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Last updated date: 30th Nov 2023
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# Predict the molecular shape of $O{F_2}$ .

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Hint :To predict the molecular shape of oxygen difluoride, $O{F_2}$ , we will use VSEPR theory. In this question, both molecular geometry and electron geometry are important.

In oxygen difluoride, $O{F_2}$ , each fluorine atom has $7$ valence electrons. Since there are two fluorine atoms, therefore, total valence electrons are $14$ .
The atomic number of oxygen is $8$ and hence its electronic configuration is $1{s^2}2{s^2}2{p^4}$ . The $2s$ and $2p$ orbitals of atoms are hybridized. This results in the formation of four hybridized orbitals namely: $2s$ , $2{p_x}$ , $2{p_y}$ , $2{p_z}$ . Hence, the molecule is $s{p^3}$ hybridized.
Hence, the total number of valence electrons is $6 + 14 = 20$ . These $20$ valence electrons form bonds in $O{F_2}$ . Here, oxygen undergoes $s{p^3}$ hybridization and its two hybrid orbitals contain the lone pair as we saw above.
As per the VSEPR theory, $O{F_2}$ has a linear geometry. The linear geometry is because of the lone pair repulsion of oxygen and fluorine electrons. The molecular arrangement of oxygen difluoride is similar to water and it entails it to a bent shape due to lone pair repulsion. However, the bond angle is a bit more than the angle between hydrogen atoms in water due to the very high lone pair repulsion of fluorine.
Therefore, we can say that the electron geometry of $O{F_2}$ is tetrahedral and the molecular geometry is linear.