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Question: Bond angle decreases except when? A) \(NH_4^ + \) is converted to \(N{H_3}\) B) \(NH_2^ - \) is ...

Bond angle decreases except when?
A) NH4+NH_4^ + is converted to NH3N{H_3}
B) NH2NH_2^ - is converted to N3{N_3}^ -
C) SO3S{O_3} is converted to SO2S{O_2}
D) CO2C{O_2} is converted to CO32CO_3^{2 - }

Explanation

Solution

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 NH4+NH_4^ + as Tetrahedral. The bond angle of the tetrahedral is 109.28{109.28^ \circ } . And that of NH3N{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{109.28^ \circ }
B) NH2NH_2^ - also has a sp3s{p^3} hybridisation and Tetrahedral Geometry but it has two Lone Pairs present which reduces the bond angle to 104 \approx {104^ \circ } . N3{N_3}^ - on the other hand, has a Linear Geometry and bond angle of 180{180^ \circ } . Hence, the Bond Angle increases. Option (B) is correct.
C) SO3S{O_3} has sp2s{p^2} hybridisation, has the shape of a Trigonal Planar with zero lone pairs. The bond angle is hence 120{120^ \circ } . SO2S{O_2} also has similar Hybridisation but has angular geometry and bond angle of 119{119^ \circ }, because of the presence of one lone pair which causes repulsion. Bond angle decreases.
D) CO2C{O_2} has linear geometry with bond angle 180{180^ \circ } whereas CO32CO_3^{2 - } has Trigonal Planar shape as well as geometry, with zero lone pairs. The Bond Angle will be 120{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=no.of LP on Central Atom+no.of Hydrogen attached+no.of Halogens attached±Formal Charge2Hybridisation = \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. NoHybridisationGeometry
12spspLinear
23sp2s{p^2}Trigonal Planar
34sp3s{p^3}Tetrahedral
45sp3ds{p^3}dTrigonal Bipyramidal
56sp3d2s{p^3}{d^2}Octahedral