Arrange $N H_{3}, P H_{3}, A s H_{3}, S b H_{3}$ in decreasing order of 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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S b H_{3} > P H_{3} > A s H_{3} > N 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, the radius of elements increases and electronegativity decreases so ionic character of compounds increases and covalent nature decreases down the group.

Therefore, the decreasing order of covalent character is:

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

Hence, option D is correct.

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}

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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