# How to calculate the hybridization of ethane with H=½[V+M-C+A]?

I wanted to calculate the hybridization of this molecule $\ce{CH3-CH3}$. I noticed a shortcut formula given in my class notes for calculating the hybridization. It was

$$H=\frac{1}{2}×[V+M-C+A].$$

With

• $H$: Hybridization
• $V$: Number of valence electrons of central atom.
• $M$: Number of monovalent atoms (i.e hydrogen or halogen) around central atom.
• $C$: Cationic Charge.
• $A$: Anionic Charge.

When I tried to calculate the hybridisation of $\ce{CH3-CH3}$ by using above formula, where for any one of the carbon atom $V=4$, $M=3$, $C=0$, $A=0$ then $H=\frac{7}{2}$. But as we all know hybridization of every carbon in ethane is $\ce{sp^3}$, i.e. $H$ should be $4$. I know other methods of calculating $H$ like $H= \text{no. of sigma bonds} + \text{no.of lone pairs}$, which gives me the right answer, but why does the above formula give something different? And also for any molecule if using that formula gives fractional answer then what does that mean?

• Are you sure you don't mean formal charge? – ringo Feb 28 '17 at 6:29
• But what difference formal charge will create as it is also zero. – Avi Feb 28 '17 at 6:55
• I have not come across this formula before and I cannot make any sense of it. Please be aware that hybridisation itself is just a model, it is just a visualisation and interpretation technique. Simple approaches to that will only give you a very approximate answer to what would mathematically be correct. In ethane the $\ce{C-C}$ bond would have more p-character than the $\ce{C-H}$ bonds. – Martin - マーチン Feb 28 '17 at 7:12
• Consider CH3 as one of the monovalent group for first C to calculate the value of H. – narendra kumar Feb 28 '17 at 9:47