Among ${N H}_{3}, {P H}_{3}, A s H_{3}$ and $S b H_{3}$ which one is a strongest reducing agent?

{N H}_{3}

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{P H}_{3}

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A s H_{3}

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S b H_{3}

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Solution

The correct option is D $S b H_{3}$
When an atom loses an electron then the element is said to be a reducing agent. The reducing power of an element depends upon how quickly it can lose electrons. In the case of electrons held very loosely by the nucleus, such element can easily lose their valence electrons and hence, higher reducing agents.

No. of shells is highest in " $S b$ " which means the outermost electron is most loosely held in Sb. Therefore, $S b H_{3}$ is the strongest reducing agent.
Hence, the correct option is D.

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}

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

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