Consider the following reactions:
`MX_(4)+X'_(2) to MX_(4)X_(2)'`
If atomic number of M is 52 and X and X' are halogens and X' is more electonegative than X. Then choose the correct statement regarding the given information:

Options

A. Both X' atoms occupy axial positions which are formed by overlapping of p and d-orbitals only
B. All M-X bond lengths are identical in both `MX_(4) and MX_(4)X'_(2)` compounds
C. Central atom `M` does not use any valence non-axial set of d-orbitals in hybridization of final product
D. Hybridization of central atom `M` remains same in both reactant and final product

Explanation

To solve the given problem, we need to analyze the reactions and the properties of the compounds involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Element M
- The atomic number of M is 52.
- The element with atomic number 52 is Tellurium (Te).

**Hint:** Look up the periodic table to find the element corresponding to the atomic number.

### Step 2: Determine the Group and Period of M
- Tellurium (Te) is in Group 16 and Period 5 of the periodic table.
- This means it has 6 valence electrons.

**Hint:** Remember that the group number indicates the number of valence electrons for main group elements.

### Step 3: Write the Electronic Configuration
- The electronic configuration of Tellurium is:
- [Kr] 4d10 5s2 5p4.
- The valence shell (5th shell) has 6 electrons (2 from 5s and 4 from 5p).

**Hint:** Use the Aufbau principle to write the electronic configuration of elements.

### Step 4: Analyze the Compound Formation
- The reaction given is: \( MX_4 + X'_2 \rightarrow MX_4X'_2 \).
- Here, \( X \) and \( X' \) are halogens, with \( X' \) being more electronegative than \( X \).

**Hint:** Consider the properties of halogens and how their electronegativity affects bond formation.

### Step 5: Determine Hybridization of \( MX_4 \)
- For \( MX_4 \), Tellurium will hybridize its orbitals to accommodate 4 bonds with \( X \).
- The hybridization will be \( sp^3d \) (5 orbitals: 1 s, 3 p, 1 d).
- The geometry will be trigonal bipyramidal, with one lone pair occupying an equatorial position according to Bent's rule.

**Hint:** Recall that hybridization depends on the number of electron pairs (bonding and lone pairs) around the central atom.

### Step 6: Determine Hybridization of \( MX_4X'_2 \)
- For \( MX_4X'_2 \), Tellurium will now form 6 sigma bonds (4 with \( X \) and 2 with \( X' \)).
- The hybridization will change to \( sp^3d^2 \) (6 orbitals: 1 s, 3 p, 2 d).
- The geometry will be octahedral.

**Hint:** The hybridization can change based on the number of bonds formed by the central atom.

### Step 7: Analyze the Statements
1. **Statement 1:** Both \( X' \) atoms occupy axial positions.
- **Incorrect:** In an octahedral geometry, all positions are equivalent.

2. **Statement 2:** All \( MX \) bond lengths are identical in both \( MX_4 \) and \( MX_4X'_2 \).
- **Incorrect:** Bond lengths differ due to different hybridizations and geometries.

3. **Statement 3:** Central atom M does not use any valence non-axial set of d-orbitals in hybridization of final product.
- **Correct:** In \( sp^3d^2 \), the d-orbitals used are axial.

4. **Statement 4:** Hybridization of central atom M remains the same in both reactant and final product.
- **Incorrect:** The hybridization changes from \( sp^3d \) to \( sp^3d^2 \).

### Conclusion
The correct statement regarding the given information is **Statement 3**: "Central atom M does not use any valence non-axial set of d-orbitals in hybridization of final product."

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