oxidation number of b in nabh4
Let the oxidation state of Boron be X
As the oxidation state of Na is +1 and Hydrogen is -1 because here hydrogen exist as hydride.
Therefore
X + (+1) + 4(-1) = 0
X = +3
Hence the oxidation state of boron is +3.
oxidation number of b in nabh4
Let the oxidation state of Boron be X
As the oxidation state of Na is +1 and Hydrogen is -1 because here hydrogen exist as hydride.
Therefore
X + (+1) + 4(-1) = 0
X = +3
Hence the oxidation state of boron is +3.