Can anyone give me any information about the Tetrahedron 6-compound? Help is greatly appreciated!?
147899
Here is a graph and some explanation:\nhttp://mathworld.wolfram.com/Tetrahedron6-Compound.html
908581
C, Al(OH)3...\nWould you like an explanation? Okay, I'm supposed to be learning about this in Chemistry right now, so, I think this is how it works...\n\nIf you study the periodic table of elements, then you will find the number of valence (or outer) electrons each element has. Al has 3, O has 6, and H has 1. Before we start, it is important to know that a filled orbital has 8 valence electrons... okay? \nO and H combine, so they have a total of 7 valence electrons. You can imagine this by thinking of the letters OH surrounded by 7 dots in a circle, 2 pairs of dots on each side and 1 alone (a filled orbital with have four pairs opposite each other). The lone pair needs an electron, so it can borrow one from aluminum, which has 3 electrons. Aluminum loses its electron, so now it has 2, and the OH is now full, because it has 8 electrons. However, the aluminum has to get rid of its other two electrons, so another OH is brought in. The aluminum gives up an electron and the second OH is now full. Finally, the last OH pair is brought in and the aluminum gives away its last electron. So now, it is empty and the OH's are full. I guess they're balanced. Anyway, you are now left with 1 Al and 3 OH's, making Al(OH)3. Do you understand now? \n\nAl (3), and OH (7) (because O=6 & H=1, which equals 7)...\nAl - 1 = (2), so now OH = (8) (now it's full)\nAl - 1 = (1), so a 2nd OH (7) is brought in, making it OH (8)\nAl - 1 = (0), so a 3rd OH (7) is brought in, making it OH (8)\n\nIt took 1 Al and 3 OH's, so the new compound is Al(OH)3\nHope this helps!!!\nP.S. sorry for any confusion earlier when I tried to "draw" them. The computer changed the spacing so all you could see was a mess of stars and stuff. : )
837515
You need to look at the central atom and determine the number of lone pairs and bonding pairs of electrons (multiple bonds in this case will not count, as the orbitals used in those bonds will be orthogonal to the bonding axis and thus won't be hybridized).  2 is sp, 3 is sp2, 4 is sp3, 5 is sp3d, 6 is sp3d2.
762916
In the proposed structure for C60(fullerenes), each vertex of the truncated icosahedron is occupied by a carbon atom, and each carbon is three-connected to other carbon atoms by one double bond and two single bonds. Carbon atoms with this kind of connectivity are usually referred to as "sp carbons" because the orbitals used to sigma-bond the three adjacent carbons are hybrids of the 2s orbital and the two 2p orbitals (2p and 2p). The remaining 2p orbital (2p) is responsible for the pi-bond.\n\nUsually, an sp carbon and its three neighboring atoms are all coplanar. This is the case, for example, in graphite where there are infinite planar sheets of sp carbons arranged in edge-sharing hexagons. The p orbitals all lie parallel to each other and perpendicular to the graphite plane, generating a sea of - electron density above and below the plane.\n\nIn graphite, the C-C bonds are all equal in length and intermediate between normal single and double bonds. This "delocalization" can be explained by noting that graphite has many different but equivalent "resonance structures". (The resonance structures differ only in the placement of the double bonds.) \nthe proposed structure for C60 is not planar! The angle between a p axis and a C-C bond vector is 101.6° (as compared to 90° in planar graphite) . The bowl-shape or concavity at each sp carbon center introduces some strain into the molecule. However, the high symmetry distributes that strain evenly across the entire structure.\n\n# Other Fullerenes\n\nThe truncated icosahedron is not the only possible fullerene-type structure. There are, in fact, many other hollow cage structures that can be constructed using only pentagons and hexagons (Kroto, 1987). Interestingly, each of these structures contains exactly twelve pentagons, while the number of hexagons is arbitrary. The pentagons are necessary for closure. (Recall that graphite, which consists only of fused hexagons, is essentially a planar sheet. While this sheet may warp and bend, it can never close.) The number of vertices in closed fullerene-type structures is necessarily even.\n\nThe smallest possible fullerene would be C20, containing twelve pentagons and zero hexagons. However, such a structure would possess a great deal of strain, because the local topology at each carbon center would be highly non-planar. Other possible fullerenes include C28, C32, C50, and C70. Because the molecular strain tends to be concentrated in the five-membered rings that are responsible for closure, structures that avoid contiguous (edge-sharing) pentagons are particularly stable. It turns out that C and C are the smallest carbon clusters for which this can be achieved.\n\n# Other Possible Structures for C60\n\nWhile the truncated icosahedron structure for C60 is elegant and aesthetically-pleasing, it is important to bear in mind that when it was proposed, there was only one piece of supporting experimental evidence - an unusually large peak at 720 in the mass spectrum. Other C60 cluster structures, including the following three, were possible.\n\n   1. Planar graphite fragments.\n      The problem with this type of structure is that it contains many unsatisfied valences, and it is not clear why C60 should be any more stable than any other cluster size.\n   2. Cyclic polyalkynes.\n      In this structure, there are no unsatisfied valences but the stability of C60 vs. other cluster sizes is not easily rationalized.\n   3. Other fullerenes, i.e., other hollow cage structures with different atom arrangements.\n      These would necessarily be less highly symmetric than the truncated icosahedron and would less effectively distribute strain evenly around the molecule. \n\nGiven these possible, albeit less attractive, alternatives, it was crucial to obtain additional experimental evidence for the structure of C60.
