Sigma Bonds Involving p orbitals  HCl?
541623
The three p orbitals of chlorine are degenerate, so it doesn't matter which of the three orbitals has the lone electron.  And no coordinate system is established for a lone atom.\n\nThe z direction is usually considered the bonding direction, so whichever orbital the hydrogen bonds to is usually considered to be the pz orbital.
497226
2NH4V03 + H2S04 ---> (NH4)2S04 + 2V02 + H20
834622
yes, they always do. if you want to have a better understanding about sigma and pi bonds, and hybrid orbitals visit this website:\n\nhttp://www.wellesley.edu/Chemistry/chem120/hybrids.html
825051
Covalent bonds, as in NH4+
1183446
the P has three bonds and all the Cl's have 4 bonds
161405
The elements Hydrogen (H+) and Chlorine (Cl-) to make HCl (aka hydrochloric acid)
956142
2Rb(s)+Cl2(g)--2RbCl(s)
910236
The H will attach to the Carbon atom which is already bonded to other hydrogen atoms, in this case, to carbon,2.  Bromine will bond to the other carbon, hence carbon,1.\n\nAns: d.
896467
OK, first of all, those little numbers at the top right aren't exponents, they're charges, so you don't need to use the ^ symbol.  It should be written "HCl(g) + H2O(l) -> Cl- + H3O+" or "HCl(g) + H2O(l) -> Cl-1 + H3O+1".\n\nThat aside, HCl and H3O+ are Lewis acids.  In fact they're very strong acids by any definition.
172774
your formula refers to several different structures which of course would give different NMR spectra. It could be 2-bromo-X-methylpyridine where X is the 3,4,5 or 6 position or Bromoaniline, with the various positional isomers. If you don't know the name go to the sigma website and find the structure/name (it may even have the NMR).
823065
Well, whether the bond is a single, double, or triple bond also determines the hybridization of the sigma bond.\n\nFor example, if you have two pi bonds in addition to the sigma bond, the sigma bond is an sp hybridized bond because two p orbitals on each atom are consumed making the two extra pi bonds.\n\nIf you have one pi bond in addition to the sigma bond, the sigma bond is an sp2 hybridized bond because only one p orbital on each atom is consumed to make the single pi bond.\n\nIf no p orbitals are used to make pi bonds all p orbitals are available to make the sp3 sigma bonds.\n\nJust count the number of bonds.  If one the sigma bond is sp3.  If two the sigma bond is sp2.  If three bonds, the sigma bond is sp.
233816
None.  The sell chemicals with alkyl groups.  In CH3Cl, CH3 is the alkyl.  They don't exist on their own.  Start with\n\nhttp://www.sigmaaldrich.com/Local/SA_Splash.html
977750
Mn(+3) has the electronic configuration of 1s(2) 2s(2) 2p(6) 3s(2) 3o(6) 3d(2)\n\nthe one before me missed the 4s2 so his is wrong
512329
Strictly speaking, the Lewis theory is not applicable to Xenon compounds (& if XeCl3 exists at all). The proper way goes by the Valence Bond theory. In that structure, Xe forms 3 single-sigma bonds with 3 Cl. That leaves out 5 valence electrons which is 2 lone pairs & 1 unpaired. This should (if possible) thus dimerise as Xe2Cl6, each segment being an octrahedron with 2 lone pairs. there should be 2 such octahedra (parallel planes) sharing 1 bond.
132325
No pi bonds. And four sigma bonds to the four fluorine atoms.
504495
Hydrogen peroxide give hydrolysis of complexes, since the hydrogens on water ligands are VERY acidic the ligands become hydroxides and the complex usually ends as a metal oxide. Ammonia acts as a lewis base with the complex accepting the electron density and most likely kicking off some chlorides. HCl just storms the complex with chloride ions. And The metal oxygen bonding is much stronger than the metal chloride bond (mainly due to differences in pi backbonding) so the chlorides shouldnt do much to the original complex. \n\n\nAs for the exact molecular formula of what should happen, try asking god. For the answer that your teacher wants try reading a book.
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.