Trends in Stability | Oxidation State | d Block elements| Lecture 8 | Line to Line NCERT | Gagan Sir
Автор: Dikki Institute
Загружено: 2022-07-14
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Highest Oxidation Number achieved with halides
Ti Tetrahalides TiX4 (+4 Oxidation state).
V Pentafluoride VF5 (+5 Oxidation State).
Cr Hexafluoride CrF6 (+6 Oxidation State).
In Mn (+7) oxidation state is not represented in simple halides. But MnO3F exists.
In Fe, FeX3 (+3) exists for (Cl, F, Br).
In Co, CoF3 (+3)
In Ni, NiX2 (+2) (F,Cl)
In Cu, CuX2 is known except for Iodides.
In Zn, ZnX2 (+2)
Question: Why is the highest oxidation state generally most stable with fluorine?
Answer: Because of higher lattice energy as in case of a CoF3 or higher bond enthalpy due to higher covalent bonds example VF5,CrF6
Note:
+5 oxidation state in ’V’ is only by VF5 but other halides of ‘V’ undergo hydrolysis to give oxoacids of +5 oxidation state (VOX3)
Fluoride are relatively unstable in lower oxidation state example VX2 (only Cl,Br,I) and CuX (Cl,Br,I), no F and it is because of instability in lower oxidation state
Question: Why do all halogens form CuX2 except Iodide?
Answer: Because in this case Cu2+ oxidises I- to I2.
2Cu2++4I-Cu2I2+I2
Question: Why Cu+(I) compounds are unstable in aqueous solution?
Answer:
Copper(I) is unstable in aqueous solution, hence undergo disproportionation
2Cu+Cu2++Cu
The Stability of Cu2+(aq) rather than Cu+(aq) is due to much more negative hydH (Hydration enthalpy) of Cu2+(aq) than Cu+ which compensates more for second ionisation enthalpy of Cu.
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