Chapter 5: Coordination Compounds

Ans: The structures of optical isomers are given below:

In total, three isomers are possible.

Ans: The structures of isomers are given below:

Trans-isomer is optically inactive 

Ans: The structures of all the isomers are given below:

Ans: There are three isomers possible:

No one will display optical isomers from the given isomers. Optical isomerization of tetrahedral compounds is uncommon. You only accomplish this in the presence of unsymmetric chelating substances.

Ans: Aqueous $\text{CuS}{{\text{O}}_{\text{4}}}$ solution exists as $\text{ }\!\![\!\!\text{ Cu(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ S}{{\text{O}}_{\text{4}}}$ which has blue color due to ${{\text{ }\!\![\!\!\text{ Cu(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$ ions.

(i) a green precipitate with aqueous potassium fluoride, and

When KF is added, the weak ${{\text{H}}_{\text{2}}}\text{O}$ ligand is replaced by ${{\text{F}}^{\text{-}}}$ ligands forming ${{\text{ }\!\![\!\!\text{ Cu}{{\text{F}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ ions which is a green precipitate. The reaction is given below:

\[{{\text{ }\!\![\!\!\text{ Cu(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\text{ + 4}{{\text{F}}^{\text{-}}}\text{ }\to \text{  }\!\![\!\!\text{ Cu(F}{{\text{)}}_{\text{4}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}\text{ + 4}{{\text{H}}_{\text{2}}}\text{O}\]

(ii) a bright green solution with aqueous potassium chloride

Explain these experimental results.

When KCl is added, $\text{C}{{\text{l}}^{\text{-}}}$ ligand replace the weak ${{\text{H}}_{\text{2}}}\text{O}$ ligands forming ${{\text{ }\!\![\!\!\text{ CuC}{{\text{l}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ ion which has bright green color. The reaction is given below:\[{{\text{ }\!\![\!\!\text{ Cu(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\text{ + 4C}{{\text{l}}^{\text{-}}}\text{ }\to \text{  }\!\![\!\!\text{ Cu(Cl}{{\text{)}}_{\text{4}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}\text{ + 4}{{\text{H}}_{\text{2}}}\text{O}\] 

Ans: The reactions are given below:\[\text{CuS}{{\text{O}}_{\text{4}}}\text{(aq) + 4KCN(aq) }\to \text{ }{{\text{K}}_{\text{2}}}\text{ }\!\![\!\!\text{ Cu(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ (aq) + }{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{(aq)}\] 

i.e., ${{\text{ }\!\![\!\!\text{ Cu(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\text{ + 4C}{{\text{N}}^{\text{-}}}\text{ }\to \text{  }\!\![\!\!\text{ Cu(CN}{{\text{)}}_{\text{4}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}\text{ + 4}{{\text{H}}_{\text{2}}}\text{O}$ 

Thus, the process-formed coordination entity is ${{\text{K}}_{\text{2}}}\text{ }\!\![\!\!\text{ Cu(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }$, which is highly stable, which does not ionise to add $\text{C}{{\text{u}}^{\text{2+}}}$ ions to water. Therefore when the ${{\text{H}}_{\text{2}}}\text{S}$ (g) passes through the solution, $\text{C}{{\text{u}}^{\text{2+}}}$ ions are not precipitated.

Ans: In ${{\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{4-}}$ the oxidation state of Fe is in +2 oxidation state, which will cause the configuration of Fe as ${{\text{d}}^{\text{6}}}$.

Since $\text{C}{{\text{N}}^{\text{-}}}$ is a strong field ligand, it causes unpaired electrons to be paired. This will lead to ${{\text{d}}^{\text{2}}}\text{s}{{\text{p}}^{\text{3}}}$ hybridization. This is shown below:

Therefore, the geometry of the complex is octahedral and it is a diamagnetic complex.

Ans: In ${{\text{ }\!\![\!\!\text{ Fe(F}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}$ the oxidation state of Fe is in +3 oxidation state, which will cause the configuration of Fe as ${{\text{d}}^{5}}$.

Since ${{\text{F}}^{\text{-}}}$ is a weak field ligand, it does not cause pairing of electrons. This will lead to $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybridization. This is shown below:

The geometry of the complex is octahedral and it is paramagnetic.

Ans: In ${{\text{ }\!\![\!\!\text{ Co(}{{\text{C}}_{2}}{{\text{O}}_{4}}{{\text{)}}_{3}}\text{ }\!\!]\!\!\text{ }}^{3-}}$ the oxidation state of Co is in +3 oxidation state, which will cause the configuration of Co as ${{\text{d}}^{\text{6}}}$.

Since oxalate is a strong field ligand, it causes unpaired electrons to be paired. This will lead to ${{\text{d}}^{\text{2}}}\text{s}{{\text{p}}^{\text{3}}}$ hybridization. This is shown below:

Therefore, the geometry of the complex is octahedral and it is a diamagnetic complex.

Ans: In ${{\text{ }\!\![\!\!\text{ Co}{{\text{F}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}$ the oxidation state of Co is in +3 oxidation state, which will cause the configuration of Co as ${{\text{d}}^{6}}$.

Since ${{\text{F}}^{\text{-}}}$ is a weak field ligand, it does not cause pairing of electrons. This will lead to $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybridization.

The geometry of the complex is octahedral and it is paramagnetic.

