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Originally posted by Flying grenade:
Having said that, there is a significant difference between CO2 and XeF2, which is : in CO2, the central partially positively charged C atom has no lone pairs, while the central partially positively charged Xe atom has 3 lone pairs. Therefore the (relative) magnitude of permanent dipole - permanent dipole attractions will be expected to be more significant for CO2 than for XeF2. And yet XeF2 has a much higher melting point than CO2, because XeF2 has a lot more electrons, has much larger molecular size, and therefore has significantly more polarizable electron charge clouds.)
Why CO2's Pd-Pd attraction is relatively stronger than XeF2's Pd-Pd attraction??
I thought XeF2 would have a larger net dipole moment than CO2, hence more significant Pd-Pd than CO2??
Both XeF2 and CO2 are linear, and hence have no overall net dipole. However, because C has no lone pairs, it is more electron-deficient compared to Xe with 3 lone pairs. Hence, based on this factor alone, you might expect C to be more partial positive compared to Xe. However, as I pointed out years ago, the most important factor remains polarizability of electron charge clouds, and Xe being in Period 5, has a much larger atomic radius than C, as well as greater shielding effect from greater number of inner shell electrons, and as such, coupled with the greater electronegativity of F over O, the magnitude of partial positive charge on Xe will still be greater than on C. Consequently, the intermolecular van der Waals interactions (all 3 types of van der Waals forces : permanent dipole - permanent dipole Keesom forces, permanent dipole - induced dipole Debye forces, and instantaneous dipole - induced dipole London dispersion forces) will still be far stronger between XeF2 molecules compared to between CO2 molecules.Based on the simplified (and often erroneous) teachings of Singapore JCs, almost all H2 Chem students (and even some JC teachers) will be instantly dismissive of the idea of any permanent dipole - permanent dipole intermolecular interactions between (overall) non-polar molecules like CO2. And yet permanent dipole - permanent dipole Keesom forces *does* exist between some non-polar molecules like CO2, On a related note, during hydration which occurs before hydrolysis, (overall) non-polar CO2 can actually even participate in hydrogen bonding. I won't elaborate or explain why here on the forum (where's the fun in that?), interested students can go figure out for themselves (or ask their school teachers or private tutors).
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When do we write Pd-Id force, i.e. force between a permanent dipole and a corresponding induced dipole (Debye force) ?
How to tell there's Pd-Id intermolecular attractions?
Sch didn't teach Pd-Id,
But i remember you teaching keesom-debye-vandawaals forces
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The relative magnitude of pdpd attractions are affected by the number of lone pairs on the central atom and also the magnitude of partial positive on the central atom ah?
"in CO2, the central partially positively charged C atom has no lone pairs, while the central partially positively charged Xe atom has 3 lone pairs. Therefore the (relative) magnitude of permanent dipole - permanent dipole attractions will be expected to be more significant for CO2 than for XeF2."
*Cries* , i feel that there's so much that idontknow
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Originally posted by Flying grenade:
When do we write Pd-Id force, i.e. force between a permanent dipole and a corresponding induced dipole (Debye force) ?
How to tell there's Pd-Id intermolecular attractions?
Sch didn't teach Pd-Id,
But i remember you teaching keesom-debye-vandawaals forces
The relative magnitude of pdpd attractions are affected by the number of lone pairs on the central atom and also the magnitude of partial positive on the central atom ah?
"in CO2, the central partially positively charged C atom has no lone pairs, while the central partially positively charged Xe atom has 3 lone pairs. Therefore the (relative) magnitude of permanent dipole - permanent dipole attractions will be expected to be more significant for CO2 than for XeF2."
