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Current Question (ID: 10620)
Question:
$\text{Which of the following is in the correct sequence?}$ $\text{[For the correct order mark (T) and for the incorrect order mark (F)]:}$ $\text{(a) Lewis acidity order: } \text{SiF}_4 < \text{SiCl}_4 < \text{SiBr}_4 < \text{SiI}_4$ $\text{(b) Melting point: } \text{NH}_3 > \text{SbH}_3 > \text{AsH}_3 > \text{PH}_3$ $\text{(c) Boiling point: } \text{NH}_3 > \text{SbH}_3 > \text{AsH}_3 > \text{PH}_3$ $\text{(d) Dipole moment order: } \text{NH}_3 > \text{SbH}_3 > \text{AsH}_3 > \text{PH}_3$
Options:
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1. $\text{(a) F, (b) T, (c) F, (d) T}$
(Correct)
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2. $\text{(a) T, (b) F, (c) T, (d) F}$
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3. $\text{(a) F, (b) F, (c) T, (d) T}$
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4. $\text{(a) F, (b) F, (c) T, (d) F}$
Solution:
$\text{Hint: Ammonia has the highest boiling point because of hydrogen bonding}$ $\text{(a) (F); As the size of the halogen atom increases crowding on the Si atom will increase, hence, the tendency of attack of Lewis base decreases.}$ $\text{(b) (T); Melting point of NH}_3 \text{ is highest due to intermolecular H-bonding in it. Next lower melting point will be of SbH}_3 \text{ followed by AsH}_3 \text{ due to high mol.wt.of SbH}_3\text{. The correct order is NH}_3 > \text{SbH}_3 > \text{AsH}_3 > \text{PH}_3\text{.}$ $\text{(c) (F); In NH}_3\text{, molecules are associated by hydrogen bonding and thus its boiling point is comparatively high in comparison to PH}_3 \text{ and AsH}_3 \text{ where no hydrogen bonding is present. In moving from PH}_3 \text{ to AsH}_3\text{, boiling points increase due to an increase in the magnitude of van der Waals' forces due to an increase in molecular size.}$ $\text{The correct order of boiling point is SbH}_3 > \text{NH}_3 > \text{AsH}_3 > \text{PH}_3\text{.}$ $\text{(d) (T); The dipole moment order is not regular and the given trend for dipole moment is correct.}$
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