Question 14 Marks
Explain colour in coordination compounds.
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Question 24 Marks
Explain magnetic property of coordination compound.
Answer
View full question & answer→$\rightarrow$ 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.
$\rightarrow$ The magnetic behaviour of these free jons and their coordination entities is similar.
$\rightarrow$ 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).
$\rightarrow$ 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.
$\rightarrow$ The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing $d^6$ ions. However, with species containing $d^4$ and $d^5$ ions there are complications.
$\rightarrow$ $\left[ Mn ( CN )_6\right]^{3-}$ has magnetic moment of two unpaired electrons while $\left[ MnCl _6\right]^{3-}$ has a paramagnetic moment of four unpaired electrons.
$\rightarrow$ $\left[ Fe ( CN )_6\right]^{3-}$ has magnetic moment of a single unpaired electron while $\left[ FeF _6\right]^{3-}$ has a paramagnetic moment of five unpaired electrons.
$\rightarrow$ $\left[ CoF _6\right]^{3-}$ is paramagnetic with four unpaired electrons while $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ is diamagnetic.
$\rightarrow$ This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities.
$\rightarrow$ $\left[ Mn ( CN )_6\right]^{3-},\left[ Fe ( CN )_6\right]^{3-}$ 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.
$\rightarrow$ 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.
$\rightarrow$ The magnetic behaviour of these free jons and their coordination entities is similar.
$\rightarrow$ 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).
$\rightarrow$ 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.
$\rightarrow$ The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing $d^6$ ions. However, with species containing $d^4$ and $d^5$ ions there are complications.
$\rightarrow$ $\left[ Mn ( CN )_6\right]^{3-}$ has magnetic moment of two unpaired electrons while $\left[ MnCl _6\right]^{3-}$ has a paramagnetic moment of four unpaired electrons.
$\rightarrow$ $\left[ Fe ( CN )_6\right]^{3-}$ has magnetic moment of a single unpaired electron while $\left[ FeF _6\right]^{3-}$ has a paramagnetic moment of five unpaired electrons.
$\rightarrow$ $\left[ CoF _6\right]^{3-}$ is paramagnetic with four unpaired electrons while $\left[ Co \left( C _2 O _4\right)_3\right]^{3-}$ is diamagnetic.
$\rightarrow$ This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities.
$\rightarrow$ $\left[ Mn ( CN )_6\right]^{3-},\left[ Fe ( CN )_6\right]^{3-}$ 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.
$\rightarrow$ 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 34 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→$\rightarrow$ In $\left[ CoF _6\right]^{3-}$ oxidation state of cobalt is $+3$
$\rightarrow$ Electronic configuration of cobalt in ground state.
$\rightarrow$ Electronic configuration of cobalt in $+3$ oxidation state.
$\rightarrow$ $F^-$ is weak ligand so $3d$ orbital does not take place in hybridization
$\rightarrow$ In this complex $4d$ orbital hybridize. So $sp^3d^2$ hybridization occurs.
$\rightarrow$ 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.
$\rightarrow$ $ \text { Magnetic Moment } (\mu) =\sqrt{n(n+2)}$
$ =\sqrt{4(4+2)}$
$ =\sqrt{24} \text { B.M. }$
$\rightarrow$ Electronic configuration of cobalt in ground state.
$\rightarrow$ Electronic configuration of cobalt in $+3$ oxidation state.
$\rightarrow$ $F^-$ is weak ligand so $3d$ orbital does not take place in hybridization
$\rightarrow$ In this complex $4d$ orbital hybridize. So $sp^3d^2$ hybridization occurs.
$\rightarrow$ 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.
$\rightarrow$ $ \text { Magnetic Moment } (\mu) =\sqrt{n(n+2)}$
$ =\sqrt{4(4+2)}$
$ =\sqrt{24} \text { B.M. }$
Question 44 Marks
Discuss the nature of bonding in $[Co(NH_3)_6]^{3+}$ coordination entities on the basis of valence bond theory
Answer
View full question & answer→$\rightarrow$ In $[Co(NH_3)_6]^{3+}$ ion oxidation number of cobalt is $+3.$
$\rightarrow$ Electronic configuration of cobalt in ground state.
