The correct decreasing order of basic strength is :

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

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S b H_{3} > A s H_{3} > P 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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P H_{3} > A s H_{3} > S b H_{3} > N 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 the size of central atom increases the lone pair of electrons occupies a larger volume. In other words electron density on the central atom decreases and consequently its tendency to donate a pair of electrons decreases along with basic character from $N H_{3}$ to $B i H_{3}$

When comparing atoms ,the larger the atom the weaker the H-X bond and the lower the electron density making it a weaker base

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}

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}

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

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