Question

What is the oxidation state of Phosphorous in $\mathrm{Ba}{\left({\mathrm{H}}_2{\mathrm{PO}}_2\right)}_2$?

Answer 1

1. The oxidation state or oxidation number is defined as the total number of electrons lost or gained through an atom, ion, or molecule to achieve stability.
2. Element, compound, or atom show variable oxidation states.
3. Phosphorous is a chemical element with the atomic number 15.

Step 1: Given the oxidation state

• The oxidation state or oxidation number of Hydrogen is +1, as losing 1 electron will stabilize the Hydrogen atom.
• The oxidation number of Oxygen is -2 because the gain in the two-electron will stabilize the Oxygen atom.
• The $\mathrm{Ba}{\left({\mathrm{H}}_2{\mathrm{PO}}_2\right)}_2$ can also be written as ${\mathrm{Ba}}^{2+}{{\left({\mathrm{H}}_2{\mathrm{PO}}_2\right)}_2}^{-1}$this implies that the whole molecule has a -1 charge.

Step 2: Calculating the oxidation number

• Let the oxidation number of Phosphorous be x.
• The oxidation number can be calculated as follows:
• $\left(\mathrm{No}.\mathrm{of}\mathrm{H}\mathrm{atoms}\right)\left(\mathrm{Oxidation}\mathrm{No}.\mathrm{of}\mathrm{H}\right)+\left(\mathrm{No}.\mathrm{of}\mathrm{P}\mathrm{atoms}\right)\left(\mathrm{Oxidation}\mathrm{No}.\mathrm{of}\mathrm{P}\right)+\left(\mathrm{No}.\mathrm{of}\mathrm{O}\mathrm{atoms}\right)\left(\mathrm{Oxidation}\mathrm{No}.\mathrm{of}\mathrm{O}\right)=0$
• $$\begin{gathered} \left(2\right)\left(1\right)+\left(1\right)\left(\mathrm{x}\right)+\left(-2\right)\left(2\right)=-1 \\ 2+\mathrm{x}-4=-1 \\ \mathrm{x}-2=-1 \end{gathered}$$
• $\mathrm{x}=1$
• Therefore, the oxidation state of Phosphorous is 1.