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Coordinated Covalent Bond

The correct s...

Question

The correct sequence of decrease in the bond angles of the following hydrides is

NH₃> PH₃> AsH₃> SbH₃

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NH₃> AsH₃> PH₃> SbH₃

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SbH₃> AsH₃> PH₃> NH₃

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PH₃> NH₃> AsH₃> SbH₃

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Solution

The correct option is A
NH₃> PH₃> AsH₃> SbH₃

$N H_{3}$ has $s p^{3}$ hybridisation with bond angle $107^{0}$ where as $P H_{3}, A s H_{3}, S b H_{3}$ has pure orbitals overlapping with bond angle around $90^{0}$

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}

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}

Out of NH₃, PH₃, AsH₃ and SbH₃ the most basic hydride is:

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