Question

(a) explain in brief about stability of co-ordination compounds.

(b) Discuss the factors affecting the magnitude of crystal splitting, $\large\rm { \Delta}$​

Answer 1

solution for (a) part

The stability of co-ordination compound $\large\rm { [ML_{n}]}$ is measured in terms of the stability constant.

A complex compound is formed by the interaction of metal and the lig@nds.

$\large\rm { M + nL ⇌ [ML_{n}]}$

$\large\rm { \therefore }$ stability constant is given by expression

β $\large\rm { = \frac{ [ML_{n}]}{[M][L]^{n}}}$

if the reaction is completed in 'n' steps then β₁, β₂, ..... βₙ are the formation constants for each step. more is the value of β, more is the stability of the complex , e.g., in the following cases:

[Cu(CN)₄]²⁻ with higher value of β is more stable. because the metal ion is same, so this also indicates that CN⁻ is a stronger lig@nd than NH₃. Further for a given metal and lig@nd and the stability is more, if more is the charge on metal ion , e.g., [Fe (CN)₆]³⁻ is more stable than [Fe (CN)₆]⁴⁻.

solution for (b) part:

$\large\rm { \bullet}$ The lig@nd field splitting depends on the arrangement and number. The value of $\large\rm { \Delta_{t}}$ is less than that of $\large\rm { \Delta_{0}}$ . it is because in case of tetrahedral complex we only have four ligands as compared to six in octahedral complex. moreover, there is a lig@nd along each axis in the octahedral complex while no lig@nd along any axis in the tetrahedral complex. for lig@nds at the same distances from the central metal atom or ion, it has been calculated that the electrostatic effect of a tetrahedral field is approximately $\large\rm { \frac{4}{9}}$ th of an octahedral field.

$\large\rm { \Delta_{t} = \frac{4}{9} \Delta_{0} = 0.45 \Delta_{0}}$

Thus, the value of ∆ is in the following order

$\large\rm { \Delta_{sp} > \Delta_{0} > \Delta_{t}}$

$\large\rm { \bullet}$ Quantum number for d-orbitals: the value of ∆ is higher for the central ion containing 4d and 5d orbitals than 3d orbitals because the 4d and 5d orbitals are larger in size as compared to 3d orbitals and extends out farther from the metal ion and hence the interaction with lig@nds is more.

Answer 2

Answer:

Quantum number for d-orbitals: the value of ∆ is higher for the central ion containing 4d and 5d orbitals than 3d orbitals because the 4d and 5d orbitals are larger in size as compared to 3d orbitals and extends out farther from the metal ion and hence the interaction with lig@nds is more.