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-1

Actually, some have voiced an opinion that FeI3 can apparently be created with difficulty, and is reputedly very unstable, decomposing into FeI2 and I2. Similarly, with CuI2, which is likewise unstable. To quote a source reference. $$FeCl3 + 3KI = FeI3 + 3KCl ; FeI3 = FeI2 + I2$$ Observation- Large anions diminish the lattice energy of the ...

0

If you go by the "ionic" counting method, we can start by looking at the charge on the ligands: -The 4 terminal methyl groups are anionic when using the ionic counting method: 4 anions = 4- charge. -For SMe2, the covalent bond to the methyl groups completes its full shell-no need to assign it a charge as it is happy being bonded with the methyls (doesn't ...

5

For organometallic complexes it's often easier to throw the "more electronegative element" idea out entirely and focus on the characteristics of the ligands. This is what @andselisk's comment is referring to. Generally ligands are partitioned into two types, $\ce{L}$ and $\ce{X}$: $\ce{L}$ refers to neutral ligands: for example $\ce{CO}$, phosphines, $\ce{... 2 There is a very similar molecule in the literature this has aryls instead of methyls. (MEYTEP) M.A.C.Lacabra, A.J.Canty, M.Lutz, J.Patel, A.L.Spek, Huailin Sun, G.van Koten, Inorg.Chim.Acta, 2002, volume 327, page 15. In the Cambridge database this compound is considered as a Pt(II) compound in the paper. We also have the dimethyl compound with bridging ... 0 All elements E appearing in the upper and right-hand part of the table are non-metals, have a high electronegativity, and their oxides are acidic. The doublet binding O and H in E-O-H is strongly attracted by the E atom. This favors the departure of the H+ ion. 2 Compounds with Mn in formal oxidation state +1 and -1 are well known. They simply are not stable on air and in water. In general, air-stable compounds of more active 3d-metals are ionic compounds. Their stability is a result of fine balance of energy of formation of relevant ions and energy of stabilization of said ions due to ion-ion and/or ion-dipole ... 0$\ce{Mn+}$has configuration$\mathrm{(4s)^0(3d)^6},$while on the other hand$\ce{Mn^2+}\mathrm{(3d)^5(4s)^0}$and also,$\ce{Mn}$in ground state is$\mathrm{(4s)^2(3d)^5}.\$ Therefore you can see in the oxidation state +2 orbitals are half-filled, and in +1 they are not half-filled. As half-filled is more stable, therefore +2 and 0 are more stable than ...

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