Question

Which of the following represents the correct order of Cl–O
bond lengths in ClO⁻ ,ClO₂⁻ ,ClO₃⁻ ,ClO₄⁻ ?
(a) ClO₄⁻ = ClO₃⁻ = ClO₂⁻ = ClO⁻
(b) ClO⁻ < ClO₂⁻ < ClO₃⁻ < ClO₄⁻
(c) ClO₄⁻ < ClO₃⁻ < ClO₂⁻ < ClO⁻
(d) ClO₃⁻ < ClO₄⁻ < ClO₂⁻ < ClO⁻

Answer 1

ClO₄⁻ < ClO₃⁻ < ClO₂⁻ < ClO⁻ is the correct trend.

Explanation:

• The chlorate ions have one thing in common.

• They have a single bonded oxygen with chlorine and the other oxygens are bonded double with chlorine.

• In hypochlorite, there is zero Cl=O and one Cl-O.

• In chlorite, there's one Cl=O and one Cl-O.

• In chlorate, there's two Cl=O and one Cl-O.

• In perchlorate, there's three Cl=O and one Cl-O.

• All these compounds except the hypochlorite is resonance stabilised.

• So the bond order of Cl-O comes as average of the total Cl-O bonds.

• Thus, bond order in ClO- = 1.

• Bond order in ClO2- = 1.5

• Bond order in ClO3- = 1.66.

• And bond order in ClO4- = 1.75.

• So,the bond length is inversely proportional to bond order.

So. Bond length trend comes as ClO₄⁻ < ClO₃⁻ < ClO₂⁻ < ClO⁻.

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Answer 2

Explanation:

Bond length∝

Bondorder

1

Bond order of ClO

−

=1,

Bond order of ClO

2

−

=1.5,

Bond order of ClO

3

−

=1.66 ,

Bond order of ClO

4

−

=1.75.

Thus the order of bond length is ClO

4

−

<ClO

3

−

<ClO

2

−

<ClO

−

, thus correct option is B.