How can you explain why the Beryllium diatomic molecule does not exist (molecular orbital theory)?
537794
Okay, to begin with; the molecular orbital theory is based on the "octet rule."  This means that, with exceptions of hydrogen and helium of course, an atom is most stable with 8 electrons (the octet) in its outer shell.  \n\nA diatomic molecule is a pair of identical atoms forming a small molecule.  The most common diatomic molecules are in the halogens; Cl, F, I, Br.  These atoms each lack one electron from a full octet, and will covalently bond in pairs to each gain a full octet.  Oxygen gas (O2) will bond in a double pair, sharing four electrons to gain an octet on each atom.  \n\nBeryllium contains 4 electrons in total.  The electronic configuration is 2:2 or if you use the other method is 1s2 2s2.  A full octet would be 2:8 or 2:0 to gain the most stability, and the Be2+ is actually in use in labs (2:0).  \n\nBecause it would need to gain electrons from or donate them to some other element with a greater number or greater need, beryllium can't form a diatomic molecule or a pair with another of itself.  \n\nIt is difficult to explain without pictures, but there are a few shell pictures available at Chemical Elements which will let you compare these elements.
426762
you know its ionic because Na is a metal while Cl isnt. a metal and nonmetal almost always from an ionic bond. to draw a lewis digram, determine how many valence electons are in each and draw the diagram.. its kinda hard to explain without actually drawing it.. consult a chem bk.
950673
no boron as with berylium among others is an exception to the octet rule. when bonding, the boron need only to share 3 bonds with (an)other element/s translating to 6 electrons for it to become stable... an example of a compound following this setup is triflouride (BF3)\n\ni hope that helps...
626546
The simplest reasonable model of the atom is the Bohr model, which does tend to look like a mini solar system. In actuality, atomic structure is not so easy to define. The further you advance in chemistry, the more complex the models become, and the less they look like "mini solar systems". But they are still only reasonable attempts to visualize a very complex system, and cannot physically confine all the expected paths of the electrons' orbits.
140269
Mecury doesn't "lack one electron".  What you would need to do is remove one proton from mercury to make it into gold, in some kind of nuclear reaction.  Now, while not impossible in theory, I don't know of any nuclear pathway to do it, and it would no doubt be prohibitively expensive to do if there were.
852561
electron does not orbit the nuleus. this idea was dismissed long back. in fact they spin in their place in either clockwise or anticlockwise direction. \n\noption (c) is correct
731446
c3h8 can not exist in the earths atmosphere
831018
http://www.chemicalelements.com/elements/be.html\n\nthis website may help you!!!!!! ( hope you get an A)
989577
There isn't enough energy at near kelvin temperatures for molecules to do all their normal movements and fluctuations.  \n\nThe contrast to this state occurs at any temperature above 'near absolute zero', take for example the nominal temperatures we are comfortable with.  At this temperature there is more than enough energy available for a molecule to bend and vibrate in literally millions of ways....these vibrations occur on the picosecond timescale and for the most part, all accessible ways to vibrate and bend.\n  As the temperature is reduced there are less vibrational modes available and the more high energy ways to vibrate (generally those involving less favourable configurations) start to not occur.  \n\nThe vibrational efficiency and capability of a molecule thus decreases to the point of a few possible vibrational modes as absolute zero is approached.
737141
The models that already exist are good enough for most chemists maybe all. Developments in molecular orbital theory on the other hand would be welcome.
835765
1) one s + three p + one d give five equivalent sp3d orbitals\n2) O=C=O, two sigma, 2 pi bonds\n3)a) 10e-; α2s2 α*2s2 π2p4 α2p2; diamagnetic\nb) 12e-; α2s2 α*2s2 π2p4 α2p2 π*2p2; paramagnetic (isoelectronic as o2)\nc) 14e-; α2s2 α*2s2 π2p4 α2p2 π*2p4; diamagnetic\nd) 10e-; same as a) above\n4) No molecules shown!\n5) The structure of graphite is planer, all sp2 carbons with an electron in the p-orbital. There are planes of carbon atoms well bonded together. Between these planes there are pi electrons and they are free to move; they are said to be aromatic pi electrons. When you apply a voltage, these electons do move - and moving electons is current - ie it conducts electricity. On the basis of MO theory:  Because the MO theory holds that electrons are not held to only one position. Instead they are spread across the entire molecule. \n6) No, it is trigonal planar, any individual dipoles from the B-F bonds are negated by the symmetry of the molecule.
369573
I think if you put that sort of question you are familliar with SCHRODINGER ecuation. It is a complex ecuation simply described by\n\nNabla (a small triangle with the basis on the upper side) of H (hamiltonian - a complex math analysis operator) equals to zero.\n\nThis equation was solved only for the Hydrogen atom (in my physics books from college the solution occupy 8 A4 pages ...) and the results were extrapolated for others atoms (helium, lithium, berylium, etc.) but is just an extrapolation. \n\nFor other atom than Hydrogen the Schrodinger equation can not be solved with the mathematical apparatus we dispose today ...\n\nSo, the Schrodinger equation describes the POSSIBILITY to find an electron in a region of space at one moment. The shape of this spatial arrangement is called orbital (wave function) and we have S orbitals (spherical shape - maximum 2 electrons), P orbitals (spatial simetry over Ox, Oy and Oz with a shape similar to mathematical sign for infinitum on every axis - maximum 6 alectrons in 3 pairs), D orbitals (maximum 10 electrons in 5 pairs) and F orbitals (maximum 14 electrons in 7 pairs)\n\nSo the Sulphur electronical configuration is:\n1S(2)2S(2)2P(6)3S(2)3P(4)\n\nFrom the last shell is probably the easier to extract an electron. The distance from the nucleus dictate the strenght of atraction ... So, an 3S electron is harder to remove than an 3P electon but easier to remove than a 2P electron\n\nDONE!
882538
Because, it only has one electron on its outermost shell. and thus is placed with group 1 atoms (which also have one electron on thier outermost 'shell')
72779
A molecule is made up of atoms.
957221
The naming is correct. But the molecule can't exist as described. The Chlor-atom must be attached to the methyl-particle, which is attached to the 3-chloro pentane molecule at the 3rd Carbon-atom.
466437
interesting answer above.\n\nCarbon has 4 valance electrons which because of the relatively small size of the atom can move about and create all sorts of hybrid orbitals and molecular bonding orbitals allowing it to suit the bonding needs of many other atoms, thus making many varied combinations.\n\nIt is also in the middle range of electronegitivities therefore resisting in most situations becoming an ion.
556249
In Be and Mg only the s subshell is full while the p is completely vacant. Moreover for every element the most stable state is tht of the nearest noble gas configuration. So for Be and Mg it will be quite easy to give to 2 electrons to get the nearest noble gas configuration to make stable cations. Hence they are not inactive/inert. Also they can gain electron pairs in the p-subshell (although it is energetically not favourable) so they can not be inert.