what is the correct Lewis structure for CBr4?
913452
start from here---\n\nhttp://en.wikipedia.org/wiki/Lewis_dot_diagrams\nhpe this helps
931837
20) B.  CH4\n24) C.  nonpolar with difference between 0 and 1.7\n25) B. pentane (more carbons to branch)\n27) C. H and Br\n28) C. 5 pairs - 3 between carbons and one on each end between carbon and hydrogen\n29) B. nylon
1163366
http://filext.com/detaillist.php?extdetail=CBR\n\nThe CBR file is Comic book reader file, as the other guy mentioned. It is in actual fact just a renamed file according to the link above. Download winrar and open it from winrar, or just change the extention to .RAR and double click.\n\nGet Winrar\nhttp://www.rarlab.com/\n\nEDIT: incase you missed it, the link for comic book reader is\nhttp://www.geocities.com/davidayton/CDisplay
839410
This compound is an amine.  You can identify it by the NH2 (amine group) at the end.  You did a fair job with the structure considering the limits of the text here.  I am sure you realize that two of your carbons do not have the correct number of bonds.
973000
C6H10 + Br2 ---> C6H10Br2.\n\nThe two Br atoms join at either end of the double bond.
390255
i presume you mean cyclohexanone (since cyclohexanol is not a carbonyl compound, it is an alcohol).\n\nBF3 is a Lewis acid, and the B has an empty p-orbital.  The lone-pair of electrons from the carbonyl will form a dative bond with the B by donating both the electrons to this bond.  This increasese the dipole on the C=O double bond making the C more available to nucleophilic attack (remember that the C in carbonyl compounds is slightly electropositive and the O slightly electronegative).  \n\nThe mechanism is the same on any carbonyl compound, so nothing specific for cyclohexanone.\n\nHope this helps
197637
Actually, there are a few things wrong with that structure. If the Red atoms are indeed oxygen then the formal charge would be (C4H8O2)2-, the double minus change would make this very unstable. \n\nOxygen is only found by when it is double bonded, which it can't do here since carbon can only have four bonds. The red could be a halide, that is something from the F, Cl, Br and I column. \n\nIf the red guys were bromine then it would be \n2,3-dibromobutane
540818
NH4Cl (note the 4 is a subscript)
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
172042
C4H9N3O2 For the structure goto http://www.cwu.edu/~chem/courses/chem388488f00/coulombe/CREATINE1.GIF
87059
You should consult a chemistry textbook, if possible, or if not, look online for an answer; I'm sure there is information or a guide describing how to do it better than I can in a few short sentences. That said, the basic way to draw a molecule (often abbreviated mlcl) is to use each element's symbol(ie sodium=Na, carbon=C, etc) to denote an atom, and to draw a single, solid line for each bond between elements. This line/bond represents 2 electrons, aka one orbital. Most elements like to have 8 electrons in their valence (outermost) shells in order to be stable, but this doesn't always happen; there are many exceptions, hence my suggestion to look at a textbook or online text/guide.\n\nIt gets more complicated when you're dealing with things like organic chemistry or complexes or chelating agents where sharing of bonds/electrons is pretty common and there are more types of bonds. For a simple example, if you had a 6-carbon atom ring, each carbon has 6 electrons in its valence; it wants 8, so what happens, does it get frustrated with its misfortune and self destruct, destroy our little universe? Of course not... there are a total of (6)(6)=36 electrons, and we need (6)(8)=48. How is this possible? Clearly, 36 is not equal to 48. We can't create more. So what happens? As it turns out, carbon, along with most other elements, passed kindergarten with flying colors- they SHARE them, and since it's a symmetrical hexagon shape of identical atoms, they are shared equally. There are all kinds of crazy combos in which the electrons are shared unequally between atoms because of something called electronegativity, but our rules for drawing those lopsided things aren't very different. However, that's for another 'lesson;' don't worry about it for now.\n\nSo, in order to make the carbon happy, we have to share 12, or 6 pairs, of electrons. ([48 needed]-[36 that we have] = 12) \nSince this carbon ring has multiple resonance structures (atom config the same but electron pairs on different atoms), we draw six carbons in a hexagon, with one solid line bond between each, and a dotted circle in the middle- this accounts for all of its resonance structures and provides us with more info, in addition to only needing one drawing to display that info, than using single bond lines only.\n\nSo in conclusion, I've said way more than I really had to... from the level of generality of your question I assume that you don't know too much about chemistry; if you are confused with what I said, then do as I said- look up the info online or get a book. It's not too easy to understand right away but with some patience you should be able to figure it out just fine; good luck.
