Question

What are the bond angles of \[P{F_3}\] and \[{H_2}Se\] ?

Answer 1

Hint: To solve this question, we have to count the number of valence electrons contributed by each atom and predict their electron geometry, then observe the bond angle which is defined as the angle between two bonds that will originate from the same atom in a covalent species. Geometrically, it is the angle formed between the two covalent bonds.

Complete answer:
First, we count the number of valence electrons in \[P{F_3}\] and \[{H_2}Se\] .
Phosphorus trifluoride , \[P{F_3}\]
Here, the atomic number of phosphorus is \[15\] and fluorine is\[9\] . Hence, their valency will be \[5\] and \[7\] respectively. There are \[3\] fluorine atoms present so, valency of \[3\] fluorine atoms will be \[21\] .
From the above, \[P{F_3}\] contains \[5 + 21 = 26\] valence electrons in which \[6\] are lone pairs for each \[F\] atoms and \[6\] electrons are required to make \[3\] bonds. Three bond pairs are formed and one lone pair is left which gives four electron groups.
This corresponds to the tetrahedral electron geometry, but the last electron is not a bond, so the molecular geometry is Trigonal pyramidal. The lone pair of electrons takes more space and it causes coulombic repulsion to the bond pairs which compresses the bond angle.
Therefore, the bond angle is less than the bond angle of the tetrahedral geometry. Hence, it is actually \[{97.7^o}\]
Hydrogen selenide, \[{H_2}Se\]
Here, the atomic number of Hydrogen is \[1\] and selenide is \[34\] . Hence, their valency will be \[1\] and \[6\] respectively. There are \[2\] Hydrogen atoms are present so, valency of hydrogen atoms will be \[2\] .
From the above, \[{H_2}Se\] contains \[2 + 6 = 8\] valence electrons in which \[2\] electron pairs form sigma bonds as hydrogen can form only one bond. The other \[2\] electron pairs will act as lone pairs.
This corresponds to the tetrahedral electron geometry, but due to the presence of \[2\] lone pairs of electrons, it has bent molecular geometry. The lone pair of electrons takes more space and it causes coulombic repulsion to the bond pairs which compresses the bond angle. So, this bond angle is very small. It is smaller than \[P{F_3}\] , it is actually \[{90.9^o}\] .
Therefore, the bond angle of \[P{F_3}\] is \[{97.7^o}\] whereas the bond angle of \[{H_2}Se\] is \[{90.9^o}\] .

Note:
The geometric angle between the two adjacent bonds originated from the same atom in a covalent bond species is called bond angle. This bond parameter is very useful as it provides an idea about the molecular geometry of a compound.