`SbF_(3)` reacts with `XeF_(4) and XeF_(6)` to form ionic compounds `[XeF_(3)^(+)][SbF_(6)^(-)] and [XeF_(5)^(+)][SbF_(6)^(-)]` then molecular shape of `[XeF_(3)^(+)]` ion and `[XeF_(5)^(+)]` ion respectively

Options

A. square pyramidal, T - shaped
B. Bent - T - shape, square pyramidal
C. See - saw, square pyramidal
D. Square pyramidal, See - saw

Answer

To determine the molecular shapes of the ions \([XeF_3^+]\) and \([XeF_5^+]\), we will follow these steps:

### Step 1: Determine the electron count for \([XeF_3^+]\)
- **Xenon (Xe)** has 8 valence electrons.
- In \([XeF_3^+]\), xenon loses one electron (due to the positive charge), leaving it with 7 electrons.
- It forms 3 bonds with fluorine atoms, using 3 electrons.
- This leaves us with \(7 - 3 = 4\) electrons, which will form 2 lone pairs.

### Step 2: Count the total electron pairs for \([XeF_3^+]\)
- Total electron pairs = Bond pairs + Lone pairs
- Here, we have 3 bond pairs (from the 3 Xe-F bonds) and 2 lone pairs.
- Therefore, total electron pairs = \(3 + 2 = 5\).

### Step 3: Determine the hybridization for \([XeF_3^+]\)
- With 5 electron pairs, the hybridization is \(sp^3d\).
- The geometry associated with 5 electron pairs is trigonal bipyramidal.

### Step 4: Determine the molecular shape of \([XeF_3^+]\)
- In a trigonal bipyramidal arrangement, lone pairs occupy equatorial positions to minimize repulsion.
- Thus, the 2 lone pairs will occupy the equatorial positions, and the 3 fluorine atoms will occupy the axial positions.
- This results in a molecular shape that resembles a "bent T" shape.

### Step 5: Determine the electron count for \([XeF_5^+]\)
- For \([XeF_5^+]\), xenon again has 8 valence electrons.
- Losing one electron gives it 7 electrons.
- It forms 5 bonds with fluorine atoms, using 5 electrons.
- This leaves us with \(7 - 5 = 2\) electrons, which will form 1 lone pair.

### Step 6: Count the total electron pairs for \([XeF_5^+]\)
- Total electron pairs = Bond pairs + Lone pairs
- Here, we have 5 bond pairs (from the 5 Xe-F bonds) and 1 lone pair.
- Therefore, total electron pairs = \(5 + 1 = 6\).

### Step 7: Determine the hybridization for \([XeF_5^+]\)
- With 6 electron pairs, the hybridization is \(sp^3d^2\).
- The geometry associated with 6 electron pairs is octahedral.

### Step 8: Determine the molecular shape of \([XeF_5^+]\)
- In an octahedral arrangement, the lone pair will occupy one position, while the 5 fluorine atoms will occupy the remaining positions.
- This results in a molecular shape that is square pyramidal.

### Final Summary
- The molecular shape of \([XeF_3^+]\) is **bent T**.
- The molecular shape of \([XeF_5^+]\) is **square pyramidal**.

---