Question

What is the oxidation number of carbon in $C_3{O}_2$ (Carbon suboxide) ?
(A) ${\displaystyle \frac{+4}{3}}$
(B) ${\displaystyle \frac{+10}{4}}$
(C) $+2$
(D) ${\displaystyle \frac{+2}{3}}$

Answer 1

Hint: Oxidation number assigned to an element in chemical combination which represents the number of electrons lost or gained.
Some oxidation states are -
(i) Oxidation number of all alkali metal ions is always $+1$
(ii) Oxidation no. of all alkaline earth metal ion is always $+2$
(iii) Oxidation no. of oxygen in Oxide $(O^{2-})$ is $-2$. In Peroxide $(O-O^{2-})$ is $-1$
(iv) Oxidation no. of hydrogen in Proton $(H^{+})$ is $+1$. In Hydride (with metal) is $-1$

Complete step-by- step answer :
The structure of $C_3{O}_2$ is shown below.
$O=C=C=C=O$
$O\to$Oxygen
$C\to$Carbon
$=\to$Double bond
In the structure, there are three carbon atoms which are SP hybridised and have the same electronegativity. So, the central carbon atom has a zero oxidation state.
Oxygen is more electronegative atom than carbon atom. In carbon-oxygen bonds, oxygen atoms attract the shared pair of electrons.
The oxidation state of oxygen is $-2$.
Hence, the oxidation state of a carbon atom linked with oxygen atoms is $+2$.
Then, $O=C=C=C=O$
Assign the oxidation state of each carbon atom in the molecule –
$$\begin{gathered} O=C=C=C=O \\ +2O+2 \end{gathered}$$
Average oxidation number of C –
$={\displaystyle \frac{sumofoxidationstatesofeachC-atoms}{Totalno.ofC-atoms}}$
$={\displaystyle \frac{+2+0+2}{3}}={\displaystyle \frac{+4}{3}}$

Hence, the correct answer is (A) ${\displaystyle \frac{+4}{3}}$

Note:There is a short method for calculating the oxidation no. $C_3{O}_2$ is neutral. So, net charge $=O$.
Oxidation state of $C_3{O}_2=$ No. of C atom X oxidation state of C-atom + No of. O-atom $\times$ Oxidation state of O-atom.
Oxidation state of $C_3{O}_2=O$
No. of C-atom $=3$
Oxidation state of $C=x$
No. of O-atom $=2$
Oxidation state of $O=-2$
$O=3\cdot x+2\times (-2)$
$O=3x+(-4)$
$3x=+4$
$x={\displaystyle \frac{+4}{3}}$