Find oxidation number of
B in Na BH4
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2
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.
Answered by
2
Hydrogen is more electronegative then boron so hydrogen have negative one(-1) oxidation number with respect to boron and sodium always have +1 oxidation number.
Let oxidation number of boron (B) in “NaBH4” is x.
+1 + x + 4(-1) = 0
=> x + 1 - 4 = 0
=> x - 3 = 0
=> x = 3
Hence, the oxidation number of B is +3.
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