HOw many pi bonds does Sulfur Tetrafluoride have?
144785
All fluorine atoms are sp3 hybridized -- so none.
851509
There are two possibilities. Follow this link for a picture of the first possibility. http://img175.imageshack.us/img175/8223/ncssh2.jpg\n\nBasically, all 3 of these elements follow the octet rule (they want 8 electrons in their outermost shell), and as you can see, they all have 8. Now, S is carrying the - sign because it's formal charge is -1. Formal charge is calculated by Formal Charge=# of electrons the element has (S brings in 6) - the # of electrons it "owns" within the bond. Sulfur "owns" 7 in this case (3 lone pairs and one in a shared covalent bond--thus (3*2)+1=7.\n\nTherefore Formal Charge of Sulfur (in this case): is -1 because\n\n6-7=-1\n\nI probably made it sound more confusing than it is. Sorry about that.\n\nBut other than that, you know that Carbon should be the central atom here because it's coming in with only 4 electrons in it's valence shell, so it's more needy than Nitrogen with it's 5 and Sulfur with it's 6. So you stick it in the middle and then pretty much trial and error. \n\nThe other possibility is here.\nhttp://img175.imageshack.us/img175/6659/ncs2wf4.jpg\n\nThis time Nitrogen is carrying the -1 formal charge, because it comes into the bond with 5 electrons in it's valence shell, but it "owns" 6 in this case (2 lone pairs and 2 electrons out of the 4 in the N=C double bond..(2*2)+2=6. So...Formal charge of -1 comes from \n\n-1=5-6\n\nEven though these two molecules look different, they are in fact the same. This is a concept called resonance. \n\nI hope this wasn't too confusing.
279021
There's nothing surprising about the normal Group oxidation state of +4.\n\nAll of the elements in the group have the outer electronic structure ns2npx1npy1, where n varies from 2 (for carbon) to 6 (for lead). The oxidation state of +4 is where all these outer electrons are directly involved in the bonding.\n\nAs you get closer to the bottom of the Group, there is an increasing tendency for the s2 pair not to be used in the bonding. This is often known as the inert pair effect - and is dominant in lead chemistry.\n\nHowever, just giving it a name like "inert pair effect" explains nothing. You need to look at two different explanations depending on whether you are talking about the formation of ionic or covalent bonds.
789677
3 is incorrect.  Three of the carbon electrons are located in the three sp2 hybrid orbitals and the remaining electron sits in the lone p orbital (orthogonal/perpendicular) to the molecular plane.  It is this electron in the p orbital which "interacts" with the oxygen lone electron in it's unhybidized p orbital to form the pi bond.\n\nThere are no electrons left in the original carbon 2s orbital because these orbitals have been hybridized to the sp2 orbitals in order to allow for the three sigma bonds (one to each hydrogen and one to the oxygen atom)
98048
H:S:S with 2 dots on top and bottom of the S. S has 6 valence electrons and H has 1, a total of 8
806618
The only difference I can see would be the molecular geometry, due to the number of H's bonded onto the anion. Also, when drawing the Lewis dot structures, NH3 (N) would have one lone pair of electrons, HF (F) and H2O (O) would have two. This would dictate their structures according to the VSEPR rules.
823065
Well, whether the bond is a single, double, or triple bond also determines the hybridization of the sigma bond.\n\nFor example, if you have two pi bonds in addition to the sigma bond, the sigma bond is an sp hybridized bond because two p orbitals on each atom are consumed making the two extra pi bonds.\n\nIf you have one pi bond in addition to the sigma bond, the sigma bond is an sp2 hybridized bond because only one p orbital on each atom is consumed to make the single pi bond.\n\nIf no p orbitals are used to make pi bonds all p orbitals are available to make the sp3 sigma bonds.\n\nJust count the number of bonds.  If one the sigma bond is sp3.  If two the sigma bond is sp2.  If three bonds, the sigma bond is sp.
962262
1) Highest electrotronegativity.In Pauling scale flourine's electronegativity is 4. Consequently F atom has a strong tendency to attract electrons from the other elements.Fluorine thus compels the other elements to combine with it.Due to its strong electronegativity & small size Fluorine has a strong tendency to accept electrons from other atoms or ions. As a result it oxidises all other substances.\n\n\n2) The bond dissociation energy of F2 is very low, due to this Fluorine easily comes to reactive atomic state. The low bond dissociation energy can be explained in the following ways\n\na] since F atom do not contain any d orbital d(pi)-p(pi) bond in F2 molecule is not possible( which is possible in other halogens). This cause was proposed by Mulliken.\n\nb] According to Coulsan, The F-F bond distance in F2 molecule is  least because of the small size of F atom.The non-bonding electrons in 2p subshell of one F atom exerts great repulsion on that of another F atom. Which causes easy bond cleavage.\n\nc] the bond existing between two F atoms is a single covalent bond. So the bond dissociation energy is minimum.\n\n3) The bond energy of Fluorine in Fluoride compound is very high & so it requires a great amount of energy to break this bond, consequently the tendency to form Fluoride compounds is much higher.
