Na[BH4] + I2 = B2H6 + NaI + H2 |

2Na[BH₄] + I₂ → B₂H₆ + 2NaI + H₂

Sodium tetrahydroxoborate(III) react with iodine to produce diborane, sodium iodide and hydrogen.

The reaction proceeds in the diglyme (MeOCH₂CH₂)₂O. The yield was 98%. ^[16]

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Sodium tetrahydroxoborate(III) react with iodine

2Na[BH₄] + I₂ → B₂H₆ + 2NaI + H₂

Picture of reaction:

Сoding to search: 2 NaBH4 + I2 = B2H6 + 2 NaI + H2

Add / Edited: 21.08.2015 /

Evaluation of information: 5.0 out of 5 / number of votes: 1

Sodium tetrahydroxoborate(III) react with iodine

2Na[BH4] + I2 → B2H6 + 2NaI + H2

Picture of reaction:

Сoding to search: 2 NaBH4 + I2 = B2H6 + 2 NaI + H2

Add / Edited: 21.08.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

2Na[BH₄] + I₂ → B₂H₆ + 2NaI + H₂

Add / Edited: 21.08.2015 /