text:
Structure that breaks InChI
I am currently testing a piece of software that generates the InChI for a given structure. I also want to test error situations as end users will also deliver the mol files that will be used as input for the software.
I have found this description about what the current version of InChI cannot represent (Technical FAQ section 4.15 at inchi-trust.org)

InChI currently does not support the representation of:

Polymers
Complex organometallics
Markush structures
Mixtures
Conformers
Excited state and spin isomers
Non-local stereochemistry/chirality
Topological isomers
Cluster molecules
Polymorphs
Unspecific isotopic enrichment
Reactions

Also, InChI is not suitable for very large compounds; technically, InChI input may not contain more than 1023 atoms.
Please note that there are IUPAC InChI subcommittee working groups currently addressing some of these matters. Details of these efforts can be found at: http://iupac.org/web/ins/802

Unfortunately I am lacking the chemical background to generate a suitable structure as molfile that meets the described features to produce an error as an InChI cannot be generated for it.

I propose you remove that bullet list of "shortcomings" from your question as it might mislead people. What you are looking for is a structure that somehow "breaks" InChI, and you probably need to be even clearer, that "breaks" the software that generates the InChI string from the input format (MOLFILE?). ||
So the domain of answers should be MOLFILEs which pose a particular challenge to the InChI generator software. ||
BTW, that InChI is not a complete specification, but the specification is the code. That software has now been extended, I believe, to allow for more than 1023 atoms.

@GuntherSchadow That bullet list is a direct quote from the technical faq as posted on the inchi-trust homepage. As I see it, it is still there. There is absolutely no sane reason to remove this list. The InChI software as you might be thinking about is a reference implementation of the description laid out in the paper. The specification is not the code, but the algorithm that has been implemented by code. You absolutely do not need a molfile to create an InChI string, if you understood the algorithm, you can do it by hand. You cannot create an InChI for any of the mentioned points.

See also https://chemistry.stackexchange.com/questions/151072/can-geometric-isomers-have-same-inchi for pairs of compounds differing in double bond config that the current implementation of InChi fails to distinguish.

Okay, let's tackle at least one problem here. Consider the rotation of 1,2-dichloroethane (BP86/cc-pVDZ):

These conformational changes can be further rationalised:

C and C' are the same conformation, since these are mirror images.
The same applies to B and B'.
A and C are local minima and can be referred to as conformes in the above given way
B and D are transition states - they are not stable.
If you consider the rotation A to C as a reaction, this state would mark the maximum energy needed for the conversion. I believe with reactions it is meant, that an accurate description of the transition state by InChI is not possible.

For the term non-local chirality I suggest you look at BINAP or BINOL.

In this case I have used a slightly abridged version of BINAP for educational purposes, where $\ce{Ph -> Me}$.

Other wonderful examples of non-local chirality include helicenes, which have helical chirality. Here is one example of a fancy azonia derivative of hexahelicene (parent compound and links). The structure is available at the Cambridge Crystallographic Data Centre (CCDC) Database. I removed the anion and converted it to xyz coordinates which I include in the appendix. Here are the structures, left P-chirality, right M-chirality:

On another account of non/local chirality, there is the planar chirality, as can be seen in trans-Cyclooctene (the bottom two). For further reading I suggest Neuenschwander et.al. "The Conformations of Cyclooctene: Consequences for Epoxidation Chemistry."

Markush Structures are a kind of summary structure for certain compound classes. A very simple example would be a monosubstituted biphenyl system, where the upper structure is a representation of all three below. However, it is noteworthy, that each instance of a Markush structure can be well explained by InChI.

More complicated structures are often used in patents and in other scientific publications, usually specifically referring to a very general procedure or mechanism. I found one on wikipedia commons, that could well represent a million molecules:

And another way to break the InChI are clusters. Although the definition is not very specific, the $\ce{[KO{}^{\mathit{t}}Bu]_4}$ tetramer (wikipedia, chemspider) certainly contains one. While the monomer can be described via InChI, the tetra- or Polymer should fail. It was published by Chisholm et.al. in Polyhedron and the *.cif file can be obtained from CCDB. I cleaned it up a little bit and included the *.xyz in the appendix. You can quite nicely see the $\ce{K4}$ tetrahedral cluster in the middle (hydrogens omitted for clarity).

Last but not least, I'd like to cover the matter of complicated organometallic molecules. As such I'd like to introduce the Grubbs' catalysts. Grubbs obtained the Nobel Prize in Chemistry 2005 for his research in conjunction with Yves Chauvin and Richard R. Schrock. One of the key publications was "Ring-opening metathesis polymerization (ROMP) of norbornene by a Group VIII carbene complex in protic media". The crystal structure can also be obtained at CCDC. Another example stems from "Synthesis and Applications of $\ce{RuCl2(=CHR')(PR3)2}$:  The Influence of the Alkylidene Moiety on Metathesis Activity" and is also available via CCDC. This structure can also be found in the appendix. It is generally known as a 1st generation Grubbs catalyst (Hydrogens omitted for clarity).

Appendix
The following section contains the nuclear coordinates in *.xyz(XMol) file format and angstroms as unit. (If not denoted otherwise, structures are obtained at BP86/cc-pVDZ.)
1,2-dichloroethane A
C        0.000000000      0.000000000      0.000000000
C        0.000000000      0.000000000      1.521307000
Cl       1.723641000      0.000000000     -0.593797000
H       -0.486731000     -0.904743000     -0.410753000
H       -0.486731000      0.904743000     -0.410753000
Cl      -1.723641000      0.000000000      2.115104000
H        0.486731000     -0.904743000      1.932060000
H        0.486731000      0.904743000      1.932060000

1,2-dichloroethane B
C        0.030242000      0.085780000     -0.007620000
C       -0.030242000      0.085780000      1.528927000
Cl       1.518312000     -0.758527000     -0.631286000
H       -0.837191000     -0.442465000     -0.444148000
H        0.070555000      1.115212000     -0.412109000
Cl      -1.518312000     -0.758527000      2.152593000
H        0.837191000     -0.442465000      1.965455000
H       -0.070555000      1.115212000      1.933416000

1,2-dichloroethane C
C        0.066377000      0.127741000      0.004558000
C       -0.066377000      0.127741000      1.516749000
Cl       1.083384000     -1.241726000     -0.615213000
H       -0.919786000      0.050600000     -0.492443000
H        0.568277000      1.063384000     -0.317636000
Cl      -1.083384000     -1.241726000      2.136520000
H        0.919786000      0.050600000      2.013750000
H       -0.568277000      1.063384000      1.838943000

1,2-dichloroethane D
C        0.164887000      0.141535000     -0.000240000
C       -0.164887000      0.141535000      1.521547000
Cl       0.337146000     -1.473219000     -0.795160000
H       -0.633932000      0.666019000     -0.558099000
H        1.123149000      0.665665000     -0.177529000
Cl      -0.337146000     -1.473219000      2.316468000
H        0.633932000      0.666019000      2.079406000
H       -1.123149000      0.665665000      1.698836000

R-BINAP
C       -0.086423000     -0.055821000     -0.031546000
C       -0.090885000     -0.047699000      1.480085000
C        1.148913000     -0.158515000      2.237313000
C        1.064686000     -0.139089000      3.676951000
C       -0.131097000     -0.014450000      4.357670000
C       -1.357362000      0.116010000      3.652970000
C       -1.345263000      0.107319000      2.211098000
P        2.768926000     -0.321256000      1.451350000
H        1.988600000     -0.233144000      4.265654000
H       -0.135982000     -0.008251000      5.459401000
C       -2.590782000      0.269421000      4.370415000
C       -2.623021000      0.305506000      1.559779000
C        0.437392000      1.073619000     -0.788457000
C        0.410661000      0.991852000     -2.228054000
C       -0.079670000     -0.105834000     -2.908992000
C       -0.579162000     -1.233570000     -2.204534000
C       -0.579999000     -1.219208000     -0.762676000
P        1.107602000      2.557133000     -0.001640000
H        0.790955000      1.839345000     -2.816335000
H       -0.077540000     -0.112552000     -4.010726000
C       -1.063317000     -2.378355000     -2.921905000
C       -1.047368000     -2.424579000     -0.111239000
C       -3.792005000      0.431032000      3.705083000
H       -2.558632000      0.263827000      5.471798000
H       -4.735477000      0.552772000      4.258523000
C       -3.797353000      0.457087000      2.279022000
H       -2.662047000      0.340912000      0.466043000
H       -4.750761000      0.602046000      1.748556000
C       -1.503463000     -3.517255000     -0.830391000
H       -1.036094000     -2.475958000      0.982488000
H       -1.854657000     -4.415344000     -0.299810000
C       -1.523650000     -3.499569000     -2.256520000
H       -1.052652000     -2.347534000     -4.023279000
H       -1.890198000     -4.377463000     -2.809877000
C        4.061075000     -0.509159000      2.730645000
C        2.844045000     -1.846744000      0.446017000
C        1.585856000      3.772639000     -1.280334000
C       -0.173490000      3.383258000      1.008127000
H        0.254332000      4.289560000      1.485098000
H       -1.067700000      3.703930000      0.418133000
H       -0.532498000      2.717241000      1.815764000
H        1.997887000      4.670081000     -0.777298000
H        2.377005000      3.362349000     -1.938832000
H        0.749130000      4.118465000     -1.941170000
H        5.044249000     -0.601334000      2.227532000
H        4.100503000      0.383637000      3.385840000
H        3.944572000     -1.404262000      3.395364000
H        3.842692000     -1.931312000     -0.030903000
H        2.674387000     -2.779210000      1.039384000
H        2.087489000     -1.826871000     -0.361580000