*Cries* , i feel that there's so much that idontknow
By definition of electronegativity (which Cambridge can test you in the Singapore A levels), electronegativity only affects bond pairs, not lone pairs. Hence, no matter how electronegative F is, only the electron density of the bond pairs will be polarized towards F. The 3 lone pairs on the Xe atom remains unpolarized.On a related note, here's a killer Cambridge Pre-U question (that most Singapore JC students aren't able to answer if it came out for Singapore A levels) :
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Intermolecular permanent dipole - induced dipole Debye van der Waals forces exists between the polar part of a molecule (eg. polar carbonyl group of a large ketone), and the non-polar part of a neighboring (but same species) molecule (eg. non-polar alkyl group of a large ketone). This is why all 3 types of van der Waals forces usually exist between most molecules (and many Singapore JCs erroneously teach their students that dipole - dipole interactions isn't the same as van der Waals interactions, and in doing so mislead and confuse their students ; in fact all 3 types of dipole - dipole interactions *are* the 3 types of van der Waals forces).
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[Al(H2O)6]3+ is formed from AlCl3 in excess water, because Al is in period 3, there's energetically accessible vacant d-orbitals , so Al can expand it's octet, is it?
Why doesnt it form [Al(H2O)4]3+?
[Be(H2O)4]2+ is formed instead of [Be(H2O)6]2+ from BeCl2 in excess water, because period 2 elements are unable to expand their octet is it?
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Hi Ultima, may i ask you,
AJC /2014/P3/2(b)
State the shape and bond angle of [Pt(NH3)4]2+
I think the challenge to this question is, to determine the number of valence electrons of centre atom Pt right?
This is what i did : i do not know how to determine valence e- of Pt, as we did not learn how to write electronic configuration beyond period 3 (is it?) .
So i just assume Pt has 2 valence e- , so lost 2 valence e- , and 4 NH3 ligands datively bonded to Pt, i get a tetrahedral shape, bond angle 109°
The answer wrote :
Shape : square planar
Bond angle : 90°
But, accept tetrahedral 109°
How did they get sq planar or assumed pt has 6 valence e-?
Did a google search for electronic config of Pt , it's complex and their still debating abt it
May i ask u for guidance for the thought process towards this qn?
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Originally posted by Flying grenade:
[Al(H2O)6]3+ is formed from AlCl3 in excess water, because Al is in period 3, there's energetically accessible vacant d-orbitals , so Al can expand it's octet, is it?
Why doesnt it form [Al(H2O)4]3+?
[Be(H2O)4]2+ is formed instead of [Be(H2O)6]2+ from BeCl2 in excess water, because period 2 elements are unable to expand their octet is it?
Yes correct. Even if Al can accept just 4 H2O ligands, but it'll prefer 6, because bond forming (including dative bonds) are exothermic and hence thermodynamically favorable. -
Originally posted by Flying grenade:
Hi Ultima, may i ask you,
AJC /2014/P3/2(b)
State the shape and bond angle of [Pt(NH3)4]2+
I think the challenge to this question is, to determine the number of valence electrons of centre atom Pt right?
This is what i did : i do not know how to determine valence e- of Pt, as we did not learn how to write electronic configuration beyond period 3 (is it?) .
So i just assume Pt has 2 valence e- , so lost 2 valence e- , and 4 NH3 ligands datively bonded to Pt, i get a tetrahedral shape, bond angle 109°
The answer wrote :
Shape : square planar
Bond angle : 90°
But, accept tetrahedral 109°
How did they get sq planar or assumed pt has 6 valence e-?
Did a google search for electronic config of Pt , it's complex and their still debating abt it
May i ask u for guidance for the thought process towards this qn?
For transition metal cations, you can ignore any lone pairs from the metal atom itself, and deduce its geometry solely based on the coordinate dative bonds accepted from its ligands. If the ligands are huge, eg. Cl- ligands, then to avoid excessive steric strain from van der Waals repulsions between the ligands, there will only be 4 instead of 6 ligands. When you know there are 4 ligands, then whether the geometry of the coordination complex is tetrahedral or square planar depends on the Jahn–Teller effect (which is of course way beyond the A level H2 syllabus), as well as other conditions that favor either stereoisomer (some coordination complexes can exist as an equilibrium between tetrahedral and square planar, with the position of equilibrium affected by other factors beyond A levels).For A level purposes, unless the coordination complex is well-known and student familiarity is expected (eg. such as Tetraamminecopper(II) sulfate and Cisplatin), Cambridge (and AJC here) will accept either square planar or tetrahedral, as long as the corresponding bond angles are correct.
