what happens to the d-orbitals when ligands are attached to the central ion?
963808
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.
922867
Check out that link.  It has a picture that shows how it looks like: the grey Al atom is in the center, and 3 green Cl atoms are around it in a triangular arrangement.
272575
let me explain this with an example.take Cl atom which has 17 protons in the nucleus and equal no. of electrons(17)revolving around the nucleus.this means tht the 17 positive charges are attracting the 17 negative charges.\nnow let us consider the chloride ion Cl(-1). the protons are again 17 here but one negative charge is extra ie there are 18 electrons.so,in this case the 17 positive charges are pulling the 18 negative charges.\nnow u can imagine in which case is the pull less efficient atom or anion?\nobviously it is less efficient in case of anion because of the inequality of charges thus making it bigger in size.\nhope ,i ve made it clear with an example...
15546
Interesting scenario first of all i would like to comment. Well actually the bond is due to electronic forces. Hence mass has litle effect on the bond strength. Thus i thikn the effect of mass change should be a small one!! unless you change the charge of proton too!!
848068
ICl2 + should be bent like water since I will have two lone pairs and two Cl's on it.\n\nI3- will be linear\n\nSbCl6- will be octahedral
766145
It depends on how far away from the noble gases they are. Like anything in gourp IA would be 1+, IIA 2+ etc, then when theyre closer to the right then its like VIIA 1-, VIA 2-\n\nTransition metals are different because they form multiple ions. It would indicate what the charge is.
235116
The metal should be absorbed by the ligand.  It's called chelation.  What about it?
634552
an ion with both acidic and basic part
978327
When a ligand binds to Cu, it splits the 5 d orbitals into two groups (off the top of my head I cant remember the names but I think they are t2g and eg).  The color results from electrons transitioning between the two d orbital levels.  Different ligands split the orbitals different amounts so you get different wavelengths of light emitted.  \nAlso, since it is technically forbidden to have an electron transfer within an orbital level, the d-d transition peaks are much smaller than other electron transitions.
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.
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
781553
a typical metalic atom takes a positive charge when it becomes an ion.  that means that it loses its valence electrons.  that means the radius is smaller in the ion than in the atom.
709101
Because there will always be interactions between all the elements in a compund except for a bridging ligand, we can asusme that we need to take into account the charge on each atom. \n\na element in group 2 is always +2\na element in group 4 is +2,3,4\nOxygen is is -2\n\nknowing we have a neutral charge we can immedialty eliminate C and as we know that we must use all of our atoms the only answer that makes sense is D
616395
This compund has an octahedral geometry and is coordinated with 3 bidentate oxalate (C2O4) ligands.\nReaction of potassium oxalate hydrate and ferric chloride\n\n3K2C2O4ˇH2O + FeCl3 --> K3[Fe(C2O4)3]ˇ3H2O + 3KCl\n\nWater molekules do not act as ligands and potassium ions act as counter-ions
739878
I am not sure what you are asking but the answer could be a monoatomic ion. The positive being a cation and the negative being an anion.\n\nYou might ask over at http://forums.mychemistrytutor.com though.
979578
An ion is formed when one electron completely transfers over from one atom to another atom.  The one losing the electron acquires a positive charge and the one losing the electron acquires a negative charge.  But then these two ions are attracted to each other forming an ionic compound.  After coming together, the bonding electrons spend most of the time around the atom with the negative charge, that is, the more electronegative ion.  However, there are also moments when the electrons get closer to the electropositive atom.  Though it's rare to find the bonding electrons near the electropositive atom (since it's orbitals are higher in energy), calculations find that ~5% of the time the electrons are closer to the electropositive atom than to the electronegative atom.  (Don't trust me on the ~5%... this is based on only a few examples from my Inorganic Chemistry chapter I read a few weeks ago).  A popular example is that of HF.  If you use molecular orbital diagrams, you see that the 2s orbital is so low in energy (around -40eV) that it doesn't share it's electrons.  The 2p(z) orbital bonds with the hydrogen's 1s orbital but you'll see that the resulting bonding orbital is so close to the 2p(z) orbital in energy that you can sort of consider it somewhat nonbonding.  The point is that there is little influence from the H 1s orbital, but there is still a small amount of influence.  This means the bonding electrons spend a fraction of the time around the H atom, meaning that the original electron in question has no completely transferred over to the F atom.
697038
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