activity 4: find the oxidation state of each element: h2o po4^-3 mg kmno4

Answer

Explanation:

Step1: Determine oxidation state rules

Oxidation state of hydrogen is +1 (except in metal hydrides), oxygen is -2 (except in peroxides and super - oxides), and for a free element it is 0. For a poly - atomic ion, the sum of oxidation states equals the charge of the ion, and for a neutral compound it equals 0.

Step2: Oxidation state in $H_2O$

Let the oxidation state of oxygen be $x$ and hydrogen be $y$. In $H_2O$, since it is a neutral compound, $2y + x=0$. Given $y = + 1$, then $2\times(+1)+x = 0$, so $x=-2$. Oxidation state of $H$ is +1 and of $O$ is -2.

Step3: Oxidation state in $PO_4^{3 - }$

Let the oxidation state of $P$ be $a$ and of $O$ be $b=-2$. Since the charge of the ion is - 3, $a + 4b=-3$. Substituting $b=-2$ gives $a+4\times(-2)=-3$, so $a = + 5$. Oxidation state of $P$ is +5 and of $O$ is -2.

Step4: Oxidation state of $Mg$

$Mg$ is a free element. So the oxidation state of $Mg$ is 0.

Step5: Oxidation state in $KMnO_4$

Let the oxidation state of $K$ be $c = + 1$, $O$ be $d=-2$ and $Mn$ be $e$. Since it is a neutral compound, $c+e + 4d=0$. Substituting $c = + 1$ and $d=-2$ gives $1+e+4\times(-2)=0$, so $e = + 7$. Oxidation state of $K$ is +1, $Mn$ is +7 and $O$ is -2.

Answer:

In $H_2O$: $H$: +1, $O$: -2
In $PO_4^{3 - }$: $P$: +5, $O$: -2
In $Mg$: $Mg$: 0
In $KMnO_4$: $K$: +1, $Mn$: +7, $O$: -2