Chapter 5: Coordination Compounds

Ans: The formula of Tetraamminediaquacobalt (III) chloride is $\text{ }\!\![\!\!\text{ Co(}{{\text{H}}_{\text{2}}}\text{O)(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ C}{{\text{l}}_{\text{3}}}$.

Ans: The formula of Potassium tetracyanonickelate(II) is ${{\text{K}}_{\text{2}}}\text{ }\!\![\!\!\text{ Ni(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }$. 

Ans: The formula of Tris(ethane-1,2-diamine) chromium(III) chloride is $\text{ }\!\![\!\!\text{ Cr(en}{{\text{)}}_{\text{3}}}\text{ }\!\!]\!\!\text{ C}{{\text{l}}_{\text{3}}}$ 

Ans: The formula of Amminebromidochloridenitrito-N-platinate(II) is ${{\text{ }\!\![\!\!\text{ Pt(N}{{\text{H}}_{\text{3}}}\text{)BrCl(N}{{\text{O}}_{\text{2}}}\text{) }\!\!]\!\!\text{ }}^{\text{-}}}$ 

Ans: The formula of Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate is $\text{ }\!\![\!\!\text{ PtC}{{\text{l}}_{\text{2}}}{{\text{(en)}}_{\text{2}}}\text{ }\!\!]\!\!\text{ (N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$ 

Ans: The formula of Iron(III) hexacyanoferrate(II) is $\text{F}{{\text{e}}_{\text{4}}}{{\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}_{\text{3}}}$

Ans: The IUPAC name of the compound is Hexaamminecobalt(III) chloride.

Ans: The IUPAC name of the compound is Pentaamminechloridocobalt(III) chloride.

Ans: The IUPAC name of the compound is Potassium hexacyanoferrate (III).

Ans: The IUPAC name of the compound is Potassium trioxalatoferrate(III).

Ans: The IUPAC name of the compound is Potassium tetrachloridopalladate(II).

Ans: The IUPAC name of the compound is Diamminechloride (methylamine) platinum(II) chloride.

Ans: Both geometrical isomers (cis and trans) are possible for this compound. These are given below:

The trans-isomers of this compound is optically inactive but the cis-isomer is optically active. Therefore, it can also show optical isomerism. This is given below:

Ans: This compound can show optical isomerism. The structure is given below:

Its optical isomers are given below:

Ans: It can show three types of isomerism.

A pair of optical isomer, these are given below:

It can show linkage isomerism. These are given below:

$\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{(N}{{\text{O}}_{\text{2}}}\text{) }\!\!]\!\!\text{ (N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ and  }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{(ONO) }\!\!]\!\!\text{ (N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$ 

It can also show ionization isomerism. These are given below:

$\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{(N}{{\text{O}}_{\text{2}}}\text{) }\!\!]\!\!\text{ (N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ and   }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{(N}{{\text{O}}_{\text{3}}}\text{) }\!\!]\!\!\text{ (N}{{\text{O}}_{\text{3}}}\text{)(N}{{\text{O}}_{\text{2}}}\text{)}$ 

Ans: Geometrical isomerism of this compound is possible. Both cis-form and trans-form are given below:

Ans: If isomers of ionization are ionized into water, various ionization isomers are ionized. These ions react differently to various products with different reagents. The compounds given in the questions are tested with $\text{AgN}{{\text{O}}_{\text{3}}}$ solution and $\text{BaC}{{\text{l}}_{\text{2}}}$ solution. The reactions are given below:\[\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{Cl }\!\!]\!\!\text{ S}{{\text{O}}_{\text{4}}}\text{(aq) + BaC}{{\text{l}}_{\text{2}}}\text{(aq) }\to \text{ }\underset{\text{White precipitate}}{\mathop{\text{BaS}{{\text{O}}_{\text{4}}}\downarrow }}\,\]

\[\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{Cl }\!\!]\!\!\text{ S}{{\text{O}}_{\text{4}}}\text{(aq) + AgN}{{\text{O}}_{\text{3}}}\text{(aq) }\to \text{ No reaction}\] \[\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\!\!]\!\!\text{ Cl(aq) + BaC}{{\text{l}}_{\text{2}}}\text{(aq) }\to \text{ No reaction}\]

\[\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\!\!]\!\!\text{ Cl(aq) + AgN}{{\text{O}}_{\text{3}}}\text{(aq) }\to \text{ }\underset{\text{White precipitate}}{\mathop{\text{AgCl}\downarrow }}\,\]

Ans: In ${{\text{d}}^{\text{8}}}$ configuration, Ni is in +2 oxidation state. This is shown below:

As we can see that in the question, there are 4 ligands in the compound which means it is a square planar complex.

Since, $\text{C}{{\text{N}}^{\text{-}}}$ is a strong field ligand and it will cause the pairing of the 3d electrons. This will lead to $\text{ds}{{\text{p}}^{\text{2}}}$ hybridization. It is given below:

Since, all the electrons are paired, ${{\text{ }\!\![\!\!\text{ Ni(CN}{{\text{)}}_{4}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ is diamagnetic complex.

Since, $\text{C}{{\text{l}}^{\text{-}}}$ is a weak field ligand and it will not cause the pairing of the 3d electrons. This will lead to $\text{s}{{\text{p}}^{3}}$ hybridization. It is given below:

Since, two electrons are unpaired, ${{\text{ }\!\![\!\!\text{ NiC}{{\text{l}}_{4}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ is paramagnetic complex.

Ans: As we can see that in the question, there are 4 ligands in the compound which means it is a tetrahedral complex.

Since, $\text{C}{{\text{l}}^{\text{-}}}$ is a strong field ligand and it will not cause the pairing of the 3d electrons. This will lead to $\text{s}{{\text{p}}^{3}}$ hybridization. It is given below:

Since, two electrons are unpaired, ${{\text{ }\!\![\!\!\text{ NiC}{{\text{l}}_{4}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ is paramagnetic complex.

In $\text{ }\!\![\!\!\text{ Ni(CO}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }$, Ni is in the zero oxidation state i.e., it has a configuration of $\text{3}{{\text{d}}^{\text{8}}}\text{ 4}{{\text{s}}^{\text{2}}}$.

But CO is a strong field ligand. This leads unpaired 3d electrons to be coupled with one another. It also leads the 4s to go to the 3d orbital, which results in a hybridization of $\text{s}{{\text{p}}^{\text{3}}}$. Since in this situation, $\text{ }\!\![\!\!\text{ Ni(CO}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }$ is diamagnetic as no unpaired electrons are present.

Ans: In the given complexes, Fe is the central metal ion and in both the cases the oxidation state of Fe is +3. Therefore, it will be in ${{\text{d}}^{\text{5}}}$ configuration.

Since $\text{C}{{\text{N}}^{\text{-}}}$ is a strong field ligand, it causes unpaired electrons to be paired. There is just one unpaired electron in the d-orbital thus remaining.

So, we can calculate the magnetic moment as:

\[\text{ }\!\!\mu\!\!\text{  = }\sqrt{\text{n(n+2)}}\] 

\[\text{ }\!\!\mu\!\!\text{  = }\sqrt{\text{1(1+2)}}\] 

\[\text{ }\!\!\mu\!\!\text{  = }\sqrt{3}\]

\[\text{ }\!\!\mu\!\!\text{  =1}\text{.732 BM}\] 

Since ${{\text{H}}_{\text{2}}}\text{O}$ is a weak field ligand, it does not cause unpaired electrons to be paired. There are five unpaired electrons in the d-orbital.

So, we can calculate the magnetic moment as:

\[\text{ }\!\!\mu\!\!\text{  = }\sqrt{\text{n(n+2)}}\] 

\[\text{ }\!\!\mu\!\!\text{  = }\sqrt{\text{5(5+2)}}\] 

\[\text{ }\!\!\mu\!\!\text{  = }\sqrt{35}\]

\[\text{ }\!\!\mu\!\!\text{  =5}\text{.91 BM}\]

Therefore, ${{\text{ }\!\![\!\!\text{ Fe(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$ is strongly paramagnetic while ${{\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$ is weakly paramagnetic complex.

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

Since $\text{N}{{\text{H}}_{\text{3}}}$ 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, it is an inner orbital complex.

In ${{\text{ }\!\![\!\!\text{ Ni(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$ the oxidation state of Ni is in +2 oxidation state, which will cause the configuration of Ni as ${{\text{d}}^{8}}$.

Since $\text{N}{{\text{H}}_{\text{3}}}$ is a strong field ligand, it causes unpaired electrons to be paired. There is only one d-orbital left and it cannot form the ${{\text{d}}^{\text{2}}}\text{s}{{\text{p}}^{\text{3}}}$. Therefore, it will act as a weak field ligand. This will lead to $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybridization. This is shown below:

Therefore, it is an outer orbital complex.

Ans: The given complex is ${{\text{ }\!\![\!\!\text{ Pt(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$. In this complex the central metal ion is platinum and its oxidation state is +2. It forms a square planar structure. This means that it will have the hybridization of $\text{ds}{{\text{p}}^{\text{2}}}$. So, the electronic configuration Pt (+2) of is $\text{5}{{\text{d}}^{\text{8}}}$.

In the complex, $\text{C}{{\text{N}}^{\text{-}}}$ is a strong field ligand which will cause the pairing of all the unpaired electrons. Therefore, there will be no unpaired electrons.

Ans: In ${{\text{ }\!\![\!\!\text{ Mn(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$, the Mn is in +2 oxidation state and its electronic configuration is ${{\text{d}}^{\text{5}}}$. The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in ${{\text{ }\!\![\!\!\text{ Mn(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$ will be ${{\text{t}}_{\text{2}}}{{\text{g}}^{\text{3}}}\text{ e}{{\text{g}}^{\text{2}}}$.

In ${{\text{ }\!\![\!\!\text{ Mn(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$, the Mn is in +2 oxidation state and its electronic configuration is ${{\text{d}}^{\text{5}}}$. The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in ${{\text{ }\!\![\!\!\text{ Mn(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$ will be ${{\text{t}}_{\text{2}}}{{\text{g}}^{5}}\text{ e}{{\text{g}}^{0}}$.