Chemistry Help !
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Hi,
I would like to ask how charges for the polyatomic ion such as SO4 2- is derived. Based on what I have found on the internet, there are 2 explanations :
1. Formal Charges
To draw a Lewis structure for sulfate ion, there are 6 + 6(4) + 2 = 32 electrons for bonding
After doing the bonding and stuff, the two oxygen atoms that has a single bond with sulfur has both 6-6-(2/2) = -1 charge each, thus giving SO4 a negative charge of 2.
But there's something I don't understand. How do you derive a charge of 2- in the first place without even adding the 6 + 6(4) + 2 = 32 ? That's not valid. Cause if I simply give you SO4 how do you even know you have to add another two more electrons to the structure and derive the formal charges of each atom ?
That brings me to the next point,
2. Bonding
30 - 8 (4 bondings between each oxygen and sulfur atoms) = 22
Putting the 22 electrons in place, I find out that the last oxygen atom lacks 2 more electrons to fulfil a stable octet.
Thus I derive that SO4 has a negative charge of 2-
And I get the following diagram
My question is, which is right and which is wrong ?
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Originally posted by H4x0ru5:
Hi,
I would like to ask how charges for the polyatomic ion such as SO4 2- is derived. Based on what I have found on the internet, there are 2 explanations :
1. Formal Charges
To draw a Lewis structure for sulfate ion, there are 6 + 6(4) + 2 = 32 electrons for bonding
After doing the bonding and stuff, the two oxygen atoms that has a single bond with sulfur has both 6-6-(2/2) = -1 charge each, thus giving SO4 a negative charge of 2.
But there's something I don't understand. How do you derive a charge of 2- in the first place without even adding the 6 + 6(4) + 2 = 32 ? That's not valid. Cause if I simply give you SO4 how do you even know you have to add another two more electrons to the structure and derive the formal charges of each atom ?
That brings me to the next point,
2. Bonding
30 - 8 (4 bondings between each oxygen and sulfur atoms) = 22
Putting the 22 electrons in place, I find out that the last oxygen atom lacks 2 more electrons to fulfil a stable octet.
Thus I derive that SO4 has a negative charge of 2-
And I get the following diagram
My question is, which is right and which is wrong ?
This topic is something best dealt with in person during tuition, and is difficult to explain over an online forum. I'll give brief comments, do the best you can to understand them, and anything further (eg. if you still do not fully comprehend this topic) ask your school teacher or tuition teacher if you have one.
I'll comment directly on your self-drawn structure, which should go some way to auto-answer your previous queries.
First of all, these are more correctly called Kekule structures, rather than Lewis structures (unsurprisingly, different chemists, schools, examiners, use different terms).
Secondly (and you'll find some school teachers disagreeing on this point; (unlke fussy school teachers) Cambridge actually doesn't care about petty little details like this, and will accept either presentation or method), you should always show all formal charges within the square brackets, as well as simultaneously show the ionic charge outside the square brackets. Afterall, the square brackets symbolize summation (for physical and inorganic chem purposes; we do not use square brackets for organic chem purposes).
Accordingly, your drawing of the sulfate(VI) ion is erroneous in that you omitted the formal charges on your atoms : a dipositive formal charge on the S atom, and a uninegative formal charge on each of the singly bonded O atoms.
Also you neglected to show dative bonds. Everytime you see positive and negative formal charges on adjacent atoms, you should suspect the presence of dative bonds.
(Interestingly, dative bond presentation is different for all three : physical chemistry, inorganic chemistry and organic chemistry. Again, different JC teachers will disagree with each other on the best presentation for these, and again Cambridge doesn't care so much, and will accept a variety of presentations. Nonetheless, I'll briefly indicate the best presentations :
physical chem : show formal charges after dative bond formation. And there should not be any lone pair at the base of the dative bond arrow.
inorganic chem (complex ions) : show formal charges before dative bond formation. And there should be a lone pair at the base of the dative bond arrow, on the ligand.
organic chem : show a curved arrow (for which the lone pair must be shown at the 'base' of the curved arrow) to illustrate electron flow, and in the next stage of the mechanism, a straight line (*no* dative bond arrowhead should be shown) is used to indicate the bond formed.
