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Electronic Configuration and Oxidation States

Choose the co...

Question

Choose the correct order for $H O C l, H C l O_{2}, H C l O_{3}$ and $H C l O_{4}$ :

Oxidation state of halogen atom:

H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}

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Oxidising power:

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4}

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Acid strength:

H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}

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Thermal stability:

H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}

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s-character in hybrid state:

H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}

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Solution

The correct options are
A Thermal stability: $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
C Oxidation state of halogen atom: $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
E Acid strength: $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
(a) $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
Oxidation states of Cl in $H C l O_{4} - 7$
Oxidation states of Cl in $H C l O_{3} - 5$
Oxidation states of Cl in $H C l O_{2} - 3$
Oxidation states of Cl in $H O C l - 1$
(b) $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
Acid with higher oxidation number is more capable of oxidising other species.
(c) $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
Higher the oxidation number more is the acidity.
(d) $H O C l < H C l O_{2} < H C l O_{3} < H C l O_{4}$
(e) Equal (25%) each Cl-atom is $s p^{3}$ hybridised.

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Similar questions

H O C l, H C l O_{2}, H C l O_{3}

H C l O_{4}

Arrange the following compounds according to the increasing order of oxidation state of chlorine.

$H C l O_{4}, H C l O_{3}, H C l O_{2}, H C l O$

{H C l O}_{4}

{H C l O}_{3}

{H C l O}_{2}

H O C l

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

(x^{2} + y^{2})

x

y

H C l O_{4} < H C l O_{3} < H C l O_{2} < H C l O

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