214724
take a look at Pascal's Triangle:\n\n1...1\n1...2...1\n1...3...3...1\n1...4...6...4...1\n1...5..10..10...5...1\n1...6..15..20..15...6...1\n1...7..21..35..35..21...7...1\n1...8..28..56..70..56..28...8...1\n\nYou see how your tetrahedal numbers are in the fourth column?  Just follow it down to whatever you mean by "5th layer."  These numbers can be computed directly as nC3 on a TI-83.
760405
well family 13 has B and Al.  they are metaloids that are great lewis acids.  they have interesting bonding strategies that defy the normal octet rule.
872523
You need to specify a compound before we can tell you what its molecular geometry is!
40125
(this is a confusing problem because resonance structures are supposed to show how electrons configure themselves to satisfy its most stable structure--the question didn't really do a good job of representing this idea, most likely that the teacher rather have you do a challenging problem than having to learn how to do a problem)\n1) The formal charge on N=N=O appears to be N=-1, N=+1, and O=0. \n2) The formal charge on N(triple bond)N-0 is N=0, N=+1, and O=-1. \n3) The formal charge on N-N(trible bond)O is N=-2, N=+1, and O=+1\nFirst of all, Nitrogen likes to make 3 bonds with another element and Oxygen only likes to make 2 bonds with another element.\nIt also appears obvious that the higher the formal charge, the less stable the compound is. Therefore, the 3rd structure appears least frequently.\nAs a conclustion, it is plausible to say that the overall resonance structure of N-N bonding is 112 picometer since N(triple bond)N = 110 picometer and that most of the compound likes to remain in a N(triple bond)N structure, yet some also retain N=N or even N-N structure. The average averages to a number very close to 112pm.\nSimilarly, Oxygen only likes to form 2 bonds. Therefore, the N-O bond lenght comes down to 119 since it is the average of the more stable N=O with the less stable N-O.\n\nHope that helps,\nUltimeaciax
325629
It might help if you spelled "naphthalene" correctly. \nhttp://www.madsci.org/posts/archives/may98/894550073.Gb.r.html \nThis website has the structural pictures you can use to do your Lewis dot structure.  \nNaphthalene is an aromatic compound.  It's two aromatic rings sharing a side.
592711
What you describe is not a chemical compound.  tetrachloro- means 4 chlorines and -ol implies an alcohol.  Carbon atoms have tetrahedral bond angles.  Could you have mixed these terms up?
883229
PCl5, or phosphorus pentachloride, has a trigonal bipyramidal structure. A photo of the structure of this molecule is shown below:\n\nhttp://en.wikipedia.org/wiki/Phosphorus_pentachloride\n\nBrF5, or bromine pentafluoride, has a square pyramidal structure: a link to this molecule is shown below as well:\n\nhttp://www.chemistry.ccsu.edu/crundwell/Galleries/VSEPR/brf5.html\n\nBecause PCl5 has a trigonal bipyramidal structure, there is no net dipole moment since the molecule is symmetric around the central phosphorus atom.\n\nBrF5, on the other hand, does have a dipole moment due to the asymmetric structure. The absence of a sixth fluorine atom prevents the structure from becoming octahedral, which would be symmetrical in the structure of BrF6.\n\nSo in short, the differences between the two are as follows:\n\nPCl5=no dipole moment\nBrF5=dipole moment\n\nHope this helps!
115249
Try this URL... it shows some diagrams.  Although it does not have the full set of configurations for d and f orbitals.
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
848068
ICl2 + should be bent like water since I will have two lone pairs and two Cl's on it.\n\nI3- will be linear\n\nSbCl6- will be octahedral
729289
what are you supposed to prove?\n\nedit:\nyou mean why the three vertices of an equilateral triangle are concyclic? i am unable to make out theproblem
613727
try:\nhttp://lb.chemie.uni-hamburg.de/static/data/50_zzk46n8n.html
664669
Yeah, there are some special cases with some transition metals, and you're right with groups 6 and 11.  \nYou'd expect for the Cu (and the other elements in that column) to have the electron configuration [Ar]4s2 3d9.  \nInstead, the configuration is [Ar]4s1 3d10.  This configuration is more stable because the d sublevel is completely filled.  Cr's valence electron configuration (and the rest of the column's) is expected to be 4s2 3d4, but it is actually 4s1 3d5.  There is added stability in the exactly half-filled d sublevel.