronegative atom such as oxygen, or nitrogen, or fluorine.

$O {H}_{2}$, $N {H}_{3}$, and $H F$, all have disproportionately high boiling points, certainly compared to $S {H}_{2}$, $P {H}_{3}$, and $H C l$, and this must be attributed to hydrogen bonding. The $H - O$ dipole is strong, and is thus responsible for a strong intermolecular interaction. Hydrogen bonding, and charge separation is not so significant for the lower group hydrides.

$\left(i i i\right)$ We gots $S {F}_{6}$ versus $O {F}_{2}$; sulfur is a third period atom, i.e. it is LARGER than second period oxygen. The larger atom certainly should certainly support a greater number of coordinating ligands.