Which of the following shows geometrical isomers?
$(a)$ $CH_2=CHCl$
$(b)$ $CH_3CH=CH_2$
$(c)$ $CHCl=CHCl$
$(d)$ $(CH_3)_2C=CHC_2H_5$

(C) For a molecule to exhibit geometrical isomerism,each carbon atom of the double bond must be attached to two different groups.
In $(a)$ $CH_2=CHCl$,the first carbon is attached to two identical $H$ atoms.
In $(b)$ $CH_3CH=CH_2$,the first carbon is attached to two identical $H$ atoms.
In $(c)$ $CHCl=CHCl$,each carbon atom is attached to one $H$ atom and one $Cl$ atom,which are different. Thus,it shows geometrical isomerism (cis and trans forms).
In $(d)$ $(CH_3)_2C=CHC_2H_5$,the first carbon is attached to two identical $CH_3$ groups.
Therefore,only $(c)$ shows geometrical isomers.