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. Correct order of `P-P` bond length in the following compound is :

Options

A. `P_(2)F_(4)ltP_(2)(CH_(3))_(4)ltP_(2)(CF_(3))_(4)ltP_(2)H_(4)`
B. `P_(2)F_(4)ltP_(2)(CF_(3))_(4)ltP_(2)(CH_(3))_(4)ltP_(2)H_(4)`
C. `P_(2)F_(4)ltP_(2)H_(4)ltP_(2)(CH_(3))_(4)ltP_(2)(CF_(3))_(4)`
D. `P_(2)F_(4)ltP_(2)(CH_(3))_(4)ltP_(2)H_(4)ltP_(2)(CF_(3))_(4)`

Answer

To determine the correct order of P-P bond lengths in the given compounds (P2F4, P2(CH3)4, P2(CF3)4, and P2H4), we can follow these steps:

### Step 1: Analyze the Compounds
We have four compounds to analyze:
1. P2F4
2. P2(CH3)4
3. P2(CF3)4
4. P2H4

### Step 2: Understand the Effect of Substituents on Bond Length
The bond length between two phosphorus atoms (P-P) can be influenced by the electronegativity of the substituents attached to phosphorus. More electronegative substituents tend to pull electron density away from the P-P bond, leading to a shorter bond length.

### Step 3: Determine Electronegativity of Substituents
The electronegativity order of the substituents is as follows:
- F (Fluorine) > CF3 (Trifluoromethyl) > CH3 (Methyl) > H (Hydrogen)

### Step 4: Analyze Each Compound
1. **P2F4**: The presence of four fluorine atoms, which are highly electronegative, will lead to a significant decrease in P-P bond length.
2. **P2(CF3)4**: The trifluoromethyl group is also electronegative, but less so than fluorine. Therefore, the P-P bond length will be longer than in P2F4 but shorter than in P2(CH3)4 and P2H4.
3. **P2(CH3)4**: The methyl group is less electronegative than CF3 and F, leading to a longer P-P bond length than in P2F4 and P2(CF3)4.
4. **P2H4**: Hydrogen is the least electronegative substituent here, resulting in the longest P-P bond length among the four compounds.

### Step 5: Establish the Order of P-P Bond Lengths
Based on the analysis:
- P2F4 has the shortest P-P bond length.
- P2(CF3)4 has a longer P-P bond length than P2F4 but shorter than P2(CH3)4.
- P2(CH3)4 has a longer P-P bond length than P2(CF3)4.
- P2H4 has the longest P-P bond length.

Thus, the correct order of P-P bond lengths from shortest to longest is:
**P2F4 < P2(CF3)4 < P2(CH3)4 < P2H4**

### Final Answer
The correct order of P-P bond lengths is:
**P2F4 < P2(CF3)4 < P2(CH3)4 < P2H4**