The number of lone pair(s) of electrons on the central atom in `[BrF_4]^(-), XeF_6 and [SbCl_6]^(3-)` are, respectively

Options

A. 2, 0 and 1
B. 1, 0 and 0
C. 2, 1 and 1
D. 2, 1 and 0

Explanation

To determine the number of lone pairs of electrons on the central atom in the given species `[BrF_4]^(-)`, `XeF_6`, and `[SbCl_6]^(3-)`, we will analyze each molecule step by step.

### Step 1: Analyze `[BrF_4]^-`
1. **Identify the central atom**: The central atom is Bromine (Br).
2. **Count the valence electrons**: Bromine has 7 valence electrons. Fluorine (F) has 7 valence electrons, and there are 4 fluorine atoms. Therefore, the total number of valence electrons contributed by fluorine is \(4 \times 7 = 28\).
3. **Account for the negative charge**: The negative charge adds 1 extra electron. Thus, the total number of valence electrons is:
\[ 7 \, (\text{Br}) + 28 \, (\text{from F}) + 1 \, (\text{negative charge}) = 36 \, \text{electrons} \]
4. **Determine bonding and lone pairs**: Bromine forms 4 bonds with fluorine atoms, using 8 electrons (4 bonds). The remaining electrons are:
\[ 36 - 8 = 28 \, \text{electrons} \]
Since each lone pair consists of 2 electrons, the number of lone pairs on Br is:
\[ \frac{28}{2} = 14 \, \text{lone pairs} \]
However, since we only need to consider the lone pairs on the central atom, we find that there are 2 lone pairs remaining on Br after forming 4 bonds.

### Step 2: Analyze `XeF_6`
1. **Identify the central atom**: The central atom is Xenon (Xe).
2. **Count the valence electrons**: Xenon has 8 valence electrons. Each fluorine contributes 7 valence electrons, and there are 6 fluorine atoms. Therefore, the total number of valence electrons is:
\[ 8 \, (\text{Xe}) + 6 \times 7 \, (\text{from F}) = 8 + 42 = 50 \, \text{electrons} \]
3. **Determine bonding and lone pairs**: Xenon forms 6 bonds with fluorine atoms, using 12 electrons (6 bonds). The remaining electrons are:
\[ 50 - 12 = 38 \, \text{electrons} \]
Since each lone pair consists of 2 electrons, the number of lone pairs on Xe is:
\[ \frac{38}{2} = 19 \, \text{lone pairs} \]
However, since we only need to consider the lone pairs on the central atom, we find that there is 1 lone pair remaining on Xe after forming 6 bonds.

### Step 3: Analyze `[SbCl_6]^(3-)`
1. **Identify the central atom**: The central atom is Antimony (Sb).
2. **Count the valence electrons**: Antimony has 5 valence electrons. Each chlorine contributes 7 valence electrons, and there are 6 chlorine atoms. Therefore, the total number of valence electrons is:
\[ 5 \, (\text{Sb}) + 6 \times 7 \, (\text{from Cl}) = 5 + 42 = 47 \, \text{electrons} \]
3. **Account for the negative charge**: The 3 negative charges add 3 extra electrons. Thus, the total number of valence electrons is:
\[ 47 + 3 = 50 \, \text{electrons} \]
4. **Determine bonding and lone pairs**: Antimony forms 6 bonds with chlorine atoms, using 12 electrons (6 bonds). The remaining electrons are:
\[ 50 - 12 = 38 \, \text{electrons} \]
Since each lone pair consists of 2 electrons, the number of lone pairs on Sb is:
\[ \frac{38}{2} = 19 \, \text{lone pairs} \]
However, since we only need to consider the lone pairs on the central atom, we find that there is 1 lone pair remaining on Sb after forming 6 bonds.

### Final Answer
The number of lone pairs on the central atom in `[BrF_4]^(-)`, `XeF_6`, and `[SbCl_6]^(3-)` are:
- `[BrF_4]^-`: 2 lone pairs
- `XeF_6`: 1 lone pair
- `[SbCl_6]^(3-)`: 1 lone pair

### Summary
The final answer is:
- `[BrF_4]^-`: 2 lone pairs
- `XeF_6`: 1 lone pair
- `[SbCl_6]^(3-)`: 1 lone pair