Question

The bond angle in $(C{H}_3{)}_{2}S$ is close to:

• A. ${90}^{\circ }$
• B. ${109}^{\circ }$
• C. ${120}^{\circ }$
• D. ${105}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

$(C{H}_3{)}_{2}S$ is a Drago compound and it does not have any hybridisation.

According to Drago's Rule, a p-block element of the $3^{rd}$ or any higher period does not allow hybridisation if they have lone pairs on them, in molecule when they form bond with less electronegative elements such as hydrogen and carbon (in methyl group).

So, $C{H}_3{)}_{2}S$ will have an angle of ${90}^o$

Therefore, $(C{H}_3{)}_{2}S$ do not have hybridisation i.e bond angle is close to ${90}^{\circ }$.