Question

Part A:

Though it did not play a role in the video that you watched, the oxidation state of the metal cation in a coordination complex can impact the difference in energy of the d orbitals. Consider aqueous solutions of two coordination complexes of chromium: [Cr(H2O)6]2+ is blue and [Cr(H2O)6]3+ is violet. Based on this observation, make a statement about the relationship between oxidation state and the difference in energy of the d orbitals (the d orbital splitting).

Part B:

The transition metals are often colored as solids and solutions. Rationalize this on an atomic level.

Part C:

Predict the product of the reaction between CuSO4 and Na2Se and write the balanced equation below. Do you expect the product to be soluble or insoluble in water? Justify your answer

Answer 1

Part A : The splitting energy of d Orbitals is higher in [Cr(H2O)6]³⁺, because Cr is in higher Oxidation State in this complex. Thus, it absorbs at the Lower wavelength Region, and is Violet.

On the other hand, since Cr is in lower oxidation state in the complex [Cr(H2O)6]²⁺, The splitting energy is relatively lower, resulting Absorbance at higher Wavelength Region than Violet. Hence, it is Blue.

Part B : Since, the transition metals contain partially filled d Electrons, they are easily excited, by The visible light. And, again the excited electrons relax their energy by emitting Light at the visible wavelength. Thus, most of the Transition Metals are coloured.

Part C : The Balanced Equation is :

CuSO4(aq) + Na2Se(aq) = CuSe(s) + Na2SO4(aq)

Thus, the Product is Copper(II) selenide. It will be Precipitated, because there's a Considerable Amount of covalent character introduced into the Compound. Thus, CuSe will be insoluble in water.