The difference between physical and inorganic chem presentations, arise from the fact that the electronegativity difference between non-metals is small, but between metals and non-metals is large. As such, formal charges after dative bond formation is more accurate for species in which all atoms are non-metals, but formal charges before dative bond formation is more accurate for complex ions in which ligands are significantly more electronegative than the metal ions they donate dative bonds to)
What you've drawn (ie. all singly bonded O atoms) is favoured by US chemists, while UK chemists favour the expanded octet version (ie. with some doubly bonded O atoms). If you could shrink yourself to the size of an atom and observe the sulfate(VI) ion, which structure would you see? The US version or the UK version?
Tis a trick question. Both versions are equally correct, because they are really resonance contributors of each other. The actual structure (that you would see, if you could shrink yourself to the size of an atom) is known as the resonance hybrid, and has partial double bond character for all 4 bonds with all 4 O atoms.
Cambridge will accept either the UK or the US presentation, but of course, the UK presentation is recommended for Cambridge 'A' level students. Note however, for period 2 elements (since they do not have vacant, energetically accessible 3d orbitals to use to expand their octet), both UK and US structures are the same.
For the UK presentation, this is how you draw the Kekule structure of the sulfate(VI) ion :
Ionic charge is the sum of formal charges.
Since the ionic charge is dinegative, we assume two of the O atoms have a uninegative formal charge.
This means these two O atoms must have 1 bond pair and 3 lone pairs.
The other two O atoms must have no formal charge, ie. 2 bond pairs and 2 lone pairs.
The S atom thus has 6 bond pairs, and being in Group VI, thus has no lone pairs, because we do not wish the S atom to have any formal charge (since the condition or formula, that ionic charge = sum of formal charges, is already satisfied by the two singly bonded uninegative formal charge O atoms).
That's how you arrive at the UK presentation of the Kekule structure of the sulfate(VI) ion :
For the US presentation, this is how you draw the Kekule structure of the sulfate(VI) ion :
Since we do not wish to expand the central S atom's octet, thus all O atoms must be singly bonded, and thus have 1 bond pair and 3 lone pairs, and therefore a uninegative formal charge each.
Since ionic charge is the sum of formal charges, it means we require the central S atom to have a dipositive formal charge, to balance out the tetranegative formal charges of the O atoms. Accordingly, we deduce the S atom has no lone pairs, because only when it has 4 bond pairs (which it already does, based on the O atoms' bonding) and zero lone pairs, will it have a dipositive formal charge, as S belongs to Group VI.
And that's how you arrive at the US presentation of the Kekule structure of sulfate(VI) ion :
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Wikipedia shows both (UK and US preferred) structures of the sulfate(VI) ion :
as well as all the common resonance contributors of the sulfate(VI) ion :
It is recommended that Singapore-Cambridge 'A' level students draw the UK presentation as follows, for the sulfate(VI) ion :
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Originally posted by UltimaOnline:
This topic is something best dealt with in person during tuition, and is difficult to explain over an online forum. I'll give brief comments, do the best you can to understand them, and anything further (eg. if you still do not fully comprehend this topic) ask your school teacher or tuition teacher if you have one.
I'll comment directly on your self-drawn structure, which should go some way to auto-answer your previous queries.
First of all, these are more correctly called Kekule structures, rather than Lewis structures (unsurprisingly, different chemists, schools, examiners, use different terms).
Secondly (and you'll find some school teachers disagreeing on this point; (unlke fussy school teachers) Cambridge actually doesn't care about petty little details like this, and will accept either presentation or method), you should always show all formal charges within the square brackets, as well as simultaneously show the ionic charge outside the square brackets. Afterall, the square brackets symbolize summation (for physical and inorganic chem purposes; we do not use square brackets for organic chem purposes).
Accordingly, your drawing of the sulfate(VI) ion is erroneous in that you omitted the formal charges on your atoms : a dipositive formal charge on the S atom, and a uninegative formal charge on each of the singly bonded O atoms.
Also you neglected to show dative bonds. Everytime you see positive and negative formal charges on adjacent atoms, you should suspect the presence of dative bonds.
