Question

The oxidation number of platinum in $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$ is:
a.) 2
b.) 3
c.) 4
d.) 5

Answer 1

Hint: The oxidation number of $N{H_3}$ is $0$ , the oxidation number of $Cl$ is -1, the oxidation number of $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$ is $0$ , and let the oxidation number of platinum be $x$ . Then calculate the oxidation number of platinum by using the equation - Oxidation number of platinum + (Oxidation number of $N{H_3} \times 5$ ) + Oxidation number of $Cl \times 4 = $ Oxidation number of $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$ .

Complete step by step answer:
The oxidation number of an atom is defined as the number of electrons that an atom shares (gains or losses) with other atoms to form a molecule. So, in simple words oxidation number is used to describe the transfer of an electron between atoms in a compound.
Take the example of $HF$ . In this compound the oxidation state of ${H^ + }$ is +1 and the oxidation state of ${F^ - }$ is -1.
You can also notice very peculiar information from the example stated above, i.e. the oxidation number of $HF$ is zero, meaning the oxidation number of the compound will be equal to the net charge on the compound.
In the problem given to us, we have to calculate the oxidation number of platinum in $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$ .
The oxidation number of $N{H_3}$ be $0$
The oxidation number of $Cl$ be -1
The oxidation number of $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$ is $0$
Let the oxidation number of platinum be $x$
So, for the compound $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$
The oxidation number of platinum + (Oxidation number of $N{H_3} \times 5$ ) + Oxidation number of $Cl \times 4 = $ Oxidation number of $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$
So, $x + \left( {0 \times 5} \right) + \left( { - 1 \times 4} \right) = 0$
$x - 4 = 0$
$x = 4$
So, the oxidation number of $Pt$ in $\left[ {Pt{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_3}$ is $ + 4$
So, the correct answer is “Option C”.

Note: We discussed the oxidation number in the solution above, and a very important fact about the oxidation number of an element is that the same element can have two different oxidation states in different compounds. In the entire periodic table fluorine is the only element that shows a single oxidation state in all of its compounds, i.e. -1.