You visited us 3 times! Enjoying our articles? Unlock Full Access!

Borax

When NaBH4 ...

Question

When $N a B H_{4}$ reacts with $H I$ ,the product formed is:

B_{2} H_{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B_{2} H_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{3} B O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $B_{2} H_{6}$
First iodine reacti with one $N a B H_{4}$ , we get borance , NaI and HI .
Again, HI is react with another $N a B H_{4}$ , we get another borane
so,the answer is $B_{2} H_{6}$

Suggest Corrections

Similar questions

B_{2} H_{6}

B_{2} H_{6}

m L

I f k_{a} o f H_{3} B O_{3} = 8 \times 10^{- 10}, p K_{a} = 9.1; A t o m i c w e i g h t o f B = 10.8 g; M o l a r m a s s o f B_{2} H_{6} = 27.6 g m o l^{- 1}, l o g 0.8 = - 0.1

N a B H_{4}

B_{2} H_{6}

(B_{2} H_{6})

O_{2}

H_{2} O

What are the products formed when anisole reacts with HI?

Join BYJU'S Learning Program