The covalency and oxidation state respectively of boron in `[BF_4]^-`, are

Options

A. `4` and `4`
B. `3` and `4`
C. `3` and `5`
D. `4` and `3`

Answer

To determine the covalency and oxidation state of boron in the compound \([BF_4]^-\), we can follow these steps:

### Step 1: Determine the oxidation state of boron
1. Let the oxidation state of boron be \(X\).
2. Each fluorine (F) has an oxidation state of \(-1\).
3. Since there are 4 fluorine atoms, their total contribution to the oxidation state is \(4 \times (-1) = -4\).
4. The overall charge of the ion \([BF_4]^-\) is \(-1\).

Using the equation for the oxidation state:
\[ X + 4(-1) = -1 \]
\[ X - 4 = -1 \]
\[ X = -1 + 4 = +3 \]

### Step 2: Determine the covalency of boron
1. Covalency refers to the number of electrons that an atom shares with surrounding atoms.
2. Boron has an atomic number of 5, which means its electronic configuration is \(1s^2 2s^2 2p^1\). Therefore, boron has 3 valence electrons (2 in the \(2s\) and 1 in the \(2p\)).
3. In \([BF_4]^-\), boron forms bonds with 4 fluorine atoms.
4. Each bond involves the sharing of one electron from boron with one electron from each fluorine atom.
5. Since boron forms 4 single bonds with 4 fluorine atoms, it shares all 3 of its valence electrons and can accommodate an additional electron through bonding.

Thus, the covalency of boron is 4.

### Final Answer
- The oxidation state of boron in \([BF_4]^-\) is +3.
- The covalency of boron in \([BF_4]^-\) is 4.

### Summary
- Covalency: 4
- Oxidation State: +3