Arrange the given compound in increasing oder of reducing nature:

$N H_{3}, P H_{3}, A s H_{3}, S b H_{3}$

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

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

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

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

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Solution

The correct option is C $N H_{3} < P H_{3} < A s H_{3} < S b H_{3}$
As we move down the group, radius of elements increases and electronegativity decreases so ionic character of compounds increases and covalent nature decreases.

As ionic character increase reducing nature increase so order is,

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

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}

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}

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