Question

for the [cr(H2O)6]2+ and [cr(CN)6]4- ion has energy diffrence value(del zero ) are 17830cm- and 26280cm- .The pairing energy is 23520 cm- .State which coplex is high spin or spin. And calculate magnetic moment of each

Answer 1

Question #47974 – Chemistry - Inorganic Chemistry

Question:

For the $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}$ ion has energy difference value (del zero) are $17830 \mathrm{~cm}^{-1}$ and $26280 \mathrm{~cm}^{-1}$. The pairing energy is $23520 \mathrm{~cm}^{-1}$. State which complex is high spin or spin. And calculate magnetic moment of each.

Answer:

The energy of the ion in cyanide complex is lower ($26280 \, \text{cm}^{-1}$) than the pairing energy, so it is low-spin complex. Hydro complex is a high-spin, as chromium ion has enough energy to make its electrons unpaired.

The diagram for $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}$ relatively:

$$\begin{array}{c c c} \mathrm{Cr}^{2+}, & \mathrm{d}^{4} & \uparrow - \\ & & \downarrow \\ & \uparrow & \uparrow \\ & \text{high spin} & \text{low spin} \\ \end{array}$$

The magnetic moment is defined as:

$$\mu = \sqrt{n(n + 1)} (B.M.)$$

where $n$ is the number of unpaired electrons, B.M. is Bohr magneton.

For $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$:

$$\mu = \sqrt{4(4 + 1)} (B.M.) = \sqrt{20} (B.M.)$$

For $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}$:

$$\mu = \sqrt{2(2 + 1)} (B.M.) = \sqrt{6} (B.M.)$$

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