Chemistry, asked by Koyyadadhanik29, 1 year ago

Oxidation state of K2[Cr(CN)2(O)2(NH3])

Answers

Answered by Anonymous

Oxidation state of Cr in given compound is.....
(1×2)+x+(-1×2)+(-2×2)+(0×2)=0
2+x-2-4+0=0
x=4
so the oxidation state of Cr in given compound is 4

Hope it helps you

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Answered by skyfall63

The oxidation state is +4

Explanation:

The oxidation state of $\mathrm{K}_{2}\left[\mathrm{Cr}(\mathrm{CN})_{2} \mathrm{O}_{2}\left(\mathrm{NH}_{3}\right)\right]$ :

The breakdown of charges of the various elements is:

K = +1 × 2 = +2

CN = -1 × 2 = -2

O = -2 × 2 = -4

NH_3 = 0 because it is a neutral ligand

We have to calculate the oxidation state of Cr. Let it be x because Cr has positive ions.

Since the compound formed is a neutral one, so the total charges = 0

Now equating them we get,

(+2) + x + (-2) + (-4) + 0 = 0

x - 4 = 0

x = 4

Chromium has a charge of +4

From this we can say that the compound is electrically neutral.

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