Question

Draw the structure of the following molecules:
(i) $XeO{F_4}$ (ii) ${H_3}P{O_3}$

Answer 1

First try and find out the valence shell electronic configuration of the central atom. Then try to give structure to the compound in a way that all the atoms forming the bonds complete their octet.

Complete answer:
In order to draw the structure of the given molecules, we need to know the bonding between the atoms. We will see their structure one by one.
i) $XeO{F_4}$:
- We know that Xe (Xenon) is a noble gas having an atomic number of 54. However, it makes complexes.
- The valence shell electronic configuration of Xe is $[Kr]4{d^{10}}5{s^2}5{p^6}$. So, there are 8 electrons in the valence shell of Xe. There are 6 and 7 electrons in the valence shell of oxygen and fluorine respectively.
- So, Xenon should make four single bonds with four fluorine atoms in order to satisfy the octet of fluorine. Xenon will make a double bond with oxygen as it will complete the octet of oxygen atoms. There will be one lone pair on the Xenon atom.
- Thus, we can say that the hybridization of this atom will be $s{p^3}{d^2}$. So, its shape should be square pyramidal. Its structure can be drawn as below.

ii) ${H_3}P{O_3}$:
The name of this acid is phosphorus acid. This acid is a diprotic acid. This means that this acid gives two protons upon its full dissociation into ions in water.
- We know that the electronic configuration of P is $[Ne]3{s^2}3{p^3}$.
- From the electronic configuration, we can say that it can make three sigma bonds. It also has a lone pair.
- So, in its structure, it forms a double bond with an oxygen atom in order to complete its octet. It forms a sigma bond with two –OH groups and one H atom. Thus, this structure can explain its dibasic nature. The structure can be drawn as

Note:
Remember that as the formula of phosphorus acid is ${H_3}P{O_3}$, it does not have three –OH groups singly bonded to one P atom. Actually, this structure is not stable and thus ${H_3}P{O_3}$ exists in the structure given just above.