Which is not true about bond order?
966470
answer four does not appear to be correct
728949
They are (Listed in order of decreasing strength)\nIonic interactions \nHydrogen bonds \ndipole-dipole interactions \nLondon dispersion forces (Van der Waals force).\n\nA very approximate strength order would be: \nBond type/Relative strength \nIonic bonds/1000 \nHydrogen bonds/100 \nDipole-dipole/10 \nLondon (Van der Waals) Forces/1
950288
1)  Only the valence orbitals matter, and all the valence orbitals count (more complicated version: all valence orbitals which are of the correct symmetry to hybridize to form MO.  If you don't know what that means then you're in a lower level class and it doesn't matter).  # of atomic orbitals = # of molecular orbitals, and the molecular orbitals may be bonding, nonbonding, or antibonding.  Not depending on the particular class you're taking your teacher may or may not want you to draw the non-valence orbitals, but they don't matter at all.\n\n2)  You start filling all the MOs with electrons, starting with the lowest energy and moving up.\n\n3)  You wouldn't determine that for C2 because that's not correct.  The bonding and antibonding orbitals formed from the 2s atomic orbitals would both be filled (net bond order of zero) the lowest bonding (sigma) MO from the 2p AO would be filled, and each of the higher bonding (pi) MO from the 2p AO would have one electron.  None of the antibonding orbitals from the 2p AO would be filled.\n\nThis predicts that C2 would have 2 unpaired electrons, whereas a Lewis structure would predict a quadruple bond between the carbons.  One guess as to which one is right....
287953
actually, it's arranged by the number of protons in the nucleus of an atom of each element, though the list in this song: http://www.privatehand.com/flash/elements.html\nis not.
934267
u do, ionic bonds results not from a sharing of electrons but from a transefer of one of more electrons from one atom to another
218795
Cannot be answered unless you specify the redemption value of the bond and the time to maturity.
753982
You have some errors there:\n\nSO4 with the -2 charge has 4 bonded pairs and no lone pairs (unbonded pairs) on the sulfur (but three on each of the four oxygens, which I assume you are not counting).\n\nH2 is correct with 1,0\n\nH2O2 is correct with 3, 4 (two lone pair on each oxygen)\n\nCH2O has 4 bonded pairs (four between the carbon and the oxygen for a double bond) and 0 non bonded pair.\n\nCO2 has 4, 0, with double bonds between each oxygen and the central carbon\n\nHF, is 1,3\n\nCl2 is 1, 6, if you count the lone pairs on the two chlorine like you are doing with F in HF but not the O in SO4 with the -2 charge.\n\nSO2 has a resonance structure where three bonded pair are smeared over two bonds between the S and O.  There are six unbonded pair (3 per oxygen) on the two oxygen.\n\nCH4O, (methyl alcohol?, CH3OH?) has four bonded pair between the carbon and three hydrogen and the one oxygen.  The oxygen has another bonded pair for its hydrogen and two unbonded pair.  This is 5, 2 as you have it.
169339
Because the future cash flow (the repayment) for a long-term zero coupon bond is discounted over a longer period of time. This will result in a greater fluctuation in price market interest rates change.\n\nFor example, assume bond A is a $100,000 zero coupon bond that matures in one year. Assume the bond is sold when market rates are 5%. This would mean that the bond price is about $95,238 (100,000/1.05). If, a few minutes after it was sold interest rates jumped to 10%, the bond price would fall to to $90,909 (100,000/1.10).\n\nAssume that bond B is a $100,000 zero coupon bond that matures in 10 years. Assuming the same rates above, the bond would have a selling price of $61,391 (100,000/(1.05)^10). However, when rates rise to 10%, the bond's price would plummet to $38,554 (100,000/(1.10)^10).\n\nNotice how much more sensitive the long-term bond is to changes in interest rates. Its price changes dramatically.
953238
It is true for individual bonds, but not for molecules.\nThe dipole moment of the molecule will depend on the geometry of the polar covalent bonds. E.g. CF4 has 4 polar bonds but they arranged so that the sum of their dipole moments is zero and the molecule has actually no dipole moment.
819263
Absolutely not.  That is their fee for posting your bond.  When the bonding company post's your bail they essentially say that they will guarantee your return to court.  What I can't believe is that you paid 40% when the going rate is around 15% - 20%!!!  You got screwed.
769354
Not at all. Scalar or vector addition is not affected by order.
75751
if there is no double bond between two carbon atoms then it is saturated
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.
523415
http://www.bondsonlinequotes.com/index.cfm?CFID=16970449&CFTOKEN=12afc9b-ce49873d-62e5-41c4-a8db-c5ed72b4a415\n\nSign up and use
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.
280123
no, not in general.  Example:  Z(2)+Z(2) and Z(4) all are of order 4 but are not issomorphic because the element 3 of Z(4) has order 4 but there does not exist an element of order 4 in Z(2)+Z(2).
11021
covalent bonds are called the directional bonds.because the orbitals involved in bonding have specific direction where they bond.