how many unpaired electrons are present in O2 according to the molecular orbital theory?
546769
unpaired electrons are 2 according to mollecular orbital (2) theory. that's why oxygen is paramagnetic. but from the VSEPR theory, this charactoristics of O2 can not be recognised. \n\nI can not draw the diagram of MO theory otherwise I would have shown you how there is 2 unpaired electrons. In short, in the pie 2P orbital, there are 2 unpaired electrons. So, the answer is 2.
291050
each sulfur adds 6 electrons, each carbon 4 - that's a total of 20 electrons.  Logically this puts a double bond between the carbons and double bonds between each carbon and one of the sulfurs.  S=C=C=S\nH2O2 has single bonds between the oxygens and single bonds between each oxygen with one of the H. H-O-O-H\nfor the thiocyanate ion this should be a 1- charge - this means the S is double bonded to the C which is double bonded to the N (S=C=N) there is a total of 16 electrons.  However - I really think the N should have a triple bond - this means that S will be singly bonded to the C and a triple obnd between the C and N.  Cannot really show this but go the the chemistry.mcmaster link below and check it out.
747161
Answer:  SIX electrons\n\nI drew a diagram for you: http://server6.theimagehosting.com/image.php?img=untitled.34c.JPG\n\nElectron Configuration of Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 \n\nExplanation:  As you can see in the electron configuration above, there are 6 electrons in 3d orbitals of Fe in its ground state.
791559
Let's see, this may be a bit difficult to explain, but since no one else has so far, I figured I'd give it a shot.\n\nI don't have a periodic table in front of me, so I'll do my best to explain.  There is a double bond between N and one of the O's because of something called "in-bonds."  You determine the number of "in-bonds," or bonds in between elements by subtracting the number of valence electrons that you have all together from the number that you need for everything to be stable.  In this case, forget about H for the time being, and you have N03.  Each element needs 8 electrons to fill its valence shell right?  Multiply 8 by 4, since there are 4 elements, and you get 32.  However, you only have 24 electrons to go 'round.  The difference is 8, there for you can have a total of 8 electrons, or 4 individual bonds in between the elements.  It doesn't matter which O gets to share the double bond with N, as they are all the same.\n\nThe double bond was used in HNO3 and not NF3 because I believe you had enough electrons to go around, therefore there were no "in-bonds."\n\nI hope this helps to clear up your questions.
839048
a.  Expanded shell/with 6 pairs of electrons around the central molecule.\n\nb.  Expanded octet with four shared pairs and one unshared pair.\n\nc.  Not sure but free radical with one double bond between an oxygen and and Cl.
222974
Two hydrogen atoms, one oxygen atom.  H2O
362329
Consider that one electron will repel another electron (-ve does not attract -ve), each bond between a carbon and hydrogen is cased by a pair of electrons (2 electrons) so you have a total of eight electrons... this results in neutralising the charge on all the atoms.\n\nCarbon has 2s1, 2s2, 2p2 electrons which means it is theoretically 4 electrons short of its closest nobel gas (nobel gases are neutral).\nHydrogen has 1s1 electrons which means it is theoretically 1 electron short. Note that there is no p orbital in first row of periodic table.\n\nAs such each hydrogen shares its sole electron with the carbon, donating a total of 4 electrons, hence 4 hydrogens.\nThe Hydrogens get their electrons from the carbon, the carbon has 2s2 and 2p2 electrons, totalling 4 to share, 1 to each hydrogen... we do not count the s1 orbital because it is unlikely to operate in forming bonding.\n\nNow each pair repels eachother so they are furthest apart, if we do the mathematics...\nblah blah...\nWe end up with an optimum angle of 107.5 degrees between eah hydrogen, any more than that and a different hydrogen will repel it back, and likewise if less. this forms a 'tetrahedral' shape.\n\nIf you were to switch a hydrogen with another element... say chlorine, to produce CH3Cl (Chloro-methane) then the hydrogens (being the smaller atoms with less electrons) will be pushed further from the chlorine which will tend to determine the molecule's appearance. Ch3Cl is not a perfect tetrahedral as its angles will not equal 107.5, but we still call it that for simplicity's sake.\n\nI recommend that for further information do a search on the following phrases:\n\nVSEPR\nValence Shell Electron Pair Repulsion theory\nMolecular geometry\nMethane\nOrganic Chemistry
867500
O-1 will have 9 electrons and 7 valence electrons (1s2 2s2 2p5 : the outer (valence) shell has 2 from 2s and 5 from 2p)
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
162059
Knowing the number of electrons present in the outer shell and  valence of each oxygen molecule.\n\nBesides there are techniques which actually helps us to visualize electrons like electron density map.\n\nHope this helps.
