How are FEMO and SHMO related to different linear conjugated carbon chain lengths?
113800
I hate to admit it, but only one of these theories existed when I was in school.  SHMO is in the realm of computer assisted quantum chemistry that has sprung up in the last few decades.  I'm too old of a dog for this level of stuff anymore.\n\nPlease accept my apologies!
884639
The bonding will indicate both of these. \na) CH3OH - sp3 O - H single bond\nb) CO32- sp2 O-H 1/3 double bond character through resonance\nc) CO2- sp C - sp2 O, double bond\nd) CO - sp C - sp O, triple bond\nThen longest to shortest is: a) b) c) d)\nWeakest to strongest is: a) b) c) d)
912679
PCl3 has the same shape as NH3 (tetrahedral with P in the center, the 3 Cl at the 3 corners and a lone pair at the fourth corner) So the polar covalent bonds don't cancel out each other\n\nSO3 is planar trigonal and thus the dipole moments cancel out each other and it is non-polar\n\nCF4 has a tetrahedral shape and since all bonds are the same they cancel each others dipole moment;the molecule is nonpolar\n\nCH2F2 has the approximately the same shape as CF4, but since not all bonds are the same their angle is a bit different and the sum of the dipole moments is not zero; thus the molecule is polar\n\nSO2 is a bent molecule (because of the sulphur's lone pair), think of it in a way as water with S in the place of O and O in the place of H so it is polar.\n\nCO2 is a linear molecule (O=C=O) so the dipole moments of the two identical polar bonds cancel out each other; the molecule is nonpolar\n\nO=C=S has different polar bonds, so although they have opposite directions the result will be a slight dipole moment, so the molecule is polar.\n\nIn all your questions you have to compare a polar and a nonpolar molecule. Of course the one with the strongest dipole-dipole interactions in each case is the polar molecule.
860814
In Carbon Carbon bonding:\n\nC-C is  sp3 hybridization, C=C is sp2 and C triple bond C is sp hybridisation. more s character is more strong covalent nature. Covalent character is increasing when moving multiple bonds in carbon. Covalent bond means sharing of pair of electrons (common pair) in bonding by both atoms(mutual) and hence both atoms come closer, that is lowering of bond length.
746884
if we use carbon as the example-there are four electron in the outer shell.  these would normally consist of one s shell electron and 3 p shell electrons.  instead of doing that, the 4 electrons become hybrids in that they all act the same and take on some s shell and some p shell character, so instead of having 1 and 3 of s and p type, you end up with 4 electrons of sp shell.\n\nif you kept the 1-s and 3-p configuration, the s would act one way and the p's would act another.  since they all become sp electrons, they act the same way, so in the case of something like methane (CH4), each of the hydrogens are equally spaced away from each other, kind of looking like a pyramid (they each are 109.5 degrees away from each other)
852150
My OChem teacher explains it this way, looking at carbon.\n\nCarbon has four valence electrons, 1 2s orbital and  3 2p orbitals. If you look at them separately, you see that the first two electrons go into 2s because it's lower energy, the next one goes into 2px, and the last into 2py. And we see that carbon forms two bonds. There's a technical term for this: WRONG! :)\n\nSo, imagine you have 4 guests in your house, and you have 1 tomato and 3 pieces of celery. How do you make sure that everyone gets *exactly* the same thing?\n\nYou have to blend them all together. You would put your ingredients, tomato (1 of them) and celery (3 of them), in a blender, pour into four glasses, serve chilled.\n\nSo with carbon, we take our s (1 of them) and our p (3 of them), blend them together and we have 4 s1p3 (ie. sp3) orbitals. 4 single bonds, everything even.\n\nWe could also take s (1 of them) and p (2 of them), and we'll have 3 s1p2 (sp2) orbitals. This is the case with ethylene -- the sp2 orbitals make 3 sigma bonds (2 for the single, 1 for the double), and the un-blended p orbital makes the pi bond.\n\nLikewise, sp is 1 s and 1 p orbital (s1p1).\n\nHope this helps. I've found this subject troublesome, too.
441962
At first glance you might make this assumption - but pyrene is an exception to Huckel's rule.  Huckel's rule of 4N + 2 is valid for mono cyclic ring structures (not in the case of the 4-fused benzene structure of pyrene).
