Coordination Compounds · Question 8

The total number of all possible isomers for the square planar complex with formula K[M(NCS)(NO2)(gly)] is . (M = metal ion and gly = NH2CH2COO-)

Consider that the coordinating atoms of the ligands in cis-[Co(NH3)4Cl2]Cl and mer-[Co(NH3)3Cl3] octahedral complexes are at the vertices of an octahedron. The sum of total number of the triangular faces in both the complexes having one N atom and two Cl atoms at their corners is .

The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

1. [Co(CN)6]3- < [Co(NH3)6]3+ < [Co(NH3)5(H2O)]3+ < [Co(NH3)5(Cl)]2+
2. [Co(NH3)5(Cl)]2+ < [Co(NH3)5(H2O)]3+ < [Co(NH3)6]3+ < [Co(CN)6]3-
3. [Co(CN)6]3- < [Co(NH3)5(Cl)]2+ < [Co(NH3)5(H2O)]3+ < [Co(NH3)6]3+
4. [Co(NH3)6]3+ < [Co(CN)6]3- < [Co(NH3)5(Cl)]2+ < [Co(NH3)5(H2O)]3+

The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br)6]3- and [Mn(CN)6]3- is .

Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other. [en = H2NCH2CH2NH2]

1. Set-I: [Ni(CO)4] and [PdCl2(PPh3)2] Set-II: [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl
2. Set-I: [Co(en)(NH3)2Cl2] and [PdCl2(PPh3)2] Set-II: [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]
3. Set-I: [Co(NH3)3(NO2)3] and [Co(en)2Cl2] Set-II: [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl
4. Set-I: [Cr(NH3)5Cl]Cl2 and [Co(en)(NH3)2Cl2] Set-II: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2 H2O

Among V(CO)6, Cr(CO)5, Cu(CO)3, Mn(CO)5, Fe(CO)5, [Co(CO)3]3-, [Cr(CO)4]4-, and Ir(CO)3, the total number of species isoelectronic with Ni(CO)4 is . [Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]

Among the following complexes, the total number of diamagnetic species is . [Mn(NH3)6]3+, [MnCl6]3-, [FeF6]3-, [CoF6]3-, [Fe(NH3)6]3+, and [Co(en)3]3+ [Given, atomic number: Mn = 25, Fe = 26, Co = 27; en = H2NCH2CH2NH2]