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best way to find the shape of a molecule?
in AS i was told that if you are given a molecule
then look at the group no. of a the central molecule...then add that to the number of surround atoms and divide by 2 to get the ammount of bonds present in the molecule.
ie. BF3
Boron in group 3
plus the no. on surround atomes (3)
add them together (which is 6) then divide by 2 which gives 3
so there are 3 bond pairs
but in my A2's that method just dont work so how do you find the shape of all molecules easier?
then look at the group no. of a the central molecule...then add that to the number of surround atoms and divide by 2 to get the ammount of bonds present in the molecule.
ie. BF3
Boron in group 3
plus the no. on surround atomes (3)
add them together (which is 6) then divide by 2 which gives 3
so there are 3 bond pairs
but in my A2's that method just dont work so how do you find the shape of all molecules easier?
sometimes you can't. There is no clear cut rule to it really. You have to sketch the basic orientation of all the atoms based on the molecular structure and depending on lone pairs of electrons causing repulsion towards other atoms, a general structure will be distorted and shape altered slightly.
or you could just memorize them...cos they do have the distinct shape of each molecule. You have to remember the valence electron pairs and how the repulsion pushes it into tetrahedron shapes...or v.e pairs below and above the central atom to form square planars and all that.
my teacher makes us remember all that plus the angles of the molecules.
my teacher makes us remember all that plus the angles of the molecules.
papz_007
in AS i was told that if you are given a molecule
then look at the group no. of a the central molecule...then add that to the number of surround atoms and divide by 2 to get the ammount of bonds present in the molecule.
ie. BF3
Boron in group 3
plus the no. on surround atomes (3)
add them together (which is 6) then divide by 2 which gives 3
so there are 3 bond pairs
but in my A2's that method just dont work so how do you find the shape of all molecules easier?
then look at the group no. of a the central molecule...then add that to the number of surround atoms and divide by 2 to get the ammount of bonds present in the molecule.
ie. BF3
Boron in group 3
plus the no. on surround atomes (3)
add them together (which is 6) then divide by 2 which gives 3
so there are 3 bond pairs
but in my A2's that method just dont work so how do you find the shape of all molecules easier?
Well, it's not quite the way you put it. In VSEPR (valence-shell electron pair repulsion), there are two basic approaches:
1) see how many valence electrons the central atom has; then add the number of accepted electrons of the surrounding atoms (and NOT just the number of the surrounding atoms); then divide by two and you get the number of electron pairs, both bonding and lone, ON THE CENTRAL ATOM
2) you can also count the full number of valence electrons (add up the number of valence electrons of ALL atoms in the molecule), and then shape your molecule so that the number of electrons you place in your bonding as well as lone pairs of ALL atoms adds up to the number you obtain
Incidentally, boron fluoride is quite special in that it has a planar structure. This is due to p(pi)-p(pi) intramolecular bonding made possible by fluorine's lone electron pairs (ie its HOMO electrons). Such bonding is only possible when planar, so planarity is favoured (similarly N(SiH3)3 is planar due to donation of HOMO electrons on N into LUMO d-shells on Si, ie p(pi)-d(pi) bonding). Pi bonding is only effective at close distances, so the stability of the BHal3 halides decreases down the group, and acidity increases. In contract, BH3 doesn't have this bonding contribution and dimerises to diborane, B2H6, with bridging hydrogen atoms to give 3-centre-2-electron bonds. (You probably don't need to know that in A-level exams though.)
So, using VSEPR, you can easily construct fairly complex molecules. Say, XeO4 (xenon tetroxide)
Xe - 8 valence electrons
4 * O - in oxidation state 2-, so 4*2 = 8 electrons
Hence you have (8+8)/2=16/2=8 negative centres, but in fact you have 4 four-electron bonding centres. XeO4 has a tetrahedral structure with double bonds to each of the oxygens.
Or, IF7:
I - 7 valence electrons
7 * F - 7*(1-) = 7
Overall 7 negative centres, all of which are bonding.
Hence the structure is pentagonal bipyramidal.
ICl2^(-):
I - 7 valence electrons
2 * Cl - 2 electrons
negative charge on the molecule - 1 electron
Overall 10/2 = 5 negative centres, of which two bonding (to the two chlorine atoms) and three lone pairs.
The best way for the lone pairs to be maximally apart is to place them on the equatorial positions in a trigonal bipyramid. As they are then symmetrical, the molecule is linear.
But VSEPR doesn't work in many cases; it would be helpful if you could give examples of which molecules you're having problems with.
Do a dot-and-cross diagram (including any lone pairs). From this work out the number of electron pairs (a lone pair is one and so is a double bond IIRC). Memorise what shape different numbers of electron pairs result in (e.g. 4 electron pairs without any lone pairs = tetrahedral; 3 without any electron pairs = trigonal; 4 electron pairs including 1 lone pair = pyramidal; 2 electron pairs = linear.) It helps to picture the shape in your head that you get when the electron pairs repel each other and spread out as much as possible. You should also memorise the different bond angles (tetrahedral = 109.5 degrees etc). The reason why lone pairs affect the shape of the molecule is that they repel more strongly and push the other electron pairs closer together, so you minus about 2.5 degrees per lone pair present from the bond angle you'd expect if there weren't any lone pairs.
That's about it. The key is to draw a dot-and-cross diagram first!
That's about it. The key is to draw a dot-and-cross diagram first!
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