
What is the oxidation number of carbon in ${C_3}{O_2}$ (Carbon suboxide) ?
(A) $\dfrac{{ + 4}}{3}$
(B) $\dfrac{{ + 10}}{4}$
(C) $ + 2$
(D) $\dfrac{{ + 2}}{3}$
Answer
581.1k+ views
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 –
$
O = C = C = C = O \\
\,\,\,\,\,\,\, + 2\,\,\,\,\,\,O\,\,\,\, + 2 \\
$
Average oxidation number of C –
$ = \dfrac{{sum\,of\,oxidation\,states\,of\,each\,C - atoms}}{{Total\,no.\,of\,C - atoms}}$
$ = \dfrac{{ + 2 + 0 + 2}}{3} = \dfrac{{ + 4}}{3}$
Hence, the correct answer is (A) $\dfrac{{ + 4}}{3}$
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 $O = - 2$
$O = 3 \cdot x + 2 \times ( - 2)$
$O = 3x + ( - 4)$
$3x = + 4$
$x = \dfrac{{ + 4}}{3}$
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 –
$
O = C = C = C = O \\
\,\,\,\,\,\,\, + 2\,\,\,\,\,\,O\,\,\,\, + 2 \\
$
Average oxidation number of C –
$ = \dfrac{{sum\,of\,oxidation\,states\,of\,each\,C - atoms}}{{Total\,no.\,of\,C - atoms}}$
$ = \dfrac{{ + 2 + 0 + 2}}{3} = \dfrac{{ + 4}}{3}$
Hence, the correct answer is (A) $\dfrac{{ + 4}}{3}$
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 $O = - 2$
$O = 3 \cdot x + 2 \times ( - 2)$
$O = 3x + ( - 4)$
$3x = + 4$
$x = \dfrac{{ + 4}}{3}$
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