What is the effective atomic number of tetraaquadichlorochromium(II)?

Aug 24, 2017

$\text{EAN} = 34$

See here, here, and here for extra practice.

The effective atomic number is the number of electrons in the central transition metal, plus the number of contributed electrons by the ligands.

You can choose an electron-counting method, but I'll use the donor pair method, which assumes the ligands donate electron pairs and assigns the charges accordingly.

Thus, for the aqua and chloro ligands, we get:

$\text{H"_2"O} :$ $\to$ $2$ donated electrons, neutral

$\text{Cl} {:}^{-}$ $\to$ $2$ donated electrons, $- 1$ charge

In this method, we then take the central metal to have whatever oxidation state leaves it with the charge of the complex. As the aqua ligands are neutral, chromium then has a $+ 2$ oxidation state.

Its neutral electron configuration was

$\left[A r\right] 3 {d}^{5} 4 {s}^{1}$,

so its $+ 2$ configuration is

$\left[A r\right] 3 {d}^{4} 4 {s}^{0}$

and it thus has $22$ electrons to contribute. Overall, that means the effective atomic number is:

$\textcolor{b l u e}{\text{EAN}} = {\overbrace{22}}^{C {r}^{+ 2}} + {\overbrace{4 \times 2}}^{4 \times {H}_{2} O} + {\overbrace{2 \times 2}}^{2 \times C {l}^{-}} = \textcolor{b l u e}{34}$