S-BINAP
C       -1.383946000     -3.448543000     -2.256996000
C       -1.065989000     -2.266144000     -2.910129000
C       -0.652376000     -1.110469000     -2.179491000
C       -0.563545000     -1.177237000     -0.738280000
C       -0.899026000     -2.413534000     -0.098700000
C       -1.299620000     -3.519323000     -0.838209000
C       -0.154302000     -0.008671000      0.004167000
C        0.166467000      1.182512000     -0.673969000
C        0.072046000      1.221328000     -2.098779000
C       -0.323252000      0.113168000     -2.831301000
C       -0.061735000     -0.093981000      1.507135000
C       -1.110430000      0.372113000      2.322674000
C       -0.989013000      0.267385000      3.742090000
C        0.132224000     -0.289631000      4.336601000
C        1.210137000     -0.777916000      3.542877000
C        1.113248000     -0.682564000      2.103804000
C        2.206553000     -1.174312000      1.320794000
C        3.329975000     -1.726929000      1.922868000
C        3.419059000     -1.820539000      3.340084000
C        2.377998000     -1.353835000      4.130094000
P       -2.590653000      1.186493000      1.502738000
C       -3.750658000     -0.293139000      1.402088000
P        0.636241000      2.692004000      0.339668000
C        2.511918000      2.555623000      0.253510000
C       -3.385046000      2.079953000      2.947654000
C        0.388676000      4.078351000     -0.898992000
H       -1.802841000      0.635802000      4.383638000
H        0.202504000     -0.360978000      5.434128000
H        0.319305000      2.150696000     -2.632672000
H       -0.386992000      0.166196000     -3.930250000
H        2.436377000     -1.419019000      5.228476000
H        4.313241000     -2.262708000      3.805535000
H        2.146133000     -1.110484000      0.224695000
H        4.157224000     -2.097887000      1.298347000
H       -0.834575000     -2.478936000      0.997042000
H       -1.552289000     -4.458551000     -0.322477000
H       -1.128190000     -2.199481000     -4.008215000
H       -1.700498000     -4.331984000     -2.832408000
H        2.950535000      3.473538000      0.698290000
H        2.842563000      1.687658000      0.858736000
H        2.884117000      2.439555000     -0.786611000
H        0.678948000      5.023670000     -0.395042000
H        0.990357000      3.982867000     -1.828352000
H       -0.686746000      4.150262000     -1.160750000
H       -4.302108000      2.573768000      2.564446000
H       -3.673500000      1.421270000      3.794842000
H       -2.700063000      2.872570000      3.312044000
H       -4.742112000      0.063129000      1.051704000
H       -3.359638000     -1.014277000      0.656570000
H       -3.871382000     -0.807607000      2.379049000

M-8a-azonia[6]helicene (from crystal structure)
N        3.565811000      0.259443000     13.724558000
C        7.445232000     -1.196589000     12.564039000
C        8.629416000     -1.762757000     12.151335000
C        8.885324000     -1.978555000     10.783139000
C        7.914184000     -1.694865000      9.864763000
C        6.665024000     -1.171130000     10.265448000
C        5.584378000     -1.088690000      9.330104000
C        4.339207000     -0.800151000      9.738213000
C        4.088732000     -0.406137000     11.083608000
C        2.763235000     -0.187914000     11.543327000
C        2.522579000      0.029096000     12.830042000
C        3.246403000      0.443720000     15.064121000
C        4.198045000      0.692252000     15.970654000
C        5.531025000      0.961394000     15.546639000
C        6.527506000      1.327523000     16.488699000
C        7.769629000      1.675468000     16.086777000
C        8.044342000      1.864594000     14.697195000
C        9.259232000      2.459858000     14.281133000
C        9.466841000      2.806590000     12.982045000
C        8.445715000      2.608977000     12.035743000
C        7.265466000      1.985829000     12.396483000
C        7.061105000      1.553020000     13.720492000
C        5.833567000      0.912900000     14.172787000
C        4.884651000      0.303088000     13.284635000
C        5.151668000     -0.288539000     12.002691000
C        6.444932000     -0.841371000     11.632053000
H        7.296301000     -1.066868000     13.466331000
H        9.304643000     -2.036748000     12.840647000
H        9.760406000     -2.388329000     10.496986000
H        8.038683000     -1.964007000      8.943383000
H        5.780686000     -1.333585000      8.395468000
H        3.604979000     -0.909262000      9.084780000
H        2.029733000     -0.315211000     10.922946000
H        1.672696000      0.060618000     13.273677000
H        2.290439000      0.327335000     15.253240000
H        3.954366000      0.824398000     16.914659000
H        6.255785000      1.333585000     17.427224000
H        8.447606000      1.891266000     16.702563000
H        9.925491000      2.630800000     14.988120000
H       10.286033000      3.200604000     12.619714000
H        8.559741000      2.909640000     11.135042000
H        6.570317000      1.891266000     11.711236000

M-8a-azonia[6]helicene (from BP86/STO-3G)
N        0.241946000      3.510759000     13.746203000
C       -1.207087000      7.492952000     12.512135000
C       -1.801472000      8.692275000     12.089683000
C       -2.102492000      8.912745000     10.707738000
C       -1.853647000      7.899258000      9.772330000
C       -1.278056000      6.643678000     10.182732000
C       -1.146616000      5.553783000      9.229249000
C       -0.770948000      4.283969000      9.646130000
C       -0.392477000      4.045979000     11.023979000
C       -0.161103000      2.688046000     11.484550000
C        0.042219000      2.428893000     12.821687000
C        0.400500000      3.183323000     15.136378000
C        0.639392000      4.176721000     16.059802000
C        0.949043000      5.527644000     15.625469000
C        1.361398000      6.519481000     16.597330000
C        1.808770000      7.766658000     16.182869000
C        1.981223000      8.066019000     14.770425000
C        2.627560000      9.287588000     14.362758000
C        2.913363000      9.530579000     13.012425000
C        2.578886000      8.546652000     12.028110000
C        1.916409000      7.363991000     12.392255000
C        1.565834000      7.097902000     13.763105000
C        0.907348000      5.851096000     14.211296000
C        0.281492000      4.885356000     13.292158000
C       -0.309154000      5.144896000     11.968134000
C       -0.894790000      6.445456000     11.575101000
H       -1.007504000      7.339193000     13.585554000
H       -2.048406000      9.472285000     12.834293000
H       -2.558542000      9.868716000     10.388465000
H       -2.123797000      8.035037000      8.708340000
H       -1.422274000      5.738729000      8.172836000
H       -0.763274000      3.430969000      8.943112000
H       -0.251900000      1.843110000     10.776492000
H        0.087281000      1.420761000     13.274365000
H        0.295196000      2.107892000     15.372408000
H        0.697253000      3.916258000     17.133249000
H        1.320711000      6.254708000     17.669775000
H        2.110372000      8.530075000     16.926174000
H        2.921715000     10.016221000     15.141242000
H        3.423497000     10.463796000     12.708003000
H        2.853573000      8.716110000     10.969950000
H        1.691114000      6.611596000     11.618453000

$\ce{KO{}^{\mathit{t}}Bu}$ tetramer (from crystal structure)
K       -1.309381000     -1.309381000     -1.309381000
O       -1.313567000      1.313567000     -1.313567000
C       -2.107232000      2.107232000     -2.107232000
C       -3.502008000      2.298114000     -1.478495000
H       -3.842748000      1.465100000     -1.465100000
H       -4.060420000      2.821364000     -2.059512000
H       -3.516240000      2.821364000     -0.619528000
O       -1.313567000     -1.313567000      1.313567000
O        1.313567000     -1.313567000     -1.313567000
K       -1.309381000      1.309381000      1.309381000
K        1.309381000      1.309381000     -1.309381000
C       -1.478495000      3.502008000     -2.298114000
C       -2.298114000      1.478495000     -3.502008000
K        1.309381000     -1.309381000      1.309381000
C       -2.107232000     -2.107232000      2.107232000
C        2.107232000     -2.107232000     -2.107232000
O        1.313567000      1.313567000      1.313567000
H       -1.465100000      3.842748000     -1.465100000
H       -2.059512000      4.060420000     -2.821364000
H       -0.619528000      3.516240000     -2.821364000
H       -1.465100000      1.465100000     -3.842748000
H       -2.821364000      2.059512000     -4.060420000
H       -2.821364000      0.619528000     -3.516240000
C       -1.478495000     -3.502008000      2.298114000
C       -3.502008000     -2.298114000      1.478495000
C       -2.298114000     -1.478495000      3.502008000
C        3.502008000     -1.478495000     -2.298114000
C        1.478495000     -2.298114000     -3.502008000
C        2.298114000     -3.502008000     -1.478495000
C        2.107232000      2.107232000      2.107232000
H       -1.465100000     -3.842748000      1.465100000
H       -2.059512000     -4.060420000      2.821364000
H       -0.619528000     -3.516240000      2.821364000
H       -3.842748000     -1.465100000      1.465100000
H       -4.060420000     -2.821364000      2.059512000
H       -3.516240000     -2.821364000      0.619528000
H       -1.465100000     -1.465100000      3.842748000
H       -2.821364000     -2.059512000      4.060420000
H       -2.821364000     -0.619528000      3.516240000
H        1.465100000     -1.465100000     -3.842748000
H        3.842748000     -1.465100000     -1.465100000
H        4.060420000     -2.059512000     -2.821364000
H        3.516240000     -0.619528000     -2.821364000
H        2.059512000     -2.821364000     -4.060420000
H        0.619528000     -2.821364000     -3.516240000
H        1.465100000     -3.842748000     -1.465100000
H        2.821364000     -4.060420000     -2.059512000
H        2.821364000     -3.516240000     -0.619528000
C        1.478495000      2.298114000      3.502008000
C        3.502008000      1.478495000      2.298114000
C        2.298114000      3.502008000      1.478495000
H        1.465100000      1.465100000      3.842748000
H        2.059512000      2.821364000      4.060420000
H        0.619528000      2.821364000      3.516240000
H        3.842748000      1.465100000      1.465100000
H        4.060420000      2.059512000      2.821364000
H        3.516240000      0.619528000      2.821364000
H        1.465100000      3.842748000      1.465100000
H        2.821364000      4.060420000      2.059512000
H        2.821364000      3.516240000      0.619528000

