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Ion Electron Method

Which of the ...

Question

Which of the following order(s) is/are incorrect?

H_{2} 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_{5}

(oxidation state on 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 $H_{2} P O_{4} > H_{3} P O_{3} > H_{3} P O_{2}$ (reducing character)
C $S b H_{3} > N H_{3} > A s H_{3} > P H_{3}$ (reducing character)
$H_{3} P O_{4} > H_{3} P O_{3} > H_{3} P O_{2}$ (oxidising character)
+5 +3 +1
$N_{2} O < N O < N_{2} O_{5}$ (oxidation state on nitrogen atom)
+1 +2 +5
$N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$ (basicity)
because as we move down the group, basecity increases.
$S b H_{3} > N H_{3} > A s H_{3} > P H_{3}$ (oxidising character)

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}

{N H}_{3}, {P H}_{3}, A s H_{3}

S b 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}

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