The colour of $C r_{2} O_{7}^{2 -}$ changes from orange to blue-green when added to compound (A). The blue-green colour is due to the formation of :

C r^{2 +}

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C r^{3 +}

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C r O_{4}^{2 -}

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C r O_{5}

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Solution

The correct option is C $C r^{3 +}$
DU in $A = \frac{(2 n_{C} + 2) - 2_{H}}{2} = \frac{(2 \times 9 + 2) - 12}{2} = 4^{\circ}$
(It shows a benzene ring which is also confirmed by the oxidation of (A) to benzoic acid (PhCOOH).)
i. The reaction of Na with (A) shows that it contains (-OH) group.
(A) is chiral, so the possible structures of (A) can be
(II) will give iodoform test. But compound (A) does not give the iodoform test. So the structure of (A) is (I).

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The colour of Cr2O2−7 changes from orange to blue-green when added to compound (A). The blue-green colour is due to the formation of :

The colour of $C r_{2} O_{7}^{2 -}$ changes from orange to blue-green when added to compound (A). The blue-green colour is due to the formation of :