2.7 The periodic table — groups 2 and 7
Posted by: intechemistry on: September 23, 2010
Flame test video for some group 1 ions: Li+, Na+ and K+
Flame Tests – Identifying Metal Ions
(you may like to mute or lower the volume for the following video)
ions shown = Li+, Na+, K+, Ca2+, Ba2+
The apple green of Ba2+ often is hard to see and is not persistent.
Ca2+ flame:
Sr2+ flame (Crimson red)
Ba2+ (video title: baryum flame)
K+
potassium flame
30 Responses to "2.7 The periodic table — groups 2 and 7"
November 5, 2010 at 12:17 am
The reaction is actually pretty complicated. In reality this reaction is very messy: Lots of things are produced and lots of reactions happen. I think it’s OK to write it like this as a kind of summary of the overall reaction:
8HI(g) + H2SO4(l) –> H2S(g) + 4H2O(l) + 4I2(s)
Ref: New Understanding Chemistry for A-level. Lister and Renshaw 3rd Edn, p211.
But you can write other equations like equations that yield sulfur if you insist. It’s not likely they will ask you for the HI reaction (or NaI or KI) reaction with sulfur for the reason above – it’s complicated. They are far more likely to ask you for equations for the HCl (or NaCl or KCl) reaction and HBr (or NaBr or KBr) reaction.
December 12, 2010 at 11:05 pm
salam sir.
can we regard radium hydroxide as the most soluble hydroxide in group 2 hydroxides? or is it the barium hydroxide?
December 13, 2010 at 1:41 am
Salam.
Sorry, don’t have the specific figures available, so I can’t say for certain.
It’s not likely you will have to know for certain which one is more soluble between Ba(OH)2 and Ra(OH)2, but you would be expected to predict that Ra(OH)2 is more soluble than Ba(OH)2 as the trend is: Going down the gp, the solubility increases, and Ra in below Ba in the p.table.
But Ra is different from Ba in that Ra has f-orbitals. Sometimes these orbitals can alter the properties of an element in a way we wouldn’t expect (unless you have studied f-block of course, and sadly, I can’t remember much about that 
)
So there is a ‘chance’ that Ra might break the trend, but I honestly don’t know. Even if it does do something unexpected, I don’t think it matters. the important thing was you identified the major trend within the group.
The important thing about learning chem, is learing the rules that 95% of molecules obey. The 5% exceptions are just ‘the icing on the cake’. It would be a sad day if edexcel examined you disproportionately on the 5% at the expense of the 95%. So expect discussions of unusual ‘rule breaking’ things to only happen about 5% of the time. They want to grade you on what you know, not on what you don’t know.
December 13, 2010 at 11:02 pm
oh.ok.
I understand the trend, it’s just that when I read, the compounds in f-orbital are not mentioned. so, just to make sure that I don’t really have to put really much attention on that.
this “f-orbital rule” generally apply to other group as well is it? since we dont learn about them, suppose edexcel wont probably ask on that 5% thingy.
so, generally, I just have to understand the trend and the explanation behind it right?
December 13, 2010 at 11:27 pm
Don’t worry, just forget about the f-orbitals.
In the old syllabus, the chemistry of Gp IV was a learning requirement. Lead, Pb, is amongst Gp IV, and for that element, the f-orbitals very briefly made an ‘appearance’ in discussions as they were involved in Pb’s unusual behaviour of having +II as its most table Oxidation State (O.S.), in contrast to the other Gp IV elements which have +IV as their most stable O.S.
Yes, just know the trend, and no need to have to explain it in topic 2.7. Specification says:
“2.7.1.d Recall the trends in solubility of the hydroxides and sulfates of group 2 elements”
No mention of explain 
December 26, 2010 at 6:16 pm
Salam sir…
what actually affecting the solubility of G2 hydroxides and sulfates?
Is it due to the size of anion and/or cation?
=D
December 27, 2010 at 8:03 pm
The specification says: “2.7.1.d Recall the trends in solubility of the hydroxides and sulfates of group 2 elements”
It’s not a “Recall and explain” learning requirment, so you don’t need to be able to explain it. Are you sure you want an explanation? It’s a bit tricky to understand.
One didn’t have to explain it on the old syllabus either, even though we (some of us) often gave explanations at that time.
December 28, 2010 at 8:55 am
So it’s about memorizing the trend after all…
=D
thank you sir..
April 15, 2011 at 5:29 pm
Salam sir..
In Edexcel Chemistry Unit 2 June 2009, for question 20(d),
can I say that the difference in the rate of reaction between 2-fluoro-2-methylpropane and 2-chloro-2-methylpropane is due to the difference in the bond strength?
My answer was the bond strength of C-F bond is much more stronger thatn that of C-Cl bond in the molecule as flourine atom is much more electronegative than C in C-F than chlorine atom in C-Cl bond.
And the second part of this question is asking about which reagent would be the best to replace water and the reason for my choice.
