why do the bonding pairs in CH4 arrange themselves in a tetrahedral shape?
362329
Consider that one electron will repel another electron (-ve does not attract -ve), each bond between a carbon and hydrogen is cased by a pair of electrons (2 electrons) so you have a total of eight electrons... this results in neutralising the charge on all the atoms.\n\nCarbon has 2s1, 2s2, 2p2 electrons which means it is theoretically 4 electrons short of its closest nobel gas (nobel gases are neutral).\nHydrogen has 1s1 electrons which means it is theoretically 1 electron short. Note that there is no p orbital in first row of periodic table.\n\nAs such each hydrogen shares its sole electron with the carbon, donating a total of 4 electrons, hence 4 hydrogens.\nThe Hydrogens get their electrons from the carbon, the carbon has 2s2 and 2p2 electrons, totalling 4 to share, 1 to each hydrogen... we do not count the s1 orbital because it is unlikely to operate in forming bonding.\n\nNow each pair repels eachother so they are furthest apart, if we do the mathematics...\nblah blah...\nWe end up with an optimum angle of 107.5 degrees between eah hydrogen, any more than that and a different hydrogen will repel it back, and likewise if less. this forms a 'tetrahedral' shape.\n\nIf you were to switch a hydrogen with another element... say chlorine, to produce CH3Cl (Chloro-methane) then the hydrogens (being the smaller atoms with less electrons) will be pushed further from the chlorine which will tend to determine the molecule's appearance. Ch3Cl is not a perfect tetrahedral as its angles will not equal 107.5, but we still call it that for simplicity's sake.\n\nI recommend that for further information do a search on the following phrases:\n\nVSEPR\nValence Shell Electron Pair Repulsion theory\nMolecular geometry\nMethane\nOrganic Chemistry
792627
True - symmetry elements cancel dipoles. Take for example CCl4, a tetrahedral molecule. Each C-Cl bond has a dipole associated with it, the negative part is the Cl (Electronegativity (EN) of C= 2.5, EN Cl =3.5). However the symmetry due to the shape of the molecule cancels out all dipole so NET dipole of molecule is zero.\n\nTrue- ex, SO4 or NO3\n\na) 180°, b) 120°, c) less than 109° (eg 104.5° for H-O-H), d) 109' 28" (or 109.5°). The detailed whys of these are due to hybridization of the atoms involved. linear will have sp hybridization and no lone pairs. trigonal planar will have sp2 with no lone pair. Bent can be many combinations, eg sp3 hybridazation with 2 lone pairs of electrons (eg water), tetrahedral will be sp3.
554160
A molecule is polar when it has a dipole moment that does not cancel out. For example, consider the molecule CH4 (methane) which is nonpolar. It has a carbon atom bonded to 4 hydrogen atoms. Bonding is the overlap of the electron density. So you have an electron from a hydrogen and an electron from the carbon forming each of the bonds. Carbon is just very slightly more electronegative (tendency of pulling electrons closer to itself) than hydrogen. This means that the electrons are pulled by carbon away from hydrogen, but not by much (the difference is very very small). This shifting of electron density allows for the carbon to be slightly more negatively charged and hydrogen slightly electron deficient. This makes the C-H bond just slightly polar. However, because there are 4 hydrogen atoms that are bonded to the carbon, this bond polarity cancels each other out...meaning that the bonds are slightly polar, but the overall molecule is nonpolar. \n\nTo determine if the bonds are polar you just have to know the electronegativity differences. To know if a molecule is polar, you must know the correct geometry. Since electrons are negatively charged and repel each other, they want to be as far away from each other as possible. If you have a carbon bonded to 4 hydrogens, you will have a tetrahedral geometry with the hydrogens 109.5 degrees apart from each other creating the best possible arrangement. Also, if you have a lone pair of electrons, then the molecule will always be polar because of that charge separation.
800333
It is trigonal bipyramidal in shape.                                                    try this & you would understand better                                            take a cardboard or a  paper.cut it in da shape of a triangle. suspend threads from da vertices of da triangle both upward & downward.                                                                                       now catch-hold of da threads.now 1 hand holds da threads from da top. The triangle remains in da mid position & the other hand holds da other 3 threads at the bottom                                            NOW the 2 hands are Cl atoms .THE 3 vertices of da triangle r also Cl atoms                                                                                   Phosphorous is at centroid of the triangle
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.
139994
Is its structure a ring with double bonds?
731348
C3H8 is CH3-CH2-CH3\n\nthere is no way that this will fragment into CH4.\n\nyou will get fragments of\nCH3 (m/e = 15)\nCH2CH3 (m/e = 29)\n\ncan you see any way to split the above molecule into CH4?\n\nremember that fragmenting means destroying a bond, thus if you were to get CH4, you'd have to have started with CH4 bonded to something, but that breaks the octet rule...
779140
Look at the VSEPR/Lewis structure:\n\nO = C = O\n\nThe molecule is linear with no net dipole moment, because the polar bonds cancel each other out. It also has no hydrogen bond donor or acceptor sites.\n\nIt's also relatively light, with a molar mass of 44 g/mol.\n\nBecause of these molecular factors, it ends up being a gas at room temperature.
