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In order to find the d-orbital splitting in a square planar complex, it is easier to start with an octahedral compound where splitting patterns are known and then imagine slowly moving two trans ligands away from the metal. This would at first distort the molecule which is known as tetragonally distorted octahedron. Moving the ligands further to such an extent I that the interaction between the metal ion and the ligands is practically nil would result in a square planar compound. This situation is best described its square planar complex because the other two ligands are so far off that they can be considered to be unattached to the metal ion and are thus unable to form a true chemical bond.
Why do zinc hydroxide, aluminium hydroxide and lead (II) hydroxide dissolve in excess aqueous sodium hydroxide?
why methane does not absorb IR radiation?
what is alternative form of cholesterol formed in plants and prokaryotes ?
#sulphuric acid can be made with concentration of 99%.however for diferent industrial requirement ,62% h2so4 is used which has density of 1.55 g/ml.calculate the molarity,molality
brief notes &uses
Metal ion configurations in high and low spin octahedral complexes(with)table
freundlich adsorpion
Why the current flow is always oppositte in direction to the conventional current direction? Ans) Availability of free electrons and application of potential difference are the c
The ratio of area covered by second orbital to the first orbital is: (1) 1 : 2 (2) 1 : 16 (3) 8 : 1 (4) 16 : 1 Ans: 16 : 1
Which of the following statements is not correct for an electron that has the quantum numbers n=4 and m =2: (1) The electron may have the quantum number s= +1/2 (2) The e
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