Question

Question: A reference book lists the chemical formula of one form of vermiculite as$\left[\left({\mathrm{Mg}}_{2.36}{\mathrm{Fe}}_{0.48}{\mathrm{Al}}_{0.16}\right)\left({\mathrm{Si}}_{2.72}{\mathrm{Al}}_{1.28}\right){\mathrm{O}}_{10}{\left(\mathrm{OH}\right)}_2\right]\left[{\mathrm{Mg}}_{0.32}{\left({\mathrm{H}}_2\mathrm{O}\right)}_{4.32}\right]$. Determine the oxidation state of the iron in this mineral.

Short answer

The oxidation state of Fe is +3.

Step-by-step solution

Step 1:

The removal of hydrogen or the addition of oxygen is known as Oxidation.

Step 2:

The oxidation state of Mg is +2, Al is +3, Si is +4, O is -2, OH is -1, ${H_2}O$ is 0

To find the oxidation state of Fe we have to calculate with all the above molecules as,

$$\begin{gathered} \left(2.36\times 20\right)+\left(0.48\times x\right)+\left(3\times 0.16\right)+\left(4\times 2.72\right)+\left(3\times 1.28\right)+\left(10\times -2\right)+\left(-1\times 2\right)+\left(0.32\times 2\right)+\left(4.32\times 0\right)\backslash \mathrm{hfill} \\ =+3 \end{gathered}$$

Hence, the oxidation state of Fe is +3.