In which of the following molecules,all atoms | vedclass

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(C) For a molecule to be coplanar,all its atoms must lie in the same plane.$(A)$ Biphenyl: Due to steric hindrance between the ortho-hydrogens of the

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$1,1-$dicyano$-2,2-$dimethylethene

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(C) For a molecule to be coplanar,all its atoms must lie in the same plane.$(A)$ Biphenyl: Due to steric hindrance between the ortho-hydrogens of the two phenyl rings,the rings are twisted relative to each other,making the molecule non-coplanar.$(B)$ Cyclohexane: It exists in a chair conformation,which is non-planar.$(C)$ $1,1-$dicyano$-2,2-$dimethylethene: The central $C=C$ bond is $sp^2$ hybridized. The two $CH_3$ groups and two $CN$ groups are attached to these $sp^2$ carbons. While the $C=C$ bond and the atoms directly attached to it are planar,the $CH_3$ groups have $sp^3$ hybridized carbons,which are tetrahedral and not coplanar with the rest of the molecule.$(D)$ Bicyclohexyl: It consists of two cyclohexane rings,which are non-planar.Correction: Upon re-evaluating the options,none of the provided structures are strictly coplanar in their most stable conformations. However,if we consider the planar representation of the $C=C$ system in option $(C)$,it is the closest to being planar,but strictly speaking,the methyl hydrogens prevent total coplanarity. If the question implies a planar structure like $1,1-dicyanoethene$,it would be planar. Given the standard options,this question is often flawed in competitive exams. If we must choose,$1,1-dicyano-2,2-dimethylethene$ has a planar $C=C$ core,but the methyl groups are not. If the question intended $1,1-dicyanoethene$,it would be planar.

List-$I$	List-$II$
$(P). \ IF_7$	$(a). \ \text{Planar} \ \& \ \text{Polar}$
$(Q). \ SO_2$	$(b). \ \text{Non-planar} \ \& \ \text{Polar}$
$(R). \ SF_4$	$(c). \ \text{Planar} \ \& \ \text{Non-polar}$
$(S). \ CS_2$	$(d). \ \text{Non-planar} \ \& \ \text{Non-polar}$

In which of the following molecules,all atoms are coplanar?

Similar Questions

Which of the following statements is correct?

The incomplete Lewis representation of $CO_3^{2-}$ is given below. The formal charges on atoms marked as $a$,$b$,and $c$,respectively,are

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$(P). \ IF_7$ $(a). \ \text{Planar} \ \& \ \text{Polar}$

$(Q). \ SO_2$ $(b). \ \text{Non-planar} \ \& \ \text{Polar}$

$(R). \ SF_4$ $(c). \ \text{Planar} \ \& \ \text{Non-polar}$

$(S). \ CS_2$ $(d). \ \text{Non-planar} \ \& \ \text{Non-polar}$

Which of the following compounds is not hypovalent but completes its octet?

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