The covalency and oxidation state respectively of boron in [BF4]-, are
- 4 and 4
- 4 and 3
- 3 and 4
- 3 and 5
The Correct Option is B
Solution and Explanation
The structure of the [BF\(_4\)]\(^{-}\) ion is as follows:
Covalency of Boron: The covalency of boron is the number of covalent bonds it forms. In the [BF\(_4\)]\(^{-}\) ion, boron is surrounded by four fluorine atoms, forming four covalent bonds. Thus, the covalency of boron is \( 4 \).
Oxidation State of Boron: The oxidation state of boron can be calculated as follows: - Each fluorine atom has an oxidation state of \(-1\). - Let the oxidation state of boron be \( x \). The total charge on the [BF\(_4\)]\(^{-}\) ion is \(-1\), so: \[ x + 4(-1) = -1. \] Simplify: \[ x - 4 = -1 \implies x = +3. \] Thus, the oxidation state of boron is \( +3 \). ### Final Answer: The covalency and oxidation state of boron are \( 4 \) and \( +3 \), respectively.
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