Discussion:
Oxidation number of atoms in thiocyanate ion
(too old to reply)
p***@woelen.nl
2004-10-22 18:35:46 UTC
Permalink
Hello,

Recently I posted a question about oxidation products of thiocyanate
and from this we got a discussion on the oxidation state of the atoms
in thiocyanate ions. There already was a thread on this subject in
spring 2001 (google thiocyanate oxidation at groups:sci.chem). Up to
now the answers are inconsistent and somewhat confusing.

So my question is: what are the oxidation numbers of S, C and N in
thiocyanate? Even more interesting to me, can you describe, how these
oxidation numbers are derived?

Thanks,

Wilco
Uncle Al
2004-10-22 18:58:40 UTC
Permalink
Post by p***@woelen.nl
Hello,
Recently I posted a question about oxidation products of thiocyanate
and from this we got a discussion on the oxidation state of the atoms
in thiocyanate ions. There already was a thread on this subject in
spring 2001 (google thiocyanate oxidation at groups:sci.chem). Up to
now the answers are inconsistent and somewhat confusing.
So my question is: what are the oxidation numbers of S, C and N in
thiocyanate? Even more interesting to me, can you describe, how these
oxidation numbers are derived?
s(-2)
C(+4)
N(-3)

Derived as usual. Pnictides and chalcogenies are more electronegative
than tetralides.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
Mohammed Farooq
2004-10-23 11:58:10 UTC
Permalink
***@woelen.nl wrote
Further discussion on oxidation numbers is pasted here for
[Text snipped]
EN(S) > EN(C)
EN(N) > EN(C) ??
Yes.
S normally has 6 outer electrons, now it has 8, so OxNum(S) == -2.
C normally has 4 outer electrons, now it has none, so OxNum(C) == +4.
N normally has 5 outer electrons, now it has 8, so OxNum(N) == -3.
In this whole exercise, I put the single negative charge on the N-atom,
otherwise I get no octets around the atoms. I doubt whether this is
100% correct, isn't that single electron delocalized and 'spread out'
over the entiry SCN-unit?
There are many resonance structures of [SCN](-), the electron *pair* (
equivalent to a single pi-bond) rather than a single electron that is
delocalized, and it does not matter which resonance structure is used.
Ofcourse you have to use one resonance form at a time for calculating
oxidation number of each.
Why is there so much discussion on the OxNum of S? This method
unambiguously gives -2 for the S, while the value 0 also is proposed
for the OxNum of S.
If Allred-Rochow electronegativity table is used carbon is more
electronegative than sulfur, then sulfur has 0 oxidation state. Carbon
has the value 2.5 and sulfur is 2.44.
For CO, I don't get a full octet around the C if I assume that O does
not go further than a double bond. Is this correct? I get
:C::O
The OxNums I get are, however, +2 for C and -2 for O.
Carbon monoxide has a triple bond [:C:::O:], which still gives the
same result.
For molecules like NO and NO2 I have the same problem when making Lewis
dot-structures, I even end up with odd number of electrons around the
atoms.
I try to do them in my spare time.
Things get even more complex for resonance stabilized structures
like benzene. How do you solve these situations with Lewis
dot-structures?
Use one snapshot picture at a time. Have a look at
http://www.unomaha.edu/~etisko/Chem1180/Lecture_Handouts/PDF_Files/Chapter10.PDF

