Zn ( II ) ion first gives a white precipitate with NaOH which dissolves in excess of NaOH. This is due to the formation of a complex. The oxidation sate of zinc is this complex will be.

Zero

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+ I I

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+ I V

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+ V I

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Solution

The correct option is B $+ I I$
Reaction:
$Z n_{2} + + 2 O H^{\to} Z n (O H)_{2} .$ (white ppt.).
In excess of NaOH, it dissolves and form complex.
$Z n (O H)_{2} + 2 O H^{\to} Z n (O H)_{4}^{2 -}$ .
In this complex, oxidation state of Zn is +2.

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