What is the Lewis structure for nitrous oxide?
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Go to http://rutchem.rutgers.edu/Courses_f04/Chem161/Lecture19k.pdf#search='lewis%20structure%20for%20nitrous%20oxide' and scroll down until you see it
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
77405
Boron, Carbon, Nitrogen, Oxygen
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http://www.webelements.com/webelements/elements/text/O/key.html\nhttp://www.chemicalelements.com/elements/o.html\nhttp://en.wikipedia.org/wiki/Oxygen
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The Lewis dot structure would be that N in the middle with branches off to oxygen and 2 Fs and then one single electron will be on nitrogen and it is a radical I think. It is very reactive because of this one electron. The electron geometry will be something with sp3 orbital, I am not sure since it only has one electron. I would check up on that. Hope that helps
896818
G. Yes\n\nH. HNO3 (Nitric acid)\n\nI. Because they do not have carbon\n\nJ. Ba(NO 3)2 (I am not sure what is it called....barium nitroxide??!!)\n\nIf I get these right, I am a genius.....haven't touched chemistry for 10 years now maybe!! OR my teachers were really cool.................\n\n\nI am sure they are right!
373250
C3H7NO2 \n\nwww.chemfinder.com is a useful resource.  \n\nI hope this helps :)
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C4H9N3O2 For the structure goto http://www.cwu.edu/~chem/courses/chem388488f00/coulombe/CREATINE1.GIF
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There are several.  Most are non-metalic oxides.\n\nCarbon dioxide:\nCO2 + H2O --> H2CO3 (carbonic acid)\n\nSulfur trioxide\nSO3 + H2O  -->  H2SO4 (sulfuric acid)\n\nSulfur diodixe\nSO2 + H2O  -->  H2SO3 (sulfurous acid)\n\nNitrogen dioxide\n3NO2 + H2O --> 2HNO3 (nitric acid) + NO\n\nThere are many others.
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Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
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alright...\nconsider the R.H.S.\nh2o has two o-h bonds which account for 2x467 and since thre are 2 mol. u get 2x2x467--------(1)\nco2 has two c=o bonds which account for 2x799 and since u hav 4 mol. u get\n4x2x799-------------(2)\naddin 1 and 2 u get 1868+6392=8260\nconsider the L.H.S.\nu hav in c2h2, two c-h bonds which account for 2x413--------(3)\nand one triple bond accounting for 1x839---------(4)\nnow u hav 2x(3+4)=2x(826+839)=3330------(5)\nand u hav 5x(o=o) which acc for 5x(495)=2475----(6)\naddin 5 and 6 u get 5805\ntherefore ur ans is 5805-8260=-2455\nand since u want for one mol of acetylene u get -1227.5 kj/mole
843792
Be very careful:\nFirst, look at the electronegativity differences, the presence of groups like OH- or cations (like Na+ of soap)\nThen, you must know molecular geometries, it is highly important:\nFor BF3 this is a  trigonal planar compound, all differences of electronegativity B<F are canceled due to the position of the F atoms (every 120º on the same plane). It is not polar.\n[SnCl3]- this is a trigonal pyramid compound.\nCl\n I\n:Sn-Cl\n I\nCl\nAS you can see, Sn has a lone pair of electrons, making Cl atoms accommodate in the base corners of an imaginary pyramid. Even if Cl cancel each others dipolar momentum on the sides, they all point ↓. This is a polar compound.
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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
452173
the short answer is valence shell electrons.
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In the Lewis structure, dots left alone want to pair up with those of another atom.  If there are 5-6-7 dots all around, the atom will accept electrons to fill it last layer.  If there are 1-2-3 dots, the atom will want to give those electrons away.\n\nSo you look at your Lewis structures, and try to pair up alone dots from one giving atom to a receiving atom. \n\nFor your peroxyde, you would get H.   and :.O.: twice so one link between and H and a O, another link between the other O and the other H, and then the two O's link together.\n\nFor you chlorine, there are 7 dots all around each atom, so they will share their un-matched electron in a covalent bond.  \n\nFor the CH4O, their are 4 dots all around the carbon, so you would think that one H will bond to each (that's 4 H then) but to add the oxygen, you have to insert it between a H and the C (that the hydroxyde radical).  It is written CH3H usually to show the -OH polyatomic ion.\n\nFor the C2H4, the carbon has 4 dots all around it, and they will bond to each other.  Each will link to two hydrogen (to make up the four in the molecule), leaving two dots on each C, so this means that there will be one double bond between the two carbon atoms.\n\nFinally, the CH2O os trickyer.  That carbon, with its four dots, will have one H bond to each one dot, so there are two free electrons left.  That oxygen has six dots, two pairs and two free.  This means that each free electron of the O will link up to one of the free electrons on the C, which means that there is one double bond between the carbon and the oxygen.  You'll have H-C-H for the first part and C=O for the second part.\n\nIt's kinda hard to type these :-)  Hope this helps anyway
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thank you for this question (I love hybridization questions)\nFirst: ClO2 ??? Actually this is an ion → ClO2 -\nO= 6e-*2 = 12e-\nCl= 7e-*1= 7e-\n+1e-\ntotal= 20e-\nHowever, they all need an octet\n3*8e- = 24e-\n24e- - 20 e- = 4e-/2 = 2 simple bonds\nCl is the central atom, therefore, it must bond twice (one with each O)\ntherefore you require two semifilled orbitals\nNeutral Cl:\n[Ne]-(↑↓)-(↑↓)(↑↓)(↑_)\n-----3s2-///----3p5-----\nCl atom has 7 valence shell electrons by itself, so when it bonds it does one coordinated bond of the two it does, this means Cl looses one of its valence shell electron and gives it to one O atom: now it has 6 and not 7 valence electrons. So we have to check the orbitals again (with Cl atom without one 1 electron)\n[Ne]-(↑↓)-(↑↓)(↑_)(↑_)\n-----3s2-///----3p4------\nNow we have the 2 semifilled orbitals we needed!!\nNow hybridize\n[Ne]-(↑↓)(↑↓)(↑_)(↑_)\n----///--------sp3--------\nThe 2 filled orbitals are the 2 non-bonding pairs of electrons of Cl atom\nThen, your answer is sp3, with a an angular molecular geometry (just like water)
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CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle