Why is $N H_{3}$ more basic than $P H_{3}$ ?

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Solution

Basically $N H_{3}$ is a base due to the lone pair of electrons present on the Nitrogen atom which is available for donation. On the other hand $P H_{3}$ is also a base but its basic character is less than that of $N H_{3}$ because the nitrogen in phosphorous have lone pair of electrons that are not readily available for donation.

Also the P-H bonds are weaker than N-H bonds, so there’s not as much compensation for protonating $P H_{3}$ . This is a result of the inert pair effect, seen in heavier elements.

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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}

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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}

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