Question

The oxidation state of boron in potassium tetrafluoroborate is:
(A) $+ 2$
(B) $+ 3$
(C) $+ 4$
(D) $- 3$

Answer 1

Hint : First, try to dissociate the potassium tetrafluoroborate molecule into its ionic form of ${K^ + }$ and $BF_4^ -$ . Check the charge of the ions and try to formulate an algebraic equation to find the oxidation state of boron. Be careful not to include those ions which are not part of the dissociated $BF_4^ -$ ion.

Complete Step By Step Answer:
The chemical formula of potassium tetrafluoroborate is $KB{F_4}$ and it undergoes the dissociation reaction as follows:
$KB{F_4} \to {K^ + } + BF_4^ -$
The $KB{F_4}$ breaks into two ions namely, potassium ion $({K^ + })$ and tetrafluoroborate ion $(BF_4^ - )$ .
Since, we have to find the oxidation state of boron, therefore, we will neglect any other element except $BF_4^ -$ .
To find the oxidation state of boron, let us consider the oxidation state of boron be $x$ . Since, the charge on $BF_4^ -$ is $- 1$ therefore, the algebraic equation for the same will be:
$x + 4 \times ( - 1) = - 1$
Here, we have multiplied $4$ with $- 1$ because there are $4$ atoms of fluorine and each one has a charge of $- 1$ . We have added $x$ because as we know, the total charge on an ion will be the sum of the individual charges of each element. The total charge of the ion is equal to $- 1$ .
On further solving,
$x - 4 = - 1$
$x = - 1 + 4$
Which results in,
$x = + 3$
Thus, the oxidation state of boron $(B)$ is $+ 3$ .
Therefore, option B is correct.

Note :
Always check the equation of dissociation carefully before moving ahead. Any mistake in the chemical equation can result in an incorrect answer. Also keep in mind the charge and the sign of the elements involved in the reaction. Also, formulate your algebraic equation carefully.