Complex ions are not the only ions that have color in solution. For example, Cu2+ is blue in aqueous solutions but is not a complex ion.
That's correct, but you haven't answer the question. Mn and Cu are transition metals that have d-orbitals available. One of the properties of the d-orbitals is that they can unfold and allow electrons to move freely in those unfolded orbitals. The energy the electrons lose is in the visible range of the spectrum, so we see it as colour.
The oxidation number of Mn in MnO4- is +7. This is determined by balancing the charge of the whole ion (-1) with the charges of the oxygen atoms (-8) and solving for the oxidation number of Mn.
The oxidation number for Mn in MnO4^- is +7. This is found by balancing the charges on the formula MnO4^- where O has an oxidation number of -2.
The oxidation number of Mn in MnO4 is +7. This is because oxygen is almost always assigned an oxidation number of -2, so the total oxidation number for the entire MnO4 ion must be -1. Since there is one Mn atom in MnO4, the oxidation number of Mn must be +7 to make the total -1 charge balance out.
The oxidation number of oxygen (O) in MnO4 is -2. To find the oxidation number of manganese (Mn) in MnO4, we use the fact that the sum of the oxidation numbers in a compound is zero (since MnO4 is a polyatomic ion). Therefore, Mn must have an oxidation number of +7 in this compound.
-2 for each O, +7 for Mn
To find the moles of Mn in MnO4, you need to consider the molar ratio between Mn and MnO4. In MnO4, there is one Mn atom. Therefore, the moles of Mn will be the same as the moles of MnO4, which is 4.18x10^4 moles.
The oxidation number of Mn in MnO4- is +7. This is determined by balancing the charge of the whole ion (-1) with the charges of the oxygen atoms (-8) and solving for the oxidation number of Mn.
The oxidation number for Mn in MnO4^- is +7. This is found by balancing the charges on the formula MnO4^- where O has an oxidation number of -2.
The oxidation number of Mn in MnO4 is +7. This is because oxygen is almost always assigned an oxidation number of -2, so the total oxidation number for the entire MnO4 ion must be -1. Since there is one Mn atom in MnO4, the oxidation number of Mn must be +7 to make the total -1 charge balance out.
The oxidation number of oxygen (O) in MnO4 is -2. To find the oxidation number of manganese (Mn) in MnO4, we use the fact that the sum of the oxidation numbers in a compound is zero (since MnO4 is a polyatomic ion). Therefore, Mn must have an oxidation number of +7 in this compound.
manganese is a metal element. It shows +4 in Mno4 ompouund.
-2 for each O, +7 for Mn
K has an oxidation number of +1 O has an oxidation number of (-2) x 4 So... the oxidation number for Mn is whatever is needed to make 1-8 equal to zero. Therefore, the oxidation number for Mn is +7
Permanganate is a poly atomic ion, not an element. It contains two elements: manganese (Mn) and oxygen (O). The formula of the permanganate ion is MnO4-
For the MnO4- ion, the oxidation number of oxygen is typically -2. Since the overall charge of the ion is -1, the oxidation number of manganese (Mn) must be +7 to balance out the charges.
To balance the oxygens, you add H2O to one side, and the H^+ to the other (acid). Example:MnO4^- ===> Mn^2+ add 4 H2O to the right side MnO4^- ===> Mn^2+ + 4H2O this balances the oxygens. Now add 8H^+ to the left side MnO4^- + 8H^+ ===> Mn^2+ + 4H2O Now this is balanced for Mn, H and O. You now only have to balance the electrons (charge).
I assume you mean the oxidation number of Mn in the permanganate ion , MnO4- The sum of the oxidation numbers is the charge on a polyatomic ion so Mn has an oxidation number of +7 as each O is assigned -2.