Grubbs I catalyst
Ru      1.923022    1.782458    4.279841
Cl      3.508375    2.054525    6.061616
P       0.620071    3.495015    5.336792
C       2.967497    2.881791    3.239789
Cl      0.101984    1.197998    2.838894
P       2.855052   -0.401312    3.741075
C       2.954465    3.343954    1.839333
Cl      3.310061    4.678711   -2.473972
C       1.880873    3.257547    0.960743
C       1.973325    3.667349   -0.352875
C       3.185197    4.168660   -0.812833
C       4.252303    4.292067    0.012884
C       4.145603    3.890748    1.341393
C      -0.637087    4.207002    4.175769
C      -1.979932    4.694567    4.692950
C      -2.932151    4.983356    3.538285
C      -2.339331    5.946324    2.517809
C      -0.957803    5.505405    2.056512
C      -0.041546    5.252113    3.235271
C       1.483967    4.856669    6.241316
C       0.518279    5.817396    6.934517
C       1.256971    6.738015    7.899778
C       2.346867    7.495106    7.161401
C       3.302557    6.547168    6.465690
C       2.565538    5.630984    5.497585
C      -0.333586    2.698072    6.719178
C      -1.014762    1.418485    6.241149
C      -1.859827    0.796208    7.358000
C      -1.043744    0.536540    8.601177
C      -0.361483    1.805588    9.075860
C       0.493782    2.427234    7.979254
C       2.628515   -0.975242    1.991593
C       3.236690   -0.059043    0.937820
C       2.687448   -0.416553   -0.435697
C       2.971677   -1.856587   -0.778868
C       2.466883   -2.799363    0.290130
C       2.999176   -2.423218    1.675026
C       4.622054   -0.540862    4.248552
C       5.523290    0.453308    3.517538
C       6.887074    0.518241    4.175936
C       7.549339   -0.842629    4.233326
C       6.649166   -1.859830    4.903939
C       5.271941   -1.927013    4.232992
C       1.995357   -1.737789    4.704830
C       2.010202   -1.452357    6.212537
C       1.292567   -2.535071    7.008639
C      -0.120796   -2.764155    6.507353
C      -0.110733   -3.109699    5.031603
C       0.567738   -2.010334    4.218268
H       3.716836    3.154747    3.670969
H       1.120535    2.917815    1.276640
H       1.189061    3.632874   -0.962082
H       5.017596    4.645026   -0.262690
H       4.792512    4.003484    1.897392
H      -0.815388    3.399122    3.615754
H      -1.902150    5.510235    5.223685
H      -2.378700    3.974732    5.228704
H      -3.726326    5.342882    3.821555
H      -3.181986    4.184074    3.125510
H      -2.984577    6.066660    1.748479
H      -2.250423    6.781889    2.909669
H      -1.054116    4.674447    1.537657
H      -0.576082    6.145292    1.467384
H       0.723372    4.965838    2.939787
H       0.125317    6.138803    3.761321
H       1.921040    4.375890    6.908583
H       1.676706    6.235725    8.636983
H      -0.145301    5.369077    7.433963
H       0.091950    6.422548    6.224250
H       0.676388    7.347621    8.295653
H       2.791319    8.070082    7.723424
H       1.896364    8.099751    6.522077
H       3.914233    7.048872    6.008409
H       3.698100    6.058287    7.055823
H       3.116098    5.021349    5.094850
H       2.158660    6.180195    4.786983
H      -1.006024    3.336226    6.935353
H      -1.559257    1.598311    5.456258
H      -0.325083    0.765200    6.003389
H      -2.608220    1.404611    7.524315
H      -2.214635    0.029761    7.044111
H      -0.418035   -0.117788    8.397718
H      -1.627822    0.182306    9.277813
H       0.176244    1.645756    9.829964
H      -1.053680    2.460437    9.333028
H       0.877440    3.225032    8.287287
H       1.235327    1.881817    7.763580
H       1.720632   -0.869158    1.853889
H       4.184273   -0.146643    0.965428
H       3.012517    0.820454    1.136093
H       1.847900   -0.200343   -0.409930
H       3.096540    0.151437   -1.100956
H       3.904789   -1.982340   -0.878422
H       2.584028   -2.105280   -1.617970
H       2.763571   -3.703670    0.105411
H       1.556168   -2.778368    0.317905
H       4.607903   -0.216537    5.166797
H       3.990062   -2.477871    1.686571
H       2.642313   -3.025155    2.339113
H       5.156005    1.289122    3.512016
H       5.590006    0.129866    2.598457
H       6.737696    0.838429    5.032942
H       7.431919    1.106030    3.739569
H       8.390108   -0.830548    4.679900
H       7.729962   -1.148361    3.309561
H       6.514020   -1.632968    5.797587
H       7.041226   -2.668142    4.887375
H       4.662063   -2.600521    4.676553
H       5.418404   -2.225128    3.326293
H       2.516573   -2.545817    4.572816
H       1.564734   -0.621934    6.319621
H       2.866067   -1.356878    6.528769
H       1.774146   -3.375710    6.968817
H       1.285356   -2.270001    7.937592
H      -0.502310   -3.449358    7.037418
H      -0.746615   -1.928105    6.670990
H       0.328488   -3.937948    4.917492
H      -1.013735   -3.235955    4.696631
H       0.558651   -2.294513    3.264385
H       0.034321   -1.214512    4.228140

I see you're from Germany too, funny how all Chem-Informatics people seem to be either Germans or Russians. And are you a radio amateur? Your emblem looks like saying "polyluxus" in morse code. Sorry, this site has no PM, else I'd have asked that way.

@Gunther Head on over to [mattermodeling.se] you'll probably find a whole bunch of other nationalities working the field. I am not a radio amteur, but it is morse code; it's really nice to know that it can be picked up. ;)

BTW, I still like your answer as an interesting text. And I am not one of these self-appointed SE sheriffs who like do downvote, nag, vote to close, and such useless things. But actually your answer doesn't really address the asker's intent to find ERROR situations to test. What you seem to be talking about is where InChI fails to detect isomerism. But it still will create an InChI without balking, doesn't it?

@Gunther Well, I am one of these community appointed sheriffs, and I think (down-, close-) voting are necessary steps in accomplishing what this community expected of me when they chose to cast their vote for me. I have to admit, in hindsight, this answer doesn't really address what you describe. However, in the cases I have provided (xyz instead of mol because of the character limit), an InChI wouldn't be helpful, even though it could be generated. This post is more about the documented shortcomings of the InChI algorithm, but I understand your concern.


text:
Entropies of Solvation for Chlorate and Nitrate Anions
Can anyone find any data on the entropies of solvation for these two anions in water solution? I can't find any.

Look at the graph on p. 65, it puts the entropy of solvation (298 K) for chlorate and nitrate at about 25 and 21 kJ/mol respectively.  This reference suggests that the entropy term for nitrate is 116 J/(K-mol), multiplying that by 298 K yields 34.6 kJ/mol.

Would you say that the difference between 25 and 21 kJ/mol is a big difference?

No, as a rule of thumb 5.5 kJ/mol (1.4 kcal/mol) shifts an equilibrium by a factor of 10.  So if two molecules A and B were in equilibrium and the free energy difference between the two favored A by 5.5 kJ/mol, then you have an [A]/[B] equilibrium of about 91/9, relatively speaking, a small difference.  The difference your looking at is even less than this.

Should have pointed out that the above comment is a "room temperature" statement.  While I'm at it, let me add another point of comparison to demonstrate that 4 kJ/mol is a small difference: the height of the rotational barrier in ethane is around 12 kJ/mol.

@Dissenter  Just looking for feedback - what would have constituted an answer that you could have "accepted"?

I don't think your answer is inadequate; I was just holding out for more opinions.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Molecular orbitals and "generalized" aufbau principle for heteronuclear molecular configurations
Is there any general ordering or rule for heteronuclear molecules like $\ce{AX_2}$, $\ce{AX_{n}}$, and their molecular bonding/antibonding configurations ($\sigma/\pi$ orbitals) accordingly to some order in the Molecular Orbital Theory (MOT)?
Do the molecular orbital configurations and their order depend on the scheme and the linear combination of atoms (homo and heteronuclear, di and polyatomic) or is there some "standard ordering" like the aufbau principle $\ce{1s~2s~2p~3s~3p~4s~3d~4p~5s~\dots}$ for atomic orbitals in atoms? Or are molecular configurations a totally dependent on the molecule type and geometry? That is, I want to know if there is some (general) rule for molecular orbitals (ordering and energies) just like the aufbau principle for atomic orbitals.

Do the molecular orbital configurations and their order depends on the scheme and the linear combination of atoms (homo and heteronuclear, di and poliatomic)

Not only orbitals arising depends on atom configuration, but also on atomic orbitals available (s,p,d) and their ordering depends on geometry (i.e. size) and relative energy of atomic orbitals available. Unfortunately, there is no simple rule here.
Lucky enough, there is some amount of free computational packages you can use to calculate and then visualize molecular orbitals.

I use Linux. Could you recommend me one? I am interested in the trial of some noble gas compounds...I would be happy if you could share or tell me how to find those packages.

afaik, gamess is free for personal use, though requires a registration. AFAIK, recent versions of avogadro are capable of visualizing MOs from gamess output.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What total energy (in kJ) is contained in 1.0 mol of photons, all with a frequency of 2.75 • 10¹⁴ Hz?
There is a question, that says:

What total energy (in $\mathrm{kJ}$) is contained in $1.0~\mathrm{mol}$ of photons, all with a frequency of $2.75 \cdot 10^{14}~\mathrm{Hz}$?

My answer was:
The energy of photon in hydrogen atom is given by the formula $E=h\nu$, where $h$ is Planck constant and $\nu$ is the frequency. After that I got an answer different from the correct answer which is $110~\mathrm{kJ}$.
Could anyone explain why?

what values did you use in the equation?

$2.75 \times 10^{14}$ x 6.63e-34 = 1.82e-19 KJ ??

I get the same answer as you, except your answer is in J/mol, you need to divide by 1000 to convert it to kJ/m

I did, but still far away from the right answer !

The formula $E=h\nu$ is for the energy of one photon. You have a mole of photons. You need to use a slightly modified form:
$$E=Nh\nu$$
where $N$ is the number of photons, in this case
$$\begin{align}
N&=n\cdot N_\mathrm A\\[6pt]
&=1\ \mathrm{mol}\times6.02\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]
&=6.02\times10^{23}
\end{align}$$
Note that you are being asked to report energy in kilojoules not kilojoules per mole.
If I use $E=nN_\mathrm A h \nu$, I get the correct answer you cite of $110~\mathrm{kJ}$. Do you?

Yes I do ! but I didn't know that avogadro number is used to convert mol to photons .. I know that it is used to convert mol to atoms or molecules.. thanks !

The units on Avogadro's number are $\text{mol}^{-1}$, so you can convert between number of anything and moles of anything. A mole is just a counting aid for very large numbers of things.


text:
Kinetics of interstellar chemistry: Applying steady state to formation of H₂
I am trying to solve problem 2 from the 2009 International Chemistry Olympiad (IChO), which is as follows:

If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\ce{H2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\ce{H2}$ rapidly desorbs. This question examines two kinetic models for $\ce{H2}$ formation on the surface of a dust particle.
In both models, the rate constant for adsorption of $\ce{H}$ atoms onto the surface of dust particles is $k_\mathrm a = \pu{1.4*10^-5 cm3 s-1}$. The typical number density of $\ce{H}$ atoms (number of H atoms per unit volume) in interstellar space is $[\ce{H}] = \pu{10 cm-3}$. (You may treat numbers of surface adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations).
Desorption of $\ce{H}$ atoms is first order with respect to the number of adsorbed atoms. The rate constant for the desorption step is $k_\mathrm d = \pu{1.9 * 10^-3 s-1}$. Assuming that only adsorption and desorption take place, calculate the steady-state number, $N$, of $\ce{H}$ atoms on the surface of a dust particle.

First off, what do they mean by steady-state number here? I tried searching online but I could not find anything. The solution they give has the equation:
$$k_\mathrm d N − k_\mathrm a[\ce{H}] = 0$$
from which they solve for $N$. I know this has something to do with steady state, but in this case why must the rate of absorption be equal to the rate of desorption? Is it because you are considering a single $\ce{H}$ atom which is considered to be an intermediate? Also, why is it $k_\mathrm dN$?
Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?