BTW, regarding the eleven reactants equation, how much time it would
have taken you to solve it manually using a nonprogrammable scientific
calculator? The original author said it took 2 hours to do that by
non-convetional methods.
p***@woelen.nl
2004-10-23 12:47:42 UTC
Permalink
Mohammed Farooq wrote:
<snipped>
Post by Mohammed Farooq
For CO, I don't get a full octet around the C if I assume that O does
not go further than a double bond. Is this correct? I get
:C::O
The OxNums I get are, however, +2 for C and -2 for O.
Carbon monoxide has a triple bond [:C:::O:], which still gives the
same result.
This is new to me. I thought that oxygen bonding was limited to double
bonds. Is this third bond a real bond? Both electrons come from the
oxygen, while for the other two bonds half of the electrons is
delivered by the two atoms, which are bonded together.
Post by Mohammed Farooq
For molecules like NO and NO2 I have the same problem when making Lewis
dot-structures, I even end up with odd number of electrons around the
atoms.
I try to do them in my spare time.
Things get even more complex for resonance stabilized structures
like benzene. How do you solve these situations with Lewis
dot-structures?
Use one snapshot picture at a time. Have a look at
http://www.unomaha.edu/~etisko/Chem1180/Lecture_Handouts/PDF_Files/Chapter10.PDF
Thanks for the link. I certainly will study this and see if I can grasp
the ideas, worked out there.
Post by Mohammed Farooq
BTW, regarding the eleven reactants equation, how much time it would
have taken you to solve it manually using a nonprogrammable
scientific
Post by Mohammed Farooq
calculator? The original author said it took 2 hours to do that by
non-convetional methods.
I think that solving this manually would require several hours for me.
Doing so is extremely error-prone and I really dislike doing that kind
of work. You probably noticed that yourself already with the MnO4(-) /
SCN(-) redox equation with its not-so-simple ratios.
The computer program just runs a few seconds. I first determined the
singular value decomposition of the matrix to find out its rank
reliably. This takes less than a second on a 600 MHz Pentium III
processor. Having determined the rank (and hence the dimension of its
null space), it is trivial to solve the system, taking just a fraction
of a second. Solving the problem took me approximately 30 minutes, most
of the time being the typing of the matrix and checking whether I made
no typos.
f***@hotmail.com
2004-10-23 16:00:25 UTC
Permalink
Post by p***@woelen.nl
This is new to me. I thought that oxygen bonding was limited to double
bonds. Is this third bond a real bond?
triple bond = one sigma bond + two pi bonds.
Post by p***@woelen.nl
For molecules like NO and NO2 I have the same problem when making
Lewis
dot-structures,
NO is said to have a bond order of 2.5(two double bonds + one odd
electron)
Post by p***@woelen.nl
I think that solving this manually would require several hours for me.
Doing so is extremely error-prone and I really dislike doing that kind
of work. You probably noticed that yourself already with the MnO4(-) /
SCN(-) redox equation with its not-so-simple ratios.
The computer program just runs a few seconds. I first determined the
singular value decomposition of the matrix to find out its rank
reliably. This takes less than a second on a 600 MHz Pentium III
processor. Having determined the rank (and hence the dimension of its
null space), it is trivial to solve the system, taking just a
fraction
Post by p***@woelen.nl
of a second. Solving the problem took me approximately 30 minutes, most
of the time being the typing of the matrix and checking whether I made
no typos.
Can you do me a favor if you spare few minutes to solve this problem
with a computer program to 8 decimal places, if it not so cubersome for
you. I just need to check, because neither I nor the the teacher have
been able to get the correct answer perhaps the large number of decimal
places are required and rounding off gives a significant error.

The system is:
[Ca] 146.12 + [Ba] 193.66 + [Sr] 243.36 = 0.56246
[Ca] 128.10 + [Ba] 175.64 + [Sr] 225.35 = 0.50980
[Ca] 100.09 + [Ba] 147.63 + [Sr] 197.34 = 0.42730
p***@woelen.nl
2004-10-23 18:06:17 UTC
Permalink
Post by p***@woelen.nl
This is new to me. I thought that oxygen bonding was limited to
double
Post by p***@woelen.nl
bonds. Is this third bond a real bond?
triple bond = one sigma bond + two pi bonds.
I'll look up what this means for the distribution of electrons. My
knowledge of the different bonding types is buried deeply, although
once, I grasped the idea behind it.
Post by p***@woelen.nl
For molecules like NO and NO2 I have the same problem when making
Lewis
dot-structures,
NO is said to have a bond order of 2.5(two double bonds + one odd
electron)
Can you do me a favor if you spare few minutes to solve this problem
with a computer program to 8 decimal places, if it not so cubersome for
you. I just need to check, because neither I nor the the teacher have
been able to get the correct answer perhaps the large number of decimal
places are required and rounding off gives a significant error.
[Ca] 146.12 + [Ba] 193.66 + [Sr] 243.36 = 0.56246
[Ca] 128.10 + [Ba] 175.64 + [Sr] 225.35 = 0.50980
[Ca] 100.09 + [Ba] 147.63 + [Sr] 197.34 = 0.42730
The set of equations, you supplied is singular within the given
accuracy:
The singular value decomposition (SVD) of the matrix on the left hand
is
5.3546e+02
1.2642e+01
1.9671e-03
The largest eigenvalue of sqrt(AA') is more than 10^5 times as large as
the smallest one. This means that you loose approximately 5 digits of
significance, when determining the numerical solution of this system.
Because you only have 5 digits, one can safely state that this system
is singular within the given precision. The SVD shows that you have two
independent relations, the third one is just a combination of the other
2.

For your information:
The third equation can be written as eq3 = eq2 - a*(eq1 - eq2) within
the given accuracy of 5 digits, where a = 1.554384.
E.g. 100.09 = 128.10 - a*(146.12-128.10). The same relation exists for
all other coefficients.