183009
Get the CAS number.  Look it up on Chemfinder.
446372
You proffered the example of XeF4.  Now the Lewis structure for XeF4 gives an expanded octet, viz: AX4E2, where A represents the central atom (Xe) in this example; X represents a shared pair of electrons; and E represents an unshared pair.  Thus for XeF4 we have 4 shared pairs and 2 unshared pairs, i.e. 6 pairs.  \n\nNow whatever the principal energy level(n), there is 1 s orbital; and for n=2 and higher, there are 3 p orbitals; and for n=3 and higher there are 5 d orbitals. For n=4 and higher there are 7 f orbitals.\n\nNow considering XeF4 hybridization, for the 6 pairs of electrons there will be 6 orbitals [each orbital holds a maximum of two electrons with opposite spin]. That means sp3d2 hybridization.  That is the one s orbital, the three p orbitals and two of the five d orbitals form six sp3d2 orbitals.  Of course, we are talking here about valence shell electrons.  How would these orbitals fill?  Well, electrons are added to orbitals such that electrons with the same spin are added until there are no more empty orbitals, then the orbitals with one electron fill (i.e. add a second electron) by acquiring an electron of opposite spin.\n\nThis gives XeF4 sp3d2 hybrid orbitals and an octahedron orientation. \n\nConsidering the Lewis structure for water, as another example, we obtain a AX2E2 form. Thus sp3 hydrid orbitals and a tetrahedron orientation. The 4 hybrid orbitals come from 1 s and 3 p atomic orbitals.
749482
1-C Hydroboration puts -OH on least substituted C; anti-Markownikov\n2-B Correct\n3-A Correct\n4-C Br goes on the center C; product has two identical CH3CH2- groups on center C, therefore symmetrical and achiral\n5-A Correct\n6-A Correct\n7-B Correct\n8-A Correct\n9-C Markownikov addition puts H on C next to C-CH3 and OH onto HO-C-CH3
352077
Thanks NeuroProf.\n\nThe molecule AsF5 is trigonal bipyramidal in shape. First lay out all the bond pairs in that shape and then see if the 'pulling effect' of all the bonds cancel each other. Here, there are 3 bond pairs (of Cl) at the corners of the triangle, 1 bond pair at the bottom and 1 at the top. The Cl at the top pulls electrons and the pulling effect is cancelled by the one at the bottom. Considering the 3 bond pairs left in the middle, 2 are angled at 120 degress and one is at 90 degrees to the top and bottom bond pair. So the pulling effect of the 2 pairs is cancelled by the one at 90 degrees. Overall dipole cancels out and there is no charge distribution, so the molecule is non-polar.\n\nIn BrF4- (with square planar shape), the pulling effect of the lone pair at the top is cancelled by that of the lone pair at the bottom. This leaves 4 F atoms arranged at the corners of the square in the middle. The pulling effect(dipole) of 2 diagonal F atoms will cancel because they are in the same plane and are the same atoms. Hence all the 4 dipoles cancel out and that of the lone pairs too, so the molecule is non-polar.\n\nIn GaI3, the 3 I atoms are arranged in a trigonal planar manner. The dipole of the 2 I atoms at the sides will cancel, but the I atom at the top is not paired with any other atom at the bottom, so the dipole cannot cancel out. Since overall dipole moment is not equal to zero, the molecule is polar.\n\nAre you sure the molecule is SbF6- and not SeF6? If it's SbF6 then the shape will be square bipyramidal or octahedral (similar to BrF4-). The dipole of the 2 F atoms at the top and bottom will cancel and that of 2 diagonal F atoms will cancel and overall dipole will be zero, so the molecule will be non-polar.
626516
BeCl2 disobeys. It has 4 valence electrons. XeOF4 disobeys. It has 10 valence electrons.
613727
try:\nhttp://lb.chemie.uni-hamburg.de/static/data/50_zzk46n8n.html