Question

Which element in the square brackets has the highest oxidation number among the given options?

• A. $H_2\left[C\right]O_3$
• B. $\left[Mg\right]C{l}_2$
• C. $H_3\left[P\right]O_4$
• D. $K_2\left[C{r}_2\right]O_7$

Answer 1

The correct option is D

$K_2\left[C{r}_2\right]O_7$

Let's calculate the oxidation number for all the options given to us.

(1) For $H_2\left[C\right]O_3$, let oxidation number of C be $x$.

Oxidation numbers of H and O are +1 and -2 respectively
$+2+x+3\times (-2)=0$
Hence, the oxidation number of C = +4.

(2) For $\left[Mg\right]C{l}_2$, let oxidation number of Mg be $x$.

Oxidation number of Cl is -1.
$x+2\times (-1)=0$
Hence, the oxidation number of Mg = +2.

(3) For $H_3\left[P\right]O_4$, let oxidation number of P be $x$.

Oxidation numbers of H and O are +1 and -2 respectively.
$3\times (+1)+x+4\times (-2)=0$
Hence, the oxidation number of P = +5.

(4) For $K_2\left[C{r}_2\right]O_7$, let oxidation number of Cr be $x$.

Oxidation numbers of K and O are +1 and -2 respectively
$2\times (+1)+2\times x+7\times (-2)=0$
Hence, the oxidation number of Cr = +6.

So the highest oxidation is of Cr element among the given options which is equal to +6.