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Thank you.
Yeah it's perplexing, AJ wrote the answer square planar, and then a second statement below
also accept tetrahedral,
Its like suggesting sq planar should come to mind first rather than tetrahedral.
But after your input, i feel that tetrahedral is more straightforward.
I was thinking, is it becos they just assume structure consisting of 4 bond pairs? If they did this, then there's a possibility of seesaw/distorted tetrahedral.
Why then, distorted tetrahedral not accepted/wrong?
Or is it acceptable?
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Originally posted by Flying grenade:
Thank you.
Yeah it's perplexing, AJ wrote the answer square planar, and then a second statement below
also accept tetrahedral,
Its like suggesting sq planar should come to mind first rather than tetrahedral.
But after your input, i feel that tetrahedral is more straightforward.
I was thinking, is it becos they just assume structure consisting of 4 bond pairs? If they did this, then there's a possibility of seesaw/distorted tetrahedral.
Why then, distorted tetrahedral not accepted/wrong?
Or is it acceptable?
As I said, ignore lone pairs from the transition metal itself, take it that the only electron charge clouds present are the dative bonds from the ligands. And the AJC qn was closely related to Cisplatin, which is a famous chemotherapy drug and hence you should be aware that it's square planar and has cis-trans isomerism (although it's not in the Singapore H2 syllabus, but it's in the IB and other A level Chem syllabuses, and if you're serious about A grade, you should expose yourself to stuff like this beyond your own A level syllabus). -
Wow! thank you for the interesting info! appreciate it.
Yes, i do try my best to search and research new information for chem and other subjects, to understand better.
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Ultima, may i ask you how to calculate formal charge?
for e.g. this qn
For instance, in the NO molecule, because ionic charge is the sum of formal charges, and ionic charge for a neutral molecule is zero, hence we require both the N and O atoms to have no formal charge. For the O atom to have no formal charge but has a stable octet, it will have 2 lone pairs and 2 bond pairs. Hence the N atom has 2 bond pairs. If it has 1 lone pair, it has a unipositive formal charge. If it has 2 lone pairs, it has a uninegative formal charge. Therefore, in order for the N atom to have no formal charge, it must have 1.5 lone pairs, ie. an unpaired electron.
i forgot the method u taught T__T ): , which is just something minus something, which is something like the number of valence electrons minus the number of bonds and electrons attached to the atom. similar to this https://upload.wikimedia.org/wikipedia/commons/thumb/2/25/Ls4.png/450px-Ls4.png <-- how do one calculate like this?
i also tried using this method ( Is this method good to use?
where V is the number of valence electrons of the neutral atom in isolation (in its ground state); N is the number of non-bonding valence electrons on this atom in the molecule; and B is the total number of electrons shared in bonds with other atoms in the molecule.
IM not sure why, different scenarios tried different method, sometimes works sometimes doesn't , i get very confused, i've spent alot time on trying to figure this out ): please help !
If it has 1 lone pair, it has a unipositive formal charge. If it has 2 lone pairs, it has a uninegative formal charge.
how do you get the above? ^
i got this https://www.dropbox.com/s/4n9mx1l4dgmptol/20160320_014618.jpg?dl=0
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what are the steps of drawing dot and cross? e.g. N2O , N3- etc
is it determine(guess and check) the possible combinations of formal charges first? e.g. to sum up to 0 for N2O, think 0,1,-1, and -1, +3, -2
to sum up to -1 for N3- , guess and check 0,0, -1, and -1, +1, -1 , etc?
then attempt to draw the electrons for dot and cross or the bonds for Kekule structure?
whats the thought process for drawing : https://www.dropbox.com/s/taf48afts843k5q/20160320_023721.jpg?dl=0
and pentazenium
In valence bond theory, pentazenium can be described by six resonance structures:
- [N≡N+−N−−N+≡N]+
[N−=N+=N−N+≡N]+ [N≡N+−N=N+=N−]+ [N+=N−N−−N=N+]+ [N+=N−N−−N+≡N]+ [N≡N+−N−−N=N+]+ - sorry and thank you.