In the above, how did they come up with those rates? Why does absorption have $\ce{[H]}x_m$ but desorption have $\ce{m}x_m$? Finally, for the reaction rate, where did they get $m(m-1)$? I think the $\frac{1}{2}$ comes from the fact that the rate of production of $\ce{H2}$ is half the rate of absorption/desorption, is that right?

From these, they calculate the rate of change of $x_0$, $x_1$, and $x_2$ as follows:

$$\begin{align}
\frac{\mathrm dx_0}{\mathrm dt} &= -k_\mathrm a[\ce{H}]x_0 + k_\mathrm d x_1 + k_\mathrm r x_2 \\
\frac{\mathrm dx_1}{\mathrm dt} &= k_\mathrm a[\ce{H}]x_0 - (k_\mathrm a [\ce{H}] + k_\mathrm d) x_1 + 2k_\mathrm dx_2 \\
\frac{\mathrm dx_2}{\mathrm dt} &= k_\mathrm a[\ce{H}]x_1 - (2k_\mathrm d + k_\mathrm r)x_2
\end{align}$$

I sort of understand how they got the value for rate of change of $x_0$ as it is the rate of production minus the rate of consumption, and desorption from state 1 and reaction produce state 0, while absorption consumes it. However, why does the $k_\mathrm a$ term come with $[\ce{H}]$ while the other terms do not? What happened to the $m x_m$ that was mentioned above?
Further, how did they get the rate of change for $x_1$ and $x_2$?

OK, let's step through your questions one at a time.

First off, what do they mean by steady-state number here?

Intermediates formed in reactions usually exist in low concentration.  The steady-state assumption assumes that the concentration of this (low-concentration) intermediate will not change appreciably over the course of the reaction once steady state has been established.  The rate of its formation is assumed to be equal to the rate of its consumption.  Making the steady state assumption allows us to recast the previous statement as a mathematical equation as we'll see below.

In this case why must the rate of absorption be equal to the rate of desorption

Just like in a chemical equilibrium where it takes a finite time for the equilibrium to establish itself, it takes a finite time before the steady state is established. Once there is a finite, non-zero concentration of the intermediate and we have arrived at the steady state, then the rate of absorption and desorption will be equal and the non-zero concentration of the intermediate(s) will remain constant – a steady state.
Applying the steady-state assumption to the problem at hand allows us to state that for all of the intermediates (dust particles with 1 or 2 hydrogen atoms on them), the rate of hydrogen atom adsorption on an intermediate must be equal to the rate of hydrogen atom desorption from that intermediate. This can be written as
$$k_\mathrm a [\ce{H}] = k_\mathrm d N$$
and then rearranged and solved:
$$\begin{align}
N &= \frac{k_\mathrm a}{k_\mathrm d}[\ce{H}] \\
&= \frac{\pu{1.4 * 10^-5 cm3 s-1}}{\pu{1.9 * 10^-3 s-1}} \cdot \pu{10 cm^-3} \\
&= \pu{7.4 * 10^-2}
\end{align}$$

Why is it $k_\mathrm d N$?

Because $k_\mathrm d N$ represents the total rate of desorption of hydrogen atoms from dust particles, the right side of our verbal equation. It represents desorption from $x_1$ and $x_2$ - the only two species that can desorb, and it is equal to
$$k_\mathrm d N = k_\mathrm d x_1 + 2k_\mathrm d x_2$$
the factor of "2" arising because there are 2 hydrogen atoms on each dust particle in $x_2$, and both are viable candidates for desorption.

Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?

It is the same rate, that's what $k_\mathrm a [\ce{H}] = k_\mathrm d N$ means.  Rates are always a positive number, we are multiplying a rate constant times a concentration, both positive numbers.  However, when we write an equation and insert the rate expression into it, we can insert a minus sign before the rate expression if the rate expression represents a pathway that is causing the concentration of a product, reactant or intermediate to decrease.

How did they come up with those rates?

Adsorption rate: $r_\mathrm a = \sum_m k_\mathrm a [\ce{H}] x_m = k_\mathrm a [\ce{H}]x_0 + k_\mathrm a [\ce{H}]x_1$
This is just the typical kinetic expression – adsorption rate constant multiplied by the concentration of hydrogen atoms available to adsorb times the concentration of the various dust particles that can adsorb a hydrogen atom, which are only $x_0$ and $x_1$.
Desorption rate: $r_\mathrm d = \sum_m k_\mathrm d m x_m = (k_\mathrm d \cdot 0 \cdot x_0) + (k_\mathrm d \cdot 1 \cdot x_1) + (k_\mathrm d \cdot 2 \cdot x_2)$
Here $m$ is a concentration term analogous to $[\ce{H}]$.  It represents the number of hydrogen atoms on a dust particle, and can be 0, 1 or 2.  Note how it "removes" $x_0$ from the desorption equation as it should since a particle with no hydrogen atoms cannot desorb.  Notice too how it adjusts $x_2$ for the fact that there are two adsorbed hydrogen atoms in this case, so desorption is statistically twice as likely to happen.  It differs from $[\ce{H}]$ in its units: $[\ce{H}]$ is in $\pu{atoms/cm3}$ while "m" is in units of atoms/dust particle.
Reaction rate: $r_\mathrm r = \left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right) = k_\mathrm r x_2$

EDIT 1:

Also, after your explanation I understood how they got the absorption and desorption rates, but I'm still confused on how they got the reaction rate $\left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right)$. Could you please elaborate further on where all the numbers came from?

Note we have a factor of $1/2$ because it takes 2 hydrogen atoms to produce 1 hydrogen molecule.  The $2$ and the $1$ come from the concentration terms $m(m-1)$.  In the "Reaction" case we have to combine two hydrogen atoms.  It is a bimolecular reaction and just like in the bimolecular reaction $\ce{A + B -> C}$ where the rate is proportional to $[\ce{A}][\ce{B}]$, here the analogous concentrations are $m(m-1)$.  Initially the concentration is $m=2$, but after that first hydrogen atom is "selected" the remaining concentration from which to pick our second hydrogen atom has been decreased to $m-1$.  What we're saying is that when we pick the first hydrogen we have two atoms to choose from, when we pick the second hydrogen atom we have one atom to choose from. If we had 5 hydrogen atoms on a particle and were still going to make 2 picks the equation for our two picks would become $5(5-1)=20$.  We would expect this case to proceed $(20/2)=10$ times faster than the case with only 2 atoms on the dust particle.
There are 2 ways to form $\ce{H2}$ in our problem. Label the hydrogen atoms A and B, then we can either pick via an AB or BA sequence. $m(m-1)$ is exactly analogous to $[\ce{H}][\ce{H}]$, it's just not a square because there is a significant difference between $2\cdot 2$ and $2\cdot 1$. When dealing with a mole of atoms ($N = 6 \cdot 10^{23}$, or Avogadro's number) and they react with one another in a bimolecular reaction, there is no need to write $(6\cdot 10^{23})(6\cdot 10^{23} - 1)$, but in our case with such small numbers the difference matters.

Why does absorption have $[\ce{H}]x_m$ but desorption have $mx_m$?

As noted above, $m$ is a concentration term analogous to $[\ce{H}]$, but with different units.

Finally, for the reaction rate, where did they get $m(m−1)$?

See above.

why does the $k_\mathrm a$ term have $[\ce{H}]$ while the other terms do not?

The other terms use $m$ which, as explained above, is analogous to $[\ce{H}]$ in that they both reflect concentrations, $[\ce{H}]$ for hydrogen atoms in space and $m$ for hydrogen atoms on dust particles; they use different units as explained above.

What happened to the $mx_m$ that was mentioned above?

Since the full rate expressions have been written out, they have been replaced with integers, 0, 1, or 2, representing the concentration of hydrogen atoms on a dust particle.

Further, how did they derive the rate of change for $x_1$ and $x_2$?

The rate of change of $x_1$ is the rate of formation minus the rate of destruction.
The rate of formation is given by $k_\mathrm a[\ce{H}]x_0$ (the rate of a bare dust particle picking up a hydrogen atom) plus $2k_\mathrm d x_2$ (the rate of a dust particle with two hydrogen atoms losing one).
The rate of destruction is given by $k_\mathrm a[\ce{H}]x_1 + k_\mathrm d x_1$ (the rate of a dust particle with one hydrogen atom either picking up one more, or losing it).
For the time being, I'll leave the rate of change of $x_2$ as an exercise for you to test yourself on what you've learned.

Why wouldn't you use combination instead of permutation? If you want to choose 2 H atoms to react and you have two H atoms on a dust particle shouldn't it just be 2C2 = 1? Why do you take into account the order of choosing which would give 2*1 = 2? Don't you get the same molecule $\ce{H2}$ regardless of the order the atoms are chosen? Also, I've never seen permutations used in rate expressions. Why are there usually no permutations (just the rate constant times concentration to a power), but in this case you have one?

I do use combination instead of permutation, permutation is used when order matters, combination is used when order doesn't matter, all of the H atoms are identical so order does not matter 2) the H atoms are being picked one at a time, not two at a time 3) there is no order of choosing, you pick one atom out of two and then one out of one 4) yes, order doesn't matter, the H atoms are identical, you get the same hydrogen molecule 5) "m" is representing the "concentration" of H atoms in the reaction and desorption cases, the "concentration" is either 0, 1 or 2.

But it is ${m}P{2}$ that gives a net result of $m(m-1)$. If you have two H atoms, $m(m-1)$ gives that there are two ways to form $H_2$, but there is clearly only one way. Shouldn't you use ${m}C{2}$ instead, which would give $1$?. Also, if you were normally writing the rate expression for $\ce{H + H -> H2}$ you would have something like $k[H]^2$. Why do you not have the squared term here and instead have the permutation?

Good questions.  There are 2 ways to form $\ce{H_{2}}$.  Label the atoms A and B, then we can go via the AB or the BA route.  m(m-1) is exactly analogous to [H][H], it's just not a square because there is a significant difference between 22 and 21.  When dealing with a mole of molecules (N = Avogadro's number, 610^23) and one reacts in a bimolecular reaction with itself, no need to write (610^23)((6*10^23)-1), in our case the difference matters.  BTW, the more I think about it, the more I like the approach in point 5) above.

Taking the approach that "m" represents concentration is much easier to understand than the statistical approach.  The statistical approach is correct but uses (IMO) much more cumbersome language than the concentration approach.  When I get a few minutes I will rework the answer removing the probability approach and replacing it with the concentration analogy.  Thanks for continuing to ask questions until we arrived at a comfortable place.

