Coordination Compounds

Due to presence of strong field ligand (CN^–) pairing occurs in which two d-orbitals i.e., d_x²–y² and d_z² directly face the CN^- ligand.

Compounds that contain at least one carbon-metal bond are called organometallic compounds.

Magnetic moment = $\sqrt {n\left( {n + 2} \right)} $ BM

$ \therefore $ $\sqrt {n\left( {n + 2} \right)} $ = 3.9

$ \Rightarrow $ n = 3

$ \therefore $ no. of unpaired electron = 3

H₂O is weak field ligand so no pairing of electrons happens.

Fe²⁺ = t_2g⁴e_g²

Co²⁺ = t_2g⁵e_g²

V²⁺ = t_2g³e_g⁰

$ \therefore $ M is V, Co.

Oxalato (C₂O₄^2–) is a bidentate and H₂O is unidentate ligand.

4C₂O₄^2– creates 8 covalent bonds.

2H₂O creates 2 covalent bonds.

$ \therefore $ Around Th 10 coordinate covalent bonds will be present.

Co $ \to $ Vitamin B₁₂

Zn $ \to $ Carbonic anhydrase

Rh $ \to $ Wilkinson catalyst

Mg $ \to $ Chlorophyll

CoCl₃ + en $ \to $ [Co(en)₂Cl₂]Cl
   1      :   2

A and B are Geometrical isomers.

$ \therefore $ The difference in the number of unpaired electrons = 3 - 1 = 2

So, The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO₂ )] is 12

Stronger the ligand, absorption of light having lower wavelength is more.

Order of $\lambda $ : Red $>$ Green $>$ Blue

Hence, ligand strength is L₃ $<$ L₁ $<$ L₂

As complex K₃[Co(CN)₆] have CN^$-$ ligand which is strongfield ligand amongst the given ligands in other complexes.

NH₃ is a strong field ligand.

H₂O is weak field ligand.

Let $\Delta $_violet is the crystal field splitting energy of complex A and $\Delta $_yellow is for complex B.

$ \therefore $   $\Delta $_yellow > $\Delta $_violet.

We know, $\Delta $ = ${{hc} \over {{\lambda _{absorb}}}}$

Violet color will absorb yellow color and yellow color will absorb violet color.

$ \therefore $   $\Delta $_yellow = ${{hc} \over {{\lambda _{violet}}}}$

and $\Delta $_violet = ${{hc} \over {{\lambda _{yellow}}}}$

So, $\Delta $_o of A is calculated from energy of yellow light and $\Delta $_o of B is calculated from energy of violet light.

(a)   [Co(H₂O) Cl(en)₂]Cl  $ \to $  it shows gemetrical isomerism.



(b)   [Cr(H₂O)₅Cl]Cl₂  :   This complex is [MA₅B] type. It does not show geometrical isomerism.

(c)   K[Pt(NH₃)Cl₃]  :  Thiscomplex is [MAB₃] type. It does not show geometrical isomerism.

(d)   K₃[Cr(OX)₃]  :  This complex is [M(AA)₃] type. It does not show geometrical isomerism.

Wilkinson's catalyst is [RhCl(PPh₃)₃]

Here central atom is Rh.

Oxidation state of Rh here is = + 1

$ \therefore $   Electronic configuration of Rh⁺ = [Kr]4d⁸


And the complex is square planar

When reactant is cis isomer then following reaction takes place.


So, if reactant is cis - isomer then two isomers are produced.

When reactant is trans isomer then following reaction takes place.



So, if the reactant is trans isomer then only one isomer is produced.

Assume oxidation state of Cr in all the compounds = x

(i)$\,\,\,$ In [Cr(H₂O)₆] Cl₃ oxidation state of Cr is

x + 0 $ \times $ 6 + ($-$1 $ \times $ ) = O

$ \Rightarrow \,\,\,\,\,$ x + 0 $-$ 3 = O

$ \Rightarrow \,\,\,\,$ x = + 3

(ii)$\,\,\,$ [Cr (C₆ H₆)₂] oxidation state of Cr is

x + 0 $ \times $ 2 = 0

$ \Rightarrow \,\,\,\,$ x = 0

(iii) $\,\,\,$ In K₂ [ Cr(CN)₂ (O)₂ (O₂) (NH₃)] oxidation state of Cr is

1 $ \times $ 2 + x + ($-$ 1 $ \times $ 2) + ($-$2 $ \times $ 2) + ($-$2) + 0 = 0

$ \Rightarrow \,\,\,\,$ x = + 6

$\therefore\,\,\,$ + 3, 0 and + 6 is the correct answer.

