How would you draw Lewis Structures for Boron Trichloride, And Borane?
916374
The trichloride is a little more complicated than borane (BH3).\n\nCheck out the following:
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.
1093944
i) add NaNH2(1 eq.)  then CH3I(1 eq.) (I=iodine)then NaNH2(1 eq.)  then C2H5I then add hydrogen in presence of palladium and BaSO4.\nii)add NaNH2(1 eq.) then C3H7I(1 eq.)  then add NaNH2(1 eq.)  then again C3H7I(1 eq.) then add ZnBr2 or MgBr2.
811115
Ball and stick is more useful in displaying molecular/VSEPR geometry. Lewis diagrams are more useful in lone electron pair location and determining charges.
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.
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...
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No, we don't use the electronegativity chart; instead we draw the full structure of the molecule and then see which 'pulling of bonds' cancel each other. \nIn the molecule AsF3, since As has 5 outer most electrons(in group 5), each F is bonded to As with a single covalent bond. Hence, the shape of the molecule becomes trigonal pyramidal, with 3 bond pairs (with 3 F atoms) and 1 lone pair. The 3 bond pairs pull the electrons towards themselves in the lower plane, while the 1 lone pair pulls at the top. The pulling effect of the 2 F atoms at the 2 sides cancel out  because the pulling effect is the same for both F atoms; however, the pulling effect of the lone pair at the top and the 1 F bond pair at the bottom will not cancel, because it is not from the same atom/species. Hence the overall dipole moment( dipole moment = charge x distance) does not cancel out, so the molecule is polar, with some charge distribution remaining due to the unequal pulling effect.\n\nIn CF3Cl, the shape is tetrahedral, with the 3 Cl atoms and 1 F atom bonded by single covalent bonds to the central C atom. The bond pairs spread out with the 3 Cl atoms pulling at the bottom and the 1 F atom pulling at the top. The pulling effect of the 2 Cl atoms at the sides cancels out, but that of the F atom at the top and the Cl atom at the bottom does not, as the pullin effect of 2 different atoms is different. Hence, as before, overall dipole does not cancel and the molecule is hence polar.
770078
alcl3 is trigonal planar (bond angle 120')\nco2, straight (180')\nh20, bent (~105')\nch4 and ch2cl2 are tetrahedral. (~110')
325629
It might help if you spelled "naphthalene" correctly. \nhttp://www.madsci.org/posts/archives/may98/894550073.Gb.r.html \nThis website has the structural pictures you can use to do your Lewis dot structure.  \nNaphthalene is an aromatic compound.  It's two aromatic rings sharing a side.
956142
2Rb(s)+Cl2(g)--2RbCl(s)
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Maybe this wil lhelp.         \n\n ELECTRON DOT DIAGRAMS\n1. Let the element's SYMBOL represent the nucleus and inner energy level .\n2. Write the ELECTRON CONFIGURATION for the element (or look at the PERIODIC table).  Count the electrons in the OUTER energy level.\n3. Let the top, bottom, and two sides represent the ORBITALS.\nPlace the appropriate number of DOTS in each "orbital".\nUse ONE dot if the orbital has 1 electron and TWO dots if the orbital has 2 electrons.  The electron dot diagrams for:\n \nNa(Z=11)  1s2   2s2   2p6  3s1       Na\n \nMg(Z=12)  1s2   2s2   2p6  3s2             Mg\n \nAs(Z=33)  1s2   2s2   2p6  3s2  3p6  4s2  3d10  4p3      As\n \nThe periodic table can also be used to write electron dot diagrams.\nAll the elements in the first column have ONE outer electron(s).  They have ONE dot(s) in their electron dot diagram. All of the elements in column 2 have TWO outer electrons.  They have TWO dots in their electron dot diagram.  Most of the elements in column 18 have EIGHT outer electrons except HELIUM which only has TWO outer electrons.  All of the elements in column 18 have EIGHT dots in their electron dot diagram except HELIUM  which only has TWO.
512329
Strictly speaking, the Lewis theory is not applicable to Xenon compounds (& if XeCl3 exists at all). The proper way goes by the Valence Bond theory. In that structure, Xe forms 3 single-sigma bonds with 3 Cl. That leaves out 5 valence electrons which is 2 lone pairs & 1 unpaired. This should (if possible) thus dimerise as Xe2Cl6, each segment being an octrahedron with 2 lone pairs. there should be 2 such octahedra (parallel planes) sharing 1 bond.
