In dichromate titrations, the end point is blue colour due to the oxidation of:

C r O_{4}^{2 -}

ions

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C r_{2} O_{7}^{2 -}

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M n O_{4}^{-}

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I^{-}

ions

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Solution

The correct option is D $I^{-}$ ions
During dichromate titrations, the iodide ions are oxidized to iodine. Hence, the colour changes to blue.
The reaction is given below:
$2 I^{-} \to I_{2} + 2 e^{-}$

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Similar questions

(I) C r_{2} O_{7}^{2 -} (I I) M n O_{4}^{2 -} (I I I) C r O_{4}^{2 -} (I V) M n O_{4}^{-}

The colour of $M n O {_{4}}^{-}$ ion is due to:

Number of $C r - O$ bonds in dichromate ion $(C r_{2} O_{7}^{2 -})$ is:

({C r_{2} O_{7}}^{2 -})

C r^{3 +}

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

C O_{2}

C r O_{4}^{2 -}

C r_{2} O_{7}^{2 -}

C r O_{4}^{2 -} (a q) C O_{2} - - \to C r_{2} O_{7}^{2 -}

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