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[STPM] Chemistry reference books

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  #81 Old 22-10-2005 Default

can we write the essay questions in point form??
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infested_ysy
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  #82 Old 22-10-2005 Default

Just a quick question (lol last minute study before STPM kicks in yo):

Is there a shortcut method in knowing what color would a complex metal ion (transition metals) would show? I mean, there's like 3 dozens of elements bonded to different ligands showing different kind of colors despite having the same oxidation state, and it's a ***** to blindly memorise the list (and I don't see a pattern in the color change).

And questions that asks about the color of the complexes comes out rather frequently on Paper 1/2 (paper 2 in 1999,2001 and 2002 asked about this). So... yeah, help, plz? kthxbai.
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allison
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  #83 Old 25-10-2005 Default

Hie..i'm a little confuse between electron donating groups and activating groups. I read from the book that activating groups are also electron donating groups.

However, activating groups are such as alkyl groups, -OH, -NH2 and -OR groups but electron donating groups are mainly alkyl groups only. -OH and -NH2 groups are electron withdrawing.

So, pls help. Thanx..
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Penny
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  #84 Old 26-10-2005 Default

Quote:
Originally Posted by allison
... activating groups are also electron donating groups.

However, activating groups are such as alkyl groups, -OH, -NH2 and -OR groups but electron donating groups are mainly alkyl groups only. -OH and -NH2 groups are electron withdrawing.
It's different.

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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Daniel
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  #85 Old 01-11-2005 Default

[Al(H20)3(OH)3] and Al(OH)3

What is the different between this twocompunds ? Some book will tend to write [Al(H20)3(OH)3] in their equation but others will also use Al(OH)3 for the same reaction. What is the significant of the water molecule inside the compound?
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reign226
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  #86 Old 02-11-2005 Default

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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Daniel
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  #87 Old 04-11-2005 Default

how can a buffer solution control it's acidity?
For example NH3 and NH4Cl...

If i add OH-, the NH4Cll ---> NH3 and H20
but how can it control the acidity since it's product contains NH3 which is like 0H- a basic as well??

same thing happen with adding H+
NH3 + H+ ---> NH4 (NH4 is an conjugate acid right? so it still have acid inside the solution. So how it gonna control the acidity???)

Please enlighten me... Thanks
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reign226
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  #88 Old 04-11-2005 Default

Buffer:

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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Tasslehoff
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  #89 Old 26-11-2005 Default

^^ The main concept is explained in Le Chatelier's Principle...
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leng
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  #90 Old 22-01-2006 Default

Does anyone here come across on why non-polar molecules dissolve in non-polar solvent? Any explanation?
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