Thanks for taking the time to answer :D. With regards to the "When dealing with a mole of molecules and one reacts in a bimolecular reaction with itself ($\ce{H + H -> H2}$), no need to write (6*10^23)(6*10^23 - 1)", doesn't the step still occur one atom at a time? As is rather than one mol of H atoms reacting with another mol of H atoms all at once, it happens one atom at a time? So then why do we only take the leading order term ($[H]^2$) instead of writing the more precise $n(n-1)$?

Yes, everything you said is correct, but...that was my point above.  When working with large numbers like Avogadro's number, $6 \cdot 10^{23}$ (let's call it $N$), the difference between $N^2$ and $N(N-1)$ is sooo small, let's make life easy and just call it $N^2$.  But for very low concentrations like on the dust particles, then there is a big difference between $2\cdot 2$ (which is $m^2$) and $2\cdot 1$ (which is $m(m-1)$).  So in this low-concentration case we need to use the tedious, but accurate, $m(m-1)$ expression.


text:
Plot of Arrhenius Equation with temperature
I saw the following plot of the Arrhenius Equation on Wikipedia (cc-by Zivilverteidigung and Matthias M.).

It is a concave upwards graph, but according to the equation
$$k= A \mathrm e^\frac{-E_\mathrm a}{RT},$$
 it is evident that the graph is of the form
$$y= \mathrm e^\frac{-1}{x}.$$

I believe that this should have been the graph (please disregard the negative side of the $x$ axis). It is a concave downwards curve. Why is there a discrepancy?
This maybe also suitable in Mathematics.se, but I have a gut feeling that the solution to this lies in Chemistry and not maths.

May not explain the whole thing but A isnt necessarily constant as it is temperature dependent; it has to do with reactant "collisions" that form the products,  including steric factors as well as frequency (depends on T)

I believe that this should have been the graph...(please omit the
  negative side of x axis), so concave downwards. Why is there a
  discrepancy?

Your analysis and your graph are correct.  Let's examine why the discrepancy arises.  I've drawn two graphs below.  The top one is similar to your graph, only I'm examining a smaller range on the x-axis.  My lower graph examines an even smaller range and it is a range close to x=0.  You can see that this lower graph appears to be "concave-up" just like the Wikipedia graph you referenced.

Here's a third graph where I changed the numerator from -1 to -100, notice how it moves the curve further to the right along the x-axis.

Depending upon 1) what numbers we put in the numerator and denominator of our exponential equation and 2) depending upon how we scale our axes, we can dramatically change how our graph appears.  There will always be a relatively flat part to this curve located close to x=0, then there will be a concave-up section, followed by a concave-down section, and finally another relatively flat section further out on the x-axis.  You happened to choose a large range (up to x=60-70) for the x-axis of your graph.  This compressed your graph, so it did not show the first two parts of the curve (the relatively flat part located close to x=0 and the concave-up section) clearly.

This may indeed be the experimental data, but acc to the maths, if I differentiate twice, why is it becoming always negative?

@Saurabh Raje  The slope is given by the first derivative.  The first derivative of e^(-1/x) is (e^(-1/x)/(x^2) which is always positive.

sir , i am sure that concavity is checked using the second derivative, which is negative

the second derivative of e^(-1/x) is given by ((e^(-1/x))/(x^4))-((2*e^(-1/x))/(x^3)) which does change sign

so for x<1, this is positive, thats why?

yes, it seems to change sign around x=1.9378

@Saurabh Raje  Sorry, I should have said that it changes sign at 0.50.  Do you get the same second derivative and see the change in sign at x=.5?

The two problems are:

You must have an "power-intercept" with inverse-x as
  $$\ y(x)=e^{a-(b/x)}$$
  Important but not necessary, as to provide a limiting value $e^a$ to $y$ at $\infty$.

Also what you are worried about the derivatives is simple:
  $$y'(x)=\frac{b}{x^2}e^{a-(b/x)}$$
  $$y''(x)=\frac{b(b-2x)}{x^4}e^{a-(b/x)}$$with second derivate a root at $x=\frac b2$

An example graph with two resolutions:
$$\text{legend: }[\color{red}{y(x)}\;,\color{purple}{y'(x)}\;,\color{black}{y''(x)}]\;\text{ with }a=2.5,b=2.2$$
So the derivative changes the sign at $1.1$,

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text:
Crystal meth in Breaking Bad

Can good quality crystal meth really be made as in Breaking Bad?
Can a pharmaceutical drug really be reverse-engineered?

I've no idea what you mean by the second sentence and what it has to do with your first question. And I'm not sure if the first question is really on-topic here.

There's a lot of ambiguity in your question.  "Good quality" would need to be defined, and whether it can be "sold" is definitely outside of our scope here.  As @MadScientist mentioned, we can attempt to decipher what you mean (as the answerer has), but it's not immediately evident.

by good quality i mean to say the type of meth that is really synthesized to great purity.....as in walter does?
and as for pharmaceutical drug...... yes....as far as i know though i have not seen breaking bad ..... but walter does take some pseudoephedrine convert it to meth ..... but in general is it possible to take a drug or any thing for example coca cola ....reverse engineer it ?

Yes. It's what we call chemistry. That is literally the whole point of chemistry.

Can good quality crystal meth really be made and sold as in Breaking Bad?

Yes, most procedures are accurate, although the whole methylamine is so hard to get idea in Breaking Bad is nonsense. It is DEA List I chemical, but can be easily manufactured and no sane kingpin would go steal it from a cargo train (although it makes a good scene). The real problem is getting the P2P (phenyl-2-propanone, or phenylacetone) which the authors of breaking bad seem to ignore (probably on purpose, so that stupid kids won't try and make it at home).

Can a pharmaceutical drug really be reverse engineered?

Reversed engineered? Assuming you are talking about the transition of pseudoephedrine into methamphetamine, this is just a chemical reduction, following this scheme (if this kid wants to make meth, he would have found this procedure anyway, I'm not giving out any special information that would help anyone manufacture methamphetamine).

I believe that by "reverse-engineered" they are referring to whether one could determine the method of preparation of the product after just seeing the product. Of course this isn't feasible, given the numerous substrates, conditions and reagents that may have been used in the preparation.

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text:
How to determine the specific heat of methanol from calorimeter data?

A $\pu{25.95 g}$ sample of methanol at $\pu{35.6 ^\circ C}$ is added to a $\pu{38.65 g}$ sample of ethanol at $\pu{24.7 ^\circ C}$ in a constant pressure calorimeter. If the final temperature of the combined liquids is $\pu{28.5 ^\circ C}$ and heat capacity of the calorimeter is $\pu{19.3 J/^\circ C}$ determine the specific heat of methanol. The heat capacity of ethanol is $\pu{2.46 J/(g * ^\circ C)}$.

\begin{align}
  q_\mathrm{methanol} &= -q_\mathrm{ethanol}\\
  msT &= -msT\\
  \pu{25.95 g} \cdot  (\pu{28.5 ^\circ C} - \pu{35.6 ^\circ C}) \cdot  s
    &= -\pu{38.65 g} \cdot  (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C})
       \cdot  \pu{2.46 J/(g* ^\circ C)}\\
\therefore s &= \pu{1.96 J/(g * ^\circ C)}
\end{align}
The correct answer, however, is $\pu{2.36 J/(g * ^\circ C)}$.
I think that there is a mistake in my solution, because the heat capacity of the calorimeter is given, and I didn't use it.
Could anyone explain where I have been wrong?

It's a good idea to include units in your equation. Where does the number 38.5 come from? (I think you mean 28.5). Also, why are you multiplying the right side by 2.64?

yes i meant that.. 2.46 is the specific heat of ethanol I got it from the book

You don't mention ethanol in your problem, you only mention methanol. Could you double check the question you've written. Is it that you are given a sample of methanol and it is added to a sample of ethanol?

You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at $\pu{24.7 ^\circ C}$ before the methanol was added and then it rose $\pu{3.8 ^\circ C}$ along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is $\pu{19.3 J/^\circ C}$ then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. $\pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} = $ "Joules absorbed by the calorimeter".
Then we simply apply the heat lost equals heat gained concept:
$$\begin{multline}
\pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\
  = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C})
    \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C}
    \cdot \pu{3.8 ^\circ C}
\end{multline}$$
After rearranging, we get $S = \pu{2.36 J/(g * ^\circ C)}$.


text:
Why isn't SiC called carbon silicate?
Is there any rule why a crystal made of Si and C is named silicon carbide SiC, but never carbon silicate CSi? Both silicon and carbon are in the same group, so share many of the atomic properties, so why the asymmetry in naming? Are all such compounds named like <HigherPeriod> <LowerPeriod>?

As per the rules of inorganic chemistry nomenclature the element of the compound which is more electronegative is to be written at the end of molecular formula and its name is to be written after all the elements which have lower electronegativity are written hence the given compound is pronounced as silicon-carbide and not carbon-silicide
a more detailed view about electronegativity of elements can be found in this picture

Aha, so solutions like $\text{Ga}x\text{In}{1-x}\text{P}$ should be written with $\text{Ga}$ first, then $\text{In}$, or does this rule only work for the last element (here $\text{P}$)?

@Rusian As a general rule, it is assumed that electronegativity in main groups grows from left to right and from down to up in s-p blocks. It is extremely debatable in many cases, which element is more electronegative if we move on diagonal from left-up to down-right, say, chlorine in many cases is capable to oxidize chlorine and vice versa. As for numeric electronegativity scales, there are several and none is ... undebatably right.

@Ruslan i think permeakra cleared your doubt if you want any more explanation feel free to comment

Not quite. My comment asked if all elements preceding the most electronegative should also be sorted with increasing electronegativity. And permeakra just said about how electronegativity behaves across periodic table.

@Ruslan yeah Ga should be written first as per rules

@Rusian It highly depends, I can find examples of inverted writing, say, $\ce{CrO2Cl2}$ is usually written as I wrote, but electronegativity for oxygen is usually assumed to be higher than for chlorine =(. Further, $\ce{SOBr2}$ is also usually written as I wrote, and electronegativity of bromine is for sure lower than for oxygen.

@agharehanabbas and what it has to do with these molecules, where both oxygen and chlorine/bromine atoms are connected to central atom =)?

Here's a great answer related to exactly how electronegativity considerations enter IUPAC nomenclature.

@NicolauSakerNeto really it is a great answer

Silicates are salts of silicic acids, $\ce{xH2O * ySiO2}$ and $\ce{SiC}$ is not one. The proper term is silicide. It is usually assumed, that s/p element upper or righter from considered one is more electronegative, and "-ide" nomenclature assumes that second element has negative charge and is more electronegative. Of course, there are exceptions. For example, $\ce{RhB}$ is still written as is, despite $\ce{Rh}$ having higher electronegativity.
However, the ordering of elements in formulas and naming of compounds are subjects of tradition. For example, despite having higher negativity, nitrogen in the formula of ammonia precedes  hydrogen: $\ce{NH_3}$, while for water hydrogen precedes oxygen $\ce{H2O}$, and both may be called 'hydrides' in some cases. It may be viewed as another rule: in case of a isolated particle central atom of the molecule first, substituents later (composite substituents written in curly braces), but acidic hydrogens first. In brutto-formula indexes symbols of elements AFAIK are ordered alphabetically except that carbon and hydrogen are first two.

