1.6 Bonding
Posted by: intechemistry on: September 23, 2010
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24 Responses to "1.6 Bonding"
October 11, 2010 at 10:01 pm
Peace be upon you Pandora.
These are not these homework Q’s by any chance are they?
And you are right, they are making my head spin…already!
Because these Q’s are not that specific to the syllabus, I may move them to the “General Questions” section (which I’ve just noticed no longer has numbered comments! Aaaagh!) later and also please wait for when I have time to give them a good answering! 
October 11, 2010 at 11:23 pm
I shall SWEAR that these questions have nothing to do with my homework or tutorials… Somehow after I studied the whole chapter, I came upon in generalising these few Q’s that neither textbook nor reference book play a role in providing a further explanation. So eventually here ‘they’ are! LOL. (sorry if it does make ur head spinned for a moment, i guess) 
) Many thanks sir!
October 12, 2010 at 12:13 am
Don’t swear – it’s rude ;p
OK. I’ll try to provide answers as soon as I am able
and you provide the paracetamol.
February 4, 2011 at 10:20 am
salam sir
i wonder how can a polar molecule has dispersion force.. eg hydrogen flouride. it has hydrogen bon and also dispersion force
February 4, 2011 at 3:32 pm
Salam.
Every molecule has dispersion forces. They are the result of temporary fluctuations in electron density. All molecules have electrons (which they all do) so they have these fluctuations.
But the most significant IMF is the strongest one, so in HF we normally only discuss it’s hydrogen bonding IMF as the dispersion force is quite insignificant in comparison.
February 13, 2011 at 11:51 pm
Salam, sir.
Which one is the best explanation for the reason why alcohols above C5 are insoluble in water?
One, because the dispersion forces between the long hydrocarbon chains are too strong to break, eventhough the OH group on the alcohols can form hydrogen bonds with water.
Or, because the long hydrophobic chains disrupt the structure of water too much. (in which I don’t really understand -__-)
May you help, sir? Thank you:)
February 15, 2011 at 1:38 pm
Re: option one…
IMF’s for small molecules esp. hydrocarbons are weak, that’s why CO2, propane, Cl2, SO3 etc are gases. Slightly larger molecules still show weak dispersion forces, benzene, cyclohexane which is why they are volatile liquids.
Even molecules with additional forces as well as dispersion forces [propanone, ethanol, ether, pyridine] can be quite volatile because they are small. i.e. the TOTAL IMF’s on the molecule are quite low and background thermal radiation (i.e. heat or ‘room temperature’ is quite close to being able to overcome these forces; They need only a little bit of heat to boil them)
So your option one is a no-go 
Re: option two… It’s a much, much better explanation
Picture a water based H-bonding network. If a foreign substance is to dissolve, it must ‘fit’ into this network. Picture the foreign substance squeezing between the water molecules. When it does so, it will ‘block’ hydrogen-bonding interactions between the H2O molecules which used to be beside each other before. Energetically, if the energy of the broken hydrogen-bonds is greater than the bonds formed between the foreign substance and the surrounding water molecules, then the dissolving the foreign substance leads to an increase in the energy of the system. The smaller the bonding between the foreign substance and the water molecules, the lower the solubility will be.
Perhaps one could also explain what you mentioned as follows:
The proportion of non-polar, non-soluble hydrocarbon chains in the molecule is much more significant than the polar hydrogen bonding hence soluble OH portion hence C5 and above alcohols are insoluble (even butan-1-ol has difficulty dissolving but is regarded as being soluble). I don’t think pental-1-ol is soluble.
But in the physical universe (in the absence of miracles!) as with EVERYTHING… energy changes (as I touched upon earlier) is the reason why ANYTHING happens.
April 12, 2011 at 10:19 am
salam sir..
In GF AS Chemistry, the definition of lattice enthalpy states that it is the energy change when 1 mol of solid is formed from its constituent gaseous ions that start infinitely far apart.
So, what is exactly ‘…that start infinitely far apart’?
Can I just leave out these last few words; if I forgot to write in my answer ?
April 12, 2011 at 8:48 pm
Salam.
The words are there to indicate that there is no initial attraction between the oppositely charged ions; so when they come together, the total energy given off as they apporach counts towards the LE. Although the inverse square law (the 1/r^2 part of the equation relating to the force between two oppositely charged ions) means even if they start only a couple of metres away, the energy given off when they approach to the optimum distance, will be almost the same as whether they approaced from an infinite distance apart.
Best include it ‘cos it’s really the only valid way to define it.
May 5, 2011 at 10:48 pm
salam alayk
sir… sorry..
getting confuse of nuclear charge and effective nuclear charge…. can you explain it?(for how many times i can’t remembe……..)
~i hope that my q won’t make you worry.. i suppose to master it already~ -_-”
thank you sir
May 6, 2011 at 12:50 am
Salam. Sometimes we need to clarify earlier points that we may have put to the side
At a simple level, effective nuclear charge, Zeff, is the number of protons minus the number if inner (i.e. shielding) electrons. The value is attributed to EACH of the outer electrons of an atom.
