**Predict the Type of Hybridization in a Molecule or Ion**

Step 1: Add the number of valence electrons of all the atoms present in the given molecule/ion.

Step 2: In case of a cation, subtract the number of electrons equal to the charge on the cation and in case of an anion, add number of electrons equal to the charge on the anion.

Step 3: (i) If the result obtained in step 2 is less than 8, divide it by 2 and find the sum of the quotient and remainder.

(ii) If the result obtained in step 2 lies between 9 and 56, divide it by 8 and find the first quotient (Q_{1}). Divide the remainder R_{1} (if any) by 2 and find the second quotient (Q_{2}). Add all the quotients and the final remainder (R_{2}).

Let the final result obtained in (i) or (ii) be X. The type of hybridisation is decided by the value of X as follows:

**Example**

**(i) BeCl _{2}**

Total valence electrons = 2 + 7 × 2 = 16

16/8 = 2 (Q_{1}) + Zero (R_{1}); X = 2

so, Hybridization = sp

**(ii) BF _{3}**

Total valence electrons = 3 + 7 × 3 = 24

24/8 = 3(Q_{1}) zero(R_{1}); X = 3

so, Hybridization = sp^{2}

**(iii) NH _{3}**

Total valence electrons = 5 + 3 = 8; X = 4

so, Hybridization = sp^{3}