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how to calculate kc expression not using quadatic http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf
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Thanks A lot nd good luck for your exams!!!!notes for chem
The central atom is C and it is surrounded by 4 hydrogens. First find the number of electrons being shared. In this case, it is 4 from carbon and 1 from each hydrogen, totaling up to 8 electrons. Now, each bond uses 2 electrons. There are 4 bonds, and therefore no lone pairs. So CH4 has 4 bonds and no lone pairs.
Another one: NH3. The central atom is N and it makes 2 bonds with hydrogen. The total number of electrons being shared are 5 from Nitrogen and 3 from the hydrogens. So total electrons = 8. Nitrogen makes 3 bonds (which use 6 electrons), so 2 electrons are left making 1 lone pair. So NH3 has 3 bonds and 1 lone pair.
When you have the bonds and lone pairs decided, learn this:
2 bonds, 0 lone pair = linear and 180 degrees (e.g. CO2)
2 bonds, 1 lone pair = angular and 117 degrees. (e.g. BF3)
2 bonds, 2 lone pair = angular and 104.5 degrees. (e.g. H2O)
3 bonds, 0 lone pair = trigonal planar and 120 degrees. (e.g. AlCl3)
3 bonds, 1 lone pair = trigonal pyramidal = 107 degrees. (e.g. NH3 as I did above)
4 bonds, 0 lone pair = tetrahedral = 109.5 degrees. (CH4 as I did above)
5 bonds, 0 lone pair = trigonal bipyramidal = 90 and 120 degrees. (PCl5)
6 bonds, 0 lone pair = octahedral and 90 degrees. (SF6)
The theory is basically that the electron pairs arrange themselves around the central atom to minimize the amount of repulson, so they try to be as far apart as possible. Also, you need to know that lone pair - lone pair repulsion > lone pair - bond repulsion > bond-bond repulsion. Lone pair repulsions are stronger because they are closer to the atom and therefore exert more pressure
when ur given standard heat of formations of the reactants and products, then formula is: heat of formation of products - heat of formation of reactantshas anybody how to calculate the bond energy in Standard Entalply change in combustion and formation.... ???
The central atom is C and it is surrounded by 4 hydrogens. First find the number of electrons being shared. In this case, it is 4 from carbon and 1 from each hydrogen, totaling up to 8 electrons. Now, each bond uses 2 electrons. There are 4 bonds, and therefore no lone pairs. So CH4 has 4 bonds and no lone pairs.
Another one: NH3. The central atom is N and it makes 2 bonds with hydrogen. The total number of electrons being shared are 5 from Nitrogen and 3 from the hydrogens. So total electrons = 8. Nitrogen makes 3 bonds (which use 6 electrons), so 2 electrons are left making 1 lone pair. So NH3 has 3 bonds and 1 lone pair.
When you have the bonds and lone pairs decided, learn this:
2 bonds, 0 lone pair = linear and 180 degrees (e.g. CO2)
2 bonds, 1 lone pair = angular and 117 degrees. (e.g. BF3)
2 bonds, 2 lone pair = angular and 104.5 degrees. (e.g. H2O)
3 bonds, 0 lone pair = trigonal planar and 120 degrees. (e.g. AlCl3)
3 bonds, 1 lone pair = trigonal pyramidal = 107 degrees. (e.g. NH3 as I did above)
4 bonds, 0 lone pair = tetrahedral = 109.5 degrees. (CH4 as I did above)
5 bonds, 0 lone pair = trigonal bipyramidal = 90 and 120 degrees. (PCl5)
6 bonds, 0 lone pair = octahedral and 90 degrees. (SF6)
The theory is basically that the electron pairs arrange themselves around the central atom to minimize the amount of repulson, so they try to be as far apart as possible. Also, you need to know that lone pair - lone pair repulsion > lone pair - bond repulsion > bond-bond repulsion. Lone pair repulsions are stronger because they are closer to the atom and therefore exert more pressure
becoz its an exothermic reaction da answer u hv got in a (i) is a negative answr so da 2nd peak is lower..Assalamoalaikum wr wb!
In Nov:2009#1 Q:3 (c) why does the second peak needs to be lower than the first?
NH3 isnt it total of 3 e- shared by nitrogen and 1 by each 3 H...?and nitrogen has 1 lone pair...so total around N will be 8 e-......its shape is trianguler pyramidal..isntead of trigonal..
CO2 has 4 bonds thats 2 bond pairs which repel eachother equally but since the e-,it makes a lenear molecule and 180' angle...
Boron has 3 e- in its outermost shel so it will make 3 bonds with F total of 6 e-...it does doesnt achieve noble gas electronic configraion in its outer most shell..so it has 120' angle and is trigonal planar in shape...its on pg 35 and 39 of AS and A Level Chemistry book...and it can make a dative covalent bond with eg NH3...
PCL5 forms 120 and 180 degree angle not 90 degree...
H2O has a bend shape or V shape not angular...
SF6 has 90 and 180 degree angle...
hope iam right? if not plz correct ne as well?
sure...i have no idea somebody ...wanted info and i just copy pasted..... so you r going to xplain whn this topic comes/.....
till where did u study?i have no idea somebody ...wanted info and i just copy pasted..... so you r going to xplain whn this topic comes/.....
till where did u study?
2 first chapters...5 chps ....until states of matter wht abu
wow these notes just saved my life ..thank u so much !notes for chem
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_2.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w06_ms_2.pdf
In 4g) my answer is (ch2-ch2-ch2-ch2-ch2-ch2-)n But in marking scheme the structure is branched. Can somebdy please tell me how they made this branced structure?
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