Question

Calculate the oxidation number of \[C\] in the following:\[{C_2}{H_6}\],\[{C_4}{H_{10}}\],\[CO\],\[C{O_2}\] and \[HCO_3^ - \].

Answer 1

Hint: The oxidation number is defined as the count of electrons in which an atom present in a molecule can share with the other atom to form a chemical bond between them. The atomic number of carbon is 6. Thus, its electronic configuration will be\[1{s^2}2{s^2}2{p^2}\]. The covalency of carbon is 4.
Complete step by step answer:
\[{C_2}{H_6}\]
(i) In ethane two carbon atoms and six hydrogen atoms are present.
(ii) Each hydrogen atom contains \[ + 1\] charge. As six hydrogen atoms are present the overall charge on hydrogen will be 6.
(iii) Take carbon atom as x. As two carbon atoms are present, it will be 2x.
(iv) Add the total oxidation state of the atoms and equate with zero.
$2x + 1(6) = 0$
$ \Rightarrow 2x = - 6$
$ \Rightarrow x = - 3$
Thus, the oxidation number of carbon in \[{C_2}{H_6}\]is\[ - 3\].
\[{C_4}{H_{10}}\]
(i) In butane four carbon atoms and ten hydrogen atoms are present.
(ii) Each hydrogen atom contains \[ + 1\] charge. As ten hydrogen atoms are present the overall charge on hydrogen will be 10.
(iii) Take carbon atom as x. As four carbon atoms are present, it will be 4x.
(iv) Add the total oxidation state of the atoms and equate with zero.
$4x + 1(10) = 0$
$ \Rightarrow 4x = - 10$
$ \Rightarrow x = \dfrac{{10}}{4}$
Thus, the oxidation number of carbon in \[{C_4}{H_{10}}\] is \[\dfrac{{10}}{4}\].
\[CO\]
(i) Carbon monoxide is a neutral oxide. Therefore, the overall oxidation state will be zero.
(ii) The charge on the oxygen atom is \[ - 2\].
(iii) Take carbon atom as x.
(iv) Add the total oxidation state of the atom and equate.
$x + ( - 2) = 0$
$ \Rightarrow x = + 2$
Thus, the oxidation number of carbon in \[CO\] is \[ + 2\].
\[C{O_2}\]
(i) In carbon dioxide one carbon atom and two oxygen atoms is present.
(ii) The charge on the oxygen atom is \[ - 2\].
(iii) Take carbon atom as x.
(iv) Add the total oxidation state of the atom and equate.
$x + 2( - 2) = 0$
$ \Rightarrow x - 4 = 0$
$ \Rightarrow x = + 4$
Thus, the oxidation number of carbon in \[C{O_2}\] is \[ + 4\].
\[HCO_3^ - \]
(i) The charge on the bicarbonate ion is \[ - 1\].
(ii) The charge of a hydrogen atom is \[ + 1\].
(iii) The charge of an oxygen atom is \[ - 2\].
(iv)Take a carbon atom as x.
(v) Add the total oxidation state of the atom and equate with the overall charge.
$(1) + x + 3( - 2) = - 1$
$ \Rightarrow (1) + x - 6 = - 1$
$ \Rightarrow x = + 4$
Thus, the oxidation number of carbon in \[HCO_3^ - \] is \[ + 4\].

Note:
Oxidation number is also stated as an oxidation state. The electronegative atom with a higher electronegativity value shows a negative oxidation number while the electropositive atom with a higher electropositive value shows a positive oxidation number. The net charge shown by the neutral atom is zero.