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

$[MnO_4]^-$ has Mn in the +7 oxidation state. Manganese in this state has an electronic configuration of $[Ar]3d^04s^0$, meaning it has no 'd' electrons.

According to the spectrochemical series, $CN^-$ (cyanide) is a strong field ligand compared to $Br^-$, $HO^-$, and $F^-$. This means it can cause a larger splitting of the d-orbitals, making it the strongest ligand among the given options.

The most stable ion among the given options is $[Fe(CN)_6]^{3-}$. This is because cyanide (CN⁻) is a strong field ligand, which leads to a low-spin complex and greater stability due to the pairing energy being less than the crystal field splitting energy. In contrast, the other ligands such as oxalate (OX) and water (H₂O) are weaker field ligands, resulting in less stable complexes. Chloride (Cl⁻) is also a weaker field ligand compared to cyanide.

In the complex $[Ni(NH_3)_4]SO_4$, the nickel ion is coordinated to 4 ammonia (NH3) molecules. Ammonia is a neutral ligand, so it does not affect the oxidation state of nickel. The sulfate ion (SO4) has a charge of -2. Therefore, for the complex to be neutral, the nickel ion must have a charge of +2. Thus, the valency (oxidation state) of Ni is +2 and the coordination number (number of ligands attached to Ni) is 4.

Optical isomerism occurs when a compound can exist in two non-superimposable mirror images. In the given options, $[Cr(C_2O_4)_3]^{3-}$ can show optical isomerism because it has a chiral center due to the arrangement of the oxalate ligands around the chromium ion. The other compounds do not have such an arrangement that leads to optical isomerism.

In the extraction of gold, the roasted gold ore reacts with cyanide ions to form the complex $[Au(CN)_2]^-$ and hydroxide ions (OH^-). This complex then reacts with zinc to form $[Zn(CN)_4]^{2-}$ and releases gold (Au). Hence, the complexes are $X = [Au(CN)_2]^-$ and $Y = [Zn(CN)_4]^{2-}$.