You must have meant "hydrogen follows nitrogen" not "nitrogen follows nitrogen".

@Rusian yes, indeed. Corrected

why a crystal made of Si and C is named silicon carbide SiC, but never
  carbon silicate

To add to agha rehan abbas' and permeakra's comments, the "ate" suffix typically implies the presence of oxygen (carbonate, silicate, phosphate, etc.).  Since $\ce{SiC}$ does not contain oxygen, you would not use the "ate" suffix in this case.

Yeah, I guessed I was wrong here after reading agha's "and not carbon-silicide".


text:
The Principle of Internal Standardization (in the context of ion chromatography)
I am trying to grasp how internal standardization works in principle, and I am having some trouble wrapping my head around it. It is not the choosing of an IS I think about, orwhen it might benefit the analysis, but just the underlying principle.
I add a known amount of the IS to every sample: unknown samples, blank samples, calibration samples, and quality control samples. Each of these samples contain equal amounts of the internal standard (or as "equal" as our pipettes may give). Considering the calibration, when the calibration samples are analyzed, I see two peaks in the chromatogram: the analyte and the IS.
Let's say I ran calibration samples of 0, 1, 2, 5, 10, and 20 ppm, each one consiting 20 ppm internal standard. The analyte peaks will grow for each calibration sample, but the IS peaks remain more or less stationary. I construct the calibration curve by plotting the ratio of analyte peak area to IS peak area (or height), as a function of the analyte conentration in the calibration samples.
I have been told that there is a constant ratio involved, one that is used to correct analyte concentrations for loss of analyte, or varying sample introduction system. If the IS peak is lower, then the analyte concentration will be corrected upwards. This apparently is because this ratio I mentioned is constant, so when reducing the denominator, the nominator must increase. Mathematically, this is trivial; however, what is this mystical constant, and what is its value? Is that arbitrary? Can I choose this constant's value myself?
I am quite confused. Hopefully someone here can help my out!

This ratio has nothing mystical, and will not work in every case.
So let's first comment on the reasons to use an internal standard. As you mention, when you inject multiple times solutions containing the same concentration of IS:

but the IS peaks remain more or less stationary

More or less is a major point here: this means that repeatability of the measurement process is an issue, when multiple samples with the same concentration lead to a variation in the response. If the accuracy required for the measurement is such that these small variations might not be an issue, there is no need for an IS. In chrmatography measurement, sources of variability include repeatability in the volume injected, detector sensitivity. If there is some sample preparation steps involved, loss of analyte (in transfer from one tube to the other, in the filtration steps etc...) can also lead to variations from one analysis to the other.
So how will an IS solve this variability issue? The basic point is the assulmption that these external variability issues will affect both the analyte and the IS in a similar way. So one has to look at the sources of errors. For instance, let's consider an error in the measurement of a sample volume (let's call it $\epsilon$ and assume it is close to 1) . We have:
$$m_{analyte} = V_{real} c_{analyte} = \epsilon V_{set} c_{analyte}$$
$$m_{IS} = V_{real} c_{IS} = \epsilon V_{set} c_{IS}$$
If the measurement system accurately measures $m_{analyte}$ and $m_{IS}$, it is quite obvious that by dividing both equations we can remove this variability parameter and we get:
$$R_{analyte/IS} = {m_{analyte} \over{m_{IS}}} = {c_{analyte} \over c_{IS}}$$
This is the point were the ratio comes out: taking a ratio allows to remove the variability parameter ($\epsilon$) in this case.
Now this is a bit too simplistic, and a measurement procedure is not as simple. The general concept will still be to try to factor out the variability parameter as a proportion of the measured value, and then to take a ratio which will remove this variability from the measured values.
If the IS is correctly chosen so that it behaves as the analyte through the entire analytical procedure (both in terms of physico-chemical properties, as well as in the response of the detector), and if the chromatographic properties as well as the detector are such that peak area is proportional to concentration for the analyte as for the IS, then one can write:
$$A_{analyte} = \epsilon(analyte) R_{method}(analyte) c_{analyte}$$
$$A_{IS} =  \epsilon(IS) R_{method}(IS) c_{IS}$$
where $A_{analyte}$ and $A_{IS}$ are the measured peak areas for both analyte ans IS, $R_{method}(x)$ is the method response factor, which is dependent on each compound. In general the variability factor $\epsilon$ should depend on each compound, but under the above assumption of similar properties, one can assume that $\epsilon(analyte) \approx \epsilon(IS) \approx \epsilon$. Without IS, one would determine $R_{method}(analyte)$ through a calibration procedure. With an IS, one takes the ratio and the above equation can be written as:
$$R_{area} = {A_{analyte} \over A_{IS}} = {R_{method}(analyte) \over R_{method}(IS)} {c_{analyte} \over c_{IS}}$$
The ratio:
$$R_{analyte/IS} = {R_{method}(analyte) \over R_{method}}$$
is the ratio of the response factors of the method for analyte and IS. This is probably the ratio that is referred to. And it is not mystical at all, it is obtained through a standard calibration procedure with the IS.
One caution word to conclude this answer: I have tried to be quite precise in the answer. I have often times heard some people state that the use of an, internal standard is a general solution for analytical issues. Thus I want to insist on a warning: the solution is good as long as the assumptions stated above hold. If any one of these fails, then the whole analytical procedure will also fail. And of course, the use of an IS will only protect from errors after addition of the IS, and not from variability in the addition of the IS (you mention pipetting it in the analyte solution, which will limit the repeatability to that achieved by the pipettes used).

I'll throw in one more point. Retention time is frequently used as a means to identify a particular component in various chromatographic techniques. Having a consistent retention time for the IS shows that the chromatography setup is consistent since there are a zillion things that can cause the retention time to change.


text:
Why is the bond angle H-P-H smaller than H-N-H?
$\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.
But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference?

An answer to a closely related question is given by my answer  here http://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h%E2%82%82o-h%E2%82%82s-h%E2%82%82se/57973#57973

related http://chemistry.stackexchange.com/questions/38599/why-is-the-inversion-barrier-larger-in-ph3-than-it-is-in-nh3

Starting point: 2s orbitals are lower in energy than 2p orbitals.
The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees.  Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.
In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees.  This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.
So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.
There are two choices for the central atom:
remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another.  As a result,

substituents will be arranged closer together and destabilizing steric interactions will be increased
due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy
the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital

or
hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another.  As a result,

steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart
due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy
the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character

An example:
Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$.  Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine.  Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density.  Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital.  The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!).  Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds.  As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$.  Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!
Back to our problem:
Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2).  In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$.  Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed.  The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.
In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the  hybridized case the lowest energy.  Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.

It is not true that hybridisation costs energy, as it is only a mathematical concept, which has no real physical meaning, it is impossible to assign an energy value to hybridisation. It is also not possible, that an element chooses to hybridise. Hybridisation is a result of a certain - the thermodynamically most stable - geometric arrangement of atoms in a molecule.

@Martin  Thanks for the comment. Martin wrote 1) "It is not true that hybridisation costs energy" - I agree, see my edited answer. 2) "Hybridisation is a result of a certain - the thermodynamically most stable - geometric arrangement of atoms in a molecule" - I agree, in fact I said that in my initial version, "Whichever of these two pathways is most exothermic..."

Old post, I know, so there's a good chance I don't get an answer here, but does the decrease in H-P-H angle due to lack of hybridization necessarily mean more steric hindrance here? Yes, the cone angle is smaller, but an $sp^3$-orbital is shorter than a p-orbital, and with small atoms like hydrogen, I would think a small change in bond length would have a pretty significant effect on sterics around them phosphorus atom.

@DGS asked, "does the decrease in H-P-H angle due to lack of hybridization necessarily mean more steric hindrance"? If the bond angle was the only thing changing, then increased steric interaction would result.  But, as you note the bond length changes too (1.42 angstroms in phosphine) making it hard to answer your question without calculations.

@Martin-マーチン "It is not true that hybridisation costs energy, as it is only a mathematical concept..." I agree. Yet this might lead to some confusion. Hybridisation is conceptually taught (at secondary school level) via two mechanisms. The first option distinguishes initial promotion ($\neq$ hybridisation) from the mixing itself ($=$ hybridisation). The second way is to combine promotion and mixing by calling both together as hybridisation.

In the first line of reasoning, hybridisation does not involve an energy cost (as you said); whereas in the second it's essentially the promotion even though not called as such. This arguement assumes that both options are correct in their own right, which is what I've been lead to believe. (Please correct me if I'm wrong.)

@LinearChristmas Why would hybridisation not require an energetic cost? If an electron was initially in a 2s orbital while later on, it is assigned to be in a hybrid of 2s and 2p orbitals, then it is necessarily raised in energy since the hybrid is certainly of higher energy. I understand that the concept of "electron promotion" followed by hybridisation may not be correct as I have learnt from here: https://chemistry.stackexchange.com/questions/98910/electron-promotion-in-hybridisation-a-misconception

@Martin-マーチン Yes, the mixing of orbitals is a mathematical concept. But the re-organisation of electrons does require energy inputs, especially so when electrons of lower energy states now have to occupy higher energy states. Certainly, you would have to agree that there is a re-organisation of electrons when atoms form chemical bonds. Hybridisation describes this re-organisation. Thus, I think it is fair to say that hybridisation incurs energetic costs.

@TanYongBoon No, absolutely not. I do not agree with anything you have written in your comment. The formation of bonds (and the associated costs and gains) is completely independent of the mathematical description. A lone carbon atom would never, ever choose to have hybrid orbitals in the absence of an external field. Once the hydrogens form such a field in methane (for example), it is a completely different state, in which a description in terms hybrid orbitals may make sense, but is not necessary at all.

"It will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital. " Can you explain a bit whats rehybridization( i know whats hybridization) and why electrons do so? Its new to me (term) also why less electron density region will have more p character bonding chances? @ron

@Orion_Pax  "Rehybridization" is just a shorthand way of saying that the geometry and hybridization will be different when we make some change.  For example, if we compare ammonia and trifluoroamine the geometries and hybridizations will be different.  We can say that, as we replace the hydrogens in ammonia with fluorines, the molecule will adopt a different geometry, the bonds will involve different hybridizations in order for the molecule to adopt the lowest energy configuration possible.  We can say all of this quickly and simply by saying that the molecule will "rehybridize".

@Orion_Pax you asked, "why less electron density region will have more p character bonding chances?" Each carbon atom uses 1 2s and 3 2p orbitals to make all of its bonds.   Therefore, if a bond uses less s character, then the other bonds around this carbon will have to use more s character.

