Using the molecular orbital model write the electronic configuration for the following diatomic species?
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Bond order of CO is 3, bond order of CO+ is 2.5.  CO+ is paramagnetic.\n\nCO is (sigma)^2 (pi)^4, CO+ is (sigma)^2 (pi)^3
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Diatomic molecules contains two atoms (either same atoms or two different). Examples are, H2, O2, N2, Cl2, CO, NO etc.
5016
O\n                       ^\n                       |\n                      xx\n                  H-O-Br:->O\n                      xx\n                       |\n                       V\n                       O\n\nPlace the xx above and below Br. Br donates electron pairs to three oxygen atoms. There is also one Br-O-H bond in the molecule.
932025
You mean the Valence Shell Configuration?\n\nIt is easy to find this by simply looking at the Group # (these are the column numbers). \n\nElements in Group 1 all have 1 electron in their valence shell, Elements in group 2 have 2 and so on...\n'\nIf you mean how many shells there are then look at the Period #  this is the horizontal groups) the Period # corresponds with the number of electron shells.\n\nElements in Period 2 have 2 electron shells, elements in Period 3 have 3 and so on...
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1. There are two molecular orbitals for hydrogen, the lower energy orbital has its greater electron density between the two nuclei. This is the bonding molecular orbital - and is of lower energy than the two 1s atomic orbitals of hydrogen atoms making this orbital more stable than two seperated atomic hydrogen orbitals. The upper molecular orbital has a node in the electronic wave function and the electron density is low between the two positively charged nuclei. The energy of the upper orbital is greater than that of the 1s atomic orbital, and such an orbital is called an antibonding molecular orbital.\nNormally, the two electrons in hydrogen occupy the bonding molecular orbital, with anti-parallel spins. If molecular hydrogen is irradiated by ultra-violet (UV) light, the molecule may absorb the energy, and promote one electron into its antibonding orbital (s*), and the atoms will seperate. The energy levels in a hydrogen molecule can be represented in a diagram - showing how the two 1s atomic orbitals combine to form two molecular orbitals, one bonding (s) and one antibonding (s*). This is shown below - by clicking upon either the s or s* molecular orbital in the diagram - it will show graphically in a window to the right.\n\n2.Hybridisation\nThe Valence Bond model suggests that for a covalent bond to form, two atomic orbitals must overlap sufficiently so that electrons can be shared. However, if the hydrogen atoms are positioned tetrahedrally, not all of them will overlap with orbitals on the carbon atom. This is because the carbon's p orbitals are positioned octahedrally, whereas the hydrogen atoms are orientated tetrahedrally. How does the VSEPR model account for this problem? By inventing hybrid orbitals (N.B. these were proposed by Linus Pauling). In the carbon atom, the 2s and 2p orbitals have different energies. If the 2s orbital combines with the three 2p orbitals, four new sp3 hybrid orbitals will be formed. These new hybrid orbitals are identical - they are arranged tetrahedrally and have the same energy (they are degenerate). The 1s orbitals from the hydrogen atoms can now overlap with the sp3 hybrid orbitals, forming single covalent bonds.\n\nImportant: insert a picture of formation of hybrid orbitals (energy levels)\n\nGeometries of Hybrid Orbitals\n\nsp\n sp2\n sp3\n sp3d\n sp3d2\n \n[edit]\nLone Electron Pairs\nThe VSEPR model treats lone electron pairs in a similar way to bonding electrons. In ammonia (NH3) for example, there are three hydrogen atoms and one lone pair of electrons surrounding the central nitrogen atom. Because there are four groups, ammonia has a tetrahedral shape but unlike methane, the angle between the hydrogen atoms is slightly smaller, 107.3°. This can be explained by theorising that lone electron pairs take up more space physically than bonding pairs. This is a reasonable theory: in a bond, the electron pair is distributed over two atoms whereas a lone pair is only located on one. Because it is bigger, the lone pair forces the other electron pairs together.\n\nTesting this assumption with water provides further evidence. In water (H2O) there are two hydrogen atoms and two lone pairs, again making four groups in total. The electron pairs repel eachother into a tetrahedral shape. The angle between the hydrogen atoms is 104.5°, which is what we expect from our model. The two lone pairs both push the bonds closer together, giving a smaller angle than in ammonia. The scientific method has succeeded again!
