Co-ordination Compounds MCQs for NEET — Chemistry Questions with Answers

$ [Cr(C_2O_4)_3]^{3-} $ shows optical isomerism. This is because the oxalate ligands (C_2O_4) can form chiral complexes with chromium, leading to non-superimposable mirror images (optical isomers).

The coordination compound $[Cr(NH_3)_6]^{3+} [Co(CN)_6]^{3-}$ is an example of coordination isomerism. Coordination isomerism occurs when there is an interchange of ligands between cationic and anionic entities of different metal ions within the same complex. Here, $[Co(NH_3)_6]^{3+}$ and $[Cr(CN)_6]^{3-}$ can interchange their ligands to form $[Cr(NH_3)_6]^{3+} [Co(CN)_6]^{3-}$.

According to Werner’s theory, primary valencies correspond to the oxidation state of the central metal ion and are ionizable, whereas secondary valencies correspond to the coordination number and are non-ionizable. Thus, primary valencies are ionizable.

The complex formation with $Ni^{2+}$ depends on the nature of the ligands. Chloride ions ($Cl^-$) typically form tetrahedral complexes with $Ni^{2+}$. Cyanide ions ($CN^-$) form square planar complexes due to the strong field effect. Water ($H_2O$) forms octahedral complexes because it is a weak field ligand. Thus, the correct geometrical shapes for the complexes $[NiCl_4]^{2-}$, $[Ni(CN)_4]^{2-}$, and $[Ni(H_2O)_6]^{2+}$ are tetrahedral, square planar, and octahedral respectively.

The complex $[Cr(NH_3)_6]Cl_3$ involves $d^2sp^3$ hybridization, making it an inner orbital complex (using inner d-orbitals). The shape of the complex is octahedral. It is paramagnetic due to the presence of unpaired electrons. The complex also gives a white precipitate with silver nitrate solution due to the presence of chloride ions. Therefore, the incorrect statement is that it is an outer orbital complex.

The magnetic moment of a complex can be calculated using the formula $ ext{Magnetic Moment} = ext{BM} = ext{n(n+2)}$, where $n$ is the number of unpaired electrons. For $[NiCl_4]^{2-}$, Ni is in the +2 oxidation state with a configuration of $3d^8$. In a tetrahedral field, this results in 2 unpaired electrons. Thus, the magnetic moment is $ ext{BM} = ext{2(2+2)} = 2.82$ BM.

The field strength of ligands is determined by their ability to split the d-orbitals of the central metal ion. In the spectrochemical series, the increasing order of field strength for the given ligands is: $NH_3 < en < CN^- < CO$. This order is based on the fact that $CO$ is a strong field ligand, followed by $CN^-$, then $en$ (ethylenediamine), and $NH_3$ is the weakest field ligand among the given options.

The magnetic moment is related to the number of unpaired electrons in the complex. The formula for spin-only magnetic moment (μ) is given by: $ ext{μ} = ext{BM} = ext{n(n+2)}^{1/2}$, where n is the number of unpaired electrons. For [NiCl_4]^{2-}, Ni(II) has a 3d^8 configuration with 2 unpaired electrons, leading to a magnetic moment of approximately 2.82 BM.

In $Cr(CO)_6$, the chromium is in the zero oxidation state (Cr^0) and has a 3d^6 4s^0 configuration. The strong field ligand CO causes pairing of all the electrons in the d-orbitals, resulting in no unpaired electrons. Therefore, the spin-only magnetic moment is 0 BM.