What is the structure of soluble complex formed when AgCl is treated with ethylamine?

[A g (C H_{3} C H_{2} N H_{2})_{2}]^{+} C l^{-}

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[A g (C H_{3} C H_{2} N H_{2})_{4}]^{+} C l^{-}

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[A g_{2} (C H_{3} C H_{2} N H_{2})_{2}] C l_{2}

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[A g_{2} (C H_{3} C H_{2} N H_{2})_{4}] C l_{2}

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Solution

The correct option is A $[A g (C H_{3} C H_{2} N H_{2})_{2}]^{+} C l^{-}$
The structure of soluble complex formed, when AgCl is treated with ethylamine, is $[A g (C H_{3} C H_{2} N H_{2})_{2}]^{+} C l^{-}$
$A g C l + 2 C H_{3} C H_{2} N H_{2} \to [A g (C H_{3} C H_{2} N H_{2})_{2}]^{+} C l^{-}$

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When 0.2 mol $C r C l_{3} .6 H_{2} O$ is treated with excess of $A g N O_{3}$ , 0.4 mol of AgCl are obtained. The formula of the complex is

C r C l_{3} .6 H_{2} O

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[A g (N H_{3})_{2}] C l

[C o (e n)_{2} {C l}_{2}] C l

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