Question

The oxidation number of underlined element: $Ca$ in \[Ca{(Cl{O_2})_2}\] the compound is:
A. \[ + 3\]
B. \[ + 5\]
C. \[ - 2\]
D. \[ + 2\]

Answer 1

Hint: To answer this question, you should recall the concept of oxidation numbers. The oxidation state of an atom is defined as the number of electrons lost and, therefore, describes the extent of oxidation of the atom. For example, the oxidation state of carbon in\[{\text{C}}{{\text{O}}_{\text{2}}}\] would be \[ + 4\] since the hypothetical charge held by the carbon atom if both of the carbon-oxygen double bonds were completely ionic would be equal to \[ + 4\].

Complete step-by-step answer:
The most frequent terms Oxidation state and oxidation number are terms frequently used interchangeably. They are defined and described as the number of electrons lost in an atom. The values can be zero, positive, or negative. The formula of calcium hypochlorite is \[Ca{\left( {Cl{O_2}} \right)_2}\] Let \[x\] be the oxidation number of $Ca$ in \[Ca{(Cl{O_2})_2}\]. We know that the oxidation number of oxygen = \[ - 2\] but for chlorine the oxidation no. is unknown. So, first, we need to find out the oxidation number of chlorine. The structure of this compound can be drawn as:

From the structure, we can see that $Ca$ is attached with \[2Cl{O_2} - \] ions.
Let the oxidation number of chlorine be $y$.
This means that \[y + 2 \times ( - 2) = - 1\].
Solving this equation we will get
\[ \Rightarrow y = + 3\].
Now using this value to calculate the oxidation number of calcium:
\[ \Rightarrow x + 2( + 3 + 2( - 2)) = 0\].
We will get \[x = + 2\].
Thus, the oxidation number of $Ca$ in \[Ca{\left( {Cl{O_2}} \right)_2}\]​ is \[ + 2\].

Hence, option D is correct.

Note: In most of the compounds, the oxidation number of oxygens is \[ - 2\]. There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of \[ - 1\]. Example, \[N{a_2}{O_2}\]
Superoxide- Every oxygen atom is allocated an oxidation number of \[ - \dfrac{{{\text{ }}1}}{2}\] Example, \[K{O_2}\]
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of 1.