HBr - HI reduce sulphuric acid- HCl can reduce KMnO 4 - HF can reduce-A- H 2 SO 4B- KMnO 4- K 2 Cr 2 O 7D- None

The correct option is D NoneFlourine being the most electronegative element amp; small in size does not lose electrons. Hence, F^ cannot act as reducing ag ...

204

You visited us 204 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Heavy Water

HBr & HI redu...

Question

HBr & HI reduce sulphuric acid; HCl can reduce $K M n O_{4}$ & HF can reduce:

H_{2} S O_{4}

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

K M n O_{4}

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

K_{2} C r_{2} O_{7}

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

None

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

Open in App

Solution

The correct option is D None
Flourine being the most electronegative element & small in size does not lose electrons. Hence, $F^{-}$ cannot act as reducing agent.

Suggest Corrections

Similar questions

H B r

and

H I

reduce sulphuric acid,

H C l

can reduce

K M n O_{4}

and

H F

can reduce :

H B r

and

H I

reduce sulphuric acid.

H C l

can reduce

K M n O_{4}

and

H F

can reduce:

Q. HBr and HI reduce sulphuric acid, HCl can reduce

K M n O_{4}

and HF can reduce

H B r

and

H I

reduce

H_{2} S O_{4}

H C l

can reduce

K M n O_{4}

and

H F

reduces:

Q. Arrange in decreasing order of acid strength.

H F, H C l, H B r, H I

Join BYJU'S Learning Program

HBr & HI reduce sulphuric acid; HCl can reduce KMnO4 & HF can reduce:

The correct option is D NoneFlourine being the most electronegative element & small in size does not lose electrons. Hence, F− cannot act as reducing agent.

HBr & HI reduce sulphuric acid; HCl can reduce $K M n O_{4}$ & HF can reduce:

The correct option is D None
Flourine being the most electronegative element & small in size does not lose electrons. Hence, $F^{-}$ cannot act as reducing agent.