# What is M in the following equation? 4 M + 8 CN^- + 2 H2 O + O2 -> 4[M(CN)2] + 4 OH^- A Cu B Fe C Au D Zn

## Question

What is $\mathrm{M}$ in the following equation?

$$ \begin{array}{l} 4 \mathrm{M} + 8 \mathrm{CN}^{-} + 2 \mathrm{H}_2 \mathrm{O} + \mathrm{O}_2 \rightarrow 4\left[\mathrm{M}(\mathrm{CN})_{2}\right] + 4 \mathrm{OH}^{-} \end{array} $$

A $\mathrm{Cu}$ B $\mathrm{Fe}$

C $\mathrm{Au}$ D $\mathrm{Zn}$

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## Answer

**:**

To determine the identity of $\mathrm{M}$ in the equation:

$$
4 \mathrm{M} + 8 \mathrm{CN}^{-} + 2 \mathrm{H}*{2} \mathrm{O} + \mathrm{O}*{2} \rightarrow 4\left[\mathrm{M}(\mathrm{CN})_{2}\right] + 4 \mathrm{OH}^{-}
$$

we can compare it with the known reaction involving gold ($\mathrm{Au}$):

$$
4 \mathrm{Au}+8 \mathrm{CN}^{-}+2 \mathrm{H}*{2} \mathrm{O}+\mathrm{O}*{2} \rightarrow 4\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-} + 4 \mathrm{OH}^{-}
$$

In this reaction, **$\mathrm{Au}$ forms a complex ion** with cyanide ($\mathrm{CN}^{-}$), resulting in the complex $\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}$. It matches the structure of the required complex in our given reaction.

Thus, the identity of the metal $\mathrm{M}$ is **Gold ($\mathrm{Au}$)**.

**Final Answer:** C

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