My answer was aqueous NaOH solution and because it would fully-dissociate to give Na+ ions and OH- ions more than OH- ions provided by the water itself.
April 17, 2011 at 6:22 pm
Salam Arif.
The first part seems fine. One mark.
The second part is a bit shakey though, as the [OH-] ions in water is so small (1×10^-7 moles in 1000 cm3) their contribution to the rate with R-X is very small compared to the much higher [H2O], despite the fact H2O is a worse nucleophile than OH-.
A similar thing came up with the rxn of NH3 with R-X. We used a large excess of NH3 to stop the amine produced in the reaction (which is a stronger nucleophile than NH3) from reacting with the remaining R-X
You’d get one mark for choice of NaOH but lose the third mark for the reason.
April 22, 2011 at 2:28 am
Thus, by using NaOH solution, we actually increasing the [0H-] in the solution mixture. By this, the rate of reaction would significantly be higher than by just using H2O as the source of nucleophile.
Is this the correct way of answer the second part of this question sir?
April 22, 2011 at 3:34 am
I wouldn’t say “actually increasing the [0H-] in the solution” as you are still making reference to OH ions from water undergoing a reaction, but your rest is OK. So dump the connection to OH- ions from the autoionisation.
i.e. something like this (which is almost what you’ve said):
“OH- ions are a better nucleophile than water so will react more quickly, hence have a greater reaction rate than the reaction of R-X with water alone.”
May 7, 2011 at 2:23 pm
Salam sir…
I can’t draw the link between the electronegativity of a halogen atom in H-X molecule and the bond strength of the molecule..
How does the electronegativity of the halogen contribute to the bond strength of H-X molecule?
Thank you…
May 7, 2011 at 5:43 pm
For H-X. the greater the electronegativity of the halogen, the stronger the bond.
This also applies to halogenoalkanes/halogen alkanes /alkyl halides /alkyl halogens (too many words describing the same thing!) i.e. C-X bonds.
This is yet another situation where only a simple explanation of what’s happening is given in discussions explaining the effect.
Electronegativity = attraction for an electron pair WITHIN A COVALENT BOND.
Simply: The greater the attraction by X for the covalently bonded e- pair in H-X (hence the greater the polarization), the stronger the bond. The H atom has no e- in orbitals repelling with the e- in orbitals of the F atom (the F-F bond is v. weak due to this repulsion) resulting in an increasingly delta positive H atom attracted to an increasingly negatively charged F atom.
Although some people may call the following ‘chemical blasphemy’, I think it helps to think of it taking on partially ionic character. The strength of the ‘partially ionic’ bond is related to the size of the ‘ions’ via the internuclear (or inter ionic) distance. As both species are very small, the d+ H and d- F, the attraction is very strong, as we think about when discussing lattice enthalpy.
Gong down the gp, the electronegtivity of X decreases and the size of X increases, both contributing to the attraction between H and X decreasing.
This can also be used to explain (up to a point) that as you go down the gp, the H-X’s become stringer acids.
May 8, 2011 at 2:50 pm
I think it’s safe to say, if it’s not mentioned in any particular detail on the syllabus, then such detail is unlikely to come up on the exams. The chemistry in the edexcel revision guides is on the whole not very ‘deep’ and generally quite simple.
E.g. h__ttp://www.amazon.co.uk/Edexcel-Chemistry-Revision-Guide-Sciences/dp/1846905974
and
h__ttp://www.amazon.co.uk/Edexcel-Chemistry-Revision-Guide-Sciences/dp/1846905966
and
h__ttp://www.amazon.co.uk/Get-Grade-Edexcel-AS-Chemistry/dp/0340991879 – with it’s rather large pentagram !
Remove the double underscore i.e. __
May 10, 2011 at 5:44 pm
salam
sir, what should we write when the question asked the observation when aqueous bromine is added to a solution of potassium iodide?
do we say colourless solution turns brown (because iodine is brown in potassium iodide solution) or grey-black solid is formed (as iodine is formed)?
which one does exactly form?the aqueous iodine or solid iodine?
May 10, 2011 at 5:55 pm
another related question, in any reaction that produces iodine, (given that no state symbols are given in the question), how do we know whether an aqueous iodine or solid iodine is formed?
the state symbols will obviously affect our observations, so do edexcel accept observations for both solid and solution of iodine in the same answer given?
May 11, 2011 at 10:26 am
@19, hidayah, May 10, 2011 at 5:44 pm
Salam.
In this redox reaction (in this case sometimes called a displacement reaction) the I2 continually formed will be in the presence of I-(aq). the I- comes from the KI and the (aq) comes from the bromine water.
I2 in the presence of I- forms the (I3)- anion.
The (I3)- anion is considered to be the same as I2(aq) because it will ‘give off’ I2 which can be consumed in reactions.