So for Na, 11p+ and 1s2 2s2 2p6 3s1 the Zeff is +11 – (2+2+6) = +1, meaning EACH electron (in the 3s subshell) experiences 1 proton worth of charge.
for Mg, 12p+ and 1s2 2p6 3s2, the Zeff is +12 – (2+2+6) = +2
Meaning EACH outer e- of an Mg atom experience two protons worth of charge.
Because the outer e- of Mg atoms experience a greater Zeff (i.e. a greater attraction) compared to Na, we predict the Mg atom is smaller than the Na atom and we can predict the ionisation energy is greater for Mg also.
When we discuss Zeff within a group (say group 1) instead of a period as we did above, then the simple model of Zeff doesn’t yield much useful info. All gp 1 elements have Zeff = +1 which doesn’t help predict properties of those elements. At this stage an improved version of Zeff can be calculated but isn’t within the scope of the syllabus (as far as I can remember)
Nuclear charge is simply the number of protons in the nucleus.
May 11, 2011 at 3:33 pm
hi sir. i came across lots of questions about comparing the melting points of different elements and compounds. one of them is:
Why the melting point of krypton is higher than neon?
The answer given is : more number of electrons in krypton, so STRONGER (not more) Van der Waal’s forces in krypton.
So is greater number of electrons causes STRONGER or MORE Van der Waal’s forces? I’m confused.
May 11, 2011 at 5:58 pm
Hi “me”
How would you measure the NUMBER of van der Waals? Even if you could measure the number of them, wouldn’t their strength not count?
The correct way to describe them is strong/stronger and weak/weaker (stronger and weaker being used in comparisons)
May 11, 2011 at 8:13 pm
Salam sir..
Is phosphor trihalide polar molecule?eg PCl3, PI3 etc…
May 12, 2011 at 12:37 am
@15, amirul arif, May 11, 2011 at 8:13 pm
Salam.
Those slightly tricky molecules are very unlikely to come up, but if they did, you are most likely to be expected to answer yes, even though PI3 is classified as being essentially non-polar. PCl3 however is polar.
You are not expected to be able to do the maths to find the exact dipole moment of each molecule, which is why tricky molecules are unlikely to come up. Symmetrical molecules like CCl4, CH5, AlCl3, CO2, N2 are obviously non-polar. Things like water, ammonia. PH2, PCl3 and PI3 will be expected to be classified as polar based on their bond dipoles not cancelling each other out. But that’s a lazy and very incomplete method. Really, one must also consider the lone pair but at that point, things get quite a bit more tricky and beyond A-level.
Some data:
PCl3′s polarity is 0.97 D (Debye units), CO2 [non polar]= 0 H2O(vapour) = 1.82
May 20, 2012 at 4:56 pm
hello sir. i have a big confusion sir. really need ur help sir. in pearson pg87 for AS. its stated that polarisation of the bond, distortion in shape of ion and increasing covalent character in the bond will all reduce the lattice energy and show up in experimental values. but why isit that when lattice energy calculated thru Bond-Haber cycle, the value is HIGHER than the theoretical value.?
May 20, 2012 at 5:02 pm
sir if charge greater and ionic radius stronger, the ionic character is stronger ryte sir? but why isit that for aluminium chloride, it is covalent and not ionic? aluminium ion has great charge of 3+ and a small ionic radius but why instead of having strong ionic bond, they say it distorts the electron cloud on chloride ion and cause it to become covalent molecule? what determines strength of ionic bond then sir?
May 20, 2012 at 6:05 pm
smaller anions and large, low charged cations give have ionic character, e.g. CsF
Al has a high charge and is very small size. It’s highly polarizing hence AlCl3 is covalent.
Ionic bond strength is measured but the lattice enthalpy, as in the Born–Landé equation: http://en.wikipedia.org/wiki/Born%E2%80%93Land%C3%A9_equation
May 20, 2012 at 6:11 pm
#21, Kosher, May 20, 2012 at 4:56 pm
The Born-haber cycle is the experimental real world thing that gives the experimental LE.
The Born–Landé equation gives the theoretical LE.
The experimental LE will be more exo (n all cases as far as I aware) than the theoretical ones becasue of the real world polarization.
P.S. Don’t say higher or lower.
Do say “more exo”, “more endo”, “less exo”, “less endo” as appropriate.
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October 11, 2010 at 9:09 pm
Salam sir! I wish these little doubts of mine wouldn’t make ur head spins as well~ LOL.
1) What’s the definition for ‘Born energy’? Is it related to the enthalpy of atomisation?
2) What does it signifies when it is stated that some ionic compounds that possess covalent characteristics have lower melting & boiling point compared to purely ionic compounds, eg. silver halides hv lower melting point than sodium halides. Why there’s such difference? wasn’t that both ionic bond & covalent bond hv equal bond energy?
3) Why we should use the values of enthalpy change of formation to compare the stability of MgCl, MgCl2, and MgCl3 instead of using lattice energy?
TQ!