MCQ
Assertion : Sigma $(\sigma )$ is a strong bond, while pi $(\pi )$ is a weak bond.
Reason : Atoms rotate freely about pi $(\pi )$ bond.
  • A
    If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
  • B
    If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
  • If the Assertion is correct but Reason is incorrect.
  • D
    If both the Assertion and Reason are incorrect.

Answer

Correct option: C.
If the Assertion is correct but Reason is incorrect.
c
Sigma $(\sigma )$ bond is formed by axial overlap of atomic orbitals while pi $(\pi )$ bond is formed by lateral overlap. Since axial overlapping takes place to a greater extent than the lateral overlapping, former $(\sigma )$ bond is stronger than pi bond. Atoms attached to doubly bonded atom can't rotate freely around the double bond.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Benzamide on reaction with $POC{l_3}$ gives
$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution.

Given : $K_f (H_2O)$ = $2\ K\ kg\ mole^{-1}$, $K_{sp} (PbCl_2)$ = $4 × 10^{-6}$

(Assume molarity to be equal to molality) .....$^oC$

$N{H_4}Cl{O_{4\,}}\, + \,HN{O_3}(dil)\, \to \,HCl{O_4}\, + \,(X)\,\xrightarrow{\Delta }Y(g),$ $(X)$ and $(Y)$ are respectively
Which one of the following is an optically active compound
Tautomerism is exhibited by
At ${80\,^o}C,$ distilled water has $[{H_3}{O^ + }]$ concentration equal to $1 \times {10^{ - 6}}\,\,mole/litre.$ The value of ${K_w}$ at this temperature will be
$\begin{matrix}
   O\,\,\,  \\
   ||\,\,\,  \\
   C{{H}_{3}}-C-C{{H}_{2}}-  \\
\end{matrix}\begin{matrix}
   O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   ||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   C{{H}_{2}}-C-C{{H}_{3}}\xrightarrow[\Delta ]{{{(N{{H}_{4}})}_{2}}C{{O}_{3}}}\xrightarrow[\Delta ]{CC{{l}_{3}}C{{O}_{2}}Na}\underset{(major)}{\mathop{(B)}}\,  \\
\end{matrix}$

Product $(B)$ of above reaction is

$+ CH_3I$ (excess) $\to$ product; The product is 
If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then
The oxidation number of $Mn$ in $KMn{O_4}$ is