6.
To determine this the CrO4 ion must be 2- as it is balanced by 2 potassium ions which can only have a +1 charge.
The oxygen atoms are considered to be present as O2- so all four contribute 8-. As the net charge is only 2- then chromium must have charge (oxidation number or state) of 6+.
The oxidation number of Cr in K2CrO4 is +6. This is because oxygen is typically assigned an oxidation number of -2, and the overall charge of the compound is 0. By setting up the equation 2(+1) + Cr + 4(-2) = 0, we can solve for Cr to find it has an oxidation number of +6.
In KβCrOβ, the oxidation number of potassium (K) is +1, and the oxidation number of oxygen (O) is -2. Since the compound is neutral, the oxidation number of chromium (Cr) can be calculated as follows: 2(+1) + Cr + 4(-2) = 0. Solving for chromium, the oxidation number of chromium in KβCrOβ is +6.
The oxidation number of potassium (K) is +1 and the oxidation number of oxygen (O) is -2. To find the oxidation number of chromium (Cr) in K2CrO4, we can set up an equation: 2(+1) + x + 4(-2) = 0. Solving this equation, we find that the oxidation number of chromium (Cr) is +6.
+2 for Ca, +6 for Cr, -2 for each O
The oxidation number for Cr in Cr2O7^2- is +6.
The change in oxidation number of Cr depends on the specific reaction or compound involved. For example, in the reaction from Cr(III) to Cr(VI), the oxidation number of Cr changes from +3 to +6, indicating an increase in oxidation state.
The oxidation number of potassium (K) is +1 and the oxidation number of oxygen (O) is -2. To find the oxidation number of chromium (Cr) in K2CrO4, we can set up an equation: 2(+1) + x + 4(-2) = 0. Solving this equation, we find that the oxidation number of chromium (Cr) is +6.
The oxidation number of chromium in Cr2(SO4)3 is +3. This is because the overall charge of the sulfate ion (SO4) is -2, and there are 3 sulfate ions in the compound, resulting in a total charge of -6. Since the compound is neutral overall, the oxidation number of chromium must be +3 to balance out the negative charge.
+2 for Ca, +6 for Cr, -2 for each O
The oxidation number of Cr in CrBr is +3. Since Br has an oxidation number of -1 and there is only one Br atom in the compound, the oxidation number of Cr must be +3 to balance the charges to zero.
The change in oxidation number of Cr depends on the specific reaction or compound involved. For example, in the reaction from Cr(III) to Cr(VI), the oxidation number of Cr changes from +3 to +6, indicating an increase in oxidation state.
The oxidation number of Cr in Cr2O7^2- ion is +6. Each oxygen atom has a -2 charge, and the overall charge of the ion is 2-, so combining the charges gives -2*(-7) = -14. Since the overall charge is 2-, the oxidation number of Cr must be +6 to balance the charges.
As with any other element, the oxidation number of Cr depends on whether and how it is chemically bonded. The oxidation number of pure elements is arbitrarily defined to be 0. In compounds, Cr has oxidation numbers of +2, +3, and +6, depending on the compound.
The oxidation number of Cr in HCr2O7 is +6. This is because each oxygen atom has an oxidation number of -2, and there are four oxygen atoms in the dichromate ion (Cr2O7)^2-. The overall charge of the ion is -2, which means the two chromium atoms must have a total oxidation number of +12 to balance the charge, resulting in an individual oxidation number of +6 for each chromium atom.
The oxidation number of Cr in MgCrO4 is +6. This is because oxygen is typically assigned an oxidation number of -2, and the overall charge of the compound is 0, so the oxidation number of magnesium (+2) and oxygen (-2) must be balanced by the oxidation number of Cr (+6).
The oxidation number for Cr in CrO3 is +6. Each oxygen atom has an oxidation number of -2, and since there are three oxygen atoms in CrO3, the total oxidation number contributed by oxygen is -6. This means that the oxidation number of chromium must be +6 to balance the overall charge of the compound, which is neutral.
+6 for Cr
The oxidation number of chromium (Cr) in CrO4^2- is +6. Since each oxygen atom has an oxidation number of -2, and the overall charge of the polyatomic ion is -2, the oxidation number of chromium can be determined by solving the equation: (oxidation number of Cr) + 4(-2) = -2.