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lewis dot structure shows the molecule's valence e-s and the bonding of the molecule \n\nit is very easy to draw \nhere are some hints for drawing lexis structure\nfirst you need to find out the valence e-s of that molecule i.e. h2O have 8 VALENCE e-s 2 from the hydrogens abd 6 from oxygen \nthere is is Carbon present in a molecule than carbon should be placed in the middle  if there is no carbon present than least electronegative atom should be in the middle \nput one or two or three pair(s) of electrons in the middle depending on the octet rule which states that all atoms want 8 valence e-s other than H, Be, and S. H bond to get 2 valence electron, Be bonds to get 6 valence e-s and S can get more than 8 valence electrons \nif you are confused or need help with any of the models just email me i will try my best to reply
142302
Resonance in the pi-electron orbitals.
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Write down what you know about the complex:\n\n1) it's tetrahedral (sq planar)  --> ∆t  (t2 and e levels)\n2) CN   --> low spin\n3) d electrons of Ni(2+)  --> [Ar],d8\n\nBefore drawing the CFSE, state that ∆ is large for a square planar complex.  Since ∆ is large, electrons will always pair up, rather than occupying the dx²-y² level.  Since there will be no unpaired electrons, the complex will always be diamagnetic.  I cant draw the diagram here, but show the z²,x²-y² and xz,yz,xy levels, fill the diagram with 8electrons and pair them.
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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
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Electron configuration is related to the number of protons in the nucleus, hence, the element. In theory, all isotopes of a certain element should have the same electron configuration.\n\nWhat u were asking for is an electron configuration for oxygen anion, it has the same electron configuration as Neon\n\nelectron configuartion of O^2- = [Ne] (Neon core) = 1s^2 2s^2 2p^6
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H\n |    \nC=O:  (You need 2 lone pairs on Oxygen)\n |   \nH\n\nIt doesnt get much better than that :)
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Huge order. Let's see how far it can be served.\n\n1. Monoatomic negative ions: take the element name (sometimes shortened) and add "-ide" (as in hydride, nitride, oxide, sulphide, phosphide, chloride...)\nPositive ions: element-name, charge (if several charges are possible), ion (sodium ion, iron II ion...)\n\n2. Polyatomic ions:\nComplexes with various ligands list the ligands and the central ion/atom.\nOxygen as ligands has its own nomenclature, depending on oxidation state. E.g. Sulphate, sulphite, perchlorate, chlorate, chlorite, hypochlorite...\n\n3. Binary molecular compounds: Halogen, oxygen and sulphur compounds usually have the correspondent "-ide" as if you had ionic compounds. Usually the more electropositive element is used as center.\n\n4. Acids: Oxygen acids have a similar nomenclature to oxygen ions.
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C                    1s2 2s2 2p2\nSi                   1s2 2s2 2p6 3s2 3p2\nGe                   (Ar) 4s2 3d10 4p2\nSn                   (Kr) 5s2 4d10 5p2\nPb                   (Xe) 6s2 4f14 5d10 6p2\n\nConsult a good chemistry book.
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H2 because it will be electronically more like Helium. It will have an electron with 1/2 spin up and 1/2 spin down. It will therefore have a happy orbital. You can also say that the bonding orbitals are occupied, and the antibonding orbitals aren't.
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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
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first of all write the molecular orbital configuration.then count the number of electrons with sigma bonds i.e., no. of bonding orbital electrons and then count the number of electrons with pi bonds i.e., no. of antibonding orbital electrons .Then\n\n bond order = 1/2[(No. of  bonding orbital  electrons) -(No.of antibonding orbital electrons)]
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The elements in the periodic table that have partially filled d orbitals. E.G. ytterbium