Iodide of Millon's base is

H g I_{2}

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K_{2} H g I_{4}

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N H_{2} H g O \cdot H g I

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N H_{4} I

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Solution

The correct option is C $N H_{2} H g O \cdot H g I$
When $N H_{4}^{+}$ ions are treated with Nessler's reagent (alkaline solution of $K_{2} H g I_{4})$ , a brown ppt. of $N H_{2} - H g - O - H g I$ or $H g_{2} N I \cdot H_{2} O$ (iodide of Millon's base) is formed.
$2 K_{2} H g I_{4} + N H_{2} + 2 K O H \to N H_{2} \cdot H g O \cdot H g I iodide of Millon's base + 7 K I + 2 H_{2} O$ .

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