Which statement explains why a molecule of CH4 is nonpolar?

Chemistry

2 answers:

Deffense [45]3 years ago

8 0

The answer is the option (3) The geometric shape of a CH4 molecule distributes the charges geometrically. the CH4 is a tetrahedral with the Carbon atom in the center and the four H atoms in the vertices of the tetrahedral. The C-H bond angles are all identical which makes the molecule perfectly symmetrical and so any dipolar moment will cancel making the molecule non polar.

lara [203]3 years ago

5 0

The following statement explains why a molecule of $\text{CH}_4$ is nonpolar:

$\boxed{\left( 3 \right){\text{ The geometric shape of a C}}{{\text{H}}_4}{\text{ molecule distributes the charge symmetrically}}}$ .

Further explanation:

Covalent Bond

This type of bonding takes place when the bonded atoms mutually share electron pairs between them. It is also called the molecule bond. The chemical compounds formed as a result of this bond are called chemical compounds.

The polarity of any bond is primarily governed by two factors; electronegativity difference and symmetry. A bond is said to be polar if the bonded atoms have a considerable molecular bondelectronegativity difference between them. But if there is very small or no electronegativity difference between the bonded atoms, the resulting bond is nonpolar in nature.

Symmetry is another factor responsible for the polarity of the bond. The charge distribution is symmetrical in case symmetry is present in the molecule and as a result, the molecule is said to be nonpolar. But if there is asymmetry in the shape of the molecule, it will be polar in nature.

The electronegativity difference between carbon and hydrogen atoms in $\text{CH}_4$ molecule is very small or negligible so the bonds between the atoms in $\text{CH}_4$ are nonpolar in nature. But the hydrogen atoms are placed in symmetrical positions with respect to the central carbon atom, imparting symmetry to $\text{CH}_4$ molecule and therefore it is nonpolar (For structure, refer to the attached image).

Learn more: