Question

The structures of two compounds (I) and (II) are given below. Which statement is correct ?
$F_{2}C=C=C{F}_2\left(I\right)$ and $F_{2}C=C=C=C{F}_2\left(II\right)$

• A. Compound (I) shows dipole moment but (II) does not.
• B. Compound (II) shows dipole moment but (I) does not
• C. Both show dipole moment
• D. Neither of them shows dipole moment

Answer 1

Answer: (A)

Compound (I) shows dipole moment but (II) does not.

Compound (I) $F_{2}C=C=C{F}_2$ shows dipole moment but (II) $F_{2}C=C=C=C{F}_2$ does not.

(I) $F_{2}C=C=C{F}_2$ - All atoms are not in one plane because $C2$ is $sp$-hybridised and is linear, whereas $C1$ and $C3$ are $s{p}^2$−hybridized. So they will show dipole moment.

(II) $F_{2}C=C=C=C{F}_2$ - All atoms are in one plane and will not show dipole moment.

Rule : Even number of double bonds with all same four groups will show dipole moment and odd number of double bonds with same four groups will not show dipole moment.

Therefore, option A is correct.