how many orbitals of different orientation does 3p have?
153826
There are 3 orbitals: px, py, and pz. Each of the orbitals can take 2 electrons.\n\nIf you take an X-Y-z plot and draw 3 figure eights with each of three dimensions, you'll get the shape of the p orbital.\n\nThe p orbital is also symmetrical (gerade), so electrons in complex don't easily go from a p orbital to a d orbital (which is ungerade) according to LaPorte rules.
1090457
a the number of outer electron that will bond with other atoms to fill thei valence shells
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.
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.
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.
705262
They are all row three transitions metals.  So the point about 5 valence electrons is correct depending on the oxidation state of the atom.
786820
Flourines electron configuration is [He]2s2 2p5.  In the p-orbitals there are three subshells, each subshell is capable of taking only two electrons a piece.  This leaves you with one electron that is unpaired.  Because in the p-orbitals you can only have a total of 6 electrons.  So, 6-5=1.
115249
Try this URL... it shows some diagrams.  Although it does not have the full set of configurations for d and f orbitals.
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...
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
1234
n=3 means that there are s, p, and d shells involved.  The s shell can accomidate 2 electrons, the p, 6 and the d, 10, so a total of 18 electrons can be accommodated, so that's 9 pairs of electrons.
9185
Hey man..ask both the questions together.Anyways now i will explain you both. Here t means triplet. It means three similar energy orbitals. Others are\nA=Singlet\nE=Doublet\nT=triplet\n\nAlso 2 here dentoes symmetry about the center. i.e 2 are centrosymmetric while one are no. e.g p orbitals being centro symmetric fall in 2 case and sp3 hybrid orbitals fall in 1 case!!
799803
a) 36\nb) i just found the answer for b : it is 6 because its a p orbital
774231
an octet configuration with eight valence electrons each (on each atom) ?
346187
The electrons form a standing wave surrounding both nuclei in a bond. The exact shape of this three-dimensional standing wave depends on the number of electrons in either element and the main quantum number.\n\nMost common are "hybrids" of S and P orbitals, for heavier elements also S and D and P orbitals. Those hybridized orbitals usually seek maximum distance from one another, leading to a rather small set of bond symmetries.
15541
Yes.  NH3 obeys the octet rule because nitrogen shares two electrons in each of three covalent bonds to hydrogen, and has a free electron pair. 2*3+2=8.
850997
sp : 2 orbitals (1 from s & other from p; rearranging to give 2 orbitals of equivalent energey & value)\n\nsp2: 3 orbitals\n\nsp3: 4 orbitals\n\nsp3d: 5 orbitals\n\nsp3d2:  6 orbitals\n\n\nany further clarifications \n\nhttp://en.wikipedia.org/wiki/Orbital_hybridisation