Question

Which one of the following does not have a pyramidal shape?
A. ${(C{H}_3)}_{3}N$
B. $P{(C{H}_3)}_3$
C. $Si{(C{H}_3)}_3$
D. ${(Si{H}_3)}_{3}N$$$$$

Answer 1

Hint: The shape of $A{B}_3$ molecule with 4 bond pairs is tetrahedral. With 3 bond pairs and 1 lone pair is pyramidal .With 2 bond pairs and 2 lone pairs the shape will be V- shaped. We need to know about VSEPR theory for more details.

Complete step by step answer:
- VSEPR theory is defined as the electron pairs surrounding the central atom must be arranged in space as far apart as possible to minimize the electrostatic repulsion experienced between them.
- A central atom can be defined as any atom that is bonded to two or more than two other atoms. The first and the most important rule of the VSEPR theory is that the bond angles about a central atom are those that minimize the total repulsion experienced between the Electron pairs in the atom’s valence shell.
Rules:
- A lone pair of electrons occupy more space than a bonding pair of electrons because lone pair of electron is under the influence of only one nucleus of the central atom, they are expected to occupy more space with a greater electron density than the bond pair electrons which are under the influence of two nuclei. The decreasing order of repulsion is mentioned below
- Lone pair-Lone pair repulsion > Lone pair-Bond pair repulsion > Bond pair- bond pair repulsion
- Repulsion forces decrease sharply with increasing interior angle. They are stronger at 90 degree weak at 120 degree and very weak at 180 degree.
- Influence of a bonding electron pair decreases with increasing value of electronegativity of an atom forming a molecule.
- Multiple bonds behave equivalent to a single electron pair for the purpose of VSEPR bond theory.
 The two electron pair of a double bond occupies more space than one electron pair of a single bond.
  The lone pair electrons repels bond pair electrons giving rise to some distortions in the molecular shape.
- Now , Phosphorus and nitrogen have 5 valence electrons on their valence shell.
Therefore in the first option nitrogen has 3 bond pairs and one lone pair so option A is pyramidal in shape
In the second and third option also Phosphorus has three bond pairs and one lone pair so option b and c is also pyramidal in shape from VSEPR theory.
- But in option 4 , ${(Si{H}_3)}_{3}N$,Si has vacant d orbitals which provides $p\pi -d\pi$ bonding back bonding as a result nitrogen has $s{p}^2$ hybridization with trigonal planar shape.

So the correct answer is “D”:

Note:
As a result of the distortions created different types of shapes are arised :

Bond pair	Lone pair	Shape
2	0	Linear
2	1	Non linear
2	2	Non linear
3	0	Trigonal planar
3	1	Pyramidal
4	0	Tetrahedral
4	2	Square planar
5	0	Trigonal bipyramidal
6	0	Octahedral.