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 are 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 hybridized *asymmetric* molecules, we observe, that the fraction of s and p characters vary 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 hybridized 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 are

**sp**hybridized.

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

**f**

_{s}= 1 / (3.40 + 1) = 1/4.4 = 0.227.Thus, there is a 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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