Question

A. An aqueous solution of Mn(NO3)2 is very pale pink, but an aqueous solution of K4[Mn(CN)6] is deep blue. Explain why the two differ so much in the intensities of their colors.
b. Predict which of the following compounds would be colorless in aqueous solution:

a. K2[Co(NCS)4]
b. Zn(NO3)2
c. [Cu(NH3)4]Cl2
d. CdSO4
e. AgClO3
f. Cr(NO3)2

Answer 1

Answer:

See Explanation

Explanation:

The colour of many transition metal complexes stem from transitions of electrons between energy levels. These transitions are governed by the spin selection rules and the colour is determined by the magnitude of crystal field splitting.

According to the spin selection rules, transitions in which ΔS = 0 are forbidden. Hence, a Mn^2+high spin compound is expected to be colourless. However, contrary to the spin selection rules Mn^2+high spin compounds do exhibit transitions in which the intensity is only about one-hundredth of the intensity of the spin allowed transitions. Thus many Mn^2+  high spin compounds such as Mn(NO3)2 are very pale pink or off white.

Note also that the crystal field stabilization energy of Mn^2+ which is a d^5 low spin ion is zero hence the very pale colour observed.

K4[Mn(CN)6] is deep blue as a result of charge transfer. Also, the compound exhibits an observed crystal field stabilization energy because it is a d^5 low spin compound hence the observed colour. Its low spin nature is because the cyanide ion is a strong field ligand hence it causes a greater magnitude of crystal filed splitting.

The following compounds are colourless;

Zn(NO3)2

CdSO4

AgClO3

One thing that is common to all the compounds listed above is that they are all d^10 compounds. This means that they all possess completely filled d-orbitals hence they are colourless.