Nevertheless, I can supply you with a solution, computed at 15 digit
accuracy internally, but I'm afraid this solution is meaningless:
[Ca] = 0.04362447
[Ba] = -0.08224782
[Sr] = 0.04156873

I'm wondering, don't you have computer hardware and software, with
which you can solve this type of problems? There are quite some good
open source products out there, such as Octave, pari/gp and libraries
like GSL.
f***@hotmail.com
2004-10-23 18:32:12 UTC
Permalink
Post by p***@woelen.nl
Post by p***@woelen.nl
This is new to me. I thought that oxygen bonding was limited to
double
Post by p***@woelen.nl
bonds. Is this third bond a real bond?
triple bond = one sigma bond + two pi bonds.
I'll look up what this means for the distribution of electrons. My
knowledge of the different bonding types is buried deeply, although
once, I grasped the idea behind it.
Post by p***@woelen.nl
For molecules like NO and NO2 I have the same problem when
making
Post by p***@woelen.nl
Lewis
dot-structures,
NO is said to have a bond order of 2.5(two double bonds + one odd
electron)
Can you do me a favor if you spare few minutes to solve this
problem
Post by p***@woelen.nl
with a computer program to 8 decimal places, if it not so cubersome
for
you. I just need to check, because neither I nor the the teacher have
been able to get the correct answer perhaps the large number of
decimal
places are required and rounding off gives a significant error.
[Ca] 146.12 + [Ba] 193.66 + [Sr] 243.36 = 0.56246
[Ca] 128.10 + [Ba] 175.64 + [Sr] 225.35 = 0.50980
[Ca] 100.09 + [Ba] 147.63 + [Sr] 197.34 = 0.42730
The set of equations, you supplied is singular within the given
The singular value decomposition (SVD) of the matrix on the left hand
is
5.3546e+02
1.2642e+01
1.9671e-03
The largest eigenvalue of sqrt(AA') is more than 10^5 times as large as
the smallest one. This means that you loose approximately 5 digits of
significance, when determining the numerical solution of this system.
Because you only have 5 digits, one can safely state that this system
is singular within the given precision. The SVD shows that you have two
independent relations, the third one is just a combination of the other
2.
The third equation can be written as eq3 = eq2 - a*(eq1 - eq2) within
the given accuracy of 5 digits, where a = 1.554384.
E.g. 100.09 = 128.10 - a*(146.12-128.10). The same relation exists for
all other coefficients.
Nevertheless, I can supply you with a solution, computed at 15 digit
[Ca] = 0.04362447
[Ba] = -0.08224782
[Sr] = 0.04156873
I'm wondering, don't you have computer hardware and software, with
which you can solve this type of problems? There are quite some good
open source products out there, such as Octave, pari/gp and libraries
like GSL.
Thank you Wilco. I will show this solution to him (he said the he never
got the correct answer...perhaps he didn't think of eq3 = eq2 - a*(eq1
- eq2). This was a problem of a mixture of calcicum , barium and
strontium oxalates where were heated and the weight loss was recorded
at three different temperatures and hence three equations. This is
called thermogravimetric analysis.
Wilco, you would be surprised (but shouldn't be for someone from
Pakistan) that I am hearing of those names for the first time. I lag
behind atleast 20 years :-(.
p***@woelen.nl
2004-10-24 21:16:59 UTC
Permalink
Post by p***@woelen.nl
Post by p***@woelen.nl
This is new to me. I thought that oxygen bonding was limited to
double
Post by p***@woelen.nl
bonds. Is this third bond a real bond?
triple bond = one sigma bond + two pi bonds.
I'll look up what this means for the distribution of electrons. My
knowledge of the different bonding types is buried deeply, although
once, I grasped the idea behind it.
Post by p***@woelen.nl
For molecules like NO and NO2 I have the same problem when
making
Post by p***@woelen.nl
Lewis
dot-structures,
NO is said to have a bond order of 2.5(two double bonds + one odd
electron)
Can you do me a favor if you spare few minutes to solve this
problem
Post by p***@woelen.nl
with a computer program to 8 decimal places, if it not so
cubersome
Post by p***@woelen.nl
for
you. I just need to check, because neither I nor the the teacher
have
Post by p***@woelen.nl
been able to get the correct answer perhaps the large number of
decimal
places are required and rounding off gives a significant error.
[Ca] 146.12 + [Ba] 193.66 + [Sr] 243.36 = 0.56246
[Ca] 128.10 + [Ba] 175.64 + [Sr] 225.35 = 0.50980
[Ca] 100.09 + [Ba] 147.63 + [Sr] 197.34 = 0.42730
The set of equations, you supplied is singular within the given
The singular value decomposition (SVD) of the matrix on the left hand
is
5.3546e+02
1.2642e+01
1.9671e-03
The largest eigenvalue of sqrt(AA') is more than 10^5 times as
large
as
Post by p***@woelen.nl
the smallest one. This means that you loose approximately 5 digits of
significance, when determining the numerical solution of this system.
Because you only have 5 digits, one can safely state that this system
is singular within the given precision. The SVD shows that you have
two
Post by p***@woelen.nl
independent relations, the third one is just a combination of the
other
Post by p***@woelen.nl
2.
The third equation can be written as eq3 = eq2 - a*(eq1 - eq2) within
the given accuracy of 5 digits, where a = 1.554384.
E.g. 100.09 = 128.10 - a*(146.12-128.10). The same relation exists
for
Post by p***@woelen.nl
all other coefficients.
Nevertheless, I can supply you with a solution, computed at 15 digit
[Ca] = 0.04362447
[Ba] = -0.08224782
[Sr] = 0.04156873
I'm wondering, don't you have computer hardware and software, with
which you can solve this type of problems? There are quite some good
open source products out there, such as Octave, pari/gp and
libraries
Post by p***@woelen.nl
like GSL.
Thank you Wilco. I will show this solution to him (he said the he never
got the correct answer...perhaps he didn't think of eq3 = eq2 - a*(eq1
- eq2). This was a problem of a mixture of calcicum , barium and
strontium oxalates where were heated and the weight loss was recorded
at three different temperatures and hence three equations. This is
called thermogravimetric analysis.
Wilco, you would be surprised (but shouldn't be for someone from
Pakistan) that I am hearing of those names for the first time. I lag
behind atleast 20 years :-(.
Farooq,

