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

When acetic acid and potassium ferrocyanide (K₄[Fe(CN)₆]) are added to a copper salt, a chocolate precipitate of copper ferrocyanide (Cu₂[Fe(CN)₆]) is formed. This reaction is used in qualitative inorganic analysis to test for the presence of copper ions.

Sodium nitroprusside reacts with sulphide ions in an alkaline solution to produce a purple color. The reaction involved is:

$$ [Fe(CN)_5NO]^{2-} + S^{2-} ightarrow [Fe(CN)_5NOS]^{4-} $$

This complex ion $[Fe(CN)_5NOS]^{4-}$ is responsible for the purple coloration observed in the solution.

The ion $[Co(NH_3)_6]^{3+}$ is diamagnetic because cobalt in this complex is in the +3 oxidation state, which means it has an electronic configuration of $[Ar] 3d^6$. The ammonia ligands are strong field ligands, which cause pairing of all the electrons in the 3d orbitals, resulting in no unpaired electrons.

The compound $[CrCl_2(H_2O)_4]Cl.2H_2O$ is the most likely structure because when 1/3 of the total chlorine is precipitated, it suggests that one out of the three chlorine atoms is outside the coordination sphere and hence can react with $AgNO_3$ to form a precipitate of $AgCl$.

The complex $M(AA’)_2$ will be able to show cis-trans isomerism because it contains two different bidentate ligands (AA’), which can arrange themselves in different positions around the central metal atom, leading to the formation of cis and trans isomers.

The complex $K_3CoF_6$ is a high spin complex. Cobalt in this complex is in the +3 oxidation state, represented as $Co^{3+}$. In a high spin complex, the ligands (F- in this case) are weak field ligands, leading to minimal pairing of electrons. The electron configuration of $Co^{3+}$ is $[Ar] 3d^6$. The hybridization state of $Co^{3+}$ in such a high spin octahedral complex is $sp^3d^2$.

The complexes $[Co(en)_2(NCS)_2]Cl$ and $[Co(en)_2(NCS)Cl]NCS$ exhibit ionization isomerism. Ionization isomerism occurs when there is an exchange of anions between the coordination sphere and the counter ion. In this case, the anions NCS- and Cl- are exchanged between the inside and the outside of the coordination sphere.

For the complex $[X(SO_4)(NH_3)_4]Cl$, the coordination number is determined by the number of ligand donor atoms attached to the central metal ion. Here, $SO_4^{2-}$ is a bidentate ligand and $NH_3$ is monodentate. Thus, we have $1 imes 2 + 4 imes 1 = 6$. The oxidation number of X can be calculated as: Let the oxidation number of X be x. The overall charge of the complex ion $[X(SO_4)(NH_3)_4]^+$ is +1. Therefore, x + (-2) + 4(0) = +1, solving this gives x = +3. Hence, the coordination number is 6 and the oxidation number is 3.