A [M(H_2O)_6]^{2+} complex typically absorbs | vedclass

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(B) The energy of absorption is given by $E = \frac{hc}{\lambda}$, where $\lambda$ is the wavelength. Since $NH_3$ is a stronger ligand than $H_2O$ in

Vedclass · Accepted Answer

$580$

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(B) The energy of absorption is given by $E = \frac{hc}{\lambda}$, where $\lambda$ is the wavelength. Since $NH_3$ is a stronger ligand than $H_2O$ in the spectrochemical series, the crystal field splitting energy $(\Delta_o)$ for $[M(NH_3)_6]^{2+}$ is greater than that for $[M(H_2O)_6]^{2+}$. Because $E \propto \frac{1}{\lambda}$, a higher splitting energy corresponds to a smaller wavelength of absorption. Therefore, the absorption wavelength for $[M(NH_3)_6]^{2+}$ must be less than $600 \ nm$. Among the given options, $580 \ nm$ is the only value smaller than $600 \ nm$ and reasonably close to it, as the difference in splitting power between $NH_3$ and $H_2O$ is not extremely large.

$A$ $[M(H_2O)_6]^{2+}$ complex typically absorbs at around $600 \ nm$. It is allowed to react with ammonia to form a new complex $[M(NH_3)_6]^{2+}$ that should have absorption at ....... $nm$.

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