Question

3N2H4 + 2H2BO3- + 2H2O2B + 6NH2OH + 2OH- In the above reaction, the oxidation state of boron changes from -7 to +3. How many electrons are transferred in the reaction?

Name: 3n2h4 2h2bo3 2h2o2b 6nh2oh 2oh in the above reaction the oxidation state of boron changes from 7 6 5 4 3 2 101234567 to 7 6 5 4 3 2 101234567 how many electrons are transferred in th 13978
Uploaded: 2023-04-20T05:14:10-08:00
Duration: 2 min 20 s
Channel: Madhur L
Description: 3n2h4 2h2bo3 2h2o2b 6nh2oh 2oh in the above reaction the oxidation state of boron changes from 7 6 5 4 3 2 101234567 to 7 6 5 4 3 2 101234567 how many electrons are transferred in th 13978

          3N2H4 + 2H2BO3- + 2H2O2B + 6NH2OH + 2OH-
In the above reaction, the oxidation state of boron changes from -7 to +3.
How many electrons are transferred in the reaction?

Added by Corey D.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

04/20/2023

Step 1

In H2BO3-, boron is bonded to three oxygen atoms and one hydrogen atom. Oxygen has an oxidation state of -2, and hydrogen has an oxidation state of +1. So, we have: B + 3(-2) + 1 = -1 Solving for the oxidation state of boron, we get: B = +3 In H2O2B, boron is Show more…

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3N2H4 + 2H2BO3- + 2H2O2B + 6NH2OH + 2OH- In the above reaction, the oxidation state of boron changes from -7 to +3. How many electrons are transferred in the reaction?