Coordination Chemistry

This option is correct because this complex is called as fischer salt. It is a yellow powder which decomposes at melting point 200 ⁰C and used in medicine as yellow pigment.

This option is correct because when sulphur dioxide is present in the air, a layer of basic sulphate is formed and it is called as bronchanite.

In octahedral complexes, CFSE is given by the formula: CFSE = value of t²g (n) + value of eg (n) Dq, Now, CFSE for low-spin d⁴ = -4 (4) + 6 (0) = -16 + 0 = -16 Dq CFSE for high-spin d⁴ = -4 (3) + 6 (1) = -12 + 6 = -6 Dq CFSE for low-spin d⁵ = -4 (5) + 6 (0) = -20 + 0 = -20 Dq CFSE for high-spin d⁵ = -4 (3) + 6 (2) = -12 + 12 = 0 Dq CFSE for low-spin d⁷ = -4 (5) + 6 (2) = -20 + 12 = -8 Dq Hence, low-spin d⁵ will give maximum CFSE in an octahedral complex.

In a weak field ligand, the filling of electrons is in accordance with Hund’s rule. The CFSE is given by the formula  CFSE = value of t²g (n) + value of eg (n) Dq, Fe³⁺ corresponds to the d⁵ configuration. Now, CFSE for Fe³⁺ (d⁵) = - 4 (3) + 6 (2) = - 12 + 12 = 0 Dq CFSE for Fe²⁺ (d⁶) = - 4 (4) + 6 (2) = - 16 + 12 = - 4 Dq CFSE for Cr²⁺ (d⁴) = - 4 (3) + 6 (1) = - 12 + 6 = - 6 Dq CFSE for Mn⁺ (d⁶) = - 4 (4) + 6 (2) = - 16 + 12 = - 4 Dq CFSE for Mn³⁺ (d4) = - 4 (3) + 6 (1) = - 12 + 6 = - 6 Dq Hence, the metal ion with zero CFSE is Fe³⁺ when it is associated with weak field ligand.

This option is correct, because the octahedral complex of the Co possess the magnetic moment 4.9 BM. It would be concluded that the Co is in a bivalent state and occupies 3d⁷ and the bonding is free spin types, corresponding to n = 3.

This option is correct because these complexes show the feature of the ionization isomers.

This option is correct, because the Co(II) octahedral complex has 4.0 BM magnetic moment when the Co(II) act as the high spin in nature. Therefore 3 unpaired electrons contribute in t_2g and one unpaired electron in e_g set. Therefore the magnetic moment  is 4.0 BM.

This option is correct, because the complexes have same nature but ligand themselves are capable of existing isomers.

This option is correct, because the Cu(I) and Hg(II) metals ions act as the minimum repulsion in between ligand to formed linear geometry of the complex.

This option is correct,  because in the both complexes ion Ni(28) is present as Ni⁺⁺ ions. The Ni atom has valence shell configuration 3d⁸4s² but Ni⁺⁺ ions indicate 3d⁸configuration. The magnetic measurement is complex of the ions paramagnetic it has two unpaired electron. They act as high spin in nature  and it gives rise to sp³d² hybridization.

This option is correct, because the octahedral high spin complex Ti exhibits trivalent state. It belongs to the 3d¹ orbital which split in to t_2g and ^eg set. The electron is istributed in ¹t_2g and ⁰e_g. Hence,  CFES = [- 0.4p + 0.6q] Dq + mP, here p = Number of electron in t_2g, q = Number of electron in e_g, set Dq = 20000 cm^-1 m = Number of paired electron, mean pairing energy. Calculation - [- 0.4 x 1 + 0.6 x 0] 20000 cm^-1 + 0 = - 0.4 x 20000 cm^-1 = - 8000 cm^-1

Fe²⁺: [Ar] 3d⁶; 4 unpaired electrons Among the given first row transition metal ions, Fe²⁺ has the highest number of unpaired electrons.

The magnetic moment increases with increasing number of unpaired electrons. Ti²⁺: [Ar] 3d²; 2 unpaired electrons V²⁺ : [Ar] 3d³; 3 unpaired electrons Cr²⁺: [Ar] 3d⁴; 4 unpaired electrons Mn²⁺: [Ar] 3d⁵; 5 unpaired electrons Hence, the correct order of magnetic moment is Ti²⁺ < V²⁺ < Cr²⁺ < Mn²⁺.

Sc³⁺ has d⁰ configuration. Because of lack of unpaired electrons, Sc³⁺ does not show paramagnetism. Zn²⁺ has d¹⁰ configuration. Because of lack of unpaired electrons, Zn²⁺ does not show paramagnetism.

V³⁺: [Ar] 3d²; 2 unpaired electrons Cr⁴⁺: [Ar] 3d²; 2 unpaired electrons