Question

What are the oxidation states of the elements in the following compounds?
$Cu{{(N{{O}_{3}})}_{2}}$; $KN{{O}_{3}}$

Answer 1

Oxidation state of any compound is its ability to gain or lose electrons. It is represented as a charge on an atom, maybe positive or negative. The oxidation state of any atom is dependent on the other atoms that are bonded in a molecule. Electronegativity also affects the oxidation state of any atom, as it is the ability to attract a shared pair of electrons.

Complete answer:
The oxidation number or oxidation state of any atom is the ability of that atom to donate or gain electrons. The oxidation state is the charge on an atom when it is in its ionic form, or when in a molecule. The charge is negative when an atom has the tendency to take electrons (or when it is more electronegative), while the charge is positive when an atom has an ability to donate electrons.
As the oxidation state is calculated in the form of ions, so the oxidation state of any neutral molecule is zero. So, by adding the respective oxidation numbers which is equal to 0, we can take out the oxidation number of any element to be positive or negative.
So, finding the oxidation number of elements in,
-$Cu{{(N{{O}_{3}})}_{2}}$ is copper (II) nitrate, the ions in the compound are $C{{u}^{2+}}$ and $N{{O}_{3}}^{-}$, as copper is present in +2 state, so its oxidation state is +2. For nitrate ion, as we know, the oxidation state of oxygen in various compounds is -2, therefore the oxidation state for nitrogen is,
N + 3(-2) = -1
N – 6 = -1
N = -1 + 6
N = +5
Therefore, the oxidation state of Cu is +2, O is -2 and N is +5.
-$KN{{O}_{3}}$ is potassium nitrate, the ions are ${{K}^{+}}$ and $N{{O}_{3}}^{-}$, as potassium is an alkali metal it has an oxidation state of +1. While we have the oxidation states of N and O as +5 and -2 in $N{{O}_{3}}^{-}$.
Hence, the oxidation states of the elements are denoted as, $\overset{+2}{\mathop{Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}$and$\overset{+1}{\mathop{K}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}$.

Note:
Ionic compounds are formed by transfer of electrons, by neutralizing the charges, therefore for ions of one $C{{u}^{2+}}$, 2 ions of $N{{O}_{3}}^{-}$ are required as $N{{O}_{3}}^{-}$ has -1 charge and Cu has +2. For $KN{{O}_{3}}$ +1 charge of potassium ion is neutralized by only one -1 $N{{O}_{3}}^{-}$ ion. Transition metal compounds have their valences written in Roman numeral as they have variable oxidation states.