Question

The compound that has the largest $\mathrm{H}–\mathrm{M}–\mathrm{H}$bond angle $(\mathrm{M}=\mathrm{N},\mathrm{O},\mathrm{S},\mathrm{C})$is:

• A. ${\mathrm{CH}}_4$
• B. ${\mathrm{H}}_2\mathrm{S}$
• C. ${\mathrm{NH}}_3$
• D. ${\mathrm{H}}_2\mathrm{O}$

Answer 1

The correct option is A

${\mathrm{CH}}_4$

The correct option is (A):

Explanation for the correct answer:-

Bond angle:-

1. It is basically can be defined as the angle that is formed in the or between the covalent bonds that will originate from the same atom.
2. Basically the variation in the bond angle is on the account of the two reasons that are the size as well as the electronegativity of the central atom.
3. ${\mathrm{CH}}_4$ is basically known as Methane whose hybridization is ${\mathrm{sp}}^3$.
4. So when the ${\mathrm{sp}}^3$ hybridized orbitals are formed they will arrange themselves as far as possible.
5. There is the presence of tetrahedral arrangement with the angle of$109.5°$.

Explanation for the incorrect answers:

Option (B):-

${\mathrm{H}}_2\mathrm{S}$:-

1. Basically as the electronegativity of the central atom will decrease, the lone pairs of the electrons will lie further from it and so the $\mathrm{lp}-\mathrm{lp}$ repulsion will decrease.
2. So in the case of $\mathrm{H}-\mathrm{S}-\mathrm{H}$ is $92.1°$and that in $\mathrm{H}-\mathrm{O}-\mathrm{H}$ is $104.5°$.

Option (C):-

${\mathrm{NH}}_3$:-

1. ${\mathrm{NH}}_3$ is basically known as Ammonia having the hybridization as ${\mathrm{sp}}^3$ and bond angle as $107°$and having the geometry as Pyramidal or Distorted tetrahedral.
2. The central Nitrogen atom has 4 valence pairs of electrons , 3 in covalent bonds with Hydrogen atom and one in the lone pair which is actually not bonded.
3. Basically the tetrahedral geometry has bond angle equal to $109.5°$, but the lone pairs will exist closer to the nucleus than that of the bonding pairs and will have a greater repulsive effect than the three bonding pairs which will result in the decrease of $2.5°$.

Option (D):-

${\mathrm{H}}_2\mathrm{O}$:-

1. Due to the presence of the 2 lone pairs of electron on the Oxygen atom in ${\mathrm{H}}_2\mathrm{O}$ the bond angle here is $104.5°$.
2. Hybridization is ${\mathrm{sp}}^3$that means the shape is tetrahedral but because of the lone pairs present which will move further so there would be the decrease in the bond angle.

Therefore, the correct option is (A):- ${\mathrm{CH}}_4$