# What are the bond angles of "PF"_3 and "H"_2"Se"?

Mar 7, 2016

You can count the number of valence electrons each atom contributes and generate a reasonable electron geometry.

PHOSPHORUS TRIFLUORIDE

$\text{P} : 5$ ($Z = 15$)
$\text{F} : 7$ ($Z = 9$)

From the above, ${\text{PF}}_{3}$ contains $5 + 21 = 26$ valence electrons.

1. $6$ are lone pairs for each $\text{F}$.
2. $6$ are required to make three total $\sigma$ bonds. That accounts for $\setminus m a t h b f \left(24\right)$.
3. Therefore, one last lone pair accounts for the final $2$ valence electrons and gives us 4 electron groups.

That corresponds to tetrahedral electron geometry, but since one out of the four electron groups is not a bond, the molecular geometry differs from the electron geometry and is actually trigonal pyramidal.

The lone pair of electrons takes up more space than a regular bonding pair since it it is not confined to be between two atoms, so it adds coulombic repulsion to the bonding pairs and compresses the angle.

Therefore, the bond angle is less than the standard ${109.5}^{\circ}$. It is actually $\setminus m a t h b f \left({97.7}^{\circ}\right)$.

HYDROGEN SELENIDE

$\text{H} : 1$ ($Z = 1$)
$\text{Se} : 6$ ($Z = 34$)

From the above, $\text{H"_2"Se}$ contains $6 + 2 = 8$ valence electrons.

1. $2$ are used for each of two $\sigma$ bonds. Hydrogen can only make $1$ bond in normal cases, so there is no reason to put it in the center.
2. The remaining $4$ are used as two lone pairs, accounting for all $\setminus m a t h b f \left(8\right)$ valence electrons.

That corresponds to tetrahedral electron geometry, but since two out of the four electron groups is not a bond, the molecular geometry differs from the electron geometry and is actually bent.

The lone pair of electrons takes up more space than a regular bonding pair since it it is not confined to be between two atoms, so it adds coulombic repulsion to the bonding pairs and compresses the angle. Furthermore, there are two bonding pairs instead of three this time, which is easier to compress, so this bond angle is even smaller.

Therefore, the bond angle is even smaller than that of ${\text{PF}}_{3}$. It is actually $\setminus m a t h b f \left({90.9}^{\circ}\right)$.