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.

  • 29

The oxidation number of Boron in NaBH4 is 3.

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