Question

Write the oxidation number of K and B in the given species:
$K_{2}Mn{O}_4$ and $Na{BH}_4$

Answer 1

1) For $K_{2}Mn{O}_4$; The oxidation state of Mn in $KMn{O}_4$ is $+7$.

This can be obtained by taking into account the valencies of potassium and oxygen atoms.

The valency of $K$is $1$ but it loses electron so to be specific it is$+1$.

Similarly, oxygen in specific has a valency of $-2$.

2) For $NaB{H}_4$; As the oxidation state of Na is $+1$ and Hydrogen is $-1$ because here hydrogen exists as a hydride.

Hence the oxidation state of boron is $+3$.