wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The decreasing order of bond length of C=C bond in the following compounds is:

A
II > I > IV > III
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
III > I > II > IV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
IV > II > I > III
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
IV > I > II > III
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B III > I > II > IV
When a double bond involve in delocalisation of electrons, they will attain more single bond character. Thus, the double bond involve in delocalisation will have larger bond length compare to the isolated double bond.

In given compounds, compound (I) and (II) shows resonance effect whereas (II) and (IV) shows hyperconjugation effect. Since resonance effect is prior to hyperconjugation,
compound (1) and (III) has larger bond length compare to (II) and (IV).

In (III), 3 double bonds are involved in resonance while in (II) two double bonds are involved so the bond length of (III) is the largest one. Also (III) is involved in cyclic resonance which make it more stable.

Comparing (II) and (IV), II has larger bond length than (IV) because it has more α Hydrogen.

Thus, the decreasing order of bond length of C=C bond in the following compounds is,
III > I > II > IV

flag
Suggest Corrections
thumbs-up
23
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon