a. [Ni(H2O)6]^2+ (aq) is green in colour whereas [Ni(H2O)4 (en)^2+ (aq)is blue in colour , give reason in support of your answer .

Question

Answer the following questions:

a. [Ni(H₂O)₆]²⁺ (aq) is green in colour whereas [Ni(H₂O)₄ (en)²⁺ (aq)is blue in colour , give reason in support of your answer .

b. Write the formula and hybridization of the following compound:

tris(ethane-1,2–diamine) cobalt(III) sulphate

OR

In a coordination entity, the electronic configuration of the central metal ion is t_2g³ e_g¹

a. Is the coordination compound a high spin or low spin complex?

b. Draw the crystal field splitting diagram for the above complex.

Answer 1

(a) In both complexes, Ni is in +2 oxidation state with a valence shell electronic configuration of 3d⁸. In the presence of weak field water ligands, two unpaired electrons do not pair up. Hence, the complex [Ni(H₂O)₆]⁺² has two unpaired electrons which result in green colour. Due to d-d transition, red light is absorbed and complimentary light emitted is green.

However, due to the presence of (en) which is a strong field ligand, the splitting is increased. Due to the change in 
t_2g - e_g splitting the colouration of the compound changes from green to blue.

(b) The given complex, tris(ethane-1,2–diamine) cobalt(III) sulphate, contains cobalt (Co³⁺) as a central metal ion. There are three ethane-1, 2-diamine (en) ligand as the multiplicative prefix 'tris' is used. The complex ion has sulphate ion (SO₄^2-) as a counter ion. Therefore, the formula of the compound becomes [Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃.

OR

(a) The electronic configuration of the central metal ion in a coordination entity t_2g³ e_g¹ indicates that the fourth electron enters one of the e_g orbitals, since, ∆_o < P. As there are four unpaired electrons in the complex, it is a high spin complex molecule.

(b) The splitting of degenerate orbitals by the ligands into orbitals of different energy is called crystal field splitting. When the metal ion is surrounded by six ligands in octahedral directions, the d-orbitals will split into a set of two higher energy e_g (d_{x² - y²}, d_z²) orbitals and another set of three lower energy t_2g (d_xy, d_yz, d_zx) orbitals. The electronic configuration of the central metal ion in a coordination entity  t_2g³ e_g¹ indicates that the fourth electron enters one of the e_g orbitals, since, ∆_o < P. As there are four unpaired electrons in the complex, it is a high spin complex molecule. The crystal field splitting diagram is shown below:

Answer 2

(a) The colour of coordination compound depends upon the type of ligand and dd transition taking place .

H₂O is weak field ligand , which causes small splitting , leading to the d-d transition corresponding green colour , however due to the presence of (en) which ia strong field ligand , the splitting is increased . Due to the change in t_2g - e_g splitting the colouration of the compound changes from green to blue.

(b) Formula of the compound is [CO(H₂NCH₂CH₂NH₂)₃]₂ (SO₄)₃

The hybridisation of the compound is: d²sp³

OR

(a) As the fourth electron enters one of the e_g orbitals giving the configuration t₂g³ e_g¹, which indicates ∆o < P hence forms high spin complex.

(b)