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examples of cations and anions

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anion

anion , atom or group of atoms carrying a negative charge. The charge results because there are more electrons than protons in the anion. Anions can be formed from nonmetals by reduction (see oxidation and reduction ) or from neutral acids (see acids and bases ) or polar compounds by ionization. Anionic species include Cl - , SO 4-- , and CH 3 COO - . Highly colored intermediates in organic reactions are often radical anions (anions containing an unpaired electron). Salts are made up of anions and cations . See ion .

cation

cation , atom or group of atoms carrying a positive charge. The charge results because there are more protons than electrons in the cation. Cations can be formed from a metal by oxidation (see oxidation and reduction ), from a neutral base (see acids and bases ) by protonation, or from a polar compound by ionization. Cationic species include Na + , Mg ++ , and NH 4+ . The cations of the transition elements have characteristic colors in water solution. Salts are made up of cations and anions . See ion .


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Question:I have to explain why cations give off different colours when burnt in a flame. I have started but I am not sure what I am saying is correct. Please correct this and add to this with any details. Ive stopped at the bit: "How we work out the distance through the colour of the flame when burnt." What is the spectrum thing called, I forgot? Cations have a positive charge. Examples of cations include: Potassium, Lithium and Copper. Cations are usually metals. Negative ions are called "anions". Anions tend to be non metals. Metals usually form cations because they are less electronegative than nonmetals. Cations are formed when an atom loses electrons, while anions are formed when an atom gains electrons. Among the atoms, the alkali metals (Group 1) and alkaline-earth metals (Group 2) are the most reactive metals, having one and two valence electrons, respectively. Therefore, metals lose that electrons and form cations. When you heat the atoms and therefore the elections, they become excited and jump to outside rings of the atom. This causes the metal to change colour when burnt in a flame. We know and can work out the colour by the distance the electron jumps. This is measured with the Thanks for any answers.

Answers:What you're saying is basically correct but I'd change a little the language, the way you'll talk about the different colors and emissions. Let me add something: A flame is normally an exothermic reaction. The heat released, f. ex., from the combustion of butane or propane is enough to excite the electrons in an atom or ion, from its original energy level (orbital) to a higher energy level (another orbital). After excitation, many of the electrons come back down to their original electronic orbitals, or to other empty orbitals, and in this process they emit light. But the fundamental point that you should make is the following: The light is characteristics of each metal because the electrons in their electronic shells have different energy! These energy are very specific for each atom. Therefore, these emissions are like "finger prints", from these it is possible to identify which atom has emitted it. In other words, the energy of the electrons 1s in H are quite different from the energy of the 3s electrons in sodium. At the same time, the energy of the other electronic energy levels of H and Na are also quite different, f. ex., the energy of the 4p or 5p (these are empty orbitals in both of them). The light emitted will be related to the energy difference between the energy levels involved. So, the wavelenght (L) of emission by electrons of these atoms, from the same orbitals (say, 3p to 2s) would be: Del E [H (3p-2s)] = E [H (3p)] - E[H(2s)] = (h x c) / L(H) Del E [Na(3p-2s)] = E [Na (3p)] - E [Na(2s)] = (h x c) / L(Na) and the Ls would be different because the Del E would be different for each atom. The fantastic thing in all this, to me, is the fact that we can determine what is the composition of stars, just looking (in details) at the light they emit! Also, remember, things are more complicated than they look: there is also ionization and even chemical reactions in the flames, at the same time ...very busy, very confusing flames are .. Hope you got the idea. Check also the references.

Question:An unknown solution possibly containing alkali metal ions, alkaline earth metal ions, and halide ions gave a violet color when placed in a flame. The solution gave no precipitate when an ammonium carbonate solution was added. Addition of chlorine water, hexane and nitric acid produced two layers with a deep violet color in the upper layer. Identify the cations and anions present in the mixture.

Answers:cations= + NH4^+ anions= - CO3^2- Cl^- NO3^-

Question:

Answers:Anions - Negatively charged ions Cations - Positively charged ions Sulphuric acid contains H(+) and SO4(2-). It depends on what it reacts (you need to know the products to know its cation and anion). Example: 2NaOH + H2SO4 --> Na2SO4 + 2H2O So here the cations are Na(+) and H(+). The anions are SO4(2-) and OH(-).

Question:I don't recall how to do it. Is there where you use the "criss cross method"? I have a list of over 50 of them and this is the first one on the list, but before I start doing it I want to make sure I do it right. Thanks. SORRY, with anion I-1

Answers:Remember this: The sum of the oxidation numbers of the cations and anions must equal zero Take iron (III) oxide for example. The formula is Fe2O3 The Roman numeral III tells us that Fe is +3 and we know that oxygen is -2 Fe2O3 = Fe Fe O O O +3 +3 +(-2) + (-2) + (-2) = 0 Or you can "crisscross" the oxidation numbers: +3 -2 Fe O +3 -2 Fe2O +3 -2 Fe2O3 The danger in using the "crisscross method" is that sometimes the subscripts will need to be reduced. Consider lead (IV) oxide +4 -2 Pb O Pb2O4 Notice that 2 and 4 are both divisible by 2; therefore reduce PbO2 In your example, it's even easier. Since the cobalt is +3, the anions must add up to -3, so if the anion is -1, then you need three of them. Or you can crisscross. +3 -1 Co X CoX3 =========== Follow up =========== Aluminum iodide = AlI3. Yes it does look odd doesn't it.