Question

oxidation number of b in nabh4

Answer 1

oxidation no of Na is +1 (since oxidation no of alkali metal is +1) oxidation no of H is -1 (hydrogen in hydrides will have -1) sum of oxidation numbers in a compound is 0, so let oxidation no of boron be x (+1)+x+ 4(-1)=0 solving it we get x=+3 oxidation no of B is +3.

Answer 2

Answer:

The oxidation number of $B$ is +3 in $NaBH_{4}$.

Explanation:

The overall charge on the compound ($NaBH_{4}$) is zero.

The oxidation number of one $B$ is assumed to be $x$, the oxidation number of one $Na$ is $+1$ and the oxidation number of one $H$ is $-1$ (hydride).

The oxidation number of $B$ in $NaBH_{4}$ can be calculated by-

$+1 + x + 4(-1) = 0$

$+1 + x - 4 = 0$

$x - 3 = 0$

$x = +3$

The oxidation number of $B$ in $NaBH_{4}$ is +3.