Note :

O₂ molecule can have 0, $-$ 1, $-$ 2 oxidation state but in K₂ [ Cr (CN)₂ (O)₂ (O₂) NH₃ ] if we choose zero as the oxidation state of O₂ then for Cr oxidation state will be $+$ 4. But + 4 oxidation state of Cr is unstable and $+$ 6 is most stable that is why we choose $-$ 2 oxidation state of O₂.

The complexes with formula [Mabcd]^n± can have three geometrical isomers. As NO²⁻ and SCN⁻ are ambidentate ligands, therefore 4 × 3 = 12 geometrical isomers will be possible.

We know,

Spin only magnetic moment ($\mu $) = $\sqrt {n\left( {n + 2} \right)} $   B.M

Where, n is the number of unpaired electrons.

So, the complex having higher number of unpaired electrons will have higher value of spin only magnetic moment.

(1)   Zn⁺² : [Ar] 3d¹⁰
Here 0 unpaired electrons present

(2)   Ni⁺² : [Ar] 3d⁸
Here 2 unpaired electrons present.

(3)   Co⁺² : [Ar] 3d⁷
Here 3 unpaired electrons present.

(4)   Mn⁺² : [Ar] 3d⁵
Here 5 unpaired electrons present.

So, correct order is $ \to $
    [MnCl₄]^2$-$ > [CoCl₄]^2$-$ > [NiCl₄]^2$-$ > [ZnCl₄]^2$-$

[NiCl₄]^2– : Oxidation state of Ni in [NiCl₄]^2– = + 2


Cl^– is a weak field ligand and cannot take part in pairing of electrons.
Hence, the complex is tetrahedral and paramagnetic with two unpaired electrons.

[Ni(CO)₄] : Oxidation state of Ni in [Ni(CO)₄] is zero. CO is a strong field ligand thus pairing of electrons takes place in d-orbitals.
Hence, the complex is tetrahedral and diamagnetic.

1. $\left[\mathrm{FeF}_6\right]^{4-}$

(A) Let oxidation state of iron (Fe) in $\left[\mathrm{FeF}_6\right]^{4-}$

$ \begin{aligned} x+6 \times(-1) =-4 \\\\ \Rightarrow x =-4+6=+2 \end{aligned} $

$ \therefore $ Iron exist as $\mathrm{Fe}^{2+}$


Electron is $3 d$ orbitals remains unpaired as fluorine is weak ligand. One $4 s$, three $4 p$ and two $4 d$ orbitals undergo hybridisation to form six $s p^3 d^2$ hybrid orbital each of which accumulate a pair of electron from fluorine.


$ \text { Iron }(\mathrm{Fe}) \text { in }\left[\mathrm{FeF}_6\right]^{4-} \text { has } s p^3 d^2 \text { hybridisation. } $

2. $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right]$ : The arrangement of electrons in $\mathrm{Ti}^{3+}\left(3 d^1\right)$ will be


Electron in $3 d$ orbitals remains unpaired as water $\left(\mathrm{H}_2 \mathrm{O}\right)$ and chlorine $(\mathrm{Cl})$ is a weak ligands. Two $3 d$, one $4 s$ and three $4 p$ orbitals undergo hybridisation to form six $d^2 s p^3$ hybrid orbitals. These orbitals accept lone pair of electrons from each of the chlorine and water molecules.

3. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ : The arrangement of electrons in $\mathrm{Cr}^{3+}\left(3 d^3\right)$ will be


Electrons in $3 d$ orbitals remains paired as ammonia $\left(\mathrm{NH}_3\right)$ is a strong ligand. Two $3 d$, one $4 s$ and three $4 p$ orbitals undergo hybridisation to form six $d^2 s p^3$ hybrid orbitals. These orbitals accept lone pair of electrons from each of the six ammonia molecules.

4. $ \left[\mathrm{FeCl}_4\right]^{2-}: \text { The arrangement of electrons in } \mathrm{Fe}^{2+}\left(3 d^6\right) \text { will be } $


Electrons in $3 d$ orbitals remains unpaired as chlorine is a weak ligand. One $4 s$ and three $4 p$ orbitals undergo hybridization to form four $s p^3$ hybrid orbitals. These $s p^3$ hybrid orbitals accept lone pair of electrons from each of the chlorine ligand.

5. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ : The arrangement of electrons in $\mathrm{Ni}^0\left(3 d^8 4 s^2\right)$ will be


Electrons in $3 d$ orbitals remains paired and paired electron from $4 \mathrm{~s}$ orbital move to $3 d$ as carbonyl is a strong ligand. One $4 s$, and three $4 p$ orbitals undergo hybridisation to form four $s p^3$ hybrid orbitals. These $s p^3$ hybrid orbitals accept lone pairs from each of the four carbonyl ligand.

6. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}:$ The arrangement of electrons in $\mathrm{Ni}^{2+}\left(3 d^8\right)$ will be


Electrons in $3 d$ orbitals pair up as cyanide is a strong ligand. One $3 d$, one $4 s$ and two $4 p$ orbitals undergo hybridisation to form four $d s p^2$ hybrid orbitals. The $d s p^2$ hybrid orbitals accept lone pair of electrons from each of the cyanide ligand.

Option (A) : Correct. The complex [Co(en)(NH₃)₃(H₂O)]³⁺ has two following isomers:

Option (B) : Correct. When bidentate 'en' is replaced by two cyanide ligands complex [Co(CN)₂(NH₃)₃(H₂O)]⁺ will have three following isomers:

Option (C) : Incorrect. Incomplex [Co(en)(NH₃)₃(H₂O)]³⁺, Co is in + 3 oxidation, that is, [Ar]d⁶. In the presence given ligands the complex will have low spin, hence, diamagnetic.

Option (D) : Correct. In complex [Co(en)(NH₃)₄]³⁺ there is larger gap between t_2g and e_g than in complex [Co(en)(NH₃)₃(H₂O)]³⁺. Hence [Co(en)(NH₃)₃(H₂O)]³⁺ will absorb light at longer wavelength as compared to [Co(en)(NH₃)₄]³⁺.

(A) [(Fe(CO₅)] & [Ni(CO)₄ complexes have 18-electrons in their valence shell.

(B) Due to strong ligand field, carbonyl complexes are predominantly low spin complexes.

(C) As electron density increases on metals (with lowering oxidation state on metals), the extent of synergic bonding increases. Hence M-C bond strength increases.

(D) While positive charge on metals increases and the extent of synergic bond decreases and hence C-O bond becomes stronger.

The structure of [Co₂(CO)₈] is

From the structure, we can see that there is one CO−CO bond, six terminals CO and two bridging CO.

Moles of complex = ${{Molarity \times Volume\left( {mL} \right)} \over {1000}}$

= ${{100 \times 0.1} \over {1000}}$ = 0.01 mole

Moles of ions precipitated with excess of AgNO₃

= ${{1.2 \times {{10}^{22}}} \over {6.023 \times {{10}^{23}}}}$ = 0.02 mole

So 0.01 × n = 0.02

$ \Rightarrow $ n = 2

It means 2Cl^– ions present in ionization sphere.

$ \therefore $ The formula of complex is [Co(H₂O)₅Cl]Cl₂.H₂O

The pink coloured aqueous solution of $\mathrm{MCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}$ has central metal atom bonded six water molecule by co-ordinate bonds and there are two primary valences of chloride ions. The formula and structure of the complex $X$ can be represented as :

Formula : $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$

$ \left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2 \rightarrow\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{2+}+2 \mathrm{Cl}^{-} $



Here, the six non ionisable water molecules are the secondary valence and two ionisable chloride ions are primary valence. Since, the hexa coordinated aqua complex of cobalt is pink in colour; hence, central metal is cobalt.

(i) Reaction of the complex X with ammonium chloride in presence of air gives hexamminecobalt (III) chloride which is a 1: 3 electrolyte.



$ \begin{aligned} & \text { Option (A) : } \\\\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3+3 \mathrm{AgNO}_3 \rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left(\mathrm{NO}_3\right)_3+3 \mathrm{AgCl}} \end{aligned} $

Option (B) : $\mathrm{NH}_3$ is a strong field ligand, therefore, pairing of electrons will take place.



