which of the following orders is correct?

$N > F > O$ : Ionization energy

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$H_{2} T e > H_{2} S e > H_{2} S > H_{2} O$ : Reducing nature

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$H_{2} O > H_{2} T e > H_{2} S e > H_{2} S$ : Boiling point

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

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Solution

The correct options are
B
$H_{2} T e > H_{2} S e > H_{2} S > H_{2} O$ : Reducing nature

C
$H_{2} O > H_{2} T e > H_{2} S e > H_{2} S$ : Boiling point

Ionization energy: F > N > O

For grup 16 hydrides, the reducing nature increases as we go down the group. Due to hydrogen bonding, the boiling point of watre is the highest of the group 16 hydrides. The remaining 3 show an increasing trend from sulphur and downwards.

Hence the following two trends are accurate:

$H_{2} T e > H_{2} S e > H_{2} S > H_{2} O$ : Reducing nature

$H_{2} O > H_{2} T e > H_{2} S e > H_{2} S$ : Boiling point

Among the oxyacids of chlorine (given), the oxidizing capacity is just the reverse:

$H C l O_{4} < H C l O_{3} < H C l O_{2} < H C l O$ : Oxidizing nature

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Similar questions

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

H_{2} O < H_{2} T e < H_{2} S e < H_{2} S

H_{2} O > H_{2} T e > H_{2} S e > H_{2} S

H_{2} O > H_{2} S > H_{2} S e > H_{2} T e

H_{2} E

(E = O, S, S e, T e a n d P o

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

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