Match the following compounds with the oxidation number of the central atom (or specific atom in question):
Compound Oxidation number
$1$. $C_6H_{12}O_6$ (Carbon) $a$. $+1$
$2$. $CCl_4$ (Carbon) $b$. $0$
$3$. $NH_4Cl$ (Nitrogen) $c$. $+4$
$4$. $Ba(H_2PO_2)_2$ (Phosphorus) $d$. $-3$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(B) $1$. In $C_6H_{12}O_6$,the average oxidation state of $C$ is $6x + 12(+1) + 6(-2) = 0 \implies 6x = 0 \implies x = 0$.
$2$. In $CCl_4$,$x + 4(-1) = 0 \implies x = +4$.
$3$. In $NH_4Cl$,$x + 4(+1) + (-1) = 0 \implies x + 3 = 0 \implies x = -3$.
$4$. In $Ba(H_2PO_2)_2$,$Ba$ is $+2$. For $(H_2PO_2)^-$,$2(+1) + x + 2(-2) = -1 \implies x - 2 = -1 \implies x = +1$.
Thus,the correct match is $1-b, 2-c, 3-d, 4-a$.

Explore More

Similar Questions

The oxidation state of $Cr$ in $K_{2}Cr_{2}O_{7}$ is

Oxidation number of $N$ in $(NH_4)_2SO_4$ is

The oxidation number and covalency of sulphur in the sulphur molecule $S_8$ are respectively

$X + Y \rightarrow$ oleum. The sum of oxidation states of the central atom in $X$ and $Y$ is:

What is the oxidation number of $Cl$ in bleaching powder?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo