Question 14 Marks
Explain colour in coordination compounds.
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Question 24 Marks
Discuss the nature of bonding in$\left.[CoF _6\right]^{3-}$ coordination entities on the basis of valence bond theory.
Answer
View full question & answer→→ In $\left[ CoF _6\right]^{3-}$ oxidation state of cobalt is +3
→ Electronic configuration of cobalt in ground state.
→ Electronic configuration of cobalt in +3 oxidation state.
→ F- is weak ligand so 3d orbital does not take place in hybridization
→ In this complex 4d orbital hybridize. So sp3d2 hybridization occurs.
→ The paramagnetic octahedral complex, $\left[ CoF _6\right]^{3-}$ uses outer orbital ( 4 d$)$ in hybridization ( $sp ^3 d^2$ ), It is thus called outer orbital or high spin of spin free complex.
$
\begin{aligned}
\text → { Magnetic Moment } (\mu) & =\sqrt{n(n+2)} \\
&=\sqrt{4(4+2)} \\
& =\sqrt{24} \text { B.M. }
\end{aligned}
$
→ Electronic configuration of cobalt in ground state.
→ Electronic configuration of cobalt in +3 oxidation state.
→ F- is weak ligand so 3d orbital does not take place in hybridization
→ In this complex 4d orbital hybridize. So sp3d2 hybridization occurs.
→ The paramagnetic octahedral complex, $\left[ CoF _6\right]^{3-}$ uses outer orbital ( 4 d$)$ in hybridization ( $sp ^3 d^2$ ), It is thus called outer orbital or high spin of spin free complex.
$
\begin{aligned}
\text → { Magnetic Moment } (\mu) & =\sqrt{n(n+2)} \\
&=\sqrt{4(4+2)} \\
& =\sqrt{24} \text { B.M. }
\end{aligned}
$
Question 34 Marks
Discuss the nature of bonding in [Co(NH3)6]3+ coordination entities on the basis of valence
bond theory
bond theory
Answer
View full question & answer→→ In [Co(NH3)6]3+ ion oxidation number of cobalt is +3.
→ Electronic configuration of cobalt in ground state.
→ Electronic configuration of cobalt in +3 oxidation state.
→ NH3 is strong ligand so electrons get paired in d-orbital and d2sp3 hybridization occurs.
six d²sp³ hybrid orbitals
→ Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals.
→ Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron.
→ In the formation of this complex, since the inner d orbital (3d) is used in hybridization, the complex, $\left[ Co \left( NH _3\right)_6\right]^{3+}$ is called an inner orbite or low spin or spin paired complex.
→ Electronic configuration of cobalt in ground state.
→ Electronic configuration of cobalt in +3 oxidation state.
→ NH3 is strong ligand so electrons get paired in d-orbital and d2sp3 hybridization occurs.
six d²sp³ hybrid orbitals
→ Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals.
→ Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron.
→ In the formation of this complex, since the inner d orbital (3d) is used in hybridization, the complex, $\left[ Co \left( NH _3\right)_6\right]^{3+}$ is called an inner orbite or low spin or spin paired complex.
Question 44 Marks
Explain geometrical isomerism in octahedral Complex.
Answer
View full question & answer→→ This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
→ Important examples of these behaviour are found with coordination numbers 4 and 6.
→ In a square planar complex of formula [MX2L2] (X and L are unidentate), the two ligands X may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer.
Geometrical isomers (cis and trans) of Pt[(NH3)2Cl2]
→ Other square planar complex of the type MABXL (where A, B, X, L are unidentates) shows three isomers-two cis and one trans.
→ Such isomerism is not possible for a tetrahedral geometry.
→ In octahedral complexes of formula [MX2L4] in which the two ligands X may be oriented cis or trans to each other
Geometrical isomers (cis and trans) of [Co(NH3)4Cl2]+
→ This type of isomerism also arises when didentate ligands L-L [e.g., en ] are present in complexes of formula [MX2(L-L)2]
Geometrical isomers (cis and trans) of [CoCl2(en)2]
→ Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Co(NH3)3(NO2)3].
→ If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, it forms the facial (fac) isomer.
→ When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.
The facial (fac) and meridional (mer) isomers of [Co(NH3)3(NO2)3]
→ Important examples of these behaviour are found with coordination numbers 4 and 6.
→ In a square planar complex of formula [MX2L2] (X and L are unidentate), the two ligands X may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer.
Geometrical isomers (cis and trans) of Pt[(NH3)2Cl2]
→ Other square planar complex of the type MABXL (where A, B, X, L are unidentates) shows three isomers-two cis and one trans.
→ Such isomerism is not possible for a tetrahedral geometry.
→ In octahedral complexes of formula [MX2L4] in which the two ligands X may be oriented cis or trans to each other
Geometrical isomers (cis and trans) of [Co(NH3)4Cl2]+
→ This type of isomerism also arises when didentate ligands L-L [e.g., en ] are present in complexes of formula [MX2(L-L)2]
Geometrical isomers (cis and trans) of [CoCl2(en)2]
→ Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Co(NH3)3(NO2)3].
→ If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, it forms the facial (fac) isomer.
→ When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.
The facial (fac) and meridional (mer) isomers of [Co(NH3)3(NO2)3]
Question 54 Marks
What is ligand? Explain classification of ligand.
Answer
View full question & answer→→ "Ligands are atom or ions which donates electron pairs to the central metal ion."
(OR)
→ "The ions or molecules bound to the central atom/ion in the coordination entity are called ligands."
→ These may be simple ions such as Cl, small molecules such as H2O or NH3, larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules, such as proteins.
Classification of Ligands:
(1) Unidentate ligands:
→ "When a ligand is bound to a metal ion through a single donor atom, the ligand is said to be unidentate."
Example :
(a) Neutral: $H _2 \ddot{ O }$, : $NH _3,: CO ,: NO , CH _3 \ddot{N}H_2, C _5 H _5\ddot{N}$ (py)
(b) Negative ion: ${ }^{-} OH , F ^{-}, Cl ^{-}, Br ^{-}, I^{-} ,{ }^{-} CN , {}^{-}NH_2,$ $NO _3^{-}, NO _2^{-}, NCH _3 COO ^{-}\left( AcO ^{-}\right), O ^{2-}, S ^2, N^{3-}$
(2) Didentate ligands:
→ "When a ligand is bound to a metal ion through a two donor atom, the ligand is said to be didentate ".
(3) Tridentate ligands:
→ "When a ligand is bound to a metal ion through a three donor atom, the ligand is said to be Tridentate,"
Example:
(4) Hexadentate ligands:
→ When a ligand is bound to a metal ion through a six donor atom, the ligand is said to be hexadendate
Example: Ethylenediaminetetraacetate ion $\left( EDTA ^{4-}\right)$
(OR)
→ "The ions or molecules bound to the central atom/ion in the coordination entity are called ligands."
→ These may be simple ions such as Cl, small molecules such as H2O or NH3, larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules, such as proteins.
Classification of Ligands:
(1) Unidentate ligands:
→ "When a ligand is bound to a metal ion through a single donor atom, the ligand is said to be unidentate."
Example :
(a) Neutral: $H _2 \ddot{ O }$, : $NH _3,: CO ,: NO , CH _3 \ddot{N}H_2, C _5 H _5\ddot{N}$ (py)
(b) Negative ion: ${ }^{-} OH , F ^{-}, Cl ^{-}, Br ^{-}, I^{-} ,{ }^{-} CN , {}^{-}NH_2,$ $NO _3^{-}, NO _2^{-}, NCH _3 COO ^{-}\left( AcO ^{-}\right), O ^{2-}, S ^2, N^{3-}$
(2) Didentate ligands:
→ "When a ligand is bound to a metal ion through a two donor atom, the ligand is said to be didentate ".
(3) Tridentate ligands:
→ "When a ligand is bound to a metal ion through a three donor atom, the ligand is said to be Tridentate,"
Example:
(4) Hexadentate ligands:
→ When a ligand is bound to a metal ion through a six donor atom, the ligand is said to be hexadendate
Example: Ethylenediaminetetraacetate ion $\left( EDTA ^{4-}\right)$
Question 64 Marks
Explain magnetic property of coordination compound.