I've looked up the source code of my program, which solves chemical
equations. Happily I could find it back. If it helps you, I can make
the code public and give it to you. It is C-code. I compiled it on
UNIX/Linux, but it also should compile on Windows. The program has been
'sleeping' here for many years and I see no reason to keep it alone for
me. There are no legal issues in using it, it is entirely written by
myself as a hobby project and does not depend on any external library.
If compiling the program is difficult for you, then I can try to make a
binary/executable for you, but in that case I need a specification of
the hardware and OS you have. Only if I have similar hardware and OS, I
can build it for you.
Anyway, I'll cleanup the code a little bit and put some legal blurp in
the code. This will take me a few more hours. I expect the code to be
suitable for releasing after just a few days from now.
f***@hotmail.com
2004-10-25 04:52:32 UTC
Permalink
Post by p***@woelen.nl
Farooq,
I've looked up the source code of my program, which solves chemical
equations. Happily I could find it back. If it helps you, I can make
the code public and give it to you. It is C-code. I compiled it on
UNIX/Linux, but it also should compile on Windows. The program has been
'sleeping' here for many years and I see no reason to keep it alone for
me. There are no legal issues in using it, it is entirely written by
myself as a hobby project and does not depend on any external
library.
Post by p***@woelen.nl
If compiling the program is difficult for you, then I can try to make a
binary/executable for you, but in that case I need a specification of
the hardware and OS you have. Only if I have similar hardware and OS, I
can build it for you.
Would I be able to use it, I have no knowledge of programming C-code
etc?
Post by p***@woelen.nl
Anyway, I'll cleanup the code a little bit and put some legal blurp in
the code. This will take me a few more hours. I expect the code to be
suitable for releasing after just a few days from now.
p***@woelen.nl
2004-10-25 06:29:04 UTC
Permalink
<snipped>
Post by f***@hotmail.com
Would I be able to use it, I have no knowledge of programming C-code
etc?
I'll see if I can build an executable for you. Using this will not be
harder than using a webbrowser or word processor. May I assume that you
use a PC with Windows 2000, Windows XP or 2003? I can also build a
Linux binary. It is just a single small executable, which you have to
copy to the directory, where other executable are stored.
f***@hotmail.com
2004-10-25 09:41:08 UTC
Permalink
Post by p***@woelen.nl
<snipped>
Post by f***@hotmail.com
Would I be able to use it, I have no knowledge of programming C-code
etc?
I'll see if I can build an executable for you. Using this will not be
harder than using a webbrowser or word processor. May I assume that you
use a PC with Windows 2000, Windows XP or 2003?
Yes.
Post by p***@woelen.nl
I can also build a
Linux binary. It is just a single small executable, which you have to
copy to the directory, where other executable are stored.
Secondly thanks for the matrix problem and I showed my professor your
numerical solution (though the answers given in the book do not match,
and moles of barium can not negative), he appreciated it very much
since he didn't realize that the third equation is not an independent
relation. Would it be possible for you, as he has requested, just to
check whether the three equations are independent relations or not, if
they are not then then there is no solution and our approach is wrong.

