what is the oxidation state of In in In2s3?

Oh, In2s3? That sounds like a sneeze-related compound! Ah-choo! Anyway, back to your question. In the compound In2s3, the oxidation state of indium (In) is +3. So, in terms of oxidation states, it's In it to win it with a +3 charge!

To determine the oxidation state of In in In2S3, we need to assign oxidation states to sulfur first.

Sulfur is usually assigned an oxidation state of -2 in compounds. Since there are three sulfur atoms in In2S3, the total oxidation state due to sulfur is -6.

Now, let "x" be the oxidation state of Indium (In). Since the compound is neutral (no net charge), the sum of the oxidation states must be zero.

The sum of the oxidation states of In and S should give a total of zero:
2x + (-6) = 0

Simplifying the equation:
2x - 6 = 0

Adding 6 to both sides:
2x = 6

Dividing both sides by 2:
x = 3

The oxidation state of In in In2S3 is +3.

To determine the oxidation state of In (indium) in In2S3 (indium sulfide), you need to know the general rule of assigning oxidation states:

1. The oxidation state of an element in its uncombined or elemental form is always zero. So, the oxidation state of In is zero in most cases.

2. The sum of the oxidation states of all the atoms in a compound is equal to the net charge of the compound. In the case of In2S3, the net charge is zero since it is a neutral compound.

3. The sum of the oxidation states of all the atoms in a polyatomic ion is equal to the net charge of the ion. However, In2S3 is not an ion but a compound.

In2S3 contains two indium atoms (In) and three sulfur atoms (S). So, let's assume the oxidation state of indium is "x" for each indium atom.

Then, using the rule of zero net charge, we can set up the equation:

2x + 3(-2) = 0

Simplifying the equation:

2x - 6 = 0

Adding 6 to both sides:

2x = 6

Dividing both sides by 2:

x = 3

Therefore, the oxidation state of In in In2S3 is +3.

S is -2, makes Ln+3