ashwinkirtane

Any electron exchange is possible if and only if a more stable state is achieved. Now look, always remember these points: 1. A fully filled orbital is MOST stable 2. Then you have a HALF filled orbital. ANY change that involves one e- transfer and resulting configuration is more stable then it is possible. I'll give you an example then you yourself solve for Copper. Chromium has atomic number 24, its outer shell configuration should have been 3d4 4s2 but if one electron from outer s orbital jumps to the inner d orbital it gives a more stable HALF filled configuration of 3d5 4s1. So Chromium actually has this configuration. I guess this information is enough for solving your question. Do post your answer here. Good Luck!

It is nothing to worry, it is just the same. Actually this is the way you'll find in most of the books, the reason being that for the leftmost carbon when we draw on paper we make the bonds towards its left so thus they are shown by H3C- and not -CH3. However this is not the case with right most carbon.

Whenever you want to balance a redox reaction, there are essentially few steps involved. If you follow them correctly, you'll certainly arrive at the correct answer. Step 1: Write the two halves of the redox reaction (oxidation half and the reduction half) Here, Oxidation Half: I- -> I2 Reduction Half: Cu2+ -> Cu+ Step 2: Now that you've written the two halves, balance them. If required(not required in this case), add water molecules to Hydrogen deficit side, OH- if the reaction was carried out in basic medium and H+ ions if the reaction was carried out in acidic medium. Since reduction is gain of electrons and oxidation is loss of electrons, show that too. Like this, Oxidation Half: 2I- -> I2 + 2e Reduction Half: 2Cu2+ + 2e-> 2Cu+ *Make sure that the electrons on both the sides are equal (charge conservation). Step3: Add both the halves: 2Cu2+ + 2e + 2I--> 2Cu+ + I2 + 2e Now use some common sense! See, you need CuI on the product side so add two iodide ions on both side and cancel the electrons off course. What do you get? 2Cu2+ + 4I- -> 2CuI + I2