822762
Both CH3F and CF4 are tetrahedrons, but due to the substitution each has a different symmetry (C4v and Td respectively), as a rule the lower the symmetry the higher the polarity of the molecule. Actually, any Td (fully substituted tetrahedron), Oh (fully substituted octahedron), D4h (fully substituted square planar), D3h (fully substituted trigonal bi-pyramidal) will be non-polar molecules. This is due to the cancellation of the dipole moments.\n\nThus in order of the strongest dipole moment the answer is\n\nCH3F > CF4 (C3v < Td)\nXeO3 > XeO4 (C2v < D4h)\nSO2 > CO2 (C2v < D(inf)H)\nCOCl2 > BF3 (C2v < D3h)\n\nIf you are having a hard time understanding the Schoedinger group types, try a 3D modeling program (eg Spartan Build or ChemDraw 3D) to see the differences in the symmetry of the molecules.\n\nI hope that this helps. Please feel free to contact me if you have any further questions.\n\nKen
887241
Bond length is directly related to bond order, when more electrons participate in bond formation the bond will get shorter. Bond length is also directly related to bond strength and the bond dissociation energy as a stronger bond is also a shorter bond. In a bond between two identical atoms half the bond distance is equal to the covalent radius. Bond lengths are measured in molecules by means of x-ray diffraction. A set of two atoms sharing a bond is unique going from one molecule to the next. For example the carbon to hydrogen bond in methane is different from that in methyl chloride. It is however possible to make generalizations when the general structure is the same.\n\n\n\nIn chemistry, bond energy (E) is a measure of bond strength in a chemical bond. For example the carbon-hydrogen bond energy in methane E(C–H) is the enthalpy change involved with breaking up one molecule of methane into a carbon atom and 4 hydrogen radicals divided by 4. Bond energy (E) should not be confused with bond dissociation energy.
785304
bond length:\nCO  CO3(2-) CO2\n\nBond Strength:\nCO2   CO3(2-)   CO
598319
http://www.sbuniv.edu/~ggray/CHE3345/chp14.html\nhttp://web.odu.edu/sci/tisenhou/chem321/chap12/index.htm\n\nbest luck
17774
It doesn't give me room to list them all and explain them on here so let me direct you to a website I found that does a pretty good job:\nhttp://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
950288
1)  Only the valence orbitals matter, and all the valence orbitals count (more complicated version: all valence orbitals which are of the correct symmetry to hybridize to form MO.  If you don't know what that means then you're in a lower level class and it doesn't matter).  # of atomic orbitals = # of molecular orbitals, and the molecular orbitals may be bonding, nonbonding, or antibonding.  Not depending on the particular class you're taking your teacher may or may not want you to draw the non-valence orbitals, but they don't matter at all.\n\n2)  You start filling all the MOs with electrons, starting with the lowest energy and moving up.\n\n3)  You wouldn't determine that for C2 because that's not correct.  The bonding and antibonding orbitals formed from the 2s atomic orbitals would both be filled (net bond order of zero) the lowest bonding (sigma) MO from the 2p AO would be filled, and each of the higher bonding (pi) MO from the 2p AO would have one electron.  None of the antibonding orbitals from the 2p AO would be filled.\n\nThis predicts that C2 would have 2 unpaired electrons, whereas a Lewis structure would predict a quadruple bond between the carbons.  One guess as to which one is right....
284075
It's fairly easy don't worry.\n\nNo need for d level splitting diagrams to answer this since we're dealing with N and F:  use the spectrochemical series to make an assumption about the ligand.  The difference in t2g and eg levels (∆o) determines whether a complex is low or high spin.  If both ligands were the same, we would have to look at the oxidation state of the ligand in the complex.\n\nHalides <  Oxygen ligands  <  Nitrogen ligands < CN- ligands\n\n   ALWAYS HIGH SPIN    ---------------            ALWAYS LOW SPIN\n\nIf ∆o > P, then a complex is low spin.  IF ∆o < P the complex is high spin.
6203
In the formalism that separates bonds into  and  types, hyperconjugation is the interaction of -bonds (e.g. C-H, C-C, etc.) with a  network. This interaction is customarily illustrated by contributing structures, e.g. for toluene (below), sometimes said to be an example of "heterovalent" or "sacrificial hyperconjugation", so named because the contributing structure contains one two-electron bond less than the normal Lewis formula for toluene \n\n\n\nAt present, there is no evidence for sacrificial hyperconjugation in neutral hydrocarbons. \n\nThe concept of hyperconjugation is also applied to carbenium ions and radicals, where the interaction is now between -bonds and an unfilled or partially filled  or p-orbital. A contributing structure illustrating this for the tert-butyl cation is: \n\n\n\nThis latter example is sometimes called an example of "isovalent hyperconjugation" (the contributing structure containing the same number of two-electron bonds as the normal Lewis formula). \n\nBoth structures shown on the right hand side are also examples of "double bond- no-bond resonance". \n\nc yhis page, please.
859386
Due to differences in chemical structures
479762
I assume you are talking about the formula and not the structure.\n\nThe molecular formula is C6H5NO5P\n\nStructure (if in case you want it too) is as follows:\n\nDraw a benzene ring (phenyl group) minus 1 hydrogen (C6H5). To the Carbon without the hydrogen, attach a Hydroxyl group (-OH). On the other Carbon positions, whether it is the ortho positions, meta positions, or para position, erase the hydrogen and attach a nitrate group (-NO2). On the Oxygen with the double bond to Nitrogen on the Nitrate group, erase one of the bonds and attach this bond to the phosphate group (-PO3).\n\nI'm not sure if you are talking about a nitro-phospho bond or a hydroxy-phosphate bond, but the difference is only 1 hydrogen.