# Lecture 20: Order of Repulsion and Hybridization Involving d Orbitals

### NH﻿$_{3}$﻿﻿$:$﻿ Electron Configuration of N

There is no Promotion required because there is no place to promote. Only 3 bonds can be formed. There is SP﻿$^{3}$﻿ hybridization but 1 hybrid orbital is holding the lone pair. The other 3 is to hold the hydrogen bond. The molecule is derived from Tetrahedral. But one orbital is occupied by a lone pair so it is Trigonal Pyramid and the angle is 107﻿$^{\circ}$﻿

#### Order of repulsion

LP - LP > LP - BP > BP - BP

• When adding lone pairs the angle is smaller

#### Hybridization Involving d Orbitals

Promoting an electron from 3S to 3d orbital

AX﻿$_{5}$﻿ type molecule - Trigonal Bi pyramid - dSP﻿$^{3}$﻿

SP﻿$^{3}$﻿d﻿$^{2}$﻿ - 1 electron is promoted to d from 3S and the other from 3P

AX﻿$_{6}$﻿ type molecules - Octahedral Structure - d﻿$^{3}$﻿SP﻿$^{3}$﻿

#### Why isn't a double bond twice the energy of a Single bond?

Because a double bond is on ﻿$\pi$﻿ bond and one O bond and ﻿$\pi$﻿ bond is weaker.

Three unpaired electrons ﻿$\rightarrow$﻿ can form 3 bonds but it forms 5 bonds

To do this we promote the S electron to the 3d orbital then has 5 unpaired electrons.

Why can't this happen with NCI﻿$_{S}$﻿ ? Because it has 2S and 2P and there is no 2d orbital.

Now P undergoes dSP﻿$^{3}$﻿ hybridization to give 5 orbitals. Each one contains an unpaired electron. These hybridized orbital are directed to the corners of a trigonal bi pyramid (AX﻿$_{5}$﻿ )

### Valence Bond Theory

Diamagnetic - All paired electrons, repelled

Paramagnetic - Unpaired electrons, attracted to magnetic field

All Paired electrons but is Paramagnetic (By Molecular Orbital Theory)