when hybriding orbitals what is the order they fill up at?
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.
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
817827
energy levels/orbitals
178046
the s is the orbital in which 2 electrons can occupy\nthe orbitals are the s which can hold 2 electrons, the p which can hold 6 electrons, the d which can hold up to 10 electrons and so on \nhydrogen has an electron configuration of 1s1\nhelium has its elec. conf. of 1s2\nLithium has its elec. conf. of 1s2 2s1\nkeep adding \nNeon has its elec. conf of 1s2 2s2 2p6 \n\nTHOSe two 2s u add the elctrons which are 2 & 6 and u get get which makes the octect of NEON (noble conf.)
115249
Try this URL... it shows some diagrams.  Although it does not have the full set of configurations for d and f orbitals.
901447
Note that we're not taking energy differences into account.  In cases where the difference in orbital energy is greater than 15eV you get negligible overlap.  Now onto the symmetry-allowed overlaps.\n\n1s and 2s are both spherical and overlap to form a sigma bond.\n\n1s and 2p(x) form a sigma bond.  You can tell because if you draw the 2p(x) lobe you'll see it overlap with the corresponding 1s.\n\nThe 2p(x) and the 2p(y) do not overlap.  There is no possible bond.  Draw the 2p(x) extending toward the other atom as positive or shaded (remember: positive does not denote charge).  The other atom has its 2p(y) pointing up and down (perpendicular) and there is equal positive and negative overlap, thus canceling both of them out.  No net overlap.\n\n3p(y)-3p(y) form a pi bond.  Once again, draw these out and see it for yourself.  Draw the 3p(y) orbitals as rather fat so you can see how they overlap.  Since they are not cylindrically symmetric they are not a sigma bond.  Pretend you were to rotate one of the atoms about the x-axis... you would change orientation and there'd be less overlap.  For the sigma bonds above, rotation does not affect the overlap.\n\n2p(x)-2p(x) have positive overlap.  Draw them and see why.  They point toward one another.  Moreover, if you rotate one of the 2p(x) orbitals about the x-axis you keep the same structure/orientation since the 2p(x) is along the x-axis.  Thus, you can tell that this gives a sigma bond.
541274
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
955362
Violet12.. is right on both questions. Now, about the second question:\n\nFor the 4s orbital: n + l = 4 + 0 = 4.\nFor the 3d orbital: n + l = 3 + 2 = 5\n\n4 < 5, so the 4s orbital has less energy than 3d, so we put electrons first in 4s orbital and then in 3d.
748503
s p d f blocks tell which set of orbitals are the last to fill for those elements.\n\n+ or - depending on whether it will lose or gain an e-\n\ncations are smaller than atoms b/c they lose e-\nanions are larger than atoms b/c they gain e-\n\neach successive ionization energy gets higher becasue the #protons is more and more unbalanced by # e-\n\nhalogens have 7 valence e- and noble gases have 8.  by gaining one e- the halogens now have the e- configuration of a noble gas and are more stable.  gain 1 e- = 1- charge
231884
There needs to be something you are binding oxygen to.  Oxygen itself will not have any hybridized orbitals. With H2O for example there would be four sp3 hybrid orbitals. two of the orbital will have hydrogen bound to it and the other two will be taken up by the 2 lone pairs.  Once the orbitals are hybridized to sp3 there is no place left to put any electrons in a p orbital since they were all hybridized to sp3.
961493
You need to know more than just the electron configuration.  There is contraction from left to right on the periodic table, plus additional periodic trends that must be considered.   \n\nEven more importantly, you need to define whether you are looking for the largest element IN ITS NATURAL  STATE (i.e. Cl2, Hg metal, etc) or AS THE ION.   Atomic and ionic radii differ.\n\nMore clarity is needed here.
525685
You can approach this is one of two ways:\n\n1)  By analogy with H2O (because O and S are both in the same group)\n\n2) From first principles ie octet rule, Lewis diagram, hybridisation, shape. \n\nNow S has 6 electrons in outer shell, while H has 1 each, therefore, S will share 1 e- from each H to complete its octet and each H will share an e- with S so H (which is too small to have 8e-s in outer shell) will have a full outer shell. So S is now covalently bonded to to H atoms, accounting for 4 e-s, with 2 non-bonding pairs. That makes 4 pairs of electrons around S all together, so hybridisation is sp3. Shape is tetrahedral minus two vertices ie bent (just like H2O).\n\nI hope this helps.
802906
they fill the s orbitals with 2 electrons
964369
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.
354063
yes, it's the 1s orbital. The electron configuration of hydrogen is 1s1.  The electron configuration of helium is 1s2.
917414
You apparently know what the diagonal rule looks like, but the link below gives a quite nice picture of it.\n\nBasically it is a summary of the order in which electron shells will be filled up. Follow the arrows round to discover where the electrons will go, starting at the 1s level.\n\nSo for 15:\n2 electrons will go into 1s, filling that up\n1s2\nthe next electrons will go into 2s2, 2p6, 3s2 and then the remaining 3 electrons into the 3p orbital, giving us a structure of:\n1s2 2s2 2p6 3s2 3p3\nRemeber, on your diagram, the first two characters (2p, 4d etc) are the names of the orbital, and the final digit is the number of electrons in that orbital. On your diagram the final digit shows the maximum number of electrons on the orbital; when you write out the structure the first few orbitals will be full (1s2 2s2 2p6 3s2) and the last one will be part full of however many electrons you have left (3p3.)
784377
This is true.  The Px, Py, and Pz orbitals are degenerate in all atoms provided the atom is not subjected to a magnetic field. \n\nIn hydrogen and other single electron systems (He+ etc) only the n quantum number matters to determine energy.  So you can actually say that the 2s, 2px, 2py and 2pz are all degenerate.  \n\nThe s, p etc orbitals lose this degeneracy when there is more than one electron due to electrostatic repulsion between the electrons, so then you need both the n quantum number and the l quantum number to determine energy.