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Current Question (ID: 10619)
Question:
$\text{Which of the following statements is true?}$
Options:
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1. $\text{NH}_3 \text{ is less soluble than } \text{PH}_3 \text{ in water}$
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2. $\text{NH}_3 \text{ is a stronger base and stronger reducing agent than } \text{PH}_3$
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3. $\text{NH}_3 \text{ has a higher boiling point than } \text{PH}_3 \text{ and has a lower melting point than } \text{PH}_3$
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4. $\text{PH}_3 \text{ is a stronger reducing agent than } \text{NH}_3 \text{ and it has lower melting critical temperature than } \text{NH}_3$
(Correct)
Solution:
$\text{NH}_3 \text{ is highly soluble in water but other hydrides are less soluble because } \text{NH}_3 \text{ show hydrogen bonding with water.}$ $\text{Down the group reducing nature increases. This shows that bond strength M-H decreases as the electronegativity of M decreases. } \text{NH}_3 \text{ is a weak reducing agent while } \text{AsH}_3\text{, } \text{SbH}_3 \text{ and } \text{BiH}_3 \text{ are powerful reducing agents.}$ $\text{In } \text{NH}_3\text{, molecules are associated with hydrogen bonding and thus its boiling point is comparatively high in comparison to } \text{PH}_3 \text{ and } \text{AsH}_3 \text{ where no hydrogen bonding is present.}$ $\text{The basic character decreases from } \text{NH}_3 \text{ to } \text{BiH}_3\text{. The basic nature is due to the presence of lone pair of electrons on the central atom. } \text{NH}_3 \text{ is the strongest electron-pair donor due to its small size' as the electron density of the electron pair is concentrated over a small region. As the size increases the electron density gets diffused over a large region and hence the ability to donate the electron pair (basic nature) decreases.}$ $\text{Hence, option 4th is the correct answer.}$
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