Arrange $N H_{3}, P H_{3}, A s H_{3}, S b H_{3}$ in increasing covalent nature :

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