0.77680 = 146.12[Ca] + 193.66[Sr] + 243.36[Ba]

0.70468 = 128.10[Ca] + 175.64[Sr] + 225.35[Ba]
0.59269 = 100.09[Ca] + 147.63[Sr] + 197.34[Ba]

Sincerely

Farooq.
p***@woelen.nl
2004-10-25 10:04:40 UTC
Permalink
Post by p***@woelen.nl
Post by p***@woelen.nl
<snipped>
Post by f***@hotmail.com
Would I be able to use it, I have no knowledge of programming
C-code
Post by p***@woelen.nl
Post by f***@hotmail.com
etc?
I'll see if I can build an executable for you. Using this will not be
harder than using a webbrowser or word processor. May I assume that
you
Post by p***@woelen.nl
use a PC with Windows 2000, Windows XP or 2003?
Yes.
OK, I'll make a build for Windows. If I have built the program, I'll
let you know. This may take a few days, I've got to find some spare
time to do the build.
Post by p***@woelen.nl
Post by p***@woelen.nl
I can also build a
Linux binary. It is just a single small executable, which you have to
copy to the directory, where other executable are stored.
Secondly thanks for the matrix problem and I showed my professor your
numerical solution (though the answers given in the book do not match,
and moles of barium can not negative), he appreciated it very much
since he didn't realize that the third equation is not an independent
relation. Would it be possible for you, as he has requested, just to
check whether the three equations are independent relations or not, if
they are not then then there is no solution and our approach is wrong.
0.77680 = 146.12[Ca] + 193.66[Sr] + 243.36[Ba]
0.70468 = 128.10[Ca] + 175.64[Sr] + 225.35[Ba]
0.59269 = 100.09[Ca] + 147.63[Sr] + 197.34[Ba]
Sincerely
Farooq.
Farooq,

The set you gave me now also is not independent. The coefficients in
the 3x3 matrix are the same as in the previous example and that is the
reason that the equations are not independent. Only the values at the
left are different, up to some scale factor (appr. 1.38). You only
measure two independent quantities and the third measurement does not
add new independent data in the mathematical sense.
It does however add new data from a practical point of view, because it
allows you to perform a least squares solution, which averages the
measurement errors. This solution looks like inverse(A'A)*A'*b. Here A
is the 3x2 matrix of e.g. Ca, Sr and b equals the vector L - x*[Ba],
with L the left vector you gave and x the vector of coefficients for
[Ba]. Here in this example I assume that you determine the
concentration of Ba in an independent way. Of course this does not help
you, if there is no way to determine a concentration independently.
Wilco
f***@hotmail.com
2004-10-25 11:30:25 UTC
Permalink
Post by p***@woelen.nl
Post by p***@woelen.nl
Post by p***@woelen.nl
<snipped>
Post by f***@hotmail.com
Would I be able to use it, I have no knowledge of programming
C-code
Post by p***@woelen.nl
Post by f***@hotmail.com
etc?
I'll see if I can build an executable for you. Using this will
not
Post by p***@woelen.nl
be
Post by p***@woelen.nl
Post by p***@woelen.nl
harder than using a webbrowser or word processor. May I assume that
you
Post by p***@woelen.nl
use a PC with Windows 2000, Windows XP or 2003?
Yes.
OK, I'll make a build for Windows. If I have built the program, I'll
let you know. This may take a few days, I've got to find some spare
time to do the build.
Post by p***@woelen.nl
Post by p***@woelen.nl
I can also build a
Linux binary. It is just a single small executable, which you
have
Post by p***@woelen.nl
to
Post by p***@woelen.nl
Post by p***@woelen.nl
copy to the directory, where other executable are stored.
Secondly thanks for the matrix problem and I showed my professor your
numerical solution (though the answers given in the book do not
match,
Post by p***@woelen.nl
and moles of barium can not negative), he appreciated it very much
since he didn't realize that the third equation is not an
independent
Post by p***@woelen.nl
Post by p***@woelen.nl
relation. Would it be possible for you, as he has requested, just to
check whether the three equations are independent relations or not,
if
Post by p***@woelen.nl
they are not then then there is no solution and our approach is
wrong.
Post by p***@woelen.nl
0.77680 = 146.12[Ca] + 193.66[Sr] + 243.36[Ba]
0.70468 = 128.10[Ca] + 175.64[Sr] + 225.35[Ba]
0.59269 = 100.09[Ca] + 147.63[Sr] + 197.34[Ba]
Sincerely
Farooq.
Farooq,
The set you gave me now also is not independent. The coefficients in
the 3x3 matrix are the same as in the previous example and that is the
reason that the equations are not independent. Only the values at the
left are different, up to some scale factor (appr. 1.38). You only
measure two independent quantities and the third measurement does not
add new independent data in the mathematical sense.
It does however add new data from a practical point of view, because it
allows you to perform a least squares solution, which averages the
measurement errors. This solution looks like inverse(A'A)*A'*b. Here A
is the 3x2 matrix of e.g. Ca, Sr and b equals the vector L - x*[Ba],
with L the left vector you gave and x the vector of coefficients for
[Ba]. Here in this example I assume that you determine the
concentration of Ba in an independent way. Of course this does not help
you, if there is no way to determine a concentration independently.
Wilco
Thanks you very much. It is the last equation that is troublesome. The
last equation consists of the formula weights of Ca, Sr, and Ba
carbonates and I am hundred percent sure that if the weights are
replaced by the formula weights of thier oxides, this would again not
give a independent relation. Anyway, thanks again.
Let give you background of this question (if you're interested), the
question to determine Ca, Sr, and Ba simultaneously when their hydrated
oxalates are thermally decomposed. The first equation was the total
weight of oxalates MC2O4.H2O at T1. The second one was the weight at T2
weight the mixture entirely consisted of M2C2O4, and the third equation
was at the temperature T3 when the mixture entirely consisted of MCO3.
So the total weight in grams of the mixture was written in terms of the
components as
[mol CaC2O4]146.12 + 193.66[mol SrC2O4] + ...

