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 ...

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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}

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K M n O_{4}

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

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None

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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.

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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

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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.