Question

What is the oxidation state of Cr in (NH4)2Cr2O7 ?

Answer 1

Answer:

The oxidation state of Cr in (NH₄)₂Cr₂O₇(Ammonium dichromate) is +6.

Explanation:

• An atom's oxidation state (also known as its oxidation number) in a chemical compound gives information on how many electrons were lost from it and, as a result, describes the degree of oxidation of the atom.
• The imagined charge that an atom would carry if all of its links to other atoms were entirely ionic in nature can be used to determine an atom's oxidation state.

The compound given to us is (NH₄)₂Cr₂O₇(Ammonium dichromate).

The oxidation state of

Let the oxidation of "Cr" be "x".

$\mathrm{2+2x+7(-2)=0}$

$\mathrm{2+2x-14=0}$

$\mathrm{2x-12=0}$

$\mathrm{2x=12}$

$\mathrm{x=6}$

So, the oxidation state of Cr in (NH₄)₂Cr₂O₇(Ammonium dichromate) is +6.

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Answer 2

At (NH₄)₂Cr₂O₇(Ammonium dichromate), Cr is in the +6 oxidation state.

Explanation:

The oxidation state of an atom, also referred to as its oxidation number, in a chemical compound indicates how many electrons were lost from it and hence describes the level of oxidation of the atom.

It is possible to calculate the oxidation state of an atom by imagining the charge that it would have if all of its bonds to other atoms were totally ionic.

The substance we have been given is (NH₄)₂Cr₂O₇ (Ammonium dichromate).

oxidation level of

Let the oxidation of "Cr" be "x".

$2+2x+7(-2)=0$

$2+2x-14=0$

$2x-12=0$

$x=6$

In (NH₄)₂Cr₂O₇ (ammonium dichromate), Cr is therefore in the +6 oxidation state.

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