903420
In all cases the arragngement is tetrahedral and thus the ideal bond angle would be 109.5 degrees. In detail:\n\na) O has 2 unpaired e and 2 lone pairs. In H3O+, the 2 e participate in two single bonds with H and one lone pair participates in a dative bond with H+. Thus you have 3 single bonds + 1 lone pair so the arrangement is tetrahedral. lone pair-bond electron repulsion is stronger than bond electron-bond electron and thus the actual bond angles will be smaller.\nThe shape of the molecule will be trigonal pyramid (It is actually tetrahedral with O in the center and on one corner you have the lone pair but the lone pair is not "visible" in the molecular level)\n\nb) Si belongs to the same group as C. It forms 4 identical single  bonds in SiCl4 and has no lone pairs. Thus both the molecular shape and the electron arrangement are normal tetrahedral with bond angles 109.5\n\nc)N has 3 unpaired e and one lone pair. In NF3 it forms 3 single bonds through the unpaired e and so it has 3 bonds + 1 lone pair. Thus it is similar to H3O+ (tetrahedral electron arrangement, ideal bond angle 109.5, true bond angle smaller, molecular shape trigonal pyramid)
817827
energy levels/orbitals
443303
It only has 7 valence electrons :)\n\nRival
115249
Try this URL... it shows some diagrams.  Although it does not have the full set of configurations for d and f orbitals.
423134
CH4 does not have any non-bonding electron pairs.  It has 4 sp3 hybrid orbitals which are all involved in bonding.  SF2 has 4 electron domains according to it's Lewis structure.  This makes the hybridization sp3.  two of these hybrid orbitals are involved with bonding while the other two contain the non-bonding pairs of electrons.  To find the non-bonding pairs you must draw the correct Lewis structure.  a single atom in the molecule will only have a single hybridization.  S in SF2 can not be sp2 and sp3 hybrid.  it is only sp3.  Therefore the non-bonding electrons exist in the sp3 hybrid orbital.
41079
Oxygen atoms like to form two bonds with other atoms. They will be more stable that way. It is related to the outer shell of electrons(valence electrons). Individual oxygen atoms have 6 valence atoms and would prefer to have 8(it's called the octet rule!)
80472
I'm trying to work with this info:\n\nOctet Rule - atoms want to have eight electrons in their valence shell (two for H) (most will form octets, some will not)\n\nMultiple Bonds\nSingle bond - a covalent bond in which a single pair of electrons is shared by two atoms\ndouble bond - two pairs of electrons are shared by two atoms\ntriple bond  - three pairs of electrons are shared by two atoms\n-generally form with C, N, O, S\n-------\nOkay, oxygen might attach easily to one of the Cls, leaving no spaces, but I'm not sure if there is a doule bond or something weird like that. Same with C.\nIf I remember correctly, CO2 looks like this:\nO=C=O\n------\nWait, check this out- I found the following (and no further info) somewhere on the net.\n\nE) COCl2\n      (C is the \n       central atom)\n\n----------\nWait. No.\n_______\nCl- C -Cl\n___O___\nI'm going to try that...\nI don't know. It lacks the symmetry that you normally see and it's kinda weird looking. What do you think of what I got?\n__________\nI found a pic: http://www.chem.ox.ac.uk/inorganicchemistry3/C/COCl2.html\nLooks kinda like yours.\n______\nHoly Mother of God - http://www.chem.ox.ac.uk/inorganicchemistry3/C/COCl2.html\nI found the name of the compound- phosgene- and did an image search. *exclamation of glee*\n___\nLol, better be right. Later.