The author of the book Skoog and West, deleted this question long time
ago, and replaced it with a two component mixture which is easily
solved.
p***@woelen.nl
2004-10-25 13:11:10 UTC
Permalink
Farooq,

Based on your last info it is clear to me, and I can easily explain
now, why the third relation is not independent. First I introduce some
symbols.

MW(BaC2O4.H2O) = Ba1
MW(BaC2O4) = Ba2
MW(BaCO3) = Ba3

Similarly I introduce symbols Ca1, Ca2, Ca3 and Sr1, Sr2, Sr3.

The equations now become

At T1: Ca1*[Ca] * Sr1*[Sr] + Ba1*[Ba] = val1
At T2, water expelled: Ca2*[Ca] * Sr2*[Sr] + Ba2*[Ba] = val2
At T3, CO expelled: Ca3*[Ca] * Sr3*[Sr] + Ba3*[Ba] = val3

Because all compounds have the same amount of water, one can write
Ca2 = Ca1 - MW(H2O)
Sr2 = Sr1 - MW(H2O)
Ba2 = Ba1 - MW(H2O)

You see that the second matrix row is the first matrix row (Ca1 Sr1
Ba1), adjusted by a constant times the vector (1 1 1).

Finally at temperature T3 one can write

Ca3 = Ca1 - MW(H2O) - MW(CO)
Sr3 = Sr1 - MW(H2O) - MW(CO)
Ba3 = Ba1 - MW(H2O) - MW(CO)

Here you see that this also can be derived from the first matrix row,
also by adjusting it with a perturbation of the form constant*(1 1 1).

Going all the way to the oxides does not yield another independent
equation, as you remarked yourself already. Now MW(CO2) is subtracted
from all the compounds and again a multiple of the row vector (1 1 1)
is added to the first row.

In general, when this kind of methods is used, then it is important to
have perturbations, which are not linear combinations of each other.
Here your second perturbation vector for determining equation 3 is a
scalar multiple of the first perturbation vector for determining
equation 2.

Your method would have worked, if one of the compounds (e.g. the
Ba-salt) would start with 2H2O. Then the base vectors of the
perturbations would be (1 1 2) and (1 1 1), which are linearly
independent. If all salts were starting off from 2H2O then again we
would end up with a linear dependent set, based on base vectors (2 2 2)
and (1 1 1), which are linearly dependent again.
I hope you grasped the idea and understand my explanation.

Wilco
f***@hotmail.com
2004-10-25 14:07:10 UTC
Permalink
Wilco,
Your reasoning is perfectly clear. Thank you very much.
Farooq.
Bob
2004-10-23 17:05:07 UTC
Permalink
Post by Mohammed Farooq
Further discussion on oxidation numbers is pasted here for
[Text snipped]
EN(S) > EN(C)
Maybe, maybe not. In fact, they are often given as equal (2.5).


But this brings us to the larger point. Why do you care what the ON
(oxidation numbers) are? (I seem to have missed the previous thread.)
They really don't mean anything in particular. Sometimes, when very
clear they allow us to describe the molecule -- to summarize our
chemical understanding. Also, we use ON as a tool for balancing. But
for that use, you can declare the ON any way you want, so long as you
are consistent.

So, there is no "right answer" for ON for thiocyanate ion.

...
Post by Mohammed Farooq
For CO, I don't get a full octet around the C if I assume that O does
not go further than a double bond. Is this correct? I get
:C::O
The OxNums I get are, however, +2 for C and -2 for O.
Carbon monoxide has a triple bond [:C:::O:],
Note that that structure gives octets. And it agrees with the measured
dipole moment, which has O+.
Post by Mohammed Farooq
which still gives the
same result.
For molecules like NO and NO2 I have the same problem when making Lewis
dot-structures, I even end up with odd number of electrons around the
atoms.
yep. These molecules are fundamentally radicals. Any N oxide with an
odd number of N will have an odd number of valence electrons.

bob
f***@hotmail.com
2004-10-23 18:22:34 UTC
Permalink
Post by Bob
But this brings us to the larger point. Why do you care what the ON
(oxidation numbers) are? (I seem to have missed the previous thread.)
They really don't mean anything in particular. Sometimes, when very
clear they allow us to describe the molecule -- to summarize our
chemical understanding. Also, we use ON as a tool for balancing. But
for that use, you can declare the ON any way you want, so long as you
are consistent.
So, there is no "right answer" for ON for thiocyanate ion.
Bob,
The problem was that Wilco posted an experimetal stoichiometric ratio
between KMnO4 and NH4SCN in acidic medium, the proposed products were
SO4(2-) and HCN. SO I thought (just for fun) that we write a
hypothetical equation showing the reaction as follows
SCN(-) + MnO4(-) + H(+) ----> SO4(2-) + Mn(2+) + HCN + H2O

