There are two groups of compounds A and B. Groups A contains three compounds `Px_(4), Qy_(3), Rz_(2)`. Groups B also contains three compounds `Sx_(4), Ty_(3), Uz_(2)`. Hybridization of each central atom of group A compounds is same as that of iodine in `IBrCl^(-)` while in group B compounds it is same as that of iodine `IBrCl^(+)`. Substituents X, Y and Z exhibit covalency of one in ground state. Then find the value of x/y.
Where, x and y are total number of lone pair present at central atoms of compounds of group A and B respectively.

Explanation

To solve the problem, we need to determine the total number of lone pairs present at the central atoms of compounds in groups A and B, and then find the ratio \( \frac{x}{y} \), where \( x \) is the total number of lone pairs in group A and \( y \) is the total number of lone pairs in group B.

### Step 1: Determine the hybridization of the central atom in group A compounds.

The compounds in group A are \( P_xCl_4 \), \( Q_yBr_3 \), and \( R_zI_2 \). The hybridization of the central atom in group A is the same as that of iodine in \( IBrCl^- \).

1. **Find the steric number for \( IBrCl^- \)**:
- Valence electrons of Iodine (I) = 7
- Number of monovalent species (Br and Cl) = 2
- Charge = -1 (add 1 for negative charge)
- Steric number = \( \frac{(7 + 2 + 1)}{2} = \frac{10}{2} = 5 \)

2. **Determine the hybridization**:
- A steric number of 5 corresponds to \( sp^3d \) hybridization.

### Step 2: Determine the lone pairs in group A compounds.

For \( sp^3d \) hybridization (trigonal bipyramidal geometry):
- There is 1 lone pair on the central atom.

1. For \( P_xCl_4 \): 1 lone pair
2. For \( Q_yBr_3 \): 2 lone pairs
3. For \( R_zI_2 \): 3 lone pairs

Total lone pairs in group A (\( x \)):
\[ x = 1 + 2 + 3 = 6 \]

### Step 3: Determine the hybridization of the central atom in group B compounds.

The compounds in group B are \( S_xCl_4 \), \( T_yBr_3 \), and \( U_zI_2 \). The hybridization of the central atom in group B is the same as that of iodine in \( IBrCl^+ \).

1. **Find the steric number for \( IBrCl^+ \)**:
- Valence electrons of Iodine (I) = 7
- Number of monovalent species (Br and Cl) = 2
- Charge = +1 (subtract 1 for positive charge)
- Steric number = \( \frac{(7 + 2 - 1)}{2} = \frac{8}{2} = 4 \)

2. **Determine the hybridization**:
- A steric number of 4 corresponds to \( sp^3 \) hybridization.

### Step 4: Determine the lone pairs in group B compounds.

For \( sp^3 \) hybridization (tetrahedral geometry):
- There are no lone pairs for \( S_xCl_4 \).
- There is 1 lone pair for \( T_yBr_3 \).
- There are 2 lone pairs for \( U_zI_2 \).

Total lone pairs in group B (\( y \)):
\[ y = 0 + 1 + 2 = 3 \]

### Step 5: Calculate the ratio \( \frac{x}{y} \).

Now that we have \( x = 6 \) and \( y = 3 \):
\[ \frac{x}{y} = \frac{6}{3} = 2 \]

### Final Answer:
The value of \( \frac{x}{y} \) is \( 2 \).

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