
Assertion: In $N{H_3},N$ is $s{p^3}$ hybridised but bond angle is ${107^\circ}$.
Reason: Shape of the $N{H_3}$ molecule is trigonal pyramidal.
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
(C) Assertion is correct but Reason is incorrect.
(D) Both Assertion and Reason are incorrect.
Answer
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Hint: As we know that ammonia molecules generally possess three bond pairs and a single lone pair of electrons and we also know that lone pair and bond pair repulsions are greater than the bond pair and bond pair repulsions.
Complete answer:
As we know that ammonia contains the nitrogen which is present in group five and thus possesses five outer electrons where three electrons are busy in bonding with three hydrogen molecules and one lone pair is present which makes the arrangement of this molecule as tetrahedron like in methane.
Nitrogen in ammonia is present in $s{p^3}$ hybridisation where one s-orbital and three p-orbitals combines. Lone pairs in case of ammonia are present in orbitals that are shorter and rounder than the orbitals which the bonding pairs occupy. Due to this, there is more repulsion between a lone pair and a bond pair than there is between bond pair and bond pair and therefore the bond angle which is naturally ${109.5^\circ}$ in ammonia is reduced to ${107^\circ}$.
Therefore, the assertion is true.
Now, talking about the shape of ammonia, as we have already discussed that nitrogen in ammonia contains five electrons and among these five, three make bonds with three hydrogen molecules and two electrons make up the lone pair giving it a trigonal pyramidal shape involving nitrogen at the peak and hydrogens at corners.
Thus, the reason is also correct.
Therefore, the correct answer is option ‘(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion’.
Note: Always remember that the number of electrons present in the central atom will bond with the helping atoms in such a manner that lone pair and bond pair repulse each other and thus there is greater repulsion between these two than the two bond pairs and less repulsions than the two lone pairs.
Complete answer:
As we know that ammonia contains the nitrogen which is present in group five and thus possesses five outer electrons where three electrons are busy in bonding with three hydrogen molecules and one lone pair is present which makes the arrangement of this molecule as tetrahedron like in methane.
Nitrogen in ammonia is present in $s{p^3}$ hybridisation where one s-orbital and three p-orbitals combines. Lone pairs in case of ammonia are present in orbitals that are shorter and rounder than the orbitals which the bonding pairs occupy. Due to this, there is more repulsion between a lone pair and a bond pair than there is between bond pair and bond pair and therefore the bond angle which is naturally ${109.5^\circ}$ in ammonia is reduced to ${107^\circ}$.
Therefore, the assertion is true.
Now, talking about the shape of ammonia, as we have already discussed that nitrogen in ammonia contains five electrons and among these five, three make bonds with three hydrogen molecules and two electrons make up the lone pair giving it a trigonal pyramidal shape involving nitrogen at the peak and hydrogens at corners.
Thus, the reason is also correct.
Therefore, the correct answer is option ‘(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion’.
Note: Always remember that the number of electrons present in the central atom will bond with the helping atoms in such a manner that lone pair and bond pair repulse each other and thus there is greater repulsion between these two than the two bond pairs and less repulsions than the two lone pairs.
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