What is hybridisation of central atom of anionic part of `PBr_(5)` in crystalline state

Options

A. `sp^(2)`
B. `sp^(3)`
C. sp
D. not applicable

Answer

To determine the hybridization of the central atom of the anionic part of \( PBr_5 \) in the crystalline state, we need to follow these steps:

### Step 1: Identify the structure of \( PBr_5 \)
- \( PBr_5 \) consists of phosphorus (P) and bromine (Br). In the crystalline state, it can be represented as \( PBr_4^+ \) (cation) and \( Br^- \) (anion).

### Step 2: Focus on the anionic part
- The anionic part we are concerned with is \( Br^- \).

### Step 3: Determine the electron configuration of bromine
- Bromine has 7 valence electrons (as it is in group 17 of the periodic table). When it gains an electron to become \( Br^- \), it has a total of 8 electrons.

### Step 4: Analyze the hybridization of \( Br^- \)
- Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals. However, in the case of \( Br^- \), it is a single atom with a complete octet (8 electrons) and does not form bonds with other atoms in this anionic state.

### Step 5: Conclusion on hybridization
- Since \( Br^- \) does not involve the mixing of orbitals (as it is a monoatomic ion), there is no hybridization applicable to the anionic part. Therefore, the hybridization of the central atom of the anionic part of \( PBr_5 \) is not applicable.

### Final Answer
- The hybridization of the central atom of the anionic part of \( PBr_5 \) in crystalline state is **not applicable**.

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