(Interestingly, dative bond presentation is different for all three : physical chemistry, inorganic chemistry and organic chemistry. Again, different JC teachers will disagree with each other on the best presentation for these, and again Cambridge doesn't care so much, and will accept a variety of presentations. Nonetheless, I'll briefly indicate the best presentations :
physical chem : show formal charges after dative bond formation. And there should not be any lone pair at the base of the dative bond arrow.
inorganic chem (complex ions) : show formal charges before dative bond formation. And there should be a lone pair at the base of the dative bond arrow, on the ligand.
organic chem : show a curved arrow (for which the lone pair must be shown at the 'base' of the curved arrow) to illustrate electron flow, and in the next stage of the mechanism, a straight line (*no* dative bond arrowhead should be shown) is used to indicate the bond formed.
The difference between physical and inorganic chem presentations, arise from the fact that the electronegativity difference between non-metals is small, but between metals and non-metals is large. As such, formal charges after dative bond formation is more accurate for species in which all atoms are non-metals, but formal charges before dative bond formation is more accurate for complex ions in which ligands are significantly more electronegative than the metal ions they donate dative bonds to)
What you've drawn (ie. all singly bonded O atoms) is favoured by US chemists, while UK chemists favour the expanded octet version (ie. with some doubly bonded O atoms). If you could shrink yourself to the size of an atom and observe the sulfate(VI) ion, which structure would you see? The US version or the UK version?
Tis a trick question. Both versions are equally correct, because they are really resonance contributors of each other. The actual structure (that you would see, if you could shrink yourself to the size of an atom) is known as the resonance hybrid, and has partial double bond character for all 4 bonds with all 4 O atoms.
Cambridge will accept either the UK or the US presentation, but of course, the UK presentation is recommended for Cambridge 'A' level students. Note however, for period 2 elements (since they do not have vacant, energetically accessible 3d orbitals to use to expand their octet), both UK and US structures are the same.
For the UK presentation, this is how you draw the Kekule structure of the sulfate(VI) ion :
Ionic charge is the sum of formal charges.
Since the ionic charge is dinegative, we assume two of the O atoms have a uninegative formal charge.
This means these two O atoms must have 1 bond pair and 3 lone pairs.
The other two O atoms must have no formal charge, ie. 2 bond pairs and 2 lone pairs.
The S atom thus has 6 bond pairs, and being in Group VI, thus has no lone pairs, because we do not wish the S atom to have any formal charge (since the condition or formula, that ionic charge = sum of formal charges, is already satisfied by the two singly bonded uninegative formal charge O atoms).
That's how you arrive at the UK presentation of the Kekule structure of the sulfate(VI) ion :
For the US presentation, this is how you draw the Kekule structure of the sulfate(VI) ion :
Since we do not wish to expand the central S atom's octet, thus all O atoms must be singly bonded, and thus have 1 bond pair and 3 lone pairs, and therefore a uninegative formal charge each.
Since ionic charge is the sum of formal charges, it means we require the central S atom to have a dipositive formal charge, to balance out the tetranegative formal charges of the O atoms. Accordingly, we deduce the S atom has no lone pairs, because only when it has 4 bond pairs (which it already does, based on the O atoms' bonding) and zero lone pairs, will it have a dipositive formal charge, as S belongs to Group VI.
And that's how you arrive at the US presentation of the Kekule structure of sulfate(VI) ion :
Thank you for your wonderful post. I still have a question though. If I am given the ion SO4, and asked to find out the charges, how would I be able to find that out based on the Kekule structure ?
Is it by
1. Realising that 2 more electrons is need for the last oxgyen atom to achieve stable octet configuration (as indicated by the red dots) so simply given SO4 a charge of 2-
or
2. Through the formal charges of each atom.
But the problem is :
Without even knowing the charges of SO4 in the first place, a student will get 6 + 4(6) = 30 electrons.
and they will draw this
such that they will realise that the last oxygen atom has only 4 electrons. and if based on each formal charges
Sulfur would have = +2
the 3 oxygen atom = -3
and the last oxygen atom = 6 - 4 - (2/2) = +1
total charges = 0
this is,however, not valid.
3. The UK way of of drawing the structure is due to having knowledge that the ions has a 2- charge. so they know that there would be 30 + 2 and thus have this structure
and thus from the structure, having 2 oxygen atoms to form double bond with sulfur
so sulfur would have a charge of 0. and deriving this,
Without knowing the charges of SO4 in the first place, is it possible to derive it by drawing the Kekule structure ? And if so, how ?
