Question

Two oxidation states for chlorine are found in the compound:
(A) $CaOC{{l}_{2}}$
(B) KCl
(C) $KCl{{O}_{3}}$
(D) $C{{l}_{2}}{{O}_{7}}$

Answer 1

Oxidation number is also known as an oxidation state. It is defined as the total number of electrons that an atom either gains or loses, in order to form chemical bonds with another atom.

Complete step by step answer:
We will check the oxidation states in all the given molecules in the question.
i) The molecule $CaOC{{l}_{2}}$ can also be written as Ca(OCl)Cl which contains two chlorine atoms with different oxidation states in the $C{{l}^{-1}}$ ,it is -1 Oxidation State while in $OC{{l}^{-1}}$ ion is in +1 oxidation state. This is proved with the molecular structure of the compound given below.
$\text{C}{{\text{l}}^{-}}\text{ C}{{\text{a}}^{2+}}\text{ }{{\text{ }}^{-}}\text{OCl}$
Now, let’s find the oxidation states of the chlorine atom in the given ions.
Oxidation state of O in $OC{{l}^{-}}$ is -2. So, we can write that
$\begin{aligned} & -2\text{ + X = }-1 \\\ & \text{X = +1} \\\ \end{aligned}$
For $C{{l}^{-}}$ ion, we know that in $CaOC{{l}_{2}}$ , Ca is in +2 oxidation state and $OC{{l}^{-}}$ has a formal charge of -1. So, we can write that
$+2+Y+1=0$
$Y=-1$
Hence, we can say that the oxidation state of chlorine in $C{{l}^{-}}$ is -1.

ii) Oxidation state of Cl in $\text{KCl}$ can be determined as -1 for Potassium has an Oxidation state of +1. This is demonstrated by the following diagram of $\text{KCl}$ compound and with equations proving the oxidation state of chlorine to be -1.
$K-Cl$

& \text{For KCl,} \\\ & +1\text{ + Y = 0 (stable molecule)} \\\ & \text{K Cl } \\\ & \text{Y = }-1 \\\ & \text{ Oxidation number of Cl = }-1 \\\ \end{aligned}$$ Being an ionic compound, Potassium’s oxidation state is always +1, so in the compound $KCl{{O}_{3}}$, potassium has an oxidation state of +1 while chlorate that is $$\text{Cl}{{\text{O}}_{\text{3}}}^{-}$$ has an oxidation state of -1. We know that the oxidation state of oxygen is -2 and 3 oxygen atoms being present in the compound, the total oxidation number of -6 is achieved by oxygen. The following is the molecular structure of the compound inclusive of equations prove the fact that chlorine has an Oxidation state of +5 in $$\text{KCl}{{\text{O}}_{\text{3}}}$$.  So, for $KCl{{O}_{3}}$, we can write that $$\begin{aligned} & +1\text{ + Y + 3(}-2\text{) = 0 (stable molecule)} \\\ & \text{K Cl }{{\text{O}}_{3}} \\\ & \text{Y = +}5 \\\ \end{aligned}$$ So, Y= Oxidation state of Cl = +5 iv) In the compound $$\text{C}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{7}}}$$, which is also known as chlorine heptoxide, we know the oxidation number of oxygen is -2 and hence the total oxidation number of oxygen in this compound is -14 which is denoted in the equation given below and thus we can equate the oxidation number of Chlorine. Chlorine having two atoms in the compound has an oxidation number of +7. The molecular structure and the equations proving the oxidation number of Chlorine is given as follows  $$\begin{aligned} & \text{For C}{{\text{l}}_{2}}{{\text{O}}_{7}}\text{,} \\\ & \text{2Y + 7(}-2\text{) = 0 (stable molecule)} \\\ & \text{C}{{\text{l}}_{2}}\text{ }{{\text{O}}_{7}} \\\ & \Rightarrow \text{ 2Y = 14} \\\ & \Rightarrow \text{ Y = 7} \\\ & \text{Y = +7} \\\ & \text{ Oxidation number of Cl = +7} \\\ \end{aligned}$$ Hence we can see that the compound $$\text{CaOC}{{\text{l}}_{\text{2}}}$$ is the only compound given in the options which has two oxidation states of chlorine. **So, the correct answer is “Option A”.** **Note:** We should know that the oxidation number of a monatomic ion is equal to the charge of the ion. This can be shown with an example. Like the oxidation number of H is +1. But when it recombines with less electronegative elements, the oxidation state becomes -1. Similarly, the oxidation number of oxygen is usually -2, but it becomes -1, in case of the peroxides.