JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 5)
The metal complex that is diamagnetic is (Atomic number: $$\mathrm{Fe}, 26 ; \mathrm{Cu}, 29)$$
$$\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]$$
$$\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]$$
$$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]$$
$$\mathrm{K}_{4}\left[\mathrm{FeCl}_{6}\right]$$
Explanation
$$\Rightarrow \mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]$$ is diamagnetic
$$\mathrm{Cu}(\mathrm{I}) \Rightarrow \mathrm{d}^{10}$$ configuration $$\Rightarrow$$ No unpaired electrons.
$$\Rightarrow \mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right], \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]$$ and $$\mathrm{K}_{4}\left[\mathrm{FeCl}_{6}\right]$$ are paramagnetic in nature
$$\mathrm{Cu}(\mathrm{I}) \Rightarrow \mathrm{d}^{10}$$ configuration $$\Rightarrow$$ No unpaired electrons.
$$\Rightarrow \mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right], \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{4}\right]$$ and $$\mathrm{K}_{4}\left[\mathrm{FeCl}_{6}\right]$$ are paramagnetic in nature
Comments (0)
