where are the lobes of the orbitals (for e.g.of s,p,d,f) of electrons with respect to the nucleus?
115249
Try this URL... it shows some diagrams.  Although it does not have the full set of configurations for d and f orbitals.
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proton (positive) in the nucleus\nneutron (no charge) in the nucleus\nelectron (negative) orbits nucleus at a great distance
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13. And I believe you mean 13 electrons. 13 protons, 22 neutrons, and 13 electrons. The atomic number is the number of protons in the nucleus.
332147
Those ions do follow Hund's rule, though. The d orbital looks like this:\n\n\n_________ ___________\n\n__**_______ ____*____ ___*____\n\nThe electrons are only paired when absolutely necessary.
841087
Se has the electron arrangement 4p4. One of the 4p orbitals will be full holding 2 electrons.\nTwo of the 4p orbitals will have one electron each in - ie half full
1010723
First the electron would have an sp1 configuration not np1.  As to which orbital it would revolve in...it doesn't matter.  The x, y, and z are just ways to distinquish the 3 different orbitals.  the point in the center is a null point.  the electrons do not actually travel through that spot.  Orbital configurations just define a volume where an electron might spend some of its time.  it could be found anywhere in the volume.
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chemically: electrons\n\nphysically: nucleons and/or protons (alpha, beta or gamma ray for nuclear elements)
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Orbital does not mean where an electron is.  It means it is the highest probability of where the electron exist around the nucleus.  Heisenberg said it is impossible for us to know where the electron is (since it is always moving) thus we have calculated wave functions....regions of high probably of where the electron PROBABLY is.
817827
energy levels/orbitals
44394
1. The Aufbau principle: Orbitals are occupied in order of increasing energy. 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d\n\n2. The Pauli exclusion principle: Only two electrons can ocoupy an orbital and they must have opposite spin.\n\n3. Hund's rule: When there are degenerate orbitals, one electron is added to each with their spins parallel until all are half full. Then a second electron with opposite spin is added to each degenerate orbital. (See below)\n\n(Basically every space/box/circle must have an arrow pointing up before any space/box circle can have one pointing down.)\n(Everyone gets one before anyone gets two.)
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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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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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Ground State - 1s2 2s2 2p6 3s1 : 2,8,1\n\nElectron moves from the 2p orbital into the 3s orbital.\nExcited State - 1s2 2s2 2p5 3s2 : 2,7,2
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the short answer is valence shell electrons.
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The electrons in the first orbit has the least potential energy. The further from the nucleus the orbit is, the more potential energy it's electrons possess.
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An electron in an s orbital has zero angular momentum.\n\nViewed classically it is in Simple Harmonic Motion (SHM) passing through the nucleus of the atom.\n\nThe restoring force in the SHM is the electrical attraction between the electron and the positively charged nucleus.  The attraction is  an inverse square law which enables you to derive the equation of motion of the electron.\n\n  (Some "proton rich" isotopes undergo a type of radioactivity where this electron is captured by a proton in the nucleus which changes to a neutron which makes the atomic Number of the atom decrease by 1)
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1) one s + three p + one d give five equivalent sp3d orbitals\n2) O=C=O, two sigma, 2 pi bonds\n3)a) 10e-; α2s2 α*2s2 π2p4 α2p2; diamagnetic\nb) 12e-; α2s2 α*2s2 π2p4 α2p2 π*2p2; paramagnetic (isoelectronic as o2)\nc) 14e-; α2s2 α*2s2 π2p4 α2p2 π*2p4; diamagnetic\nd) 10e-; same as a) above\n4) No molecules shown!\n5) The structure of graphite is planer, all sp2 carbons with an electron in the p-orbital. There are planes of carbon atoms well bonded together. Between these planes there are pi electrons and they are free to move; they are said to be aromatic pi electrons. When you apply a voltage, these electons do move - and moving electons is current - ie it conducts electricity. On the basis of MO theory:  Because the MO theory holds that electrons are not held to only one position. Instead they are spread across the entire molecule. \n6) No, it is trigonal planar, any individual dipoles from the B-F bonds are negated by the symmetry of the molecule.