I see but Sir in hybridization all four hybrid are degenerate orbitals isnt ? So if some bond uses less s character as for formation will not it be having a different energy compared to what a hybrid should be ?

@Orion_Pax  In some cases like methane, hybridization produces 4 equivalent (degenerate) orbitals that are all at the same energy.  In other cases like methyl fluoride, the 4 hybridized orbitals are not equivalent. Bent's Rule tells us that the $\ce{C-F}$ bond will have less s character and be at higher energy relative to the 3 degenerate $\ce{C-H}$ bonds which will each have a bit more s character and consequently lie at lower energy.

First note, that hybridisation is a mathematical concept which can be applied to interpret a bonding situation. It has no physical meaning whatsoever. Instead it helps us to understand the direction of bonds better.
Second note, that the second period usually behaves quite differently from the remaining elements in a group. So in a way, ammonia behaves unnatural or anomalous.

If you compare nitrogen with phosphorus, you will note, that the former is much smaller than the latter, i.e. van der Waals radii $r(\ce{N})=155~\mathrm{pm};\ r(\ce{P})=180~\mathrm{pm}$ (ref. wikipedia), covalent radii $r(\ce{N})=71~\mathrm{pm};\ r(\ce{P})=107~\mathrm{pm}$ (ref. wikipedia). Therefore also the orbitals in nitrogen are smaller, and $\ce{s}$ and $\ce{p}$ orbitals will occupy more of the same space than in phosphorus. As a result the $\ce{N-H}$ bond distance will naturally also be shorter.
A lone pair is usually most stable in an orbital that has high $\ce{s}$ character. Bonds will most likely be formed with the higher lying $\ce{p}$ orbitals. The orientation of these towards each other is exactly $90^\circ$.
In ammonia this would lead to very close $\ce{H\cdots{}H}$ contacts, which are repulsive and therefore the hydrogen atoms are pushed away from each other. This is possible since in the second period the $\ce{s-p}$ splitting is still very small and the nitrogen $\ce{s}$ orbital is accessible for the hydrogen atoms. This will ultimately result in mixing $\ce{s}$ and $\ce{p}$ orbitals for nitrogen in the respective molecular orbitals. This phenomenon can be referred to as hybridisation - the linear combination of orbitals from the same atom. This term is therefore somewhat independent from its most common usage.
It is also very important to know, that the molecular wavefunction of a molecule has to reflect its overall symmetry. In this case it is $C_{3v}$, which means there is a threefold rotational axis and three vertical mirror planes (the axis is element of these planes). This gives also rise to degenerate orbitals. A canonical orbital picture has to reflect this property (BP86/cc-pVDZ; valence orbitals are ordered with increasing energy from left to right).

Note that the lowest lying valence molecular orbital is formed only from $\ce{s}$ orbitals (There is one additional $\ce{1s^2-N}$ core orbital.)
Now Natural Bond Orbital (NBO) Theory can be used to transform these delocalised molecular orbitals to a more common and familiar bonding picture, making use of atomic hybrid orbitals. This method is called localising orbitals, but it has the expense of losing the energy eigenvalue that may be assigned to canonical orbitals (NBO@BP86/cc-pVDZ; valence NBO cannot be ordered by energy levels).

In this theory you will find three equivalent $\ce{N-H}$ bonds, that are composed of $32\%~\ce{1s-H}$ and $68\%~\ce{s^{$0.87$}p^3-N}\approx\ce{sp^3-N}$ orbitals. Note that the lone pair orbital at nitrogen has a slightly higher $\ce{s}$ orbital contribution, i.e. $\ce{s^{1.42}p^3-N}\approx\ce{sp^3-N}$.
So the thermodynamically most favoured angle is found to be $107^\circ$ due to a compromise between optimal orbital overlap and least internuclear repulsion.

The canonical bonding picture in phosphine is very similar to ammonia, only the orbitals are larger.  Even in this case it would be wrong to assume, that there is no hybridisation present at all. However, the biggest contribution to the molecular orbitals stems from the $\ce{p}$ orbitals at phosphorus.

Applying the localisation scheme, one end up with a different bonding picture. Here are three equal $\ce{P-H}$ bonds that are composed of $48\%~\ce{1s-H}$ and $52\%~\ce{s^{$0.5$}p^3-P}$ orbitals. The lone pair at phosphorus is composed of $57\%\ce{s} + 43\%\ce{p}$ orbitals.

One can see the difference of the molecules also in their inversion barrier, while for ammonia the inversion is readily available at room temperature, $\Delta E \approx 6~\mathrm{kcal/mol}$, it is very slow for phosphine, $\Delta E \approx 40~\mathrm{kcal/mol}$.
This is mostly due to the fact, that the nitrogen hydrogen bonds have already a significant $\ce{s}$ orbital contribution, which can be easily increase, to form the planar molecule with formally $\ce{sp^2}$ hybrids.

@ martin : so hybridization is totally hypothetical and not practical,right?

@user36790 In the most common usage, this is true, but it can give you somewhat more insight into the shape of molecules. In a more genaral meaning, hybridisation just refers to the mixing of orbitals at the same atom in a molecule, and that might be observable.

Since Phosphorus has vacant 3d orbitals, the s-orbitals of hydrogen can directly overlap with the orbitals of Phosphorus. The procedure of bonding does not see much benefit in hybridisation as it is energetically expensive. Thus the hybridised character of the bonds is less dominant than in ammonia and the structure resembles the Lewis structure that we would expect instead of the tetreahedral (VSEPR) structure. This effect is called Drago's Rule and can also be observed in hydrogen Sulfide.

d-orbitals are too far removed to do anything in phosphorus and sulphur. Check out Martin's answer.


text:
Why does an acid exist when dissolved in water?
Acid is a substance that releases $\ce{H+}$ ions when mixed with water But if it releases $\ce{H+}$ ions the  acid will ultimately not exist. For e.g we take the acid $\ce{H2SO4}$. If we add water then it becomes $$\ce{H2SO4 + H2O -> H+ + SO4^{-2} + H2O}$$ So the bond between hydrogen and sulphate will be broken therefore the acid will not exist, but we see that diluted acids are possible. Why?

Your doubts are legitimates, but there are some aspects that you should take in account:
1-Not all acid are fully dissociated
Not all acids fully dissociate, so if you take acetic acid you have in fact an equilibrium with some acetic acid molecules not dissociated. So your question holds only for strong acid like sulfuric acid you mentioned.
2-You are using pseudo-Arrhenius definition, why don't you use Lewis acid definition?
If you think the acid as a 'substance that that releases $H^{+}$' this is still an acid even if it dissociate, and even here you should notice that acidity is linked with $H^{+}$. However if you want to dig deeper there are many definitions of acid if you use Lewis acid theory and so you think the acid as a specie that accepts electrons you can easily see that in fact that the acidity is linked mainly with $H^{+}$!

$BH_3$ will accept a pair of electrons, so it is a Lewis Acid.

@LDC3 Hi! what's the point? I never say that it isn't, I've said that the acidity in that case is linked with the H+ ion

Your answer seems to indicate that $H^+$ is the "only" acid. Maybe you should omit the word "only"?

@LDC3 ah I see, thank you for the suggestion! i really appreciate it!

As a user mentioned, not all acids dissociate that way. Sometimes, there is also the production of Hydronium ions.
Refer : http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/water.php
$$\ce{2 H2O~(l) <=> H3+O~(aq) + {}^{-}OH~(aq)}$$
The hydronium ions react with the sulphate radical, forming an acid like entity. Hence, the acidity is not lost.

Well, this answer is technically inaccurate, but it's not delete-worthy. The author did state their opinion; and in this case, the downvotes should indicate its quality.

Yes, exactly; only ions can't not exist. Water molecule has dipole nature. When acid is mixed with water, then because of ion-dipole interaction which is stronger than dipole -dipole interaction b/w water molecules and ion-ion interaction b/w acid molecules, opposite ends of water dipoles surround the ions of acid ( means the dipole getting delta + charge will attract negative ion of acid and vica versa). So, now ions get hydrated as process of hydration takes place. They are still free entities, but surrounded by water dipoles. So, the solution of acid and water exists.

Dipole nature ?

Yes, haven't you studied that?? See, as hydrogen and oxygen both have different electro-negativities( oxygen's is higher ) so oxygen pull the shared pair of electron towards itself. Because of this pull, oxygen gets a slight negative charge (called delta - charge) and hydrogen gets delta positive. This group of charges is called dipole. Because of the positive charge attained by oxygen it surrounds the positive end of acid ( as it is very small charge delta - charge so it can't attract the positive end like that in ionic compounds.)


text:
How does temperature affect equilibrium, according to Le Chatelier's principle??
I want to understand this concept in detail. I am getting confused with increase in temperature backward reaction takes place or forward?? And how?? Proper explanation is needed.

In a simple equilibrium we have either $$\ce{A + heat <=> B}$$ where the forward reaction is endothermic and the back reaction exothermic, or $$\ce{A <=> B + heat}$$ where the forward reaction is exothermic and the back reaction endothermic.  So in a simple equilibrium, the reaction in one direction will be exothermic and the reaction in the other direction will be endothermic.
When the temperature of an equilibrium is increased, the rate of both the forward and back reaction is increased.   However the rate of the endothermic reaction (the reaction that absorbs the added energy) is increased more.  So in the first reaction shown above, adding heat will shift the reaction to the right; in the second reaction heat will push it to the left.
The reason behind this behavior is that both the forward and back reaction can be described by the Arrhenius equation $$\ce{rate=Ae^{-E_{act}/RT}}$$  and the rate with the larger activation energy (the endothermic reaction) will always be more sensitive to changes in temperature. Therefor, the endothermic reaction will increase faster with increasing temperature than the exothermic reaction will.  Put some numbers into the equation and convince yourself.


text:
pKa explanation through empirical formula?
Suppose two compounds with, say, $\mathrm{p}K_{\mathrm{a}1}=2.8$ and $\mathrm{p}K_{\mathrm{a}2}=3.2$, and formulae $\ce{C_{$a$}H_{$b$}O_{$c$}}=\ce{C_{$p$}O_{$q$}(OH)_{$r$}}$ and $\ce{C_{$d$}H_{$e$}O_{$f$}}=\ce{C_{$u$}O_{$v$}(OH)_{$w$}}$, respectively, with given $abcdef$, $pqruvw$. Can we explain the above $\mathrm{p}K_\mathrm{a}$ with (ONLY) the knowledge of the given numbers of atoms of $\ce{C}$, $\ce{H}$ and $\ce{O}$ using the theory of Usanovich, and no knowledge of the internal chemical structure of the chemical bonds inside the two molecules?

