The correct order of reducing power of halide ions is :

C l^{-} > B r^{-} > I^{-} > F^{-}

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C l^{-} > I^{-} > B r^{-} > F^{-}

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B r^{-} > C l^{-} > I^{-} > F^{-}

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I^{-} > B r^{-} > C l^{-} > F^{-}

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Solution

The correct option is C $I^{-} > B r^{-} > C l^{-} > F^{-}$
From the reduction potential values, it can be seen that iodide is best reducing agent. It has low reduction potential value which indicates that it easily reduces others and itself gets oxidised.

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Similar questions

Q. The correct order of polarizability is

I^{-}, B r^{-}, C l^{-}, F^{-}

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy ) is
(a) F > Cl > Br > I (b) F < Cl < Br < I (c) F < Cl > Br > I (d) F < Cl < Br < I

Q. The correct order of polarizability is

I^{-}, B r^{-}, C l^{-}, F^{-}

Q. Arrange the following bonds according to their average bond energies in descending order:

C - C l, C - B r, C - F, C - I

Q. Which of the following options does not give to the correct trends indicated against it?
(I)

F_{2} > C l_{2} > B r_{2} > I_{2}

: Bond dissociation energy
(II)

F_{2} > C l_{2} > B r_{2} > I_{2}

: Oxidizing nature
(III)

C l > F > B r > I

: Electron gain enthalpy
(IV)

F > C l > B r > I

: Electronegativity

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The correct order of reducing power of halide ions is :

The correct option is C I−>Br−>Cl−>F−From the reduction potential values, it can be seen that iodide is best reducing agent. It has low reduction potential value which indicates that it easily reduces others and itself gets oxidised.

The correct option is C $I^{-} > B r^{-} > C l^{-} > F^{-}$
From the reduction potential values, it can be seen that iodide is best reducing agent. It has low reduction potential value which indicates that it easily reduces others and itself gets oxidised.