Question of ion difficult (快入)

2011-07-14 9:14 am

It is a bit long, please be patient to finish the whole question, THX!~

According to what I have learnt, atoms will donate/accept electrons to form ions, and thus, attain the electronic structures of Group 0 elements.
This enables atoms to form compounds, which is of a lower energy level, and become more stable (things about
圖片參考：http://imgcld.yimg.com/8/n/HA00656648/o/701107140006213873428440.jpg
).

So, I feel a bit puzzled :

For the the atoms of the same element of transition metals/lanthanide/actinides, of course, have the same electronic diagram,
but when they lose electron(s) to form cation, their electronic structures should same as that of Group 0 elements.
Which means that, they can only lose a specific number of electrons to form ions.

However, the truth is that, they can form more than one type of ion, take Molybdenum(atomic number = 42; [Kr] 4d5 5s1) as an example, it can form the following compounds:
MoCl2, MoCl3, MoCl4, MoCl5 and MoCl6

So, since metal and non-metal forms ionic compounds. Then, the compounds above should classified as ionic compounds.

As ion of chlorine is Cl-, so , the cation will be Mo(2+), Mo(3+), Mo(4+), Mo(5+) and Mo(6+) respectively.

And here comes my problem, isn't an ion have a electronic structure of Group 0 element? How can the Mo atom loss 2~6 electrons in different compounds and still attain an electronic structure of Group 0 element.

Is this about oxidation states, octet rule, 18-electron rule, orbitals, ligand field theory, energy level or else?

Can anyone explain these to me? THX a lot a lot!!!

If I have any mistakes/misconceptions, please point them out!!!

THX!!!

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回答 (1)

己式庚辛

2011-07-15 7:22 am

✔ 最佳答案

0.
atoms do not necessarily form ions to attain octet structure. they can form covalent bonds, or even remain non-octet (e.g. boron trifluoride, nitrogen dioxide).

1.
compounds between metals and non-metals are not necessarily ionic. aluminium chloride, dimethyl mercury and tetraethyl-lead are covalent, for example.

2.
even if they're ionic, ions do not necessarily follow octet rule; depending on their electronic structure, attaining octet structure may or may not be more stable. it's all about energetics.

3.
the ions are stable at specific conditions. for example, Cr(III) is more stable than Cr(II) under air; with absence of air, Cr(II) cannot be oxidized to give Cr(III). with stronger oxidants, Cr can be oxidized to form oxycation, chromate or dichromate.
in these cations, Cr increases in O.N. not by losing electrons, but by forming covalent bonds with more electronegative elements (i.e. oxygen).

4.
transition metal ions form complexes by accepting lone pairs from ligands, giving 18-electron species (filling up ns, np and (n-1)d orbitals). of course there're exceptions, but as a rule of thumb..

5.
molybdenum chlorides are no simple ionic compounds; they can form metal-metal bonds, forming metal clusters.
there's no Mo(5+) or so, but there's [Cl5Mo-MoCl5] complex, Mo2Cl10 or over-simplified, "MoCl5".
also, there's no MoCl6 under normal conditions.

6.
Mo has stable complex with CO , as Mo(CO)6 and has 18-electron structure. O.N. = 0 , Mo has 6 electrons, together 2x6 = 12 electrons from 6 carbon monoxide (carbonyl) ligands.

7.
THE POINT IS, when the energy loss due to losing electrons (to gain so-called stable structures) is GREATER than the gain from forming a compound, the compound isn't stable at all.
yes, Mo may lose 6 electrons to give octet, but the energy required is way too large. even the lattice energy of such ion is large, it cannot compensate the energy input for producing such an ion.

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