Find the total number of compounds having equal all bond lengths equal :
`PCl_(5),PF_(5),SF_(4),XeF_(2), XeF_(4),XeF_(6),SF_(6),ClF_(3), BrF_(5),Ni(CO)_(4),IF_(7)`

Explanation

To determine the total number of compounds from the given list that have equal bond lengths, we will analyze each compound step by step.

1. **PCl₅ (Phosphorus Pentachloride)**:
- Phosphorus has 5 valence electrons, and it forms 5 bonds with chlorine.
- The hybridization is sp³d, resulting in a trigonal bipyramidal shape.
- In this shape, the axial bonds (2) and equatorial bonds (3) have different bond lengths.
- **Conclusion**: Not all bond lengths are equal.

2. **PF₅ (Phosphorus Pentafluoride)**:
- Similar to PCl₅, phosphorus forms 5 bonds with fluorine.
- The hybridization is also sp³d, with a trigonal bipyramidal shape.
- Again, axial and equatorial bond lengths differ.
- **Conclusion**: Not all bond lengths are equal.

3. **SF₄ (Sulfur Tetrafluoride)**:
- Sulfur has 6 valence electrons and forms 4 bonds with fluorine.
- The hybridization is sp³d, and the shape is seesaw due to one lone pair.
- The bond lengths are not equal due to the presence of the lone pair.
- **Conclusion**: Not all bond lengths are equal.

4. **XeF₂ (Xenon Difluoride)**:
- Xenon has 8 valence electrons and forms 2 bonds with fluorine.
- The hybridization is sp³d, resulting in a linear shape.
- Both bond lengths are equal as both fluorine atoms occupy axial positions.
- **Conclusion**: All bond lengths are equal.

5. **XeF₄ (Xenon Tetrafluoride)**:
- Xenon forms 4 bonds with fluorine and has 2 lone pairs.
- The hybridization is sp³d², resulting in a square planar shape.
- All bond lengths are equal due to symmetry.
- **Conclusion**: All bond lengths are equal.

6. **XeF₆ (Xenon Hexafluoride)**:
- Xenon forms 6 bonds with fluorine and has no lone pairs.
- The hybridization is sp³d³, resulting in an octahedral shape.
- All bond lengths are equal due to symmetry.
- **Conclusion**: All bond lengths are equal.

7. **SF₆ (Sulfur Hexafluoride)**:
- Sulfur forms 6 bonds with fluorine and has no lone pairs.
- The hybridization is sp³d², resulting in an octahedral shape.
- All bond lengths are equal due to symmetry.
- **Conclusion**: All bond lengths are equal.

8. **ClF₃ (Chlorine Trifluoride)**:
- Chlorine has 7 valence electrons and forms 3 bonds with fluorine.
- The hybridization is sp³d, resulting in a T-shaped structure due to 2 lone pairs.
- The bond lengths are not equal due to the lone pairs affecting the geometry.
- **Conclusion**: Not all bond lengths are equal.

9. **BrF₅ (Bromine Pentafluoride)**:
- Bromine has 7 valence electrons and forms 5 bonds with fluorine.
- The hybridization is sp³d², resulting in a square pyramidal shape.
- The bond lengths are not equal due to the lone pair affecting the geometry.
- **Conclusion**: Not all bond lengths are equal.

10. **Ni(CO)₄ (Nickel Tetracarbonyl)**:
- Nickel forms 4 bonds with carbon monoxide.
- The hybridization is sp³, resulting in a tetrahedral shape.
- All bond lengths are equal due to symmetry.
- **Conclusion**: All bond lengths are equal.

11. **IF₇ (Iodine Heptafluoride)**:
- Iodine has 7 valence electrons and forms 7 bonds with fluorine.
- The hybridization is sp³d³, resulting in a pentagonal bipyramidal shape.
- All bond lengths are equal due to symmetry.
- **Conclusion**: All bond lengths are equal.

### Summary of Findings:
- Compounds with equal bond lengths: **XeF₂, XeF₄, XeF₆, SF₆, Ni(CO)₄, IF₇**
- Total number of compounds with equal bond lengths: **6**

### Final Answer:
The total number of compounds having all bond lengths equal is **6**.