Question

The central oxygen atom in ether is_______

Answer 1

Hybridization of Oxygen in ether:-

• Valence Shell Electron Pair Repulsion (VSEPR) theory is a theory that is used in order to predict the molecular shape of any compound, with the help of the electron pairs that are present around the central atom of the compound.
• The total number of bond pairs and the lone pair can be calculated with the help of the formula:$\frac{1}{2}[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}]$
• Here, V= No. of valance electron on the central atom, M= No. of monovalent atoms directly attached to the central atom, C= Charge on Cation, A= Charge on Anion.
• For ether, if we use this formula we can find out the number of lone pairs and number of bond pairs and thus predict its shape.
• Here in ether, V= 6 as Oxygen has 6 valence electrons, M= 2 as there are two alkyl group atoms attached to Oxygen, C= 0 and A= 0 as the charge of the compound is 0
• There applying the formula we get,

$$\begin{gathered} \frac{1}{2}[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}] \\ =\frac{1}{2}[6+2-0+0)] \\ =\frac{1}{2}\times 8 \\ =4 \end{gathered}$$

• From here we can see that there are 2 bond pairs $(\mathrm{O}-\mathrm{R})$and 2 lone pairs, and the hybridization of the compound is ${\mathrm{sp}}^3$.
• Hence the shape is Tetrahedral.

Therefore, the central oxygen atom in ether is ${\mathrm{sp}}^3$ hybridized