Option (C) :



Option (D) : When X and Z are in equilibrium at 0°C

$ \underset{\text{Pink}}{\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]}\rightarrow \underset{\text{Blue}}{\left[\mathrm{CoCl}_4\right]^{2-}} $

Since, formation of chlorine complex takes place in presence of excess chlorine at room temperature $\left(25^{\circ} \mathrm{C}\right)$. At lower temperature $\left(0^{\circ} \mathrm{C}\right)$, the equilibrium is more shifted towards left favouring the formation of pink coloured hexaaquacobalt (II) chloride complex.

Option (D) is correct.

Homoleptic complexes are those compounds in which all the ligands bound to the central metal are identical. Thus, these types of complexes have only one type of ligands. In the complex [Co(NH₃)₆]Cl₃ , central atom Co has ammonia ligands as all six.

Heteroleptic complexes are those compounds in which central transition metal has more than one type of ligands. So, rest of the options are examples of heteroleptic complexes.

[Cr(H₂O)₆]Cl₃ has the highest primary valency among the given complexes, that is, three, therefore, it will consume three moles of AgNO₃ and precipitate three moles of AgCl.

[Cr(H₂O)₆]Cl₃ + 3AgNo₃ $\to$ 3AgCl + [Cr(H₂O)₆](NO)₃

$\Delta$₀ of complex [Mo(H₂O)₆]²⁺ is greater than that of [Cr(H₂O)₆]²⁺. It is due to the fact that Mo²⁺ belongs to second row transition element while Cr²⁺ is first row transition element.

Magnitude of crystal field splitting energy ($\Delta$₀) depends on the charge on the metal ion. Greater the charge, greater is the crystal field splitting energy. Therefore, [Ti(H₂O)₆]³⁺ > [Ti(H₂O)₆]²⁺.

Since $(a)$ and $(b),$ each has $4$ unpaired electrons, they will have same magnetic moment.

Optical isomerism occurs when a molecule is non-superimposable with its mirror image hence the complex $cis - \left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$ is optically active.

The hybridisation of central atoms and the geometries of ammonia complex are tabulated as follows:

$ \begin{array}{|l|c|c|c|l|} \hline \text { Metal Ion } & \text { Hybridisation } & \text { Coordination Number } & \text { Complex with } \mathbf{N H}_3 & \text { Geometry } \\ \hline \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 d^8 4 s^2 & d^2 s p^3 & 6 & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} & \text { Octahedral } \\ \hline \mathrm{Pt}^{2+}:[\mathrm{Xe}] 4 f^4 5 d^9 6 s^1 & d s p^2 & 4 & {\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]^{2+}} & \text { Square planar } \\ \hline \mathrm{Zn}^{2+}:[\mathrm{Ar}] 3 d^{10} 4 s^2 & s p^3 & 4 & {\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_4\right]^{2+}} & \text { Tetrahedral } \\ \hline \end{array} $

For the given complexes:

In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ : $\mathrm{Ni}$ (at. no. 28): $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 4 s^2$

$\mathrm{CO}$ being a strong field ligand causes the electrons to pair; hence, the complex diamagnetic.


$ \text {In }\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{2+} \text { (at. No. 26): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 $

There are two unpaired electrons, so the complex is paramagnetic.


$ \begin{aligned} & \mathrm{In}\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}: \mathrm{Co}^{3+}\text {(at. No. 24): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6 \end{aligned} $


The ligand $\mathrm{NH}_3$ results in formation of low spin complex, which is diamagnetic.

$ \text {In } \mathrm{Na}_3\left[\mathrm{CoF}_6\right]: \mathrm{Co}^{3+}\text {(at. No. 24): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6 $


$\mathrm{F}$ is a weak field ligand and results in the formation of high spin complex with four unpaired electrons, which is paramagnetic.

In $\mathrm{Na}_2 \mathrm{O}_2$ : The molecular orbital configuration of $\mathrm{O}_2^{2-}$ is

$ \sigma 1 s^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2,\left\{\begin{array}{l} \pi 2 p_x^2 \\ \pi 2 p_y^2 \end{array},\left\{\begin{array}{l} \pi * 2 p_x^2 \\ \pi * 2 p_y^2 \end{array}\right.\right. $

There is no unpaired electron, so it is diamagnetic.