Can we explain the above pka with (ONLY) the knowledge of the given
  numbers of atoms of C, H and O using the theory of Usanovich

No, nobody's theory (including Usanovich's) can explain $\ce{pK_{a}}$ differences based on molecular formula coefficients.  This is best illustrated by the observation that even compounds with the same coefficients (same values for "abcdef") can have different $\ce{pK_{a}}$ values.  For example, $\ce{C3H6O}$
\begin{aligned}
\ce{&acetone &&pK_{a}=20}\\
\ce{&allyl~ alcohol &&pK_{a}=$15.5$}\\
\end{aligned}

Even if the two molecular compounds have DIFFERENT C,H and O contents? Your example is a same molecular formula with 2 differents pka, I was arguing about two different compounds (note the numbers/letters were deliberately different).

Yes, I understand.  My point is, if it fails - as it must - for the same empirical formula, then it must also fail for different empirical formulas.

And my point is, that, according to Usanovich theory, acid or base character is defined by the ability of donating/recieving positive/negative species. If you have a compound with differente number of atoms, in principle, there is no match between the possible number of positive/negative ions and possible charges. I can not see your point, ron...


text:
light Intensity and its effects on energy and emitted electron on metal surface
we have a metal surface with electrons in a specific energy state say X ,now we shine light on it of specific wavelength and intensity.
now what if i double the intensity of light , what would happen to the number of electrons emitted and the average energy of the electrons ?
my guess is that since intensity is doubled, then number of photons emitted also double and the energy also doubles .... but i am not really sure about it

what if i double the intensity of light , what would happen to the
  number of electrons emitted and the average energy of the electrons?

This is known as the "Photoelectric Effect" and is described by the following equation
$$\ce{E_{max}=h\nu- WF_{M}}$$
where $\ce{E_{max}}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\ce{WF_{M}}$ is the workfunction for the particular metal. Your guess is part right, doubling the intensity of the incoming beam doubles the number of electrons ejected from the metal surface, but their kinetic energy distribution remains unchanged.  The maximum kinetic energy of an emitted electron is still determined by the above equation and therefor still capped by $\ce{E_{max}}$ and the kinetic energies of all electrons emitted are still distributed from 0 to $\ce{E_{max}}$.  If you'd like more information, here's a good, concise reference.

The intensity of emitted light varies directly with the number of photoelectrons emitted. If the intensity of light is varied, keeping the frequency of the source and the accelerating potential constant, the resulting photo-current produced is found to increase,i.e. the photo-current vs intensity of incident light curve is linear having a constant slope. Since the number of photoelectrons emitted per second is directly proportional to photo current produced, therefore it is also proportional to the intensity of incident light.

The above equation that has been stated in the previous answer can also be written as:

E(max)=eV, where e is the basic electronic charge and V represents the stopping potential. Now it has been found that the stopping potential is dependent on the frequency of incident light and is independent of its intensity. This implies that the maximum kinetic energy of emitted photo-electrons is independent of intensity too[since, it remains proportional to stooping potential].


text:
Does nitrogen tetrafluoride act as an acid or a base in its reaction with water?
$\ce{NF4+}$ ion reacts with water as:
$$\ce{NF4+ + H2O -> NF3 + H2F+ + {1\over 2}O2}$$
My question is, does $\ce{NF4+}$ function as an acid or base in this reaction, and why?

Neither. The reaction is NOT one of ion exchange; it is a redox reaction. Thus, $\ce{NF4+}$ works as electron acceptor here, like this:
$$\ce{NF4+ + 2e- -> NF3 + F-}$$
Since the reaction mechanism is unclear, it is possible that some of its stages may have acid-base character (for example, $\ce{F-}$ sometimes acts as a base), but $\ce{NF4+}$ certainly isn't acting primarily as base or acid here.


text:
Which acids are created by the combination of a gas and water?
Which acids are created by the combination of a gas and water?  I know that carbonic acid is created by combining carbon dioxide and water, but are there others?  Is there a specific name for this kind of acid?

I don't think there is a specific name for these acids. But I can think of at least 4 more. The gases are $HCl,\ SO_2,\ SO_3,$ and $NO_2$.

To add to this, I believe all of the hydrogen halides are gases under standard conditions. As far as gaseous compounds that actually react to form acidic products (rather than just dissolving), off the top of my head I can add $\ce{X2}$ and $\ce{COX2}$ (for $\ce{X = F, Cl}$), $\ce{ClO2}$, $\ce{OF2}$. I'm sure there must be others.


text:
What is electrodialysis and is electrolysis prevented or suppressed?
Recently, I was introduced to a method of desalinization called electrodialysis in which a current was applied to salt water in such a way that the positive and negative were separated and drawn through membranes, leaving less salty water in between.  how do the membranes facilitate a one-way flow of ions?
Also, this method sounds a lot like the electrolysis.  How is electrolysis prevented or suppressed?

Could electrodialysis be the method you are searching for?

Yes the name I was looking for is electrodialysis


text:
ice cream float - what causes it?
When you mix ice cream and a soft drink (like coke,)  a reaction happens that creates a thick foam.  I am assuming the foam id composed of carbon dioxide gas and milk curd, which is produced by the milk reacting with an acid.
Is the milk curd produced by the carbonic acid or the phosphoric acid?
Is the carbon dioxide released by the added sugar, or another ingredient?
Are there other important processes going on in there?

Just a guess, but I would imagine that the foam is caused by the effervescence of carbon dioxide (which was dissolved in the soda) made more rapid by nucleation on the ice cream.


text:
Why do acids favor the oxidizer? And bases the reducer?
Simply put, why does adding acid to the cathode up the cell voltage? From what I understand, acids are proton donors. Do these positive protons attract electrons toward the cathode compartment of the electrochemical cell?

The effect that can be observed by adding acids to a solution to increase conductivity is known as the Grotthuss mechanism. Protons have a much higher mobility in solution ($3.62\cdot10^{-3}~\mathrm{cm^2~V^{-1}~s^{-1}}$) than any other cation, basically allowing it to tunnel through a solution. (A similar effect can be noted for hydroxyl anions.)
Basically this allows a much faster charge exchange through hydrogen bond networks, also resulting in a higher cell voltage.
(But then again I am not really sure I understood your question correctly.)

I donn't know what a "detailed canonical answer" is, but intutitively, I can appreciate that it is easier to take an electron away from (i.e. oxidize) something with a negative charge than something that is electrically neutral. That is because it costs energy to separate charged particles and it produces energy to reunite them. So many examples - a thiol is stable to air in the absence of a base, but in the presence of base, air will oxidize it to a disulfide. Same with phenol - stable in air but with base present, you soon get colored condensation products.


text:
Reconciling the units of pH calculation
As part of a simulation I am developing I have an acid in aqueous solution represented as a concentration field of mass density (units g/L) at various points in space (denoted $[S]$). I want to calculate the pH of the solution. I am attempting to use the method for weak acids outlined here, but am struggling with the dimensionality of the variables,
$$[H]^2 + K_a [H] - K_a C_a = 0.$$
Looking at the quadratic equation, and knowing that $K_a$ has no units, it implies that $[H]$ and $C_a$ are also dimensionless. Does this mean that $[H]$ is in fact the activity of the the solution, and not the $H^+$ concentration? And is the relationship between $C_a$ and $[S]$ $$C_a = \gamma \frac{[S]}{M(S)c^\ominus},$$
where $\gamma$ is the activity coefficient, $M(S)$ is the molar mass of the acid, and $c^\ominus = 1$M is the standard amount concentration?
Followup question:
The solution to the quadratic equation is
$$[H] = \frac{-K_a \pm \sqrt{K_a^2 + 4K_a C_a}}{2},$$
thus $[S] = 0 \implies [H] = \frac{-K_a \pm K_a}{2} = -K_a$ or $0$.
As $pH = -\log_{10}\left([H]\right)$ and $K_a > 0$  there is no real solution, but logic would dictate that $pH = 7$ in this case. Where am I going wrong?

$K_a$ is not unitless (in this case). Some equilibrium constants may have their units totally cancel, but the acidity constant has units of concentration:
$$K_a = \dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\implies\dfrac{\frac{\text{mol}}{\text{L}}\cdot\frac{\text{mol}}{\text{L}}}{\frac{\text{mol}}{\text{L}}}=\frac{\text{mol}}{\text{L}}$$
I have used moles per liter (molarity) as it is a common concentration unit.
In this case, $K_a$ and $[\ce{H+}]$ both have concentration units, as does $C_a$, so your initial equation $[\ce{H+}]^2 + K_a [\ce{H+}]+ K_a C_a =0$ is valid (at least for units.
If you recast $K_a$ and $C_a$ in terms of activities, then you need to replace $[\ce{H+}]$ with $\alpha_{\ce{H+}}$ (which is unitless).

I suggest you consider the following:
$$\ce{HA->H+ +A-}\\
\begin{array}{c|c|}\hline
&&HA&H^+&A^-\\\hline
&\text{initial}&C&0&0\\\hline
&\text{final}&C(1-\alpha)&C\alpha&C\alpha\\\hline
\end{array}$$
where $C$ is initial concentration and $\alpha$ ($\le1$) is dissociation constant of the acid.
In equilibrium:
$$K_a=\frac{[H^+][A^-]}{[HA]}=\frac{C\alpha^2}{1-\alpha}$$
As $\alpha\ll1$,
$$K_a\approx C\alpha^2\implies\alpha=\sqrt{\frac{K_a}C}\implies[H^+]=\sqrt{K_aC}$$
$$\LARGE pH=-\frac12\log{(K_aC)}\;\small\text{for weak acids}$$


text:
Does H3O+ donate a H+ to Water?
Does the hydronium ion donate a proton to water? If the collision of 2 molecules of water can cause a proton to be transferred, does hydronium 'pass' a proton to water?

$\ce{H3O^+}$ is really just a "shorthand notation" for what really occurs in solution.  Typically higher hydrates of the proton, particularly $\ce{H5O2+}$ the "Zundel" cation and $\ce{H9O4+}$ the "Eigen" cation are thought to be the main players in the proton transfer process.  As permeakra has pointed out, Grotthuss was one of the early pioneers studying this process and his name is still used to describe the mechanism.  The mechanism is still an active area of research with considerable controversy remaining over the details of the mechanism.  This reference provides some diagrams of the general process.

Basically, just as electrons are said to use a "hopping" transport mechanism, the same has been said of proton transport.  The making and breaking of hydrogen bonds, accompanied by the twisting and turning of protons completes the process.  This proton hopping mechanism between these larger water aggregates is thought to be a key step at room temperature.  However proton tunneling plays a role as well, particularly as the temperature is decreased.

Yes, it is actually the main reason for superior mobility of hydrogen ion in water solutions
see the link: http://en.wikipedia.org/wiki/Grotthuss_mechanism