Answer
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Hint: Actually we can say that, at room temperature, covalent compounds exist in all three physical states and have low boiling and melting points. So it is said to be since covalent compounds do not contain charged particles capable of carrying electrons, they do not conduct electricity.
Complete answer:
We can clearly say Option D- ammonia from hydrogen bond is the right option. Because,
\[N{H_3}\] has a much higher boiling point than \[P{H_3}\]. It's because \[N{H_3}\] has a lot of hydrogen bonding. As a result, choice D is the correct answer.
Between themselves, \[N{H_3}\] molecules form strong intermolecular H bonds. Larger \[P{H_3}\] molecules, on the other hand, cannot form hydrogen bonds between themselves due to their lower electronegativity. In\[N{H_3}\], hydrogen bonds exist, while in \[P{H_3}\], hydrogen bonds do not exist. So we can say that as a result, the boiling point of \[N{H_3}\] is higher than that of \[P{H_3}\].
\[N{H_3}\] has very polar N-H bonds, which leads to heavy hydrogen bonding, despite being a larger molecule with greater dispersion forces than ammonia. This is the most powerful intermolecular force, causing \[N{H_3}\] molecules to attract each other more strongly than \[P{H_3}\] molecules.
Intermolecular forces determine boiling points. Since nitrogen is more electronegative than phosphorus, ammonia is more polar than phosphine. In fact, it is less electronegative than carbon and is more electronegative than hydrogen.
As a result, electrons in ammonia spend more time close to nitrogen than they do near phosphorus in phosphine. Ammonia has higher partial charges on the heteroatom and hydrogen than phosphine, resulting in stronger hydrogen bonds (larger charge difference = stronger electrostatic interaction).
So this clarifies that Option D- ammonia from hydrogen bond is the right option.
Note:
To boil a liquid, the molecules must be separated. The more energy it takes to sever those bonds and form a gas, the greater the interaction.
Complete answer:
We can clearly say Option D- ammonia from hydrogen bond is the right option. Because,
\[N{H_3}\] has a much higher boiling point than \[P{H_3}\]. It's because \[N{H_3}\] has a lot of hydrogen bonding. As a result, choice D is the correct answer.
Between themselves, \[N{H_3}\] molecules form strong intermolecular H bonds. Larger \[P{H_3}\] molecules, on the other hand, cannot form hydrogen bonds between themselves due to their lower electronegativity. In\[N{H_3}\], hydrogen bonds exist, while in \[P{H_3}\], hydrogen bonds do not exist. So we can say that as a result, the boiling point of \[N{H_3}\] is higher than that of \[P{H_3}\].
\[N{H_3}\] has very polar N-H bonds, which leads to heavy hydrogen bonding, despite being a larger molecule with greater dispersion forces than ammonia. This is the most powerful intermolecular force, causing \[N{H_3}\] molecules to attract each other more strongly than \[P{H_3}\] molecules.
Intermolecular forces determine boiling points. Since nitrogen is more electronegative than phosphorus, ammonia is more polar than phosphine. In fact, it is less electronegative than carbon and is more electronegative than hydrogen.
As a result, electrons in ammonia spend more time close to nitrogen than they do near phosphorus in phosphine. Ammonia has higher partial charges on the heteroatom and hydrogen than phosphine, resulting in stronger hydrogen bonds (larger charge difference = stronger electrostatic interaction).
So this clarifies that Option D- ammonia from hydrogen bond is the right option.
Note:
To boil a liquid, the molecules must be separated. The more energy it takes to sever those bonds and form a gas, the greater the interaction.
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