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In benzene, -OH, -NH2 etc (those with lone pair) are electron donating groups. Because the lone pair of the O strengthen the C-O bond, and becomes C=O, this is known as resonance effect. But, in aliphatic case, eg. alkane, aliphatic acid etc, -OH, NH2, etc. are electron withdrawing. Because O is more electronegative. (Correct me if I'm wrong...
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Aluminium will ALWAYS tend to form compounds with six dative bonds. IE it will always try to accept six pairs of electrons from lewis bases such as NH3 and H20. It's more complete when you write without leaving out the H2O but sometimes only the (OH)3 is important because maybe you're showing how it will turn into a solid ppt when it becomes (OH)4. Anyway i guess the book omits the H2O sometimes because it's not relevant to the question.
Mmyea i need help memorizing the colours of ions, as well as what happens when you add such and such reagents to it. Failing that, I guess we can always do other questions. *sigh*
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Lets say a mixture of NH4Cl and NH3 (base and salt of base) NH4Cl -> NH4+ and Cl- ions (1) NH3 + H2O -> NH4+ and OH- ions (2) upon addition of acid, ie H+ ions, it will react with OH- to form water upon addition of OH- ions from a base, it will react with NH4+ to form NH3 and water. The NH3 won't react with the water to form back NH4 and Cl because the concentration of NH4+ is already too high (from fully disassociated ammonium chloride salt). If you want to go the route of: H+ plus NH3 -> NH4+ That's fine because even tho NH4+ is a conjugate acid, its formation does not involve creation of free OH- molecules. If you see the equation, the only products are NH4+ ions, which is in itself neutral. Acid/base is only when they react and somehow manage to release H+ and OH- into the equation. If they don't, the acidity/basicity remains constant.
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