- if too burden to explain this via forums and typing out, its ok u dont have to type out
- [N≡N+−N−−N+≡N]+
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I think dot and cross drawing structure questions can all be done via guess and check, but it would be good to learn the scientific and systematic way of drawing structures , and also guess and check at times too time consuming
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Ummm yeah, fact is, it reeaaallly is way too cumbersome, ineffective and inefficient to attempt a resolution of this topic on an online forum, it's infinitely more effective to discuss this face-to-face during tuition. PM me when you're ready to join my tuition... 7 months left to 2016 A levels.
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The purpose of orbital hybridisation is to achieve the ideal bond angles as predicted by VSEPR theory, to maximise stabilities by minimising electron pair repulsions , is it?
Does it also " useful in the explanation of molecular geometry and atomic bonding properties" ? (from wiki)
I ask this becos of this ''The electrons rearrange themselves again in a process called hybridisation. This reorganises the electrons into four identical hybrid orbitals called sp3hybrids (because they are made from one s orbital and three p orbitals)" ,
From http://www.chemguide.co.uk/basicorg/bonding/methane.html#top
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Originally posted by Flying grenade:
The purpose of orbital hybridisation is to achieve the ideal bond angles as predicted by VSEPR theory, to maximise stabilities by minimising electron pair repulsions , is it?
Does it also " useful in the explanation of molecular geometry and atomic bonding properties" ? (from wiki)
I ask this becos of this ''The electrons rearrange themselves again in a process called hybridisation. This reorganises the electrons into four identical hybrid orbitals called sp3hybrids (because they are made from one s orbital and three p orbitals)" ,
From http://www.chemguide.co.uk/basicorg/bonding/methane.html#top
Yes, they're both talking about the same or related concepts. So what's your question, exactly? -
Do we need to know how a pi bond is formed? Or just say side on overlap will suffice?
P.s. tis qn and the above qn not related
I ask that above qn becos the first paragraph is what you've told me before. So i ask if hybridisation theory also explains 'atomic bonding properties'
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Originally posted by Flying grenade:
Do we need to know how a pi bond is formed? Or just say side on overlap will suffice?
That *is* how a pi bond is formed : sideways or side-on overlap of unhybridized p orbitals (pz if double bond, py and pz if triple bond, but such details are not required unless specified by the question). -
Originally posted by Flying grenade:
Do we need to know how a pi bond is formed? Or just say side on overlap will suffice?
P.s. tis qn and the above qn not related
I ask that above qn becos the first paragraph is what you've told me before. So i ask if hybridisation theory also explains 'atomic bonding properties'
Atomic bonding (ie. covalent bonding between atoms) involves hybridization of orbitals (to achieve the ideal bond angles as predicted by VSEPR theory). These are related concepts, and testable at A level H2 Chem syllabus. Only molecular orbital theory is beyond A level H2 Chem syllabus. -
神
感�
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how a pi bond is formed : sideways or side-on overlap of unhybridized p orbitals (pz if double bond, py and pz if triple bond
U mean for double bond, both atom 's unhybridised pz orbital side on overlap, is it?
And for triple bond, both atom's side-on overlap of unhybridized py and pz orbitals , is it?
No px is it because the other 1 of the sp hybridised orbital used to bond with another atom e.g. H , the other sp hybridised orbital sigma bonded with the other Carbon atom?
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Yes, when there's a triple bond, the C atom's px orbital has already been used up to form the two 2sp hybridized orbitals (linear or 180 deg), in turn used to overlap head-on or end-on, to form the sigma bonds with the 2 adjacent atoms.
Read CS Toh's A level Study Guide's 2 pages (3rd page is on benzene) on orbital hybridization in the chapter "Intro to Org Chem", sums it up nicely for the H2 syllabus.
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Ultima, everytime u take the A levels, your paper 1 - 5 all 100% correct?
