Question

Find out the oxidation number of chlorine in the following compounds.
$NaCl{{O}_{4}},NaCl{{O}_{3}},NaClO,KCl{{O}_{2}},C{{l}_{2}}{{O}_{7}},Cl{{O}_{3}},C{{l}_{2}}O,NaCl,C{{l}_{2}},Cl{{O}_{2}}$

Answer 1

The oxidation number is defined as the charge on the atom which appears to have on forming ionic bonds with other heteroatoms. The atoms which have high electronegativity are considered to have a negative oxidation state. The net charge which is present on the neutral atom or molecules is considered to be zero. So their overall oxidation state is zero.

Complete solution:
The oxidation state of chlorine can be determined from the following compounds by summing the oxidation state of the other atom or molecules which are bonded to it and taking the unknown oxidation of chlorine as x. so now we will find the oxidation state of chlorine in $NaCl{{O}_{4}}$. The oxidation number of Na is +1, for oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms:

& 1+x+4(-2)=0 \\\ & x-7=0 \\\ & x=7 \\\ \end{aligned}$$ So the oxidation state of chlorine in $$NaCl{{O}_{4}}$$is =7. so now we will find the oxidation state of chlorine in $$NaCl{{O}_{3}}$$. The oxidation number of Na is +1, for oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & 1+x+3(-2)=0 \\\ & x-5=0 \\\ & x=5 \\\ \end{aligned}$$ So the oxidation state of chlorine in$$NaCl{{O}_{3}}$$ is +5. So now we will find the oxidation state of chlorine in $$NaClO$$. The oxidation number of Na is +1, for oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & 1+x+-2=0 \\\ & x-1=0 \\\ & x=1 \\\ \end{aligned}$$ So the oxidation state of chlorine in$$NaClO$$ is +1. So now we will find the oxidation state of chlorine in $$KCl{{O}_{2}}$$. The oxidation number of K is +1, for oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & 1+x+2(-2)=0 \\\ & x-3=0 \\\ & x=3 \\\ \end{aligned}$$ So the oxidation state of chlorine in $$KCl{{O}_{2}}$$ is =3. So now we will find the oxidation state of chlorine in $$C{{l}_{2}}{{O}_{7}}$$. The oxidation number of oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & 2x+7(-2)=0 \\\ & 2x=14 \\\ & x=7 \\\ \end{aligned}$$ So the oxidation state of chlorine in $$C{{l}_{2}}{{O}_{7}}$$is +7. so now we will find the oxidation state of chlorine in $$Cl{{O}_{3}}$$. The oxidation number of oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & x+3(-2)=0 \\\ & x-6=0 \\\ & x=6 \\\ \end{aligned}$$ So the oxidation number of chlorine in$$Cl{{O}_{3}}$$ is +6. so now we will find the oxidation state of chlorine in $$C{{l}_{2}}O$$. The oxidation number of oxygen is -2 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & 2x+-2=0 \\\ & 2x=2 \\\ & x=1 \\\ \end{aligned}$$ So the oxidation number of chlorine in $$C{{l}_{2}}O$$ is +1. So now we will find the oxidation state of chlorine in NaCl. The oxidation number of sodium is +1 and for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & x+1=0 \\\ & x=-1 \\\ \end{aligned}$$ So the oxidation number of chlorine in NaCl is -1. so now we will find the oxidation state of chlorine in $$C{{l}_{2}}$$ . The oxidation number for chlorine we will assume ‘x’. so now adding the oxidation number of all the atoms: $$\begin{aligned} & 2x=0 \\\ & x=0 \\\ \end{aligned}$$ So the oxidation number of chlorine in $$C{{l}_{2}}$$ is 0 so now we will find the oxidation state of chlorine in $$Cl{{O}_{2}}$$ . The oxidation number for chlorine we will assume ‘x’ and for oxygen it is -2. so now adding the oxidation number of all the atoms: $$\begin{aligned} & x+2(-2)=0 \\\ & x-4=0 \\\ & x=4 \\\ \end{aligned}$$ So the oxidation state of chlorine in $$Cl{{O}_{2}}$$is +4. **Note:** if the charge on the atom is high so the ionization energy of the atom will be high because it will be difficult to remove the electron from it. If the negative charge on the atom is high then it will be difficult to add the electron. Oxidation state larger than three is practically impossible.