Question

The $ C{H_3}F $ molecule has which molecular geometry, as dictated by the hybridization, and polarity?
(A) $ s{p^3} $ , non-polar
(B) $ s{p^3} $ , polar
(C) $ s{p^2} $ , polar
(D) $ s{p^2} $ , non-polar

Answer 1

Hint : First calculate the no of bonds formed by $ C $ in $ C{H_3}F $ . Next, find the no of lone pairs and calculate the steric number. After you get the steric number, you will be able to tell the hybridization of the molecule and thus its shape. Next, explain why $ C{H_3}F $ is polar in nature. Try to use diagrams to explain the theoretical aspects of the question.

Complete Step By Step Answer:
 $ C{H_3}F $ or Fluoromethane has a $ s{p^3} $ molecular geometry. However, it is a polar molecule.
To check for its molecular structure, we will first obtain the hybridization of $ C{H_3}F $ :
In $ C{H_3}F $ ,
No of $ \sigma $ bonds are $ = 4 $ ( $ 3 $ hydrogen atoms and $ 1 $ fluorine atom)
No of lone pairs $ = \dfrac{{v - b - c}}{2} $
Here, $ v \to $ no of valence electrons in free state
 $ b \to $ no of bonds formed (both $ \sigma $ and $ \pi $ included)
 $ c \to $ charge on atom that forms the bonds
In $ C{H_3}F $ , $ v = 4 $ , $ b = 4 $ , $ c = 0 $
 $ \therefore $ no of lone pairs $ = \dfrac{{4 - 4 + 0}}{2} = 0 $
We know that,
Steric number $ = $ no of lone pairs + no of $ \sigma $ bonds
 $ = 4 + 0 = 4 $
And we know that if the steric number is $ 4 $ then the hybridization of $ C{H_3}F $ is $ s{p^3} $ and thus, it has a tetrahedral shape.

As shown in above diagram, in $ C{H_3}F $ , carbon atom is central and it has four covalent bonds with hydrogen and fluorine atoms which contribute to its tetrahedral geometric shape. Since the electronegativity of fluorine is quite high as compared to carbon and hydrogen atoms, it acquires a partial negative charge and other atoms in the molecule acquire a partial positive charge thus making the $ C{H_3}F $ molecule polar. Also, due to very high electronegativity of fluorine, the dipole moment of $ C{H_3}F $ becomes non zero.
Therefore, option B is correct.

Note :
Always calculate the $ \sigma $ bonds carefully. Also keep in mind the formula of calculating lone pairs as any mistake in calculating lone pairs can lead to an incorrect steric number which ultimately leads to an incorrect hybridization and a wrong molecular structure. Try to use diagrams in explanation of theoretical aspects of the question and make them as clear as possible.