In $\mathrm{CsO}_2$ : The molecular orbital configuration of $\mathrm{O}_2^{-}$is

$ \sigma 1 s^2, \sigma * 1 \mathrm{~s}^2, \sigma 2 s^2, \sigma * 2 s^2, \sigma 2 p_z^2,\left\{\begin{array}{l} \pi 2 p_x^2 \\ \pi 2 p_y^2 \end{array},\left\{\begin{array}{l} \pi * 2 p_x^2 \\ \pi * 2 p_y^1 \end{array}\right.\right. $

There is one unpaired electron, so it is paramagnetic. Hence, there are three paramagnetic compounds.

Zn₂[Fe(CN)₆] is white in color as it does not have unpaired electrons.

Square planar complexes of type $M\left[ {ABCD} \right]\,$ form three isomers. Their position may be obtained by fixing the position of one ligand and placing at the trans position any one of the remaining three ligands one by one.

(A) [Mg (H₂O)₆ ]²⁺ + (EDTA)^4- $\buildrel {excess\,NaOH} \over \longrightarrow $ [Mg (EDTA) ]²⁺ + 6H₂O

The above equation is incorrect because the product formed would be [Mg (EDTA)]²⁻ .

(B) CuSO₄ + KCN $\to$ K₂[Cu (CN)₄] + K₂SO₄

The above equation is incorrect. Thus, the correct equation is

2KCN + CuSO₄ $\to$ K₂SO₄ + Cu + (CN)₂(Cyanogen Gas)

(C) Li₂O + 2KCl $\to$ 2LiCl + K₂O

The above equation is incorrect because K₂O, a stronger base cannot be generated by a weaker base, Li₂O.

(D) [CoCl(NH₃)₅]⁺ + 5H⁺ $\to$ Co²⁺ + $5NH_4^+$ + Cl^-

The above equation is correct because amine complexes decomposes under acidic medium. Thus, the complex [CoCl(NH₃)₅] decomposes under acidic conditions to give ammonium ions.

For a given metal ion, weak filed ligands create a complex with smaller $\Delta $, which will absorbs light of longer $\lambda $ and thus lower frequency. Conservely, stronger filled ligands create a larger $\Delta ,$ absorb light of shorter $\lambda $ and thus higher $v.i.e.$ higher energy.

$\mathop {{\mathop{\rm Re}\nolimits} d}\limits_{\lambda = 650\,nm} \,\,\mathop { < Yellow}\limits_{570\,\,nm} \,\,\mathop { < Green}\limits_{490\,nm} < \mathop {Blue}\limits_{450\,nm} $

So order of ligand strength is

${L_1} < {L_3} < {L_2} < {L_4}$

For [Cr(NH₃)₄Cl₂]Cl,

For [Ti(H₂O)₅Cl](NO₃)₂,

The complex is paramagnetic due to presence of unpaired electron.

Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Pt(en)(NH₃)Cl]NO₃,

The complex is square planar and all electrons are paired. So it is diamagnetic. Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Co(NH₃)₄(NO₃)₂]NO₃, NH₃ is a strong ligand and causes pairing

Octahedral coordination entities of the type $M{a_3}{b_3}$ exhibit Geometrical isomerism. The compound exists both as facial and meridional isomers.

The electronic configurations of the central metal ions are as follows:

The number of unpaired electrons is P = 5, Q = 3 and R = 4. From the relation $\mu = \sqrt {n(n + 2)} $, where n is the number of unpaired electrons, we have the order of spin only magnetic moment as Q < R < P.

${[Co{(N{H_3})_4}C{l_2}]^ + }$ and ${[Pt{(N{H_3})_2}({H_2}O)Cl]^ + }$ are octahedral and square planar complexes, which show geometrical isomerism. $[Pt{(N{H_3})_3}(N{O_3})]Cl$ and $[Pt{(N{H_3})_3}Cl]Br$ show ionisation isomerism.

$\left[ {Cr{{\left( {en} \right)}_2}B{r_2}} \right]Br$

dibromidobis (ethylenediamine) chromium $\left( {{\rm I}{\rm I}{\rm I}} \right)$ Bromide.

The configuration of Ni = 3d⁸ 4s² and that of Ni²⁺ = 3d⁸.

In paramagnetic state, the hybridisation is sp³ and geometry is tetrahedral.

In diamagnetic state, the hybridisation is dsp² and geometry is square planar.