Ans: We know that there are five subshells of d-orbital, i.e., ${{\text{d}}_{\text{xy}}}\text{, }{{\text{d}}_{\text{yz}}}\text{, }{{\text{d}}_{\text{zx}}}\text{, }{{\text{d}}_{{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}}}\text{ and }{{\text{d}}_{{{\text{z}}^{\text{2}}}}}$. When the energy is provided, the splitting occurs in such a way that ${{\text{d}}_{{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}}}\text{ and }{{\text{d}}_{{{\text{z}}^{\text{2}}}}}$ experience the rise in energy so, they form ${{\text{e}}_{\text{g}}}$ level, while ${{\text{d}}_{\text{xy}}}\text{, }{{\text{d}}_{\text{yz}}}\text{ and }{{\text{d}}_{\text{zx}}}$ experience the low energy level so, they form ${{\text{t}}_{\text{2}}}\text{g}$ level. This is shown below:

Ans: A spectrochemical set of ligands is the organisation of their crystal field splitting energy (CFSE) values in the ascending order. The ligands on the R.H.S are strong ligands of fields whereas L.H.S is weak ligands of field. In d'orbitals, strong field ligands are more divided than weak field ligands. The series is given below:

${{\text{I}}^{\text{-}}}$ < $\text{B}{{\text{r}}^{\text{-}}}$ < ${{\text{S}}^{\text{2-}}}$ < $\text{SC}{{\text{N}}^{\text{-}}}$ < $\text{C}{{\text{l}}^{\text{-}}}$ < ${{\text{F}}^{\text{-}}}$ < $\text{O}{{\text{H}}^{\text{-}}}$ < ${{\text{C}}_{\text{2}}}\text{O}_{\text{4}}^{\text{2-}}$ < ${{\text{O}}^{\text{2-}}}$ < ${{\text{H}}_{\text{2}}}\text{O}$ < $\text{NC}{{\text{S}}^{\text{-}}}$ < $\text{N}{{\text{H}}_{\text{3}}}$ < en < $\text{NO}_{\text{2}}^{\text{-}}$ < $\text{C}{{\text{N}}^{\text{-}}}$ < CO

Ans: D-orbitals degenerate into two levels (${{\text{e}}_{\text{g}}}$ and ${{\text{t}}_{\text{2}}}\text{g}$) in the presence of ligands in the spherical field surroundings. The separation of degenerated levels by the presence of ligands is called the splitting of the crystal field, and the energy difference between two levels (${{\text{e}}_{\text{g}}}$ and ${{\text{t}}_{\text{2}}}\text{g}$) is called the splitting energy of the crystal field. The name is ${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{0}}}$.

The electrons are filled once the orbitals have divided. The filling of the fourth electron is done in two different methods after 1 electron (each) has been filled in 3 ${{\text{t}}_{\text{2}}}\text{g}$ orbitals.

The electron can be either in the ${{\text{e}}_{\text{g}}}$ orbital (giving birth to ${{\text{t}}_{\text{2}}}{{\text{g}}^{\text{3}}}\text{, }{{\text{e}}_{\text{g}}}^{\text{1}}$ similar electrical configuration) or in the ${{\text{t}}_{\text{2}}}\text{g}$ orbital (giving rise to ${{\text{t}}_{\text{2}}}{{\text{g}}^{4}}\text{, }{{\text{e}}_{\text{g}}}^{0}$ like electronic configuration). The electrons enter the ${{\text{e}}_{\text{g}}}$ orbital when the ${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{0}}}$ value of the ligand is lower than pairing energy (p).

On the other hand, if a ligand's ${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{0}}}$ value exceeds the pairing energy (P), the electrons then enter ${{\text{t}}_{\text{2}}}\text{g}$ orbital.

Ans: In ${{\text{ }\!\![\!\!\text{ Cr(N}{{\text{H}}_{3}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$ the oxidation state of Cr is in +3 oxidation state, which will cause the configuration of Cr as ${{\text{d}}^{3}}$.

Since $\text{N}{{\text{H}}_{\text{3}}}$ is a weak field ligand, it does not cause pairing of electrons. This will lead to ${{\text{d}}^{\text{2}}}\text{s}{{\text{p}}^{\text{3}}}$ hybridization. 

The geometry of the complex is octahedral and it is paramagnetic.

In ${{\text{ }\!\![\!\!\text{ Ni(CN}{{\text{)}}_{4}}\text{ }\!\!]\!\!\text{ }}^{2-}}$ the oxidation state of Ni is in +2 oxidation state, which will cause the configuration of Ni as ${{\text{d}}^{8}}$.

Since $\text{C}{{\text{N}}^{\text{-}}}$ is a strong field ligand, it causes unpaired electrons to be paired. This will lead to $\text{ds}{{\text{p}}^{2}}$ hybridization. This is shown below:

Therefore, the geometry of the structure is square planar and it is diamagnetic.

Ans: ${{\text{ }\!\![\!\!\text{ Ni(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$, the field ligand ${{\text{H}}_{\text{2}}}\text{O}$ is weak. Unpaired electrons are therefore available in $\text{N}{{\text{i}}^{\text{2+}}}$. In this complex d electrons can be stimulated to the higher energy level from the lower energy level, i.e. there is a potential to change d to d-orbital. Therefore, it has color. The ${{\text{ }\!\![\!\!\text{ Ni(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ have a strong ligand in the form of $\text{C}{{\text{N}}^{\text{-}}}$. As a result, the transition to d-d in ${{\text{ }\!\![\!\!\text{ Ni(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ is not feasible. It's colorless, therefore.