$C_{1} - C_{2}$ bond is shortest in:

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
In option (d), there is some partial double bond character that arises in $C_{1} - C_{2}$ bond due to resonance.
Also, $C_{1}$ and $C_{2}$ carbons are $s p^{2}$ and $s p$ hybridized respectively. More the s-character, more will be the electronegativity. So $C_{1} - C_{2}$ bond is the shortest.

Suggest Corrections

Similar questions

C_{1} - C_{2}

bond is shortest in:

Q. If

C_{1} : x^{2} + y^{2} - 20 x + 64 = 0

and

C_{2} : x^{2} + y^{2} + 30 x + 144 = 0

. Then the length of the shortest line segment

P Q

which touches

C_{1}

P

and to

C_{2}

Q

is ?

If $c_{1} : x^{2} + y^{2} - 20 x + 64 = 0$ and $c_{2} :$ $x^{2} + y^{2} + 30 x + 144 = 0$ . Then the length of the shortest line segment $P Q$ which is tangent to $c_{1}$ at $P$ and $c_{2}$ at $Q$ is

C_{1} - H, C_{2} - H, C_{3} - H

and

C_{4} - H

homolytic bond dissociation energy is in order:

Q. If

C_{1} : x^{2} + y^{2} - 20 x + 64 = 0

and

C_{2} : x^{2} + y^{2} + 30 x + 144 = 0

. The line PQ touches the circle touches

C_{1}

P

and

C_{2}

Q

. Then the length of the shortest line segment

P Q

Join BYJU'S Learning Program

C1−C2 bond is shortest in:

The correct option is D In option (d), there is some partial double bond character that arises in C1−C2 bond due to resonance. Also, C1 and C2 carbons are sp2 and sp hybridized respectively. More the s-character, more will be the electronegativity. So C1−C2 bond is the shortest.

$C_{1} - C_{2}$ bond is shortest in: