Please i need help with this question below. Kindly choose the right answer, and
ID: 1029622 • Letter: P
Question
Please i need help with this question below. Kindly choose the right answer, and please explain your answer as well. show your workings.Thank you.
Which ONE of the following statements about inert and labile complexes is correct? O Tetracyanonickelate(2-) is labile because it has a large value for its formation constant, K. t has a large value for ts Tetracyanonickelate(2) is labile because it rapidly exchanges its cyanide groups O Tetracyanonickelate(2-) is inert because it has a large negative free energy of formation.Explanation / Answer
Answer is second option:- Tetracyanonickelate(2-) is albile bacause it rapidly exchanges its cyanide groups.
All the square planar complexa are labile bacause of the empty d-orbital are present in the square planer molecule. these are easy exchange ligands and some case it acept the ligand and forms to octahydral complex very fastly. And labile complex are low CFSE value.
Labile complex In these complexes, the rate of substitution of ligands is relatively high.
Inert complex In this case, ligands are not exchanged easily. The rate of exchange is very slow.
Factors affecting the labile/inert nature of complexes
1. Size of the central metal ion Smaller the size of the metal ion, greater will be the inertness because the ligands are held tightly by the metal ion.
2. Charge on the central metal ion Greater the charge on the metal ion, greater will be the inertness of the complex. 3. d-electron configuration If electrons are present in the antibonding eg* orbitals, the complex will be labile because the ligands will be weakly bonded to the metal and hence can be substituted easily. If the complex contains empty t2g orbitals, then it will be labile because ligands can approach easily without much repulsion. In short, if the complex contains less than three d-electrons, it will be labile. Or, if one or more eg electrons are present, it will be labile.