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Originally posted by H4x0ru5:
Thank you for your wonderful post. I still have a question though. If I am given the ion SO4, and asked to find out the charges, how would I be able to find that out based on the Kekule structure ?
Is it by
1. Realising that 2 more electrons is need for the last oxgyen atom to achieve stable octet configuration (as indicated by the red dots) so simply given SO4 a charge of 2-
or
2. Through the formal charges of each atom.
But the problem is :
Without even knowing the charges of SO4 in the first place, a student will get 6 + 4(6) = 30 electrons.
and they will draw this
such that they will realise that the last oxygen atom has only 4 electrons. and if based on each formal charges
Sulfur would have = +2
the 3 oxygen atom = -3
and the last oxygen atom = 6 - 4 - (2/2) = +1
total charges = 0
this is,however, not valid.
3. The UK way of of drawing the structure is due to having knowledge that the ions has a 2- charge. so they know that there would be 30 + 2 and thus have this structure
and thus from the structure, having 2 oxygen atoms to form double bond with sulfur
so sulfur would have a charge of 0. and deriving this,
Without knowing the charges of SO4 in the first place, is it possible to derive it by drawing the Kekule structure ? And if so, how ?
No, you're looking at the problem the wrong way.
For 'A' level purposes, you will either be given the formulae (including the charge) in the question, or are expected to know the formulae (including the charge), and then tasked to draw the structure.
OR (but this is not common for 'A' levels, but is even simpler to do!)
You're given the structure, and tasked to find the formal charges and hence overall ionic charge.
Simply indicate the formal charges, and then add them all up to obtain the ionic charge.
For instance, if the structure given has :
a singly bonded O atom, it means it has 1 bond pair and 3 lone pairs, and thus a uninegative formal charge (since O is in Group VI).
a doubly bonded O atom, it means it has 2 bond pairs and 2 lone pairs, and thus no formal charge (since O is in Group VI).
a triply bonded O atom, it means it has 3 bond pairs and 1 lone pairs, and thus a unipositive formal charge (since O is in Group VI).
You can extrapolate this for any atom, eg. S, N, C, etc.
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One more minor problem with your drawings :
To indicate the dipositive formal charge on S, you must write it as 2+, instead of +2. The +ve or -ve sign is behind the number.
In contrast, Oxidation States (OS) aka Oxidation Numbers (ON), are written in brackets next to the atom, and has the +ve or -ve sign in front of the number.
Oxidation State = Formal Charge + Electronegativity consideration.
Thus for the US resonance contributor, OS of S = (+2) + (+4) = +6
Thus for the UK resonance contributor, OS of S = (0) + (+6) = +6
Hence for SO4 2-, the latin name is sulfate ion, the stock name is sulfate(VI) ion (ie. the stock name has the OS of the heteroatom indicated).
Note that Ionic Charge is the Sum of Formal Charges, and also the Sum of Oxidation States.
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Originally posted by UltimaOnline:
No, you're looking at the problem the wrong way.
For 'A' level purposes, you will either be given the formulae (including the charge) in the question, or are expected to know the formulae (including the charge), and then tasked to draw the structure.
OR (but this is not common for 'A' levels, but is even simpler to do!)
You're given the structure, and tasked to find the formal charges and hence overall ionic charge.
Simply indicate the formal charges, and then add them all up to obtain the ionic charge.
For instance, if the structure given has :
a singly bonded O atom, it means it has 1 bond pair and 3 lone pairs, and thus a uninegative formal charge (since O is in Group VI).
a doubly bonded O atom, it means it has 2 bond pairs and 2 lone pairs, and thus no formal charge (since O is in Group VI).
a triply bonded O atom, it means it has 3 bond pairs and 1 lone pairs, and thus a unipositive formal charge (since O is in Group VI).
You can extrapolate this for any atom, eg. S, N, C, etc.
Ah thank you very much. Really enlighten me about this.
I have a last question relating to ions.
Iodine Tetrafluoride, IF4 - is a anion.
Iodine, the central atom, has 4 bond pairs and 2 lone pairs. The Fluorine atom has 1 bond pair, 3 lone pairs.
