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Saturday, April 16, 2022

Compare the splitting energy (CFSE) into the following compound and give appropritate reason ? [Co(NH3) 6] 3+ , [Rh(NH3) 6] 3+,[Ir(NH3) 6] 3+

In given compounds number of ligands, types of ligands and oxidation state is same for central atom belive to same group but different transition series 3d, 4d and 5d respectively. We know that as move top to bottom size of d-orbital(3d--> 4d -->5d) is increases so ligand approches to d-orbitals more closely so the repulsion between d-orbital of metal and ligand is high and splitting energy increases. 
Thus order of splitting energy 
[Co(NH3)6]3+ < [Rh(NH3)6]3+ < [Ir(NH3)6]3+

Related questions:



Which is correct statement for regarding I2Cl6 molecule?

(A) The molecule is planar
(B) 8 Number of angles of type 02 are present
(C) 03 > 01 > 02
(D) Total 4 unshared lone pairs are present
Ans key (A)
What is structure and also explain the bridge bonding of I2Cl6 ?

Related Questions:

(1) What is the difference between the structure of AlCl3 and diborane?

Thursday, April 14, 2022

Ellingham Diagram and it's features:


The changes in Gibbs energy that occur when one moleofoxygen isused may be plotted against temperature for a number of reaction of metals to form
their oxides. Such a graph is shown in Figure below and is called an Ellingham diagram for oxides. Similarly, we can plot Ellingham diagrams for halides.
The Ellingham diagram for oxides show the following important features :
(1) Ellingham diagram normallyconsist of plots of dfG° vs T for the formation of oxides of elements, i.e. for the reaction.
(2) The graphs for metal oxide all slope upwards because the change in Gibbs energybecomes less negative with increase in temperature.
(3) Each plot follows a straight line unless there is some change in phase.
(4) When the temperature is increased, a point will be reached when the line crosses dG = 0 line. Below this temperature the dfG° of oxide is negative and hence the oxide is stable.Above this temperature dfG° of the oxide is positive and hence the oxide becomes unstable and decomposes on its own into metal and oxygen.