Question

. Arrange the following in order of increasing bond angles

(i) CH₄, NH₃, H₂O, BF₃, C₂H₂ (ii) NH₃, NH₂^-, NH₄⁺

Answer 1

1) In this , C₂H₂ is a linear molecule , so its bond angle will be 180^o.
While BF_​3 has a trigonal planar structure , so its bond angle will be 120^o.
CH₄ has a tetrahedral structure , so it will have bond angle 109^o 28^'.
NH₃ does not have a perfect tetrahedral structure , one of the sp³ hybridised orbital is occupied by a lone pair , so due to the lone pair -bond pair repulsion the bond angle is less than the expected tetrahedral bond angle , i.e. it is has bond angle of 107^o.
H₂O does not have a perfect tetrahedral structure as the 2 sp³ hydridised orbitals are occupied by 2 lone pairs, so due to 2 lone pairs their is much stronger lone pair -bond pair repulsion which further decreases the bond angle and come to the value of 104.5^o.
So, the increasing order of the bond angles is , $H_{2}O<N{H}_3<C{H}_4<B{F}_3<C_2{H}_2$
2) NH₄⁺ has a perfect tetrahedral structure, so the bond angle is 109^o 28' .
NH₃ doesnot have a perfect tetrahedral structure , one of the sp³ hybridised orbital is occupied by a lone pair , so due to the lone pair -bond pair repulsion the bond angle is less than the expected tetrahedral bond angle , i.e. it is has bond angle of 107^o.
In NH₂^- doesnot have a perfect tetrahedral structure , it has two non-bonded pair of electrons which created more lone pair-bond pair repulsion and makes the hydrogen atoms come closer leading to decrease in the bond angle , making it to close to 105^o.
So, the increasing order of bond angle is : $N{H_2}^{-}<N{H}_3<N{H_4}^{+}$