The sum of the number of lone pairs of electrons on each central atom in the following species is :
`[TeBr_6]^(2-) , [BrF_2]^+ , SNF_3`, and `[XeF_3]^(-)` (A) Atomic numbers: 7, F = 9, S = 16, Br = 35, Te = 52, Xe= 54)

Answer

To find the sum of the number of lone pairs of electrons on each central atom in the species \([TeBr_6]^{2-}\), \([BrF_2]^+\), \(SNF_3\), and \([XeF_3]^-\), we will analyze each species step by step.

### Step 1: Analyze \([TeBr_6]^{2-}\)

1. **Valence Electrons Calculation**:
- Tellurium (Te) has 6 valence electrons.
- The charge of -2 adds 2 more electrons, giving a total of \(6 + 2 = 8\) valence electrons for Te.
- Each Bromine (Br) has 1 valence electron, and there are 6 Br atoms contributing \(6 \times 1 = 6\) electrons.
- Total valence electrons = \(8 + 6 = 14\).

2. **Bonding and Lone Pairs**:
- Te forms 6 bonds with 6 Br atoms, using 6 electrons.
- Remaining electrons = \(14 - 6 = 8\) electrons.
- These 8 electrons form 4 lone pairs on Te (since 2 electrons make 1 lone pair).

**Lone pairs on Te**: 4

### Step 2: Analyze \([BrF_2]^+\)

1. **Valence Electrons Calculation**:
- Bromine (Br) has 7 valence electrons.
- The positive charge (-1) reduces the count by 1, giving \(7 - 1 = 6\) valence electrons for Br.
- Each Fluorine (F) has 1 valence electron, and there are 2 F atoms contributing \(2 \times 1 = 2\) electrons.
- Total valence electrons = \(6 + 2 = 8\).

2. **Bonding and Lone Pairs**:
- Br forms 2 bonds with 2 F atoms, using 2 electrons.
- Remaining electrons = \(8 - 2 = 6\) electrons.
- These 6 electrons form 3 lone pairs on Br.

**Lone pairs on Br**: 3

### Step 3: Analyze \(SNF_3\)

1. **Valence Electrons Calculation**:
- Sulfur (S) has 6 valence electrons.
- Each Nitrogen (N) has 5 valence electrons.
- Each Fluorine (F) has 1 valence electron, and there are 3 F atoms contributing \(3 \times 1 = 3\) electrons.
- Total valence electrons = \(6 + 5 + 3 = 14\).

2. **Bonding and Lone Pairs**:
- N forms a double bond with S and 3 single bonds with 3 F atoms.
- This uses \(2 + 3 = 5\) electrons.
- Remaining electrons = \(14 - 5 = 9\) electrons.
- These 9 electrons form 0 lone pairs on S (since all are used for bonding).

**Lone pairs on S**: 0

### Step 4: Analyze \([XeF_3]^-\)

1. **Valence Electrons Calculation**:
- Xenon (Xe) has 8 valence electrons.
- The negative charge adds 1 more electron, giving \(8 + 1 = 9\) valence electrons for Xe.
- Each Fluorine (F) has 1 valence electron, and there are 3 F atoms contributing \(3 \times 1 = 3\) electrons.
- Total valence electrons = \(9 + 3 = 12\).

2. **Bonding and Lone Pairs**:
- Xe forms 3 bonds with 3 F atoms, using 3 electrons.
- Remaining electrons = \(12 - 3 = 9\) electrons.
- These 9 electrons form 3 lone pairs on Xe.

**Lone pairs on Xe**: 3

### Step 5: Sum of Lone Pairs

Now, we sum the lone pairs from each central atom:
- Lone pairs from \([TeBr_6]^{2-}\): 4
- Lone pairs from \([BrF_2]^+\): 3
- Lone pairs from \(SNF_3\): 0
- Lone pairs from \([XeF_3]^-\): 3

**Total Lone Pairs**: \(4 + 3 + 0 + 3 = 10\)

### Final Answer
The sum of the number of lone pairs of electrons on each central atom in the given species is **10**.

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