In the last post, we have seen that the bond angles (θ) , hybridization index (i) and the fraction of s and p character in a bond (f_{s} and f_{p} ) values are related to each other.

Determining %s and %p character ,bond lengths and bond angles for symmetric molecules is very easy . However, for asymmetric molecules , the composition of each hybrid orbital varies.Thus, the bond angles and bond lengths change too.We term the hybridization in asymmetric molecules as isovalent /second order hybridization.

**Isovalent/second order hybridization. **

When we study hybridised *asymmetric* molecules , we observe, that the fraction of s and p character varies in them. The bond angles vary too ( from ideal bond angles).

*e.g. – *Experimentally it is found that the bond angle **between two N-H bonds** in ammonia molecule is 107.1º.

Now let us find the amount of s and p character in hybridised ammonia orbitals.

**FOR BONDING ORBITALS**

**STEP 1** -Find the hybridization index, i.

** i = -1 / cosθ .
** ∴ i

_{NH3 }= – 1/ cos(107.1) = 3.40 – We calcuate this using a scientific calculator.

Thus, in ammonia molecule , the

**bonding orbitals of N**atom is

**sp**hybridised.

^{3.4}**STEP 2** – **f _{s}= 1 / (i + 1)
**∴

**f**

_{s}= 1 / (3.40 + 1) = 1/4.4 = 0.227.Thus, there is 22.7 % s- character in bonding hybrid orbitals of ammonia i.e the orbitals which form the N-H bonds. This percentage is *less than the expected value* ( 25%) .

**NOTE – THE ABOVE CALCULATION IS FOR THE BONDING ORBITALS WHICH HAVE INTERHYBRID ORBITAL ANGLE = 107.1 º BETWEEN THEM.**

**FOR NON – BONDING ORBITALS**

For calculating % s- character of the orbital containing non- bonding or lone pair of electrons , we have to use the formula from the previous post –

**∴∑ f _{s} = 1 **for all orbitals of NH

_{3}molecule.

∴3 × f_{s bonding)}+f_{s(non bonding)} = 1

3 (0.227) +f_{s(non bonding)} =1

∴f_{s(non bonding)} = 1 – 0.681

**f _{s(non bonding)}= 0.319.**

We also know that , **f _{s(non bonding)} +f_{p(non bonding)} = 1.**

∴f_{p(non bonding)}= 1-f_{s(non bonding) }= 1- 0.319 =** 0.681**

So, i_{(non bonding) }= f_{p(non bonding)}/f_{s(non bonding)} .

= 0.681 / 0.319

= 2.13.

∴The hybrid orbital in ammonia which contains **non bonding **pair of electrons is **sp ^{2.13 }**

**hybridised**.

This kind of hybridization , where fractional number of atomic orbitals mix to form new hybrid orbitals, is called **isovalent or second order hybridization.**

In our next post we shall study the bond angle in much detail. Till then,

Be a perpetual student of life and keep learning..

Good day !

Superb entry! I was struggling to understand this concept and your blog helped me a lot!

I´ll just be super picky and tell you I think there is a little confusion in the last sentence about sp2.13 hybridisation. I think you wanted to say “ammonia” but instead you wrote “methane”.

Regards from Brazil!

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Thank you so much for writing to me ! I am glad I could help.

Yes! You are right… I will get that corrected. Thanks again.

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