Chemistry: Post your doubts here!

[saudha]

hey anglegirl.... ive not even done anything for p2..... im so damn scarrrrrrrrred

 

[leosco1995]

As-Salaamu 'Alaykum.

Someone please explain to me the Electron-Pair Repulsion Theory; I don't get it at all. How do we determine what the bond angles and shapes of molecules are? Jazak Allah.

WA.

Please give this a read, it helped me quite a bit. Also, let me try to explain with an example, CH4.

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. SO2)
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.

 

[saudha]

WA.

Please give this a read, it helped me quite a bit. Also, let me try to explain with an example, CH4.

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.

this is really in detail.... wow thxx

 

[raamish]

PLease answer this question why does cis-trans isomerism arise in a molecule?

 

[leosco1995]

PLease answer this question why does cis-trans isomerism arise in a molecule?

They arise due to the lack of rotation around the C=C bond.

 

[Saad (سعد)]

WA.

Please give this a read, it helped me quite a bit. Also, let me try to explain with an example, CH4.

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.

O_O I understand now... almost.

Jazak Allah; thanks a lot that really made things simpler. Just one thing;

2 bonds, 1 lone pair = angular and 117 degrees. (e.g. BF3)

Isn't BF3 trigonal planar with 3 bonds and no lone pair? ;

 

[leosco1995]

O_O I understand now... almost.

Jazak Allah; thanks a lot that really made things simpler. Just one thing;




Isn't BF3 trigonal planar with 3 bonds and no lone pair? ;

You are right.. I made a mistake there. I changed it to SO2.

 

[Saad (سعد)]

You are right.. I made a mistake there. I changed it to SO2.

... Doesn't SO2 have two lone pairs?

 

[smartangel]

syllabus states: b) explain qualitatively in terms of intermolecular forces and molecular size:
(i) the conditions necessary for a gas to approach ideal behaviour
(ii) the limitations of ideality at very high pressures and very low temperatures

ans to i is high temp and low pressure...what abt ii??

 

[leosco1995]

... Doesn't SO2 have two lone pairs?

Nope, it only has 1. The oxygen's have 2 lone pairs, but sulfur has only 1.

 

[DragonCub]

As-Salaamu 'Alaykum.

Someone please explain to me the Electron-Pair Repulsion Theory; I don't get it at all. How do we determine what the bond angles and shapes of molecules are?

And whats the shape of/bond angles in an ammonium ion, NH4+?

Jazak Allah.

All electrons have negative charges so there are repulsive forces among them.
In a covlent compound, the repulsion is obvious among bonding pairs and lone pairs.
For a simple molecule, the covalent bonds stretch out from the centre in a 3D space. The relative strength of the repulsive forces determine the shape of the molecule.
N.B. Lone pairs have stronger repulsive ability than bonding pairs. The repulsive ability of single bond and double bond is almost the same.

Examples:
- methane, CH4, has four bonding pairs. The repulsive forces among them are all the same, so the shape of methane molecule is tetrahedral.
- ammonium ion, NH4(+), likewise, has four single bonds - four bonding pairs. The shape is also tetrahedral, with bond angle 109.5 deg.
- water, H2O, has two bonding pairs and two lone pairs (on oxygen). Lone pairs show stronger repulsion. They will squeeze the bonding pairs so that the bond angle is smaller (about 106 deg).
- sulphur trioxide, SO3, has three bonds (1 double and 2 single). We regard them as three bonding pairs. So the force of repulsion should be mutually equal. Here there are three bonding pairs rather than four, so the shape should be planar triangular.

To figure out the shape of a molecule, here is a useful procedure:
- find the central atom
- count the number of all electron pairs (bonding pairs and lone pairs)(count each double or triple bond as ONE pair)
- select the corresponding approximate shape:
2 as LINEAR
3 as COPLANAR TRIANGLE
4 as TRIANGULAR PYRAMID
5 as DOUBLE TRIANGULAR PYRAMID
6 as OCTAHEDRAL
- If there is no lone pair, the bonds then divide the space equally.
- The presence of each lone pair would squeeze the bonding pairs closer, making the shape more distorted.

 

[Saad (سعد)]

syllabus states: b) explain qualitatively in terms of intermolecular forces and molecular size:
(i) the conditions necessary for a gas to approach ideal behaviour
(ii) the limitations of ideality at very high pressures and very low temperatures

ans to i is high temp and low pressure...what abt ii??

At very high pressures or very low temperatures, in both cases, gas molecules are too close together and intermolecular forces have a large effect. This prevents the gas from behaving ideally, because an ideal gas has 'no' intermolecular forces.

Hope that helps.

 

[DragonCub]

syllabus states: b) explain qualitatively in terms of intermolecular forces and molecular size:
(i) the conditions necessary for a gas to approach ideal behaviour
(ii) the limitations of ideality at very high pressures and very low temperatures

ans to i is high temp and low pressure...what abt ii??

High pressure and low temperature shall lead to relatively strong intermolecular forces. This means there is potential energy among the molecules. Ideal gases don't have any molecular potential energy.

 

[Saad (سعد)]

Nope, it only has 1. The oxygen's have 2 lone pairs, but sulfur has only 1.

Sulfur has 6 electrons, forms two bonds with two Os which makes 8 shared electrons. Which makes, as you taught, 4 shared electron pairs. 2 of these are bond pairs because there are 2 oxygens; doesn't that leave behind 2 lone pairs on the sulfur?

 

[smartangel]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_21.pdf
Q3d...isnt the formula of calculating enthalpy change..Enthalpy change of products - enthalpy change of reactants...?? ms is doing the opposite..

 

[Saad (سعد)]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_21.pdf
Q3d...isnt the formula of calculating enthalpy change..Enthalpy change of products - enthalpy change of reactants...?? ms is doing the opposite..

This I need an answer to as well... I get this wrong almost every single time. When I memorize Enthalpy change of products - reactants, and solve a paper, the MS says reactants - products. When I do that in the next paper, THEN the MS goes back to Product - Reactants.

X_x

 

[smartangel]

This I need an answer to as well... I get this wrong almost every single time. When I memorize Enthalpy change of products - reactants, and solve a paper, the MS says reactants - products. When I do that in the next paper, THEN the MS goes back to Product - Reactants.

X_x

EXACTLY. i"ve posted this question so many times in different threads here but no one has answered it yet..i hope someone does soon because im really confused!

 

[leosco1995]

Sulfur has 6 electrons, forms two bonds with two Os which makes 8 shared electrons. Which makes, as you taught, 4 shared electron pairs. 2 of these are bond pairs because there are 2 oxygens; doesn't that leave behind 2 lone pairs on the sulfur?

Actually, sulfur forms a double bond. In double bond, you add 2 electrons (as opposed to adding 1 in a single bond, I forgot to mention that sorry). So sulfur has 6, each oxygen has 2 making a total of 10 electrons. Now, in each double bond, 4 electrons will be used leaving 2 behind. These 2 form a lone pair of sulfur.

Likewise, if you have a triple bond, then you add 3 electrons instead of 2. But very few questions come when you determine the shape of a molecule with a triple bond.

Also, when you have an ion like NH4+. You treat the + as a loss of electron. So for example, Nitrogen has 5, each hydrogen has 4 which means 9, but an electron is lost so there's 8. The rest of the shape is determined the same way.

 

[smartangel]

is there a formula for calculatiing enthalpy change of rxn using bond energies?

 

[Soulgamer]

is there a formula for calculatiing enthalpy change of rxn using bond energies?

= Bond Energy Of Reactants - Bond Energy Of Products