Question

${CH}_4$ has four identical $C-H$ bonds and non-polar in nature. It would have been better if it were supposed to have square planar structure but it is assigned tetrahedral geometry. This is due to the fact:

• A. tetrahedral shape leads to zero dipole moment
• B. square planar shape leads to zero dipole moment
• C. in case of square planar shape two isomers on mono substitution will be possible which is not observed in ${CH}_4$
• D. the net dipole moment of ${CH}_4$ is zero

Answer 1

The correct option is C in case of square planar shape two isomers on mono substitution will be possible which is not observed in ${CH}_4$

$C{H}_4$ has four identical $C-H$ bonds and non-polar in nature. It would have been better if it were supposed to have square planar structure but it is assigned tetrahedral geometry. This is because in case of square planar shape two isomers on mono substitution will be possible which is not observed in $C{H}_4$.It is a reason propose square planar geometry for $C{H}_4$.For square planar arrangement for $C{H}_4$, the four $C-H$ bond pairs have to lie right angle to each other.