Which of the following is the correct order of basicity of hydride compounds?

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $S b H_{3} < A s H_{3} < P H_{3} < N H_{3}$
In group 15, the hydrides of the formula $X H_{3}$ have a lone pair on the central atom. As we go down the group, size of the atom increases. And hence, electron density over the group 15 element decreases. Thus the tendency to donate electrons decreases, hence basicity decreases down the group as
$N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
option C 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}

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

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}

Join BYJU'S Learning Program