
Bond angle decreases except when?
A) $NH_4^ + $ is converted to $N{H_3}$
B) $NH_2^ - $ is converted to ${N_3}^ - $
C) $S{O_3}$ is converted to $S{O_2}$
D) $C{O_2}$ is converted to $CO_3^{2 - }$
Answer
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Hint: By knowing the Hybridisation, 3D orientation, no. of lone pairs on the central atom, we can easily identify the general bond angle of that compound. We can then modify it keeping in mind the effect of lone pairs on the lone pairs.
Complete answer:
A) From the formula we get the Hybridisation of $NH_4^ + $ as Tetrahedral. The bond angle of the tetrahedral is ${109.28^ \circ }$ . And that of $N{H_3}$ is also tetrahedral, but because of the Presence of one lone pair, and Lone Pair- Bond Pair repulsion the Bond Angle decreases to < ${109.28^ \circ }$
B) $NH_2^ - $ also has a $s{p^3}$ hybridisation and Tetrahedral Geometry but it has two Lone Pairs present which reduces the bond angle to $ \approx {104^ \circ }$ . ${N_3}^ - $ on the other hand, has a Linear Geometry and bond angle of ${180^ \circ }$ . Hence, the Bond Angle increases. Option (B) is correct.
C) $S{O_3}$ has $s{p^2}$ hybridisation, has the shape of a Trigonal Planar with zero lone pairs. The bond angle is hence ${120^ \circ }$ . $S{O_2}$ also has similar Hybridisation but has angular geometry and bond angle of ${119^ \circ }$, because of the presence of one lone pair which causes repulsion. Bond angle decreases.
D) $C{O_2}$ has linear geometry with bond angle ${180^ \circ }$ whereas $CO_3^{2 - }$ has Trigonal Planar shape as well as geometry, with zero lone pairs. The Bond Angle will be ${120^ \circ }$. The bond angle hence decreases.
The correct answer is Option (B).
Note:
The hybridisation of any compound can be easily found out by using this formula:
$Hybridisation = \dfrac{{no.of{\text{ }}LP{\text{ }}on{\text{ }}Central{\text{ }}Atom + no.of{\text{ }}Hydrogen{\text{ }}attached + no.of{\text{ }}Ha\log ens{\text{ }}attached \pm Formal{\text{ }}Ch\arg e}}{2}$.
If obtained Hybridisation is:
Complete answer:
A) From the formula we get the Hybridisation of $NH_4^ + $ as Tetrahedral. The bond angle of the tetrahedral is ${109.28^ \circ }$ . And that of $N{H_3}$ is also tetrahedral, but because of the Presence of one lone pair, and Lone Pair- Bond Pair repulsion the Bond Angle decreases to < ${109.28^ \circ }$
B) $NH_2^ - $ also has a $s{p^3}$ hybridisation and Tetrahedral Geometry but it has two Lone Pairs present which reduces the bond angle to $ \approx {104^ \circ }$ . ${N_3}^ - $ on the other hand, has a Linear Geometry and bond angle of ${180^ \circ }$ . Hence, the Bond Angle increases. Option (B) is correct.
C) $S{O_3}$ has $s{p^2}$ hybridisation, has the shape of a Trigonal Planar with zero lone pairs. The bond angle is hence ${120^ \circ }$ . $S{O_2}$ also has similar Hybridisation but has angular geometry and bond angle of ${119^ \circ }$, because of the presence of one lone pair which causes repulsion. Bond angle decreases.
D) $C{O_2}$ has linear geometry with bond angle ${180^ \circ }$ whereas $CO_3^{2 - }$ has Trigonal Planar shape as well as geometry, with zero lone pairs. The Bond Angle will be ${120^ \circ }$. The bond angle hence decreases.
The correct answer is Option (B).
Note:
The hybridisation of any compound can be easily found out by using this formula:
$Hybridisation = \dfrac{{no.of{\text{ }}LP{\text{ }}on{\text{ }}Central{\text{ }}Atom + no.of{\text{ }}Hydrogen{\text{ }}attached + no.of{\text{ }}Ha\log ens{\text{ }}attached \pm Formal{\text{ }}Ch\arg e}}{2}$.
If obtained Hybridisation is:
| Sr. No | Hybridisation | Geometry | |
| 1 | 2 | $sp$ | Linear |
| 2 | 3 | $s{p^2}$ | Trigonal Planar |
| 3 | 4 | $s{p^3}$ | Tetrahedral |
| 4 | 5 | $s{p^3}d$ | Trigonal Bipyramidal |
| 5 | 6 | $s{p^3}{d^2}$ | Octahedral |
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