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NIDOHUBIN
Answered

what is the oxidation number of Na in the compound NaBH4​

Sagot :

Y07IN

Answer:

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.

Explanation:

EMILYSANCHEZ040411IN

Answer:

As the oxidation state of Na is +1 and Hydrogen is -1 because here hydrogen exist as hydride. Hence the oxidation state of boron is +3. 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.

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