Question #608d1

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Answer 1 · 2017-11-17T10:07:38

The oxidation number of $C r$ $Cr$ is $+ 6$ $+6$

We will rewrite the formula as an equation.

$2 K + 2 C r + 7 O = 0$ $2K + 2Cr + 7O = 0$

The zero comes from there being no overall charge on the compound.

Each oxidation number is represented by the element's symbol.

Since this compound is not a peroxide or superoxide, the oxidation value for $O$ $O$ is $- 2$ $-2$ .

$K$ $K$ being in group 1 has a $+ 1$ $+1$ charge and therefore has an oxidation number of $+ 1$ $+1$

Now, by putting known values in we have:
$2 (+ 1) + 2 C r + 7 (- 2) = 0$ $2(+1) + 2Cr + 7(-2)=0$

$2 + 2 C r - 14 = 0$ $2+2Cr-14=0$

$2 C r = 0 + 14 - 2 = + 12$ $2Cr=0+14-2=+12$

$C r = + \frac{12}{2} = + 6$ $Cr=+12/2=+6$

To make it clear, we must keep the positive sign even if we know it is positive.

Answer 2 · 2017-11-17T10:07:40

$+ 6$ $+6$ .

$K$ $"K"$ has an oxidation number of $+ 1$ $+1$ . There are two $K$ $"K"$ atoms, thus $+ 2$ $+2$ oxidation number for both $K$ $"K"$ .

$O$ $"O"$ has an oxidation number of $- 2$ $-2$ . There are seven $O$ $"O"$ atoms, thus $- 14$ $-14$ oxidation number for all $O$ $"O"$ atoms.

Overall, the oxidation numbers have to equal 0 as it is an uncharted molecule.

Thus

$- 14 + 2 = - 12$ $-14 + 2 = -12$

So the $Cr$ $"Cr"$ atoms have to have a combined oxidation number equaling $+ 12$ $+12$ to cancel out the $- 12$ $-12$ number we just calculated.

There are $2$ $2$ $Cr$ $"Cr"$ atoms present, and so

$\frac{+ 12}{2} = + 6$ $(+12) / 2 = +6$