Answer
View full question & answer→→ Metal ions with upto three electrons in the d orbitals, like $Ti ^{3+}\left( d ^1\right) ; V ^{3+}\left( d ^2\right) ; Cr ^{3+}\left( d ^3\right) ;$ tuo vacant $d$ orbitals are available for octahed $d^3$ hybridization with 4 s and 4 p orbitals.
→ The magnetic behaviour of these free jons and their coordination entities is similar.
→ When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund's rule).
→ Thus, for $d ^4\left( Cr ^{2+}, Mn ^{3+}\right), d ^5\left( Mn ^{2+}, Fe ^{3+}\right), d ^6$ $\left( Fe ^{2+}, Co ^{3+}\right)$ cases, a vacant pair of d orbitals results only by pairing of 3 d electrons which leaves two, one and zero unpaired electrons, respectively.
→ The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and d5 ions there are complications.
→$\left[ Mn ( CN )_6\right]^{3-}$ has magnetic moment of two unpaired electrons while $\left[ MnCl _6\right]^{3-}$ has a paramagnetic moment of four unpaired electrons.
→ $\left[ Fe ( CN )_6\right]^{3-}$ has magnetic moment of a single unpaired electron while $\left[ FeF _6\right]^{3-}$ has a paramagnetic moment of five unpaired electrons.
→ $\left[ CoF _6\right]^{3-}$ is paramagnetic with four unpaired electrons while $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ is diamagnetic.
→ This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities.
→ $\left[ Mn ( CN )_6\right]^{3-},\left[ Fe ( CN )_6\right]^{3-}$ and $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ are inner orbital complexes involving $d ^2 sp ^3$ hybridisation, the former two complexes are paramagnetic and the latter diamagnetic.
→ On the other hand, $\left[ MnCl _6\right]^{3-},\left[ FeF _6\right]^{3-}$ and $\left[ CoF _6\right]^{3-}$ are outer orbital complexes involving $sp ^3 d^2$ hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons.
→ The magnetic behaviour of these free jons and their coordination entities is similar.
→ When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund's rule).
→ Thus, for $d ^4\left( Cr ^{2+}, Mn ^{3+}\right), d ^5\left( Mn ^{2+}, Fe ^{3+}\right), d ^6$ $\left( Fe ^{2+}, Co ^{3+}\right)$ cases, a vacant pair of d orbitals results only by pairing of 3 d electrons which leaves two, one and zero unpaired electrons, respectively.
→ The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and d5 ions there are complications.
→$\left[ Mn ( CN )_6\right]^{3-}$ has magnetic moment of two unpaired electrons while $\left[ MnCl _6\right]^{3-}$ has a paramagnetic moment of four unpaired electrons.
→ $\left[ Fe ( CN )_6\right]^{3-}$ has magnetic moment of a single unpaired electron while $\left[ FeF _6\right]^{3-}$ has a paramagnetic moment of five unpaired electrons.
→ $\left[ CoF _6\right]^{3-}$ is paramagnetic with four unpaired electrons while $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ is diamagnetic.
→ This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities.
→ $\left[ Mn ( CN )_6\right]^{3-},\left[ Fe ( CN )_6\right]^{3-}$ and $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ are inner orbital complexes involving $d ^2 sp ^3$ hybridisation, the former two complexes are paramagnetic and the latter diamagnetic.
→ On the other hand, $\left[ MnCl _6\right]^{3-},\left[ FeF _6\right]^{3-}$ and $\left[ CoF _6\right]^{3-}$ are outer orbital complexes involving $sp ^3 d^2$ hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons.
Question 74 Marks
Explain different types of Structural isomerism with example.
Answer
View full question & answer→(1) Linkage Isomerism:
→ Linkage isomerism arises in a coordination compound containing ambidentate ligand.
→ A simple example is provided by complexes containing the thiocyanate ligand, NCS-, which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.
→ Jorgenson discovered such behaviour in the complex $\left[ Co \left( NH _3\right)_5\left( NO _2\right)\right] Cl _2$, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( - ONO ), and as the yellow form, in which the nitrite ligand is bound through nitrogen $\left(- NO _2\right)$.
