Which if these three hydrogen bonds is the strongest?
931315
I think it's "a," do to the very high electronegativity of F.  I'll have to check it out...
40125
(this is a confusing problem because resonance structures are supposed to show how electrons configure themselves to satisfy its most stable structure--the question didn't really do a good job of representing this idea, most likely that the teacher rather have you do a challenging problem than having to learn how to do a problem)\n1) The formal charge on N=N=O appears to be N=-1, N=+1, and O=0. \n2) The formal charge on N(triple bond)N-0 is N=0, N=+1, and O=-1. \n3) The formal charge on N-N(trible bond)O is N=-2, N=+1, and O=+1\nFirst of all, Nitrogen likes to make 3 bonds with another element and Oxygen only likes to make 2 bonds with another element.\nIt also appears obvious that the higher the formal charge, the less stable the compound is. Therefore, the 3rd structure appears least frequently.\nAs a conclustion, it is plausible to say that the overall resonance structure of N-N bonding is 112 picometer since N(triple bond)N = 110 picometer and that most of the compound likes to remain in a N(triple bond)N structure, yet some also retain N=N or even N-N structure. The average averages to a number very close to 112pm.\nSimilarly, Oxygen only likes to form 2 bonds. Therefore, the N-O bond lenght comes down to 119 since it is the average of the more stable N=O with the less stable N-O.\n\nHope that helps,\nUltimeaciax
891589
None of the first two row elements can exceed 8 valance electrons.\n\nCarbon can make four covalent bonds. Oxygen (typically) only makes two covalent bonds since it already has two lone pairs. And hydrogen can only ever have one bond.
923446
I would say CH2O2 (I'm assuming you mean formic acid, not dioxirane), CH3CO2H, and water.  \n\n"Important" is a tough word here, as all of these compounds can participate in h-bonding to varying degreees, with water being the strongest and CH2F2 being the weakest.  That said, strong hydrogen bonding gives rise to many of the nifty properties of water and in all three cases the h-bonding is powerful enough to raise the boiling points to be significantly higher than what you would expect based on molecular weight.
335753
1. Covalent Bonds\n2. Ionic Bonds\n3. Hydrogen Bonds
910236
The H will attach to the Carbon atom which is already bonded to other hydrogen atoms, in this case, to carbon,2.  Bromine will bond to the other carbon, hence carbon,1.\n\nAns: d.
680519
yes, it does because fluorine, being such an electronegative elements allows one of its lone pairs to become a bonding pair for another hydrogen to attach (other than its already used up bonding pair). and like the first quote HF has some of the strongest H-bonds due to the fluorine being one of most electronegative elements.
103986
The Hydofluoric Acid i.e. HF has the strongest Cationic Hydrogen Bond.\nAnd I think I fogot the Binding Energy of it\nNo Problem I will tell U Later when I Read my book
496847
N2- has a bond order of 3.\n\nhttp://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html
230448
dipole-dipole\nhydrogen bonds\nlondon forces\n\nde nada
959167
HNO2...all others are strong acids.
1107546
1. There are two molecular orbitals for hydrogen, the lower energy orbital has its greater electron density between the two nuclei. This is the bonding molecular orbital - and is of lower energy than the two 1s atomic orbitals of hydrogen atoms making this orbital more stable than two seperated atomic hydrogen orbitals. The upper molecular orbital has a node in the electronic wave function and the electron density is low between the two positively charged nuclei. The energy of the upper orbital is greater than that of the 1s atomic orbital, and such an orbital is called an antibonding molecular orbital.\nNormally, the two electrons in hydrogen occupy the bonding molecular orbital, with anti-parallel spins. If molecular hydrogen is irradiated by ultra-violet (UV) light, the molecule may absorb the energy, and promote one electron into its antibonding orbital (s*), and the atoms will seperate. The energy levels in a hydrogen molecule can be represented in a diagram - showing how the two 1s atomic orbitals combine to form two molecular orbitals, one bonding (s) and one antibonding (s*). This is shown below - by clicking upon either the s or s* molecular orbital in the diagram - it will show graphically in a window to the right.\n\n2.Hybridisation\nThe Valence Bond model suggests that for a covalent bond to form, two atomic orbitals must overlap sufficiently so that electrons can be shared. However, if the hydrogen atoms are positioned tetrahedrally, not all of them will overlap with orbitals on the carbon atom. This is because the carbon's p orbitals are positioned octahedrally, whereas the hydrogen atoms are orientated tetrahedrally. How does the VSEPR model account for this problem? By inventing hybrid orbitals (N.B. these were proposed by Linus Pauling). In the carbon atom, the 2s and 2p orbitals have different energies. If the 2s orbital combines with the three 2p orbitals, four new sp3 hybrid orbitals will be formed. These new hybrid orbitals are identical - they are arranged tetrahedrally and have the same energy (they are degenerate). The 1s orbitals from the hydrogen atoms can now overlap with the sp3 hybrid orbitals, forming single covalent bonds.\n\nImportant: insert a picture of formation of hybrid orbitals (energy levels)\n\nGeometries of Hybrid Orbitals\n\nsp\n sp2\n sp3\n sp3d\n sp3d2\n \n[edit]\nLone Electron Pairs\nThe VSEPR model treats lone electron pairs in a similar way to bonding electrons. In ammonia (NH3) for example, there are three hydrogen atoms and one lone pair of electrons surrounding the central nitrogen atom. Because there are four groups, ammonia has a tetrahedral shape but unlike methane, the angle between the hydrogen atoms is slightly smaller, 107.3°. This can be explained by theorising that lone electron pairs take up more space physically than bonding pairs. This is a reasonable theory: in a bond, the electron pair is distributed over two atoms whereas a lone pair is only located on one. Because it is bigger, the lone pair forces the other electron pairs together.\n\nTesting this assumption with water provides further evidence. In water (H2O) there are two hydrogen atoms and two lone pairs, again making four groups in total. The electron pairs repel eachother into a tetrahedral shape. The angle between the hydrogen atoms is 104.5°, which is what we expect from our model. The two lone pairs both push the bonds closer together, giving a smaller angle than in ammonia. The scientific method has succeeded again!
937266
1)electron ...(also negative meson)\n2)the atom is the basic (many atoms make molecule and so on)\n3)polymer (the most famous are protiens)
549677
This is a chemistry question, the anomaly is due to Hydrogen bonding.
806205
d. 100% of the max number of hydroxide ions.\n\nIn chemistry, strong base or strong acid is not a degree of how much it would eat you alive if you got it on you, but the degree in which it disassociates in water.
992794
bp order alkanes>alcohols>carbonic acids\nbecause in alkanes there are highest hydrogen bonds and due to the H-bonds the boiling point increases thus....
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.