Question
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+ is converted to NH3
B) NH2− is converted to N3−
C) SO3 is converted to SO2
D) CO2 is converted to CO32−
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+ as Tetrahedral. The bond angle of the tetrahedral is 109.28∘ . And that of NH3 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∘
B) NH2− also has a sp3 hybridisation and Tetrahedral Geometry but it has two Lone Pairs present which reduces the bond angle to ≈104∘ . N3− on the other hand, has a Linear Geometry and bond angle of 180∘ . Hence, the Bond Angle increases. Option (B) is correct.
C) SO3 has sp2 hybridisation, has the shape of a Trigonal Planar with zero lone pairs. The bond angle is hence 120∘ . SO2 also has similar Hybridisation but has angular geometry and bond angle of 119∘, because of the presence of one lone pair which causes repulsion. Bond angle decreases.
D) CO2 has linear geometry with bond angle 180∘ whereas CO32− has Trigonal Planar shape as well as geometry, with zero lone pairs. The Bond Angle will be 120∘. 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=2no.of LP on Central Atom+no.of Hydrogen attached+no.of Halogens attached±Formal Charge.
If obtained Hybridisation is:
Sr. No | Hybridisation | Geometry | |
---|---|---|---|
1 | 2 | sp | Linear |
2 | 3 | sp2 | Trigonal Planar |
3 | 4 | sp3 | Tetrahedral |
4 | 5 | sp3d | Trigonal Bipyramidal |
5 | 6 | sp3d2 | Octahedral |