The problem was on determining the number of electron lost by SCN(-)
when it converts to SO4(2-) and HCN. The *seemingly* correct equation
shows six electron transfer, but it implicitly assumes that sulfur was
initially zero in thiocynate ion. So is it possible to determine the
no. of electron transferred some other way without determining
(absolute) value of ON of sulfur say from a Lewis structure and using
Pauling electronegativity table.
Bob
2004-10-23 19:53:22 UTC
Permalink
Post by f***@hotmail.com
Post by Bob
But this brings us to the larger point. Why do you care what the ON
(oxidation numbers) are? (I seem to have missed the previous thread.)
They really don't mean anything in particular. Sometimes, when very
clear they allow us to describe the molecule -- to summarize our
chemical understanding. Also, we use ON as a tool for balancing. But
for that use, you can declare the ON any way you want, so long as you
are consistent.
So, there is no "right answer" for ON for thiocyanate ion.
Bob,
The problem was that Wilco posted an experimetal stoichiometric ratio
between KMnO4 and NH4SCN in acidic medium, the proposed products were
SO4(2-) and HCN. SO I thought (just for fun) that we write a
hypothetical equation showing the reaction as follows
SCN(-) + MnO4(-) + H(+) ----> SO4(2-) + Mn(2+) + HCN + H2O
The problem was on determining the number of electron lost by SCN(-)
when it converts to SO4(2-) and HCN.
OK.

Let's accept that eqn. Seems reasonable. (And it explains why people
doing that reaction die.)

I presume... We agree that Mn was reduced from 7+ to 2+. And we agree
that H and O did not change.

The electrons to reduce Mn came from the SCN. That follows from the
above points.

Now, allocating those electrons among the S, C, N is tricky.

If the reason for doing it is simply to balance the eqn, why not keep
it simple. Assume that CN is unchanged (ie, assign CN together as -1),
and allocate all the electron donation to the S. That makes it easier
to balance the equation, since there is only one atom "going each
way". This choice of ON is guided by simplicity, not chemical reality.
You can assign ON other ways, and so long as you are internally
consistent (ie, conserve electrons), you can balance the eqn (and get
the same result).


If the reason to do it is insight into what the electrons did... ON
are not really the answer. As discussed already (I gather), SCN- is a
complicated beast, not easily described. The whole notion of discrete
bonds, of molecules = collection of atoms, begins to fail. Molecules
are new creations, with their own wave functions and electron
distributions. Although it may often serve us well to write electron
dots and bond lines, that is a simplification.

If you want to know more about the actual e in SCN-, go to quantum
mechanics and solve those equations, or go get some experimental data
on the species. ON is not an "explanation".

bob
Post by f***@hotmail.com
The *seemingly* correct equation
shows six electron transfer, but it implicitly assumes that sulfur was
initially zero in thiocynate ion. So is it possible to determine the
no. of electron transferred some other way without determining
(absolute) value of ON of sulfur say from a Lewis structure and using
Pauling electronegativity table.
Monika Hohlmeier
2004-10-26 07:33:55 UTC
Permalink
Post by f***@hotmail.com
Bob,
The problem was that Wilco posted an experimetal stoichiometric ratio
between KMnO4 and NH4SCN in acidic medium, the proposed products were
SO4(2-) and HCN. SO I thought (just for fun) that we write a
hypothetical equation showing the reaction as follows
SCN(-) + MnO4(-) + H(+) ----> SO4(2-) + Mn(2+) + HCN + H2O
In fact, you can avoid the use of oxidation numbers at all.
If you want to balance the equation given above, you have to set up a set of
7 equations for the seven unknown stochiometric numbers of the 7 components
(SCN-, MnO4-, H+, SO42-, Mn2+, HCN, H2O). That these are the compounds
taking part in the reaction is the only point where chemical knowledge
comes into play. The rest is mathematics.