This gives Iodine a formal charge of -1. Each fluorine atom has a charge of 0.
Why doesn't the compound IF4 exist, since without the extra electron attached to Iodine, Iodine would have a 4 bond pairs, and 3 electrons attached to it. The formal charge of Iodine would therefore be 7-3-(8/2) = 0 Why does it exists as an anion instead of a compound ?
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Originally posted by H4x0ru5:
Ah thank you very much. Really enlighten me about this.
I have a last question relating to ions.
Iodine Tetrafluoride, IF4 - is a anion.
Iodine, the central atom, has 4 bond pairs and 2 lone pairs. The Fluorine atom has 1 bond pair, 3 lone pairs.
This gives Iodine a formal charge of -1. Each fluorine atom has a charge of 0.
Why doesn't the compound IF4 exist, since without the extra electron attached to Iodine, Iodine would have a 4 bond pairs, and 3 electrons attached to it. The formal charge of Iodine would therefore be 7-3-(8/2) = 0 Why does it exists as an anion instead of a compound ?
For it to exist as a molecule (ie. ionic charge = zero), one or more atoms (to be precise, the I atom) would have an unpaired electron (to be precise, 1.5 lone pairs, and the ion would be a free radical) and is thus too unstable to exist.
IF4-
The -ve formal charge is on the central I atom (since it has a larger atomic radius and a lower charge density and is thus more stable, despite its lower electronegativity).
In addition, note that F (being in period 2) cannot expand its octet, and thus to have a stable octet, it must have 1 bond pair and 3 lone pairs, and hence no formal charge (being in Grp VII).
Since I is in Grp VII, to have a -ve formal charge, we deduce that it has 8 valence electrons, ie. 4 bond pairs, 2 lone pairs.
Hence the electron geometry is octahedral, and the ionic geometry is square planar.
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Originally posted by UltimaOnline:
For it to exist as a molecule (ie. ionic charge = zero), one or more atoms (to be precise, the I atom) would have an unpaired electron (to be precise, 1.5 lone pairs, and the ion would be a free radical) and is thus too unstable to exist.
IF4-
The -ve formal charge is on the central I atom (since it has a larger atomic radius and a lower charge density and is thus more stable, despite its lower electronegativity).
In addition, note that F (being in period 2) cannot expand its octet, and thus to have a stable octet, it must have 1 bond pair and 3 lone pairs, and hence no formal charge (being in Grp VII).
Since I is in Grp VII, to have a -ve formal charge, we deduce that it has 8 valence electrons, ie. 4 bond pairs, 2 lone pairs.
Hence the electron geometry is octahedral, and the ionic geometry is square planar.
Wow thanks alot UltimaOnline ! Really appreciate it. This forum is super useful.
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Originally posted by H4x0ru5:
Wow thanks alot UltimaOnline ! Really appreciate it. This forum is super useful.
No problem. You're welcome.Although do note (not just H4x0ru5, but to all students visiting this forum), that for 'normal' type of questions*, ie. for the usual stuff that *is* taught within JCs, that you've got to attempt the questions yourself first, and post your answers here for me to comment on, instead of posting just the questions and expect me (or others) to do your homework for you.
(*formal charges aren't usually taught in JCs, which is one of the reasons why JC students still lack a fundamental understanding of Chemistry up till the day they take the 'A' levels; and which is why I've been extra generous in providing info for this particular topic, here on this forum)
Take note (for all students visiting this forum) : for any Chemistry question that you (ie. all students visiting this forum) wanna ask here on the forum, you should always first look up the relevant info on Wikipedia and the 3 recommended 'A' level Chem websites (indicated on my webpage here), before posting your Chem question here on this forum.
Links to Rod Beavon's, Jim Clarke's, Doc Brown's 'A' level Chemistry websites.
http://infinity.usanethosting.com/Tuition/#H2_Planning_and_other_goodies
If I find the question posted to be something you could have easily checked it up yourself online (eg. on Wikipedia, or the 3 'A' level Chem websites above), I may not even reply to your post. If the question is reasonable enough, and I choose to reply, most often it'll be to give my comments and guidance, rather than direct answers (remember : this HomeWork Forum is to help you understand your homework, not do your homework for you).
Have Fun with Chemistry!