Hence, I would say:
The solution changes from an orange/light brown colour to a very dark brown colour due to the disappearance of Br2(aq) and the formation of I2(aq)
May 11, 2011 at 10:35 am
@20, hidayah, May 10, 2011 at 5:55 pm
It seems to me that then there is water present, then I2(aq) will be the likely species. In the absense of water or ‘solution chemistry’, I2 will probably be a solid. After that, give observations accordingly (just don’t say ‘purple’ solid for I2(s) !!)
The examiners will, on the whole, probably avoid any areas where any such borderline or confusing areas may occur. They are more interested to know whether you have understood the principle and are able to apply it.
When one deals at with ‘fringe’ chemistry the commonly learnet model or principles need adjusting so that’s getting away from their objective as A-Level examiners.
They may ask the occasional tricky question, but it is pretty much assured that it’ll be based on those core principles you learned.
To all: Don’t worry, they are not monstors.
May 11, 2011 at 11:03 am
ok,thank you sir.
pretty (or very much nervous), actually.
like u said, always prepared for the worst.hopefully their worst is not as horrible as monsters
May 11, 2011 at 2:16 pm
I want to say “don’t be nervous” because by saying so, I hope they will go away, but that’s not a realistic prospect is it?
Maybe however those nerves will go away (or lessen!) if you put trust in the higher power. I am sure He grants guidance(!) to those who follow His way.
January 7, 2012 at 12:32 pm
Salam sir Allan. In Alevel exam, if they asked about flame test and its only 3 marks given, should we stated bout cleaning the nichrome wire first? or should we go straight to our answer?
One more, can we used short forms for our exam as in HCl, instead of hydrochloric acid?
Hope u cn help us, as our exams are nearby. Thank you.
January 7, 2012 at 8:44 pm
Salam.
I would include the nichrome wire in with the answer. It’s always hard to say exactly how the marks are allocated. Some mark schemes for this topic may have >3 marks!, as there are more than just 3 important points in doing a flame test. My advice is to scan through the mark schemes to see what aspects of the procedure they think are important.
With abbreviations like HCl, you need to be careful. If it asks you to name some species, then “hydrogen chloride” would be the answer. You must also be careful about the nature of the species. For example, if again you were asked to name the substance used to clean the nichrome wire and also moisten the s-block salt, you would have to say “concentrated hydrochloric acid” (hydrogen chloride is a gas, only when it is dissolved in water is it called hydrochloric acid”
Generally however, e.g. when writing sentence type answers, writing things like conc. HCl, i.e. chemical symbols and formulae is fine instead of writing out the full English words.
Edexcel state on their mark schemes that examiners are to award marks in a positive sense rather than “be too strict” ( – as long as the scientific ideas gets across of course!)
May 19, 2012 at 1:18 pm
Sir, in lattice energy we learned that if more cation polarises the anion more, there is more covalency in the bonding (extra bonding), so energy released is more exo. When the lattice energy is more exo, means the bonds form are stronger.
1) So can I say polarised ionic compound is stronger than perfect/nearly perfect ionic compound?
2) If more polarised ionic comp is stronger than less polarised ionic comp because of the extra bonding, then why in group 2, we learned that heat needed to decompose nitrates and carbonates increases down the group because of less polarisation?? Shouldnt it be more polarised compound, stronger, then need more energy to decompose?
I know I got it wrong somewhere in the reasoning, I’m pretty sure. >.< THe only thing I can figure out if i'm wrong is that in the 'polarisation of ionic bond' part, I'm talking about lattice energy(exo) and lattice dissociation energy (endo). The ionic compound is formed from or dissociated into gaseous atom. and in the group 2 nitrate part i'm decomposing them into nitrates and oxygen only. But still, shouldn't decomposing lithium nitrate needs more energy than caesium nitrate since it has a little bit of covalent bonding??
Thanks
May 19, 2012 at 2:53 pm
Hi “sleep deprived bug”
1) Yes. both the polarization AND the separation of the ions need to be overcome when breaking down the lattice rather than just ‘separation of the ions’ in the purely ionic case. The different delta(H) values are actually proof of this, the more exo value indicating what we may casually call “more bonding” (bond forming releases energy) and greater stability than just a purely ionic interaction which as “less bonding”. Experimental (i.e. real world) values will almost always be more exo than purely theoretical ionic model of bonding, as in reality there will almost always be some polarization of the anions e- cloud (although fluroide anions are very difficult to polarise due to only tiny levels of shielding by 1s electron inner shell.
2) Good Q. Because in decomposition you are not simply splitting apart the ions, you are destroying the covalent bonds in the anion. Your doing a different reaction (it’s a chemical reaction, i.e. the electron distribution within species is changing) compared to just separating the ions. The more polarized the anion, the easier it is to destroy the covalent bonds in the anion.
Does that help?
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November 4, 2010 at 11:45 pm
Sir, how to write the balance equation of HI oxidised to iodine by conc. H2SO4, which itself reduced to H2S? Different books put it differently.