998521
that is wrong!!!!\nthe values should have been the other way round!\nnow look:since ethane is a parafin we cannot discuss its prop\ncome to acetone. the central carbon is double bonded to an oxygen atom\nnow when an electrophile comes to this oxygen the bond attains polar character wth the oxygen having partial negative charge(grearter electronegativity), and the carbon having partial positive charge\nnow obviously the electrophile will go towards this oxygen atom, showin its basic character\nlook at the other way(the one described above)\nwhen the oxygen atom reduces the electron cloud density, the carbon atom is more strogly attached to the hydrogens which means that it cannot be released easily which once agains shows its basic character\nhence the values should have been the othr way round
785541
1. lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. VSEPR theory. a rough rule of thumb is that with same number of electron pairs, the more lone pairs you have with respect to bond pairs, the smaller the angle between bond pairs.\n\n2. the theory i just mentioned is not obeyed exactly by formaldehyde. you would expect the molecule to be trigonal planar with 120° for all bond angles because there are only bond pairs. however, the oxygen atom is electronegative and has a tendency to pull electrons from the double bond towards itself. the other bond pairs (both C-H) thus are not repelled as much by the C=O bond electrons and so can take a wider angle.
147899
Here is a graph and some explanation:\nhttp://mathworld.wolfram.com/Tetrahedron6-Compound.html
11244
All are true except that each atom is bonded to four others.  Carbon's valence is four, and so it can only form four bonds--by having double bonds, there is no room to bond to four atoms.  Diamond, on the other hand, has each carbon atom bonded to four other atoms, but no double bonds.\n\nGraphite forms in sheets, with no covalent linking between pifferent sheets.  That is why it is so soft.\n\nthe picture on the second link does not show the double bonds, but if you pick a single direction and regard all parallel bonds in that direction to be double bonds, the valence states will make sense.
260431
R-O-CH3 \n\n R is any carbon based group. I wish I had chem draw on this computer, but the structure is actually bent due to the symmetry (this involves group theory - a linear algebra based idea in quantum mechanics) of the orbitals involved in bonding. Its bent for the same reasons that water is bent, now your teachers may say that water is bent due to electronic repulsions but if thats true how come they draw the lone pairs so close to eachother??? Its because its not true, and its part of VSEPR theory which is just a practicle method of teaching structure to people who dont know too much physics, math, and chemistry.
758383
Since the central atom, C, does not have any lone pair electrons on it, and it is bonded to 3 oxygen atoms, the general shape around the central atom is trigonal planar.  The hydrogen atom is bonded to one of the oxygen atoms.\n\nTrigonal planar indictates that all 4 atoms, C and three O, lie in a plane and that the O-C-O bond angle is 120o for all bonds.
149133
It's covalent bonding between the carbon atoms.  The whole diamond can be considered one molecule actually because of the ability of carbon atoms to covalently bond to each other.
754819
no too sure since its been a while since iv done this, but its a simple enough molecule, C2Br4  2Br to each carbon and a double bond between the carbons, its a linear moleule, not sure about the electronegativity of the elements as my notes r all packed away but the Br atoms should sit slightly of veritcal from each other toward the outside of the molecule. and im sure that the two halves sit at 90 degrees to each other, not sure if u call it this but if u do then 2 Br attached to one carbon will sit along the x- axis of the molecule and the two Br attached to the other carbon will sit along the Z-axis, ask ur tutor or teacher if this is correct as im not 100% sure\n\ntetrahedral only applies to CBr4, thats because the 4 Br are comin off the same carbon, because there are 2 carbons it changes it all together\n   (ignore the full stops, needed them to position the atoms decently)\n\nBr&#92; ........ /Br\n.......C=C\nBr/ ........ &#92;Br\n\nThis is not perfect but imagine the top Br on the left aiming toward you at 45 degrees and the top one on the right aiming away at 45 degrees, hope this helps
352077
Thanks NeuroProf.\n\nThe molecule AsF5 is trigonal bipyramidal in shape. First lay out all the bond pairs in that shape and then see if the 'pulling effect' of all the bonds cancel each other. Here, there are 3 bond pairs (of Cl) at the corners of the triangle, 1 bond pair at the bottom and 1 at the top. The Cl at the top pulls electrons and the pulling effect is cancelled by the one at the bottom. Considering the 3 bond pairs left in the middle, 2 are angled at 120 degress and one is at 90 degrees to the top and bottom bond pair. So the pulling effect of the 2 pairs is cancelled by the one at 90 degrees. Overall dipole cancels out and there is no charge distribution, so the molecule is non-polar.\n\nIn BrF4- (with square planar shape), the pulling effect of the lone pair at the top is cancelled by that of the lone pair at the bottom. This leaves 4 F atoms arranged at the corners of the square in the middle. The pulling effect(dipole) of 2 diagonal F atoms will cancel because they are in the same plane and are the same atoms. Hence all the 4 dipoles cancel out and that of the lone pairs too, so the molecule is non-polar.\n\nIn GaI3, the 3 I atoms are arranged in a trigonal planar manner. The dipole of the 2 I atoms at the sides will cancel, but the I atom at the top is not paired with any other atom at the bottom, so the dipole cannot cancel out. Since overall dipole moment is not equal to zero, the molecule is polar.\n\nAre you sure the molecule is SbF6- and not SeF6? If it's SbF6 then the shape will be square bipyramidal or octahedral (similar to BrF4-). The dipole of the 2 F atoms at the top and bottom will cancel and that of 2 diagonal F atoms will cancel and overall dipole will be zero, so the molecule will be non-polar.