The seven equations are: Conservation of the number of S, C, N, Mn, H, O
atoms (6 equations) + conservation of the number of electrons.
This recipe allways works and is unambigous, although it may not be the most
rapid to solve. Anyway it should be clear that ON are a purely formal
concept with no other meaning than a device to balance equations.
p***@woelen.nl
2004-10-26 08:07:02 UTC
Permalink
Post by Monika Hohlmeier
Post by f***@hotmail.com
Bob,
The problem was that Wilco posted an experimetal stoichiometric ratio
between KMnO4 and NH4SCN in acidic medium, the proposed products were
SO4(2-) and HCN. SO I thought (just for fun) that we write a
hypothetical equation showing the reaction as follows
SCN(-) + MnO4(-) + H(+) ----> SO4(2-) + Mn(2+) + HCN + H2O
In fact, you can avoid the use of oxidation numbers at all.
If you want to balance the equation given above, you have to set up a set of
7 equations for the seven unknown stochiometric numbers of the 7 components
(SCN-, MnO4-, H+, SO42-, Mn2+, HCN, H2O). That these are the
compounds
Post by Monika Hohlmeier
taking part in the reaction is the only point where chemical
knowledge
Post by Monika Hohlmeier
comes into play. The rest is mathematics.
The seven equations are: Conservation of the number of S, C, N, Mn, H, O
atoms (6 equations) + conservation of the number of electrons.
This recipe allways works and is unambigous, although it may not be the most
rapid to solve. Anyway it should be clear that ON are a purely formal
concept with no other meaning than a device to balance equations.
Monika,

Probably you missed the thread on oxidation products of thiocyanate,
where this option is discussed in more detail.
In fact, I use this method frequently myself. I've written a small
C-program which does something like you mentioned, although not exactly
so.

It is not a matter of simply solving the equations, because the
solution space is not a single point, but in general it is an
N-dimensional space
In the example of the equation you give, the solution space is
one-dimensional and the equation can be solved, up to a single scaling
factor for each of the coefficients of the compounds in SCN(-) +
MnO4(-) + H(+) ----> SO4(2-) + Mn(2+) + HCN + H2O. In your example
there are indeed seven equations, but only six of them are independent.
The system you get is of the form

Ax = 0.

If A were non-singular, then the only solution would be 0 for each of
the coefficients. For actual chemical equations, the matrix A always is
rank-deficient. The null-space of A is the solution space for the
coefficients of the chemical equation.
The program I've written constructs an integer basis for this null
space of A. An example of an equation with a two-dimensional solution
space is the following:
Cu + HNO3 ---> Cu(2+) + No3(-) + NO + NO2 + H2O

p***@woelen.nl
2004-10-23 18:22:42 UTC
Permalink
Post by Bob
Post by Mohammed Farooq
Further discussion on oxidation numbers is pasted here for
[Text snipped]
EN(S) > EN(C)
Maybe, maybe not. In fact, they are often given as equal (2.5).
But this brings us to the larger point. Why do you care what the ON
(oxidation numbers) are? (I seem to have missed the previous thread.)
They really don't mean anything in particular. Sometimes, when very
clear they allow us to describe the molecule -- to summarize our
chemical understanding. Also, we use ON as a tool for balancing. But
for that use, you can declare the ON any way you want, so long as you
are consistent.
So, there is no "right answer" for ON for thiocyanate ion.
Ah, this reply comes close to what I originally thought. If one knows
that thiocyanate is oxidized up to SO4(2-), with S in +6 oxidation
state, and S first has oxidation number -2, then it would give 8
electrons (going from -2 to +6). With S at zero oxidation state, it
would give 6 electrons, hence resulting in other reaction
stoichiometry. In the previous thread ("Oxidation of thiocyanate",
started last week) I also had a discussion on this. I can resolve the
issue by looking at sums of oxidation numbers, rather than looking at
the individual atoms of the SCN(-) ion.
Post by Bob
...
Post by Mohammed Farooq
For CO, I don't get a full octet around the C if I assume that O does
not go further than a double bond. Is this correct? I get
:C::O
The OxNums I get are, however, +2 for C and -2 for O.
Carbon monoxide has a triple bond [:C:::O:],
Note that that structure gives octets. And it agrees with the
measured
Post by Bob
dipole moment, which has O+.
Post by Mohammed Farooq
which still gives the
same result.
For molecules like NO and NO2 I have the same problem when making Lewis
dot-structures, I even end up with odd number of electrons around the
atoms.
yep. These molecules are fundamentally radicals. Any N oxide with an
odd number of N will have an odd number of valence electrons.
This explains their reactivity, because they "want" their sets of
electrons completed?
Post by Bob
bob
Bob
2004-10-23 19:53:24 UTC
Permalink
On 23 Oct 2004 11:22:42 -0700, ***@woelen.nl wrote:

...
Post by p***@woelen.nl
Post by Bob
For molecules like NO and NO2 I have the same problem when making
Lewis
Post by Bob
dot-structures, I even end up with odd number of electrons around
the
Post by Bob
atoms.
yep. These molecules are fundamentally radicals. Any N oxide with an
odd number of N will have an odd number of valence electrons.
This explains their reactivity, because they "want" their sets of
electrons completed?
yes.

Actually, NO and NO2 are unusually stable for radicals, esp for simple
ones.

bob
Post by p***@woelen.nl
Post by Bob
bob
Continue reading on narkive:
Loading...