If C - C (I), C = C(II) and C ≡ C(III) are three bond types, then their bond order will be respectively,

1, 2, 3

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3, 2, 1

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3, 1, 2

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2, 1, 3

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Solution

The correct option is A
1, 2, 3

Generally, we can state that number of bonds = Bond order. This is not always the case but happens to be a quick rule of approximation.

Here, $C - C \Rightarrow B o n d O r d e r = 1$
$C = C \Rightarrow B o n d O r d e r = 2$
$C \equiv C \Rightarrow B o n d O r d e r = 3$

Also, $B o n d O r d e r \propto \frac{1}{B o n d L e n g t h}$

$B o n d O r d e r \propto B o n d E n t h a l p y$

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If C - C (I), C = C(II) and C ≡ C(III) are three bond types, then their bond order will be respectively,

The correct option is A 1, 2, 3 Generally, we can state that number of bonds = Bond order. This is not always the case but happens to be a quick rule of approximation. Here, C−C⇒Bond Order=1 C=C⇒Bond Order=2 C≡C⇒Bond Order=3 Also, Bond Order∝1Bond Length Bond Order∝Bond Enthalpy