876904
The way I used to do it was you draw the structure....and then however many atoms are connected to it determines it's hybridization...thats not the "scientific" way but it does give the correct answer. For example in SO2.....Sulfur is connected to two Oxygen atoms....so the sulfur hybridization would be sp or \ns(1)p(1)  with each number representing an atom. The "s" always has 1...and p can have up to 3. Therefore in SO3 it would be sp2....and in SO4...sp3. And bonds count as one...so if there is a double bond...it counts as one..not two. Same with triple bond. Wikipedia breaks it down really good with visuals....check out the link =)\n\n\n\n\nWe recognize three basic types of hybridization: sp3, sp2 and sp. These terms specifically refer to the hybridization of the atom and indicate the number of p orbitals used to form hybrids.\n\nIn sp3 hybridization all three p orbitals are mixed with the s orbital to generate four new hybrids (all will form σ type bonds or hold lone electron pairs).  \n\nIf two p orbitals are utilized in making hybrids with the s orbital, we get three new hybrid orbitals that will form σ type bonds (or hold lone electron pairs), and  the "unused" p may participate in π type bonding.  We call such an arrangement sp2 hybridization.  \n\nIf only one p orbital is mixed with the s orbital, in sp hybridization,  we produce two hybrids that will participate in σ type bonding (or hold a lone electron pair). In this case, the remaining two p orbitals may be a part of two perpendicular π systems.
496847
N2- has a bond order of 3.\n\nhttp://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
725881
1) sp hybridized carbons always means a triple bond. Of three available p orbitals only one is hypridized to make the sp. So two unhybridized p orbitals remain. The atom therefore, can form two pi bonds, one with each unhyridized orbital. \n2) A triple bond always has one sigma bond and two pi bonds. Sigma bonds are the result of interactions between orbitals that point directly toward eachother. The sigma bond comes from the sp orbital. Pi bonds come from interaction between orbitals that are parallel to eachother. \n3) A molecule freely rotates around a sigma bond with very little energy. However in a double bond if the molecule cannot rotate because the p orbitals used to create the bond will become perpendicular -- rather than parallel. This places one p orbital in the node of the other. The node is the location of the p orbital that doesn't have any electron density which would be contridictory. It is for the same reason impossible to rotated around a triple bond.
790804
That's an easy one... Ha Ha!\n\nA pi-acid is an acid that accepts electron density into it's pi-orbital.\n\nThe term, pi molecular orbital, is a symmetry designation.  A pi molecular orbital is antisymmetric with respect to a defining plane containing at least one atom (ie. the molecular plane of ethene).  A sigma molecular orbital is symmetric with respect to the same plane.  The terms are used both in this rigorous sense (for orbitals encompassing the entire molecule) and also for localized two-centre orbitals or bonds.  In the case of two-centre bonds, a pi bond has a nodal plane that includes the internuclear bond axis, whereas a sigma bond has no such nodal plane.\nInteraction between orbitals of different symmetries (with respect to the same defining plane) is forbidden!\n\n-- pi-acids or pi-bases usually apply to  Organometallic Chemistry and Molecular Orbital Theory.
580069
Tetrahedral arrangement at the methyl carbon. (4 bonding pairs, no lone pairs)\nTrigonal planar arrangement at the carbonyl carbon. (3 bonding pairs, no lone pairs)\nBent arrangement at the single-bonded oxygen. (2 bonding pairs, 2 lone pairs)
1101405
Ionic bond, hydrogen bonds, covalent bond, ionic interactions, van der waals interaction, non-polar interaction ... the list goes on.\n\nA bit more on covalent bonds:\nThere are sigma and pi covalent bonds, mainly. Pi bonds are usually the double bonds, while sigma are single bonds. Pi bonds can have resonance!\n.\nAlso there are often non-bonding electrons and anti-bonding electrons in molecules, but they aren't really "bonds"\n.\nCheers.\n.\nP.S.: If you share "chemistry" with a girl / boy you could also call that a "chemical bond"! He he he he.
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.
114913
You have to make double bonds. These are made by moving a nonbonding electron pair from an atom that already has an octet to a position where it can bond an atom that does not have an octet.\n\nSulphur dioxide has 18 electrons. There are electron pairs below the first O, above the S, and above and below the second O.\n\n:O::S:O:\n\nClicking on "Procedure 5 Example" at this site will give you the structure for sulphur trioxide.