821051
start with carbon... carbon has 4 electrons in it's valence shell.  carbon will almost always be the central element in your compound.  \n\nfor the first one... oxygen has 6 valence electrons... a triple bond joins the 2 elements (therefore they share 6 electrons) and each element has one pair of unshared electrons... it looks like this\n    :C:::O:\nnotice that each atom has 8 electrons... and you can count the ones that carbon donated and the ones oxygen donated... maybe this will make it clearer\n    :C:;;O;    imagine that each ; is a pair of electrons from oxygen\n\nok, the 2nd one is quite easy to explain, but difficult to draw on the computer...   Carbon is the central atom, with 4 electrons.  It shares one elctron with each of the 4 other elements.  Hydrogen has 1 electron, and needs just one more to complete its shell (it is unique in this way; most need 8, but hydrogen just needs 2).  Chlorine has 7 electrons in its outer shell and needs just one more to complete its outer shell.\n\nSo, the lewis structure is drawn in a cross fashion.  \n\nHorizontally...   H:C:Cl:::\n\nVertically...   H:C:Cl:::   (where they cross at the C)\n\nSo, C has a pair of electrons on each side, and each Cl has a pair of electrons on each side (not 3 pairs on one side as I have drawn)\n\nCheck out the attached link for more help
236619
There are several steps used to do Lewis Dot Structures.\n1) Determine the number of electrons required ( in this example there are 40 electrons required)\n2) Determine the number of valence elestrons ( there are 6 +6+6+6+6 = 30 ), however since you have a -2 formal charge there are 2 extra electrons, which means you have 32\n3) Determine the number of bonds (subract the number of valence electrons from the total required and then divide by 2) ie: 40 - 32 = 8/2 = 4 bonds\n4) write the atom symbols on this page ( in this example you're going to write Se in the center with one Oxygen on every side)\n5) connect with the number of bonds that were calculated\n6) fill in electrons until all the atoms have conformed to the octet rule except for hydrogen\n7) then count the number of electrons shown to check your work\nAnswer: 1 bond connects each Oxygen to the one Selenium and each Oxygen is going to have six dots on the outside and Selenium is going to have four dots.
910312
okay. i just took the test on this and passed with a B. so i promise i know what im doing.\n\nK, look at your elements. then look them up on the periodic table. lets pick, oh say the compout Cl(subscript)2O. that is 2 chlorines, and 1 oxygen. well, chlorine is in group 17, which means that each have 7 valence electrons, and oxygen is in group 16, which means that it has 6 valence electrons. following me so far? okay, take those valence electrons, and add them up. that equals 20 valence electrons. now, position the lewis dot diagram so that the oxygen is between the two chlorines. now, distribute the 20 electrons amongst the three letters, so that each has eight. now each letter should be surrounded by exactly eight dots. keep in mind that it should look like this -- :Cl:O:Cl:, but dots on the tops and bottoms. that is the lewis dot. luckily, this one did NOT have double or triple bondage. but in the event of double bondage, it might look like this -- :Cl::O:Cl, without dots all the way around each letter.\n\nand the VSEPR theoryis quite simple. it is just talking about the potential shapes that chemicals can make. for example, there is linear, or trigonal planar, or tetrahedral. this is vital. i promise that, if you contact me, i can give you a simple cheat sheet (not really for cheating, but it helps your memorization) to remember VSEPR. it gives you the shapes, lewis dot examples for the shapes, and angle measurements.\n\nhope this helps!\n~ally
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This is kinda hard to draw on here...\n......_\nK--Br|\n......_\n\nThe lines around Br represent 2 electrons, and the -- represents a bond...ignore the periods
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what you need to do, get a SI data book, and look up their natural state. not bonded with anything else, and possibly itself.\n\nPhophorus - white: molecular (P4). red: Chain. Black: Layer\nIodine - molecular (I2)\nMagnesium - hexagonal closest-packing\nNeon - cubic closest-packing\nArsenic -grey: 3d structure - Layer formation. yellow: molecular As4\nSulfur- molecular S8\nBoron - complicated structure.\nSelenium - grey: chain, Red: molecular: Se8\nGallium - Complicated\nCesium - body-centred cubic\nOxxygen - O2, O3 molecular