Iodide of Millon's base is :

K_{2} [H g I_{4}]

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H_{2} N - H g - O - H g - I

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[H g_{2} O . N H_{2} O H] . H_{2} O

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H g (N H_{2}) I + H g

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Solution

The correct option is B $H_{2} N - H g - O - H g - I$
Reaction process:

$K_{2} H g I_{4} ⇌ 2 K I + H g I_{2}$

$H g I_{2} + 2 N H_{3} \to I - H g - N H_{2} + N H_{4} I$

$2 N H_{2} - H g - I + H_{2} O \to N H_{2} - H g - O - H g - I + N H_{4} I$

Or $2 K_{2} H g I_{4} + N H_{4} + 3 K O H \to H_{2} N - H g - O - H g - I + 7 K I + 2 H_{2} O$

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