$\begin{array}{l}\text { e.g. : }\left[ Co ( ONO )\left( NH _3\right)_5\right]^{2+} \text { and }\left[ Co \left( NO _2\right)\left( NH _3\right)_5\right]^{2+} \\ \quad \quad \quad \quad \text{Red} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\text{Yellow} \\\end{array}$
(2) Coordination Isomerism:
→ This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
e.g. : $\left[ Co \left( NH _3\right)_6\right]\left[ Cr ( CN )_6\right]$, and $\left[ Cr \left( NH _3\right)_6\right][ Co ( CN )_6$]
(3) Ionisation Isomerism:
→ This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
e.g. : $\begin{aligned} & \left.\left. [Co \left( NH _3\right)_5\right]\left( SO _4\right)\right] Br \text { and }\left[ Cr \left( NH _3\right)_5 Br \right] SO _4 \\ & {\left[ Pt \left( NH _3\right)_4 Cl _2\right] Br _2 \text { and }\left[ Pt \left( NH _3\right)_4 Br _2\right] Cl _2 } \\ & {\left[ Cr \left( NH _3\right)_4 Cl _2\right] NO _2 \text { and }\left[ Cr \left( NH _3\right)_4 Cl \cdot NO _2\right] Cl }\end{aligned}$
(4) Solvate Isomerism OR Hydrate isomerism:
→ This form of isomerism is known as 'hydrate isomerism' in case where water is involved as a solvent.
→ This is similar to ionization isomerism.
→ Three isomeric forms of $CrCl _3 \cdot 6 H _2 O$ are known
(1) $\left[ Cr \left( H _2 O \right)_6\right] Cl _3$ (violet).
(2) $\left[ Cr \left( H _2 O \right)_5 Cl \right] Cl _2 \cdot H _2 O$ (grey green).
(3) $\left[ Cr \left( H _2 O \right)_4 Cl _2\right] Cl \cdot 2 H _2 O$ (green)
→ Linkage isomerism arises in a coordination compound containing ambidentate ligand.
→ A simple example is provided by complexes containing the thiocyanate ligand, NCS-, which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.
→ Jorgenson discovered such behaviour in the complex $\left[ Co \left( NH _3\right)_5\left( NO _2\right)\right] Cl _2$, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( - ONO ), and as the yellow form, in which the nitrite ligand is bound through nitrogen $\left(- NO _2\right)$.
$\begin{array}{l}\text { e.g. : }\left[ Co ( ONO )\left( NH _3\right)_5\right]^{2+} \text { and }\left[ Co \left( NO _2\right)\left( NH _3\right)_5\right]^{2+} \\ \quad \quad \quad \quad \text{Red} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\text{Yellow} \\\end{array}$
(2) Coordination Isomerism:
→ This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
e.g. : $\left[ Co \left( NH _3\right)_6\right]\left[ Cr ( CN )_6\right]$, and $\left[ Cr \left( NH _3\right)_6\right][ Co ( CN )_6$]
(3) Ionisation Isomerism:
→ This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
e.g. : $\begin{aligned} & \left.\left. [Co \left( NH _3\right)_5\right]\left( SO _4\right)\right] Br \text { and }\left[ Cr \left( NH _3\right)_5 Br \right] SO _4 \\ & {\left[ Pt \left( NH _3\right)_4 Cl _2\right] Br _2 \text { and }\left[ Pt \left( NH _3\right)_4 Br _2\right] Cl _2 } \\ & {\left[ Cr \left( NH _3\right)_4 Cl _2\right] NO _2 \text { and }\left[ Cr \left( NH _3\right)_4 Cl \cdot NO _2\right] Cl }\end{aligned}$
(4) Solvate Isomerism OR Hydrate isomerism:
→ This form of isomerism is known as 'hydrate isomerism' in case where water is involved as a solvent.
→ This is similar to ionization isomerism.
→ Three isomeric forms of $CrCl _3 \cdot 6 H _2 O$ are known
(1) $\left[ Cr \left( H _2 O \right)_6\right] Cl _3$ (violet).
(2) $\left[ Cr \left( H _2 O \right)_5 Cl \right] Cl _2 \cdot H _2 O$ (grey green).
(3) $\left[ Cr \left( H _2 O \right)_4 Cl _2\right] Cl \cdot 2 H _2 O$ (green)