According to hybridisation theory, the % s-character in `sp, sp^(2) and sp^(3)` hybrid orbitals is 50, 33.3 and 25 respectively, but this is not true for all the species. When `theta` is the bond angle between equivalent hybrid orbitals then % s and p-character in hybrid orbitals (when only s and p-orbitals are involved in hybridisation) can be calculated by the following formula :
`costheta=(S)/(S-1)=(P-1)/(P)`
Q. Two elements X and Y combined together to form a covalent compound. If % p-character is found to be 80% in a orbital then the hybridised state of central atom X for the orbital is :

Options

A. `sp^(2)`
B. `sp^(3)`
C. `sp^(4)`
D. `sp^(5)`

Answer

To determine the hybridized state of the central atom X when the % p-character is found to be 80%, we can follow these steps:

### Step 1: Understand the relationship between s-character and p-character
The total percentage of s-character and p-character in a hybrid orbital must equal 100%. Therefore, if the % p-character is 80%, we can calculate the % s-character as follows:
\[ \text{% s-character} = 100\% - \text{% p-character} \]
Substituting the given value:
\[ \text{% s-character} = 100\% - 80\% = 20\% \]

### Step 2: Identify the hybridization state
The hybridization state can be determined from the % s-character. The hybridization states and their corresponding % s-character are as follows:
- **sp**: 50% s-character
- **sp²**: 33.3% s-character
- **sp³**: 25% s-character
- **sp⁴**: 20% s-character

From our calculation, we found that the % s-character is 20%. This corresponds to the sp⁴ hybridization state.

### Conclusion
Thus, the hybridized state of the central atom X for the orbital with 80% p-character is **sp⁴**.