Which of the following orders is/are incorrect?

H_{3} P O_{4} > H_{3} P O_{3} > H_{3} P O_{2}

(reducing character)

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N_{2} O < N O < N_{2} O_{3} < N_{2} O_{5}

(oxidation state of nitrogen atom)

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N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

(basicity)

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S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

(reducing character)

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Solution

The correct options are
A $H_{3} P O_{4} > H_{3} P O_{3} > H_{3} P O_{2}$ (reducing character)
D $S b H_{3} > N H_{3} > A s H_{3} > P H_{3}$ (reducing character)
(A) In $H_{3} P O_{2}$ , the two H atoms are bonded directly to P atom & in $H_{3} P O_{3}$ , the one H atom is bonded directly to P atom, which imparts reducing character to the acid, whereas in $H_{3} P O_{4}$ , there is no H atom bonded directly to P atom.
Correct order is $H_{3} P O_{2} > H_{3} P O_{3} > H_{3} P O_{4}$
(B) The oxidation state of nitrogen atom is +1 < +2 < +3 < +5.
(C) Down the group the availability of lone pair of electrons decreases as they are present in more concentrated s-orbital.
(D) As down the group bond energy decreases, the removal of H becomes easier. So, the correct order is $S b H_{3} > A s H_{3} > P H_{3} > N H_{3}$ .

Suggest Corrections

Similar questions

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

N H_{3}, P H_{3}, A S H_{3}, S b H_{3}, B i H_{3}

S i F_{4} < S i C l_{4} < S i B r_{4} < S i I_{4}

N H_{3} > S b H_{3} > A s H_{3} > P H_{3}

N H_{3} > S b H_{3} > A s H_{3} > P H_{3}

N H_{3} > S b H_{3} > A s H_{3} > P H_{3}

N H_{3} < P H_{3} < A s H_{3} < S b H_{3}

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