$\rightarrow$ Electronic configuration of cobalt in $+3$ oxidation state.
$\rightarrow$ $NH_3$ is strong ligand so electrons get paired in $d-$orbital and $d^2sp^3$ hybridization occurs.
six $d^2sp^3$ hybrid orbitals
$\rightarrow$ Six pairs of electrons, one from each $NH_3$ molecule, occupy the six hybrid orbitals.
$\rightarrow$ Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron.
$\rightarrow$ 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.
$\rightarrow$ Electronic configuration of cobalt in ground state.
$\rightarrow$ Electronic configuration of cobalt in $+3$ oxidation state.
$\rightarrow$ $NH_3$ is strong ligand so electrons get paired in $d-$orbital and $d^2sp^3$ hybridization occurs.
six $d^2sp^3$ hybrid orbitals
$\rightarrow$ Six pairs of electrons, one from each $NH_3$ molecule, occupy the six hybrid orbitals.
$\rightarrow$ Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron.
$\rightarrow$ 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 54 Marks
Explain different types of Structural isomerism with example.
Answer
View full question & answer→$(1)$ Linkage Isomerism:
$\rightarrow$ Linkage isomerism arises in a coordination compound containing ambidentate ligand.
$\rightarrow$ A simple example is provided by complexes containing the thiocyanate ligand, $ \text{NCS}^{-,}$ which may bind through the nitrogen to give $ \text{M-NCS}$ or through sulphur to give $ \text{M-SCN.}$
$\rightarrow$ 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 $\left(- NO _2\right)$.
$\text { e.g. : }\left[ Co ( ONO )\left( NH _3\right)_5\right]^{2+} \text { and }\left[ Co \left( NO _2\right)\left( NH _3\right)_5\right]^{2+}$
$ \text{Red} \quad\text{Yellow}$
$(2)$ Coordination Isomerism:
$\rightarrow$ 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:
$\rightarrow$ 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. : $ \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 }$
$(4)$ Solvate Isomerism $OR$ Hydrate isomerism:
$\rightarrow$ This form of isomerism is known as 'hydrate isomerism' in case where water is involved as a solvent.
$\rightarrow$ This is similar to ionization isomerism.
$\rightarrow$ 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$).$
$\rightarrow$ Linkage isomerism arises in a coordination compound containing ambidentate ligand.
$\rightarrow$ A simple example is provided by complexes containing the thiocyanate ligand, $ \text{NCS}^{-,}$ which may bind through the nitrogen to give $ \text{M-NCS}$ or through sulphur to give $ \text{M-SCN.}$
$\rightarrow$ 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 $\left(- NO _2\right)$.
$\text { e.g. : }\left[ Co ( ONO )\left( NH _3\right)_5\right]^{2+} \text { and }\left[ Co \left( NO _2\right)\left( NH _3\right)_5\right]^{2+}$
$ \text{Red} \quad\text{Yellow}$
$(2)$ Coordination Isomerism:
$\rightarrow$ 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:
$\rightarrow$ 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. : $ \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 }$
$(4)$ Solvate Isomerism $OR$ Hydrate isomerism:
$\rightarrow$ This form of isomerism is known as 'hydrate isomerism' in case where water is involved as a solvent.
$\rightarrow$ This is similar to ionization isomerism.
$\rightarrow$ 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$).$
Question 64 Marks
Explain geometrical isomerism in octahedral Complex.
Answer
View full question & answer→$\rightarrow$ This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
$\rightarrow$ Important examples of these behaviour are found with coordination numbers $4$ and $6.$
$\rightarrow$ In a square planar complex of formula $[MX_2L_2] (X$ 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 $Pt[(NH_3)_2Cl_2]$
$\rightarrow$ Other square planar complex of the type $ \text{MABXL} ($where $A, B, X, L$ are unidentates$)$ shows three isomers$-$two cis and one trans.
$\rightarrow$ Such isomerism is not possible for a tetrahedral geometry.
