Question

What is the oxidation number of chromium is NA2CR2O7?

Answer 1

Answer:

Given:

Given Compound:

$\sf Na_2Cr_2O_7$

Name of the Compound:

Sodium Dichromate

To Find:

Oxidation Number of Chromium (Cr) in the given Compound

Solution:

We know that,

Oxidation state of,

Na = $\sf +1$

O = $\sf -2$

We can clearly see that,

There are,

2 Na, 2 Cr and 7 Oxygen atoms in the given Compound,

Summing up the of oxidation states of known elements be used to find the oxidation state of the unknown element

Here We know

Na = $\sf +1$

O = $\sf -2$

And Let us take Oxidation Sate of Chromium (Cr) as $\sf x$

Hence,

Substituting these values in the Given Compound

We get,

$\sf \implies 2(+1) + 2(x) + 7(-2) = 0$

$\sf \implies 2 + 2x -14 = 0$

$\sf \implies 2x -12=0$

$\sf \implies 2x=+12$

$\sf \implies x=+6$

$\sf \boxed{x=+6}$

Hence,

The Oxidation Number of Chromium (Cr) in the given compound is

Cr = +6