how to calculate the oxidation state for 'cr' in K2CRO4 ?
You need to learn how to use the caps key. That's K2CrO4.
All compounds must add to zero with oxidation numbers.
K + +1
O = -2
(2*+1) + (Cr) + 4(-2) = 0
2+Cr-8 = 0
Cr-6 = 0
Cr = 6
To calculate the oxidation state of an element in a compound, you need to consider the known oxidation states of the other elements in the compound.
In the compound K2CrO4, we can start by assigning the oxidation state of oxygen (O) as -2. Since there are four oxygen atoms in K2CrO4, the total oxidation state contributed by oxygen is -8.
Next, we know that the oxidation state of potassium (K) is always +1 in compounds.
Now, let's assign the oxidation state of chromium (Cr) as 'x'.
Using the principle that the sum of the oxidation states of all the elements in a compound equals the overall charge of the compound, we can set up the equation:
2(K) + (x) + 4(-2) = 0
Simplifying this equation, we get:
2 + x - 8 = 0
Combining like terms, we have:
x = 6
Therefore, the oxidation state of chromium (Cr) in K2CrO4 is +6.
To calculate the oxidation state of an element in a compound, we need to assign the oxidation states for the other elements and use their charges to determine the oxidation state of the element in question.
In the compound K2CrO4, oxygen is generally assigned an oxidation state of -2. Since there are four oxygen atoms, the total charge from oxygen would be -8 (4 × -2).
Potassium (K) is an alkali metal and typically has an oxidation state of +1 in compounds. Since there are two potassium atoms, the total charge from potassium would be +2 (2 × +1).
Now, we can calculate the oxidation state for the chromium (Cr) atom:
Let's assign the oxidation state for chromium as x.
We know that the overall charge of the compound is zero since it is neutral. So, the sum of all the charges must equal to zero.
(+2) + (-8) + x = 0
Simplifying the equation, we get:
-6 + x = 0
x = +6
Therefore, the oxidation state of chromium (Cr) in K2CrO4 is +6.