what is the shape of an NF3 molecule?
440980
i think its a trigonal pyramid due to the extra lone pair on the nitrogen.\n\nrefer to this site \nhttp://www.chm.bris.ac.uk/motm/ni3/ni3c.htm
1047499
If it existed, it would be boron pentafluoride, but as far as I know, this is not a stable molecule.\n\nAre you sure you don't mean BrF5 (bromine pentafluoride), which does exist as a stable molecule?
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)
975490
NaNO3 is very soluble\nBaF2 has Ksp=1.84*10^-7\n\nBaF2 <=> Ba(+2) + 2F(-)\n\nso Ksp=[Ba(+2)][F(-)]^2 This is for the concentrations at equilibrium.\nIf the product P=[Ba(+2)][F(-)]^2 for the initial concentrations (and not the concentrations at equilibrium) is P>Ksp then you have a precipitate, if P=Ksp then you have a saturated solution and if P<Ksp you have a non saturated solution.\n\n[Ba+2]=mole/V= 5*10^-4/1=  5*10^-4\n\nNaF -> Na(+) + F(-)\nTherefore [F(-)]=[NaF]=6*10^-3\n\nP=[Ba(+2)][F(-)]^2 = (5*10^-4)*(6*10^-3)^2 =1.80*10^-8 <Ksp so no precipitate
788892
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
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.
364902
Most crystals form three basic shapes: simple cubic, body-centered cubic, and face-centered cubic.  Simple cubic has molecules at each corner of a cube, body-centered cubic has molecules at each corner of a cube and one in the center, and face-centered cubic has molecules at each corner of a cube and one in the center of each face.
750269
The shape of the molecule depends on how many electrons and bonds you have in the outer shell of the central atom.\nFor NH3, N is the central atom. It has 3 covalent bonds and a lone pair. Thus the repulsion of the lone pair electrons and the bond electrons will force them to arrange at the corners of a tetrahedral. The molecule will look as a trigonal pyramid (since the lone pair is not "visible" in the structure.\n\nFor BH3, B is the central atom and has no lone pair apart from the three bonds. So the repulsion of the bond electrons will force them to be in a planar trigonal arrangement.
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.
809957
The most polar bond will be between the atoms with the greatest difference in electronegativity.\n\nF2 has bond that is not polar since the electrons are equally shared between the two identical atoms. The electronegativity series is As<S<Br<F so the biggest difference is between As and F and the most polar bond is in AsF3.\n\nAs belongs to the same family as N. Thus in AsF3 it forms 3 single bonds and also has 1 lone pair. So the electron-group arrangement is tetrahedral but since there are only 3 identical bonds there will be a net dipole moment.\n\nI already explained why F2 has no dipole moment\n\nBrF3 has trigonal bipyramid electron group arrangment (3 bonds+2 lone pairs) but not corners are occupied by identical atoms (2 have lone pairs of Br) so the dipole moments do not cancel out and the molecule has a net dipole moment\n\nSF4 has again trigonal bipyramid electron group arrangement(4 bonds + 1 lone pair) with one lone pair at one of the corners so it has a net dipole moment\n\nSF6 has octahedral electron group arrangement (6 single bonds)\nand since all corners have F atoms the dipole moments of the bonds cancel each other and the molecule has no net dipole moment.
280076
Difluoro arsenate, AsF2O2-, should have the same general structure as H2AsO4-, with two OH- groups replaced with F-.  This will be closest to tetrahedral, but distorted toward a smaller F-As-F angle because of the resonant double bond with the oxides.
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.
436740
CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle
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.
284075
It's fairly easy don't worry.\n\nNo need for d level splitting diagrams to answer this since we're dealing with N and F:  use the spectrochemical series to make an assumption about the ligand.  The difference in t2g and eg levels (∆o) determines whether a complex is low or high spin.  If both ligands were the same, we would have to look at the oxidation state of the ligand in the complex.\n\nHalides <  Oxygen ligands  <  Nitrogen ligands < CN- ligands\n\n   ALWAYS HIGH SPIN    ---------------            ALWAYS LOW SPIN\n\nIf ∆o > P, then a complex is low spin.  IF ∆o < P the complex is high spin.
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
697397
NaOH\nSn(NO3)2\nCo2O3\nFeF2\nCuS2\nFe(OH)3\nAgBr