Question

02: A-1) What is the type of hybridization in the following complexes 2) State the electronic absorption spectra, which is expected from the following complexes: [Colen):NH刈SO4 ; K2[Ni(CN)4] 3) Give brief explanation for the following c) Bond order and bond type B- Draw the correlation diagram for d configuration as free ion and strong field. a) Lanthanide contraction b) Trans effect

Answer 1

1. a. [Zn(NH₃)₄](BF₄)₂ = [Zn(NH₃)₄]²⁺.2BF₄^-

The central metal and its oxidation state are Zn(II), which has a d¹⁰ configuration.

The coordination no. of Zn(II) = 4

So, the possible hybridization is either sp³ or dsp²

Here, the d-orbitals in Zn(II) are completely filled and that too the NH₃ ligands are of weak filed.

Hence, the type of hybridization in [Zn(NH₃)₄](BF₄)₂ = sp³

b. [Ag(NH₃)₂]₂SO₄ = 2[Ag(NH₃)₂]⁺.SO₄^2-

The central metal and its oxidation state are Ag(I), which has a d¹⁰ configuration.

The coordination no. of Ag(I) = 2

So, the type of hybridization in ​[Ag(NH₃)₂]₂SO₄ = sp

c. [Fe(CN)₅(NH₃)]^3-

The central metal and its oxidation state are Fe(II), which has a d⁶ configuration.

The coordination no. of Fe(II) = 6

So, the possible hybridization is either sp³d² or d²sp³

Here, the CN^-ligands are of strong filed.

Hence, the type of hybridization in [Fe(CN)₅(NH₃)]^3- = d²sp³

d. [Cu(en)₃]SO₄ = [Cu(en)₃]²⁺.SO₄^2-

The central metal and its oxidation state are Cu(II), which has a d⁹ configuration.

The coordination no. of Cu(II) = 6

Note: The ligands 'en' (= ethylenediamine) are bidentate N-donors, i.e. 3*2 = 6.

So, the possible hybridization is either sp³d² or d²sp³

Here, the 'en'ligands are of strong filed but there are no two empty d-orbitals.

Hence, the type of hybridization in [Cu(en)₃]²⁺ = sp³d²