characterisation of complexes
200112
See the links at \n\nhttp://www.itt.edu/staff/boyter/links/index.html
335710
View source of the that page.\nYou will see the formulas.
403781
I won't do your homework for you, but you should look here:\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1\n\nGood luck. :)
406006
So you need to prove that each element (x,y) E br(xo, yo) has a neighbourhood contained in br(xo, yo),\n\nTake a point (xa,ya) E br(xo,yo) and let d = r - ||(xa,ya) - (xo,yo)||.  d is positive since ||(xa,ya) - (xo,yo)|| < r.\nThen the neighbourhood bd(xa,ya) satisfies the condition because for any point (x,y) in bd(xa,ya) we have\n||(x,y) - (xo,yo)||\n<= ||(x,y) - (xa,ya)|| + ||(xa,ya) - (xo,yo)|| (triangle inequality)\n< d + ||(xa,ya) - (xo,yo)|| (because (x,y) E bd(xa,ya))\n= r.\n\nIn plain English: If the point (xa,ya) has distance d from the boundary of the set, then a neighbouroud of (xa,ya) whose elements are all less than d away from (xa,ya) will be contained in the set as well.
113848
Write down what you know about the complex:\n\n1) it's tetrahedral (sq planar)  --> ∆t  (t2 and e levels)\n2) CN   --> low spin\n3) d electrons of Ni(2+)  --> [Ar],d8\n\nBefore drawing the CFSE, state that ∆ is large for a square planar complex.  Since ∆ is large, electrons will always pair up, rather than occupying the dx²-y² level.  Since there will be no unpaired electrons, the complex will always be diamagnetic.  I cant draw the diagram here, but show the z²,x²-y² and xz,yz,xy levels, fill the diagram with 8electrons and pair them.
544230
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
593315
http://genchem.chem.wisc.edu/sstutorial/FunChem.htm
351568
If the ligands lie in the x-y plane, then the d-orbitals on the central atom that have all their electron density *out* of this plane will have the lowest energies.  These are the d-xz and d-yz orbitals. \n\nThe d-z^2 orbital has its density mostly out of the x-y plane, but there is a ring of electron density in the x-y plane, so this orbital will have a somewhat higher energy than the d-xz and d-yz orbitals.\n\nThe d-xy and d-(x^2-y^2) orbitals have their electron density concentrated in the x-y plane so their energies will be the highest.  They have the same energy.\n\nSummary, highest energy at the top:\n\nd-xy, x-(x^2-y^2)\nd-z^2\nd-xz, d-yz
551014
C, having both positive and negative charges on one molecule.
618258
a the number of outer electron that will bond with other atoms to fill thei valence shells
527985
CuF3, Cu(NO2)2, CoCl3
319335
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
538366
The ligand will accomodate electrons from the ligands, forming dative bonds.\n\nThe sub-shell also ceases into two groups of orbitals.  One group will be at a higher energy level as compared with the other.  The difference in energy level is acounted by the strength of the ligand, in other words, by its polarising power.  The stronger the ligand, the greater is the energy difference between the two groups of orbitals.
197608
Software programming.
167487
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
373452
It's:\n\nSum of bond enthalpies of product - Sum of bond enthalpies of reactants
474535
your question is not clear. describe positions of the objects.