Question

Among the following molecules in which chlorine shows maximum oxidation state?
(A) $C{l_2}$
(B) $KCl$
(C) $KCl{O_3}$
(D) $C{l_2}{O_7}$

Answer 1

Hint: Oxidation state or also called as oxidation number is a hypothetical number that represents the loss or gain of electrons. The oxidation state of an element, homonuclear diatomic and triatomic compounds is zero. Homonuclear means the compound is formed from the same element. Whereas heteronuclear means compound has different elements.

Complete Step by Step Solution:
to find the oxidation state, follow the following steps
Assume the oxidation to be x whose oxidation state is to be determined.
Form an equation using the known oxidation state of other elements.
If the compound is neutral then equate the oxidation state to zero. If the compound is charged then equate it equal to the charge including the sign of the charge, not only the magnitude of charge.
Oxidation state of some elements must be kept in mind.
For example, alkali metals have \[ + 1\]oxidation state. Alkaline earth metal has \[ + 2\]oxidation state. Oxygen’s most common oxidation state is\[ - 2\]. In peroxide form its oxidation state changes to\[ - 1\].
Group \[17\] has \[ - 1\]oxidation state generally.
Hydrogen shows both \[ + 1\]and \[ - 1\] oxidation state depending upon the electronegativity of the atom attached to it.
In the question we need to determine the oxidation state in various compounds of chlorine. Thus, this signifies that although chlorine belongs to group \[17\]that has \[ - 1\]as its oxidation state but it is not always that \[ - 1\]oxidation state is observed.
In$C{l_2}$, the oxidation state of$Cl$is zero. It is a diatomic molecule with both the elements as chlorine. flowing the above procedure, $2x = 0$thus, $x = 0$
In$KCl$, k has \[ + 1\]oxidation state. Therefore, $Cl$ has \[ - 1\]oxidation state.
Finding oxidation state of$KCl{O_3}$. Oxidation of k is \[ + 1\]and o is \[ - 2\]
Thus, $1 + x + (3)( - 2) = 0$
$x = 5$
Oxidation state of $C{l_2}{O_7}$ can be calculated as
$2x + 7( - 2) = 0$
$x = 7$
Thus, highest oxidation among the following compound is\[ + 7\]. The compound is $C{l_2}{O_7}$
The correct option is D.

Note: It is possible that the oxidation state comes out to be fractional. This does not mean that the electrons are gained or lost in fractions. But it means it is the average value of the oxidation state of the element. In such a case, there must be more than one element present.