$\rightarrow$ In octahedral complexes of formula $[MX_2L_4]$ in which the two ligands $X$ may be oriented cis or trans to each other
Geometrical isomers $($cis and trans$)$ of $[Co(NH_3)_4Cl_2]^+$
$\rightarrow$ This type of isomerism also arises when didentate ligands $L-L [$e.g., $en ]$ are present in complexes of formula $[MX_2(L-L)_2]$
Geometrical isomers $($cis and trans$)$ of $[CoCl_2(en)_2]$
$\rightarrow$ Another type of geometrical isomerism occurs in octahedral coordination entities of the type $[Co(NH_3)_3(NO_2)_3].$
$\rightarrow$ If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, it forms the facial $($fac$)$ isomer.
$\rightarrow$ When the positions are around the meridian of the octahedron, we get the meridional $($mer$)$ isomer.
The facial $($fac$)$ and meridional $($mer$)$ isomers of $[Co(NH_3)_3(NO_2)_3]$
$\rightarrow$ Important examples of these behaviour are found with coordination numbers $4$ and $6.$
$\rightarrow$ In a square planar complex of formula $[MX_2L_2] (X$ 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 $Pt[(NH_3)_2Cl_2]$
$\rightarrow$ Other square planar complex of the type $ \text{MABXL} ($where $A, B, X, L$ are unidentates$)$ shows three isomers$-$two cis and one trans.
$\rightarrow$ Such isomerism is not possible for a tetrahedral geometry.
$\rightarrow$ In octahedral complexes of formula $[MX_2L_4]$ in which the two ligands $X$ may be oriented cis or trans to each other
Geometrical isomers $($cis and trans$)$ of $[Co(NH_3)_4Cl_2]^+$
$\rightarrow$ This type of isomerism also arises when didentate ligands $L-L [$e.g., $en ]$ are present in complexes of formula $[MX_2(L-L)_2]$
Geometrical isomers $($cis and trans$)$ of $[CoCl_2(en)_2]$
$\rightarrow$ Another type of geometrical isomerism occurs in octahedral coordination entities of the type $[Co(NH_3)_3(NO_2)_3].$
$\rightarrow$ If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, it forms the facial $($fac$)$ isomer.
$\rightarrow$ When the positions are around the meridian of the octahedron, we get the meridional $($mer$)$ isomer.
The facial $($fac$)$ and meridional $($mer$)$ isomers of $[Co(NH_3)_3(NO_2)_3]$
Question 74 Marks
What is ligand? Explain classification of ligand.
Answer
View full question & answer→$\rightarrow$ "Ligands are atom or ions which donates electron pairs to the central metal ion."
$(OR)$
$\rightarrow$ "The ions or molecules bound to the central atom/ion in the coordination entity are called ligands."
$\rightarrow$ These may be simple ions such as $Cl,$ small molecules such as $H_2O$ or $NH_3,$ larger molecules such as $H_2NCH_2CH_2NH_2 \ or \ N(CH_2CH_2NH_2)_3$ or even macromolecules, such as proteins.
Classification of Ligands:
$(1)$ Unidentate ligands:
$\rightarrow$ "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:
$\rightarrow$ "When a ligand is bound to a metal ion through a two donor atom, the ligand is said to be didentate ".
$(3)$ Tridentate ligands:
$\rightarrow$ "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:
$\rightarrow$ When a ligand is bound to a metal ion through a six donor atom, the ligand is said to be hexadendate
Example: Ethylenediaminetetraacetate ion $\left( EDTA ^{4-}\right)$
$(OR)$
$\rightarrow$ "The ions or molecules bound to the central atom/ion in the coordination entity are called ligands."
$\rightarrow$ These may be simple ions such as $Cl,$ small molecules such as $H_2O$ or $NH_3,$ larger molecules such as $H_2NCH_2CH_2NH_2 \ or \ N(CH_2CH_2NH_2)_3$ or even macromolecules, such as proteins.
Classification of Ligands:
$(1)$ Unidentate ligands:
$\rightarrow$ "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:
$\rightarrow$ "When a ligand is bound to a metal ion through a two donor atom, the ligand is said to be didentate ".
$(3)$ Tridentate ligands:
$\rightarrow$ "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:
$\rightarrow$ When a ligand is bound to a metal ion through a six donor atom, the ligand is said to be hexadendate
Example: Ethylenediaminetetraacetate ion $\left( EDTA ^{4-}\right)$