Count the total number of unshared electrons on `I_(3)^(-)=3`
Count the total number of `ppi-dpi` bonds in `PO_(4)^(3-)=y`
Count the total number of lone pairs in `XeF_(4)=z`
Hence, find the value of `x-z-y`

Explanation

To solve the problem step by step, we will break down each part of the question and calculate the required values.

### Step 1: Count the total number of unshared electrons on \( I_3^- \) (denote this as \( x \))

1. **Determine the valence electrons of iodine**: Each iodine atom has 7 valence electrons. Since there are 3 iodine atoms, the total is \( 3 \times 7 = 21 \) electrons.
2. **Account for the negative charge**: The negative charge on \( I_3^- \) adds 1 more electron, bringing the total to \( 21 + 1 = 22 \) valence electrons.
3. **Draw the Lewis structure**: In \( I_3^- \), two iodine atoms are bonded to a central iodine atom. The central iodine forms 2 bonds (2 bond pairs), using 4 electrons.
4. **Calculate remaining electrons**:
- Total electrons used in bonding: 4 (2 bonds)
- Remaining electrons: \( 22 - 4 = 18 \) electrons.
5. **Distribute remaining electrons**: The remaining 18 electrons will be placed as lone pairs around the iodine atoms. Each lone pair consists of 2 electrons.
- Each of the two terminal iodine atoms will have 3 lone pairs (6 electrons each).
- The central iodine will have 3 lone pairs (6 electrons).
6. **Count unshared electrons**:
- Total lone pairs: 3 (from each terminal iodine) + 3 (from central iodine) = 6 lone pairs.
- Total unshared electrons: \( 6 \times 2 = 12 \) electrons.

Thus, \( x = 12 \).

### Step 2: Count the total number of \( p\pi - d\pi \) bonds in \( PO_4^{3-} \) (denote this as \( y \))

1. **Determine the valence electrons**:
- Phosphorus has 5 valence electrons.
- Each oxygen has 6 valence electrons, and there are 4 oxygens: \( 4 \times 6 = 24 \).
- The negative charge \( (3-) \) adds 3 electrons: \( 5 + 24 + 3 = 32 \) total valence electrons.
2. **Draw the Lewis structure**:
- Phosphorus will form 4 sigma bonds with the 4 oxygen atoms.
- Each oxygen forms a single bond with phosphorus, consuming 8 electrons (4 bonds).
3. **Calculate remaining electrons**:
- Remaining electrons: \( 32 - 8 = 24 \).
4. **Distribute remaining electrons**: Each oxygen atom will complete its octet by forming double bonds with phosphorus, which involves \( p\pi - d\pi \) bonding.
5. **Count \( p\pi - d\pi \) bonds**:
- Each double bond between phosphorus and oxygen contributes 1 \( p\pi - d\pi \) bond.
- Since there are 4 oxygen atoms, and phosphorus can form 4 bonds, we have 4 \( p\pi - d\pi \) bonds.

Thus, \( y = 4 \).

### Step 3: Count the total number of lone pairs in \( XeF_4 \) (denote this as \( z \))

1. **Determine the valence electrons**:
- Xenon has 8 valence electrons.
- Each fluorine has 7 valence electrons, and there are 4 fluorines: \( 4 \times 7 = 28 \).
- Total valence electrons: \( 8 + 28 = 36 \).
2. **Draw the Lewis structure**:
- Xenon forms 4 bonds with 4 fluorine atoms, using 8 electrons.
3. **Calculate remaining electrons**:
- Remaining electrons: \( 36 - 8 = 28 \).
4. **Distribute remaining electrons**: The remaining electrons will be placed as lone pairs on the xenon atom.
- Each lone pair consists of 2 electrons.
- Xenon will have 2 lone pairs (4 electrons).

Thus, \( z = 2 \).

### Final Calculation: Find the value of \( x - y - z \)

Now we can calculate \( x - y - z \):
- \( x = 12 \)
- \( y = 4 \)
- \( z = 2 \)

So,
\[ x - y - z = 12 - 4 - 2 = 6 \]

### Conclusion
The final answer is \( 6 \).