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Chemistry Finals

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Chemistry Finals Empty Chemistry Finals

Post  Oscar Thu Dec 11, 2008 4:14 pm

Hello guys,

Here we are, after months of hard work and intense despair, to the Finals.

To make the best of it, you can post any issue or concern about the CHEMISTRY FINALS here Very Happy
Oscar
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Chemistry Finals Empty Re: Chemistry Finals

Post  Bibi Sun Dec 14, 2008 3:17 pm

What's the VESPR theory?
and a brief definition of ionization energy and electronegativity would be nice! thanks
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Post  Oscar Sun Dec 14, 2008 5:02 pm

The VESPR theory states that the atoms are arranged far away from each other, repelling each others. It helps us poor mortals that we are to figure out how to draw compounds, and to give a name to that arragement. Smile

For example: CO2 will be arranged as following: O = C = O
"=" being double link
Carbon has four valences electrons, so it links with four other electons from the two oxygens , letting in total four lone pairs of electrons (on the oxygens)

This figure is said to be Linear
There are also bent, Trigonal planar, tetrahedral, pyramidal, trigonal bipyramidal, octahedral. (see handout for pictures)

-Ionization energy: this is the energy required to form a cation (therefore to lose an electron)

It decreases down the group (Li>Fr) because electrons are further away from nucleus as we go down the group.
It increases from left to right (Mg<Al)

-Electronegativity: this is the ability of an element to attract electrons when it is in a compound.

It also decreases down the group (Li>Fr)
It also increases from left to right (Mg>Al)
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Chemistry Finals Empty Atomic Theories

Post  David Sun Dec 14, 2008 5:13 pm

I need the Atomic Theories handout affraid

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Post  David Sun Dec 14, 2008 5:46 pm

and what about the isotopes calculation? pig

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Post  Bibi Sun Dec 14, 2008 6:06 pm

To calculate the atomic mass unit (a.m.u.) you will be given the isotopes with their abundance in percentages. For instance:
Nitrogen 14 has 99.63% abundance
Nitrogen 15 has 0.37% abundance
You do 14 x .9963 + 14 x .0037 = 14.0037 amu
which is equal to the atomic mass of Nitrogen
Bibi
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Post  David Sun Dec 14, 2008 6:41 pm

thanks bibi Very Happy..
I love you

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Post  John Sun Dec 14, 2008 7:07 pm

the SPDF thing...can any one help?
for example Nickel: 1s2 2s2 2p6 3s2 3p6 4s2 3d8 is this right?
if it was Arsenic would the right one be this: 1s2 2s2 2p6 3s2 3p6 4s2 4p3 4d10?
and for example how many orbital does nickel and arsenic have?

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Post  Bibi Sun Dec 14, 2008 7:25 pm

Nitrogen: 1s2, 2s2 and 2p3

Orbitals:
s = ONE
p = THREE
d = FIVE
f = SEVEN

You have to multiple these numbers x2 because there are 2 electrons per orbital.
Energy level 1 has one orbital: 1 from s
Energy level 2 has 4 orbitals in total: 1 from s and 3 from p
Energy level 3 has 9 orbitals in total: 1 from s, 3 from p and 5 from d
Energy level 4 has 16 orbitals in total: 1 from s, 3 from p, 5 from d, 7 from f

Nitrogen is located in energy level 2 so it has orbitals from energy levels 1 and 2
total 1 + 4 = 5
Arsenic is located in energy level 4 so it has orbitals from energy levels 1, 2, 3 & 4
total 1+ 4 + 9 + 16 = 30
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Post  John Sun Dec 14, 2008 7:28 pm

eu ainda n entendi o negoco de orbitals uaheuhaeuaeh
no teste ela vai fzr perguntas como sobre orbitals?

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Chemistry Finals Empty Re: Chemistry Finals

Post  Oscar Sun Dec 14, 2008 7:51 pm

John, I think you need to understand the concept of orbitals ...

Bibi presented this concept very well; what is your question?

Well, if you take a look at the periodic table, it may be easier.

The horizontal rows correspond the levels of energy
These levels of energy are divided into orbitals.

For example the first level of energy only contains one orbital, S.
So you write it: 1(first level of energy) s (the type of orbital) 2 (the number of electrons)

=> 1s2

then second row => 2s2 2p6
third row=> 3s2 3p6
fourth row => 4s2 3d10 4p6

ect ...

That's the basic, tell me your problem with orbitals, it will be easier to solve Smile
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Chemistry Finals Empty orbtals!

Post  ana Sun Dec 14, 2008 8:10 pm

Oh john... vou simplificar

cada orbital so pode ter dois electrons, entao os sublevels (s,p,d,f) tem uma quantidade de orbitals diferente.
por isso:
s tem 1 orbital e portanto 2 electrons max
p tem 3 orbitals e portanto 6 electrons max
d tem 5 orbitals e portanto 10 electrons max
f tem 7 orbitals e portanto 14 electrons max

qdo vc quer ver os orbitals no ultimo energy level, vc soma o todas as letras do level (1,2,3 etc) disponiveis.

ou seja.. se quiser saber o numero d orbitals d um sublevel (s,p,d,f) sem memorizar, divida o numero d electrons q podem caber na level por 2
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