Coordination Compounds

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.

There are two ligands in the coordination entity, that is, ammine and aqua, but according to the IUPAC rule, alphabetical order is followed, that is, ammine is named first. Now calculating the oxidation state of Co, the net charge on the coordination entity is +3 as Cl has $-$1 and there are three Cl^$-$ outside the square brackets. So the oxidation state of Co will be III as both ammine and aqua are neutral ligands. So, its correct name will be diamminetetraaquacobalt (III) chloride.

${\left[ {NiC{l_4}} \right]^{2 - }}\,\left( {{d^8}} \right)$



i.e. the number of unpaired electrons in ${\left[ {NiCl{}_4} \right]^{2 - }}\,\,$ is $2.$

$n = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( 4 \right)} $

$ = 2\sqrt 2 = 2 \times 1.41$

$ = 2.82\,BM$

$\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}\,\,$ is an inner orbital complex, because in this complex $d-$orbital used is of lower quantum number i.e. $(n-1).$ It results from ${d^2}\,\,\,$ $s{p^3}$ (inner orbital) hybridization.

(i) Nickel in +2 oxidation state $\left(\mathrm{Ni}^{2+}\right)$ prefers to form tetrahedral complexes with chlorine, e.g., $\left[\mathrm{NiCl}_4\right]^{2-}$.

The oxidation state of nickel in $\left[\mathrm{NiCl}_4\right]^{2-}$

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

Electronic configuration of $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 4 s^0$

Since, chlorine is a weak ligand, electrons in $\mathrm{Ni}^{2+}$ do not pair up. One $4 s$ and three $4 p$ orbitals undergo hybridisation to form four $s p^3$ hybrid orbitals. These vacant orbitals accept a pair of electron from each of the chloride ion.

The geometry of $s p^3$ hybridised nickel $\left[\mathrm{Ni}^{2+}\right]$ is tetrahedral

(ii) Nickel in +2 oxidation state $\left(\right.$ as $\left.\mathrm{Ni}^{2+}\right)$ forms square planner complex with cyanide ligands, e.g., $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$.

Oxidation state of $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

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

The oxidation state of nickel in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is $+2$

Electronic configuration of $\mathrm{Ni}^{2+}=[\operatorname{Ar}] 3 d^8 4 s^0$

Since, cyanide is a strong ligand, electrons in $3 d$-orbitals are paired up.



One $3 d$, one $4 s$ and two $4 p$ orbitals undergoes hybridisation to form four $d s p^2$ hybrid orbitals. These vacant orbitals accept a pair of electrons from each of cyanide ligand.



The geometry of $d s p^2$ hybridised nickel complex $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planner.



(iii) Nickel in +2 oxidation state as $\mathrm{Ni}^{2+}$ forms octahedral complex with water as ligands, e.g., $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Oxidation state of nickel in $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is:

$ \begin{aligned} x+6 \times 0 & =+2 \\\\ x & =+2 \end{aligned} $

Electronic configuration of $\mathrm{Ni}^{2+}=$



Since, water is a weak ligand, electrons in $3 d$ orbitals of Nickel remains unpaired. One $4 s$, three $4 p$ and two $4 d$ orbitals undergo hybridisation to form $s p^3 d^2$ hybrid orbitals. These vacant orbitals accept lone pair of electron from each of the water molecule.


The geometry of $s p^3 d^2$ hybridised nickel complex $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is octahedral.

The geometries of different $\mathrm{Ni}^{2+}$ complexes are as follows:

$ \begin{aligned} {\left[\mathrm{NiCl}_4\right]^{2-} } & \rightarrow \text { tetrahedral } \\\\ {\left[\mathrm{NiCN}_4\right]^{2-} } & \rightarrow \text { square planner } \\\\ {\left[\mathrm{Ni}_{(}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} } & \rightarrow \text { octahedral } \end{aligned} $

In K₃Fe(CN)₆ :

CN^$-$ being a strong field ligand, it causes pairing of electrons, so only one unpaired electron. The complex is paramagnetic.

In Co[(NH₃)₆]Cl₃ :

NH₃ being a strong field ligand it causes pairing of spins. There is no unpaired electron, hence the complex is diamagnetic.

In Na₃[Co(oxalate)₃] :

Oxalate being a strong field ligand it causes pairing of electron spins. There is no unpaired electron, hence the complex is diamagnetic.

In [Ni(H₂O)₆]Cl₂ :

Since H₂O is a weak field ligand, no pairing takes place. There are two unpaired electrons, so the complex is paramagnetic.

In K₂[Pt(CN)₄] :

CN^$-$ being a strong field ligand, it causes pairing of electrons spins, so no unpaired electron. Hence, the complex is diamagnetic.

In [Zn(H₂O)₆(NO₃)₂] : Zn²⁺ has 3d¹⁰ configuration. Therefore, it is a diamagnetic complex.

$\mathop {CoC{l_3}.6N{H_3}}\limits_{2.675g} \buildrel \, \over \longrightarrow xC{l^ - }$

$xC{l^ - } + AgN{O_3}\buildrel \, \over \longrightarrow \mathop {x\,AgCl \downarrow }\limits_{4.78g} $

Number of moles of the complex

$ = {{2.675} \over {267.5}} = 0.01$ moles

Number of moles of $AgCl$ obtained

$ = {{4.78} \over {143.5}} = 0.03$ moles

$\therefore$ $\,\,\,$ Number of moles of $AgCl$ obtained

$ = 3 \times \,\,$ No. of moles of complex

$\therefore$ $\,\,\,$ $n = {{0.03} \over {0.01}} = 3$

For a substance to be optical isomers following conditions should be fulfilled

$(a)\,\,\,$ A coordination compound which can rotate the plane of polarized light is said to be optically active.

$(b)\,\,\,$ When the coordination compounds have same formula but differ in their abilities to rotate directions of the plane of polarized light are said to exhibit optical isomerism and the molecules are optical isomers. The optical isomers are pair of molecules which are non-super-imposable mirror images of each other.

$(c)\,\,\,$ This is due to the absence of elements of symmetry in the complex.

$(d)\,\,\,$ Optical isomerism is expected in tetrahedral complexes of the type $Mabcd.$

Based on this only option $(2)$ shows optical isomerism ${\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }}$



Complexes of $Z{n^{ + + }}\,\,$ cannot show optical isomerism as they are tetrahedral complexes with plane of symmetry.

${\left[ {Co{{\left( {{H_2}O} \right)}_4}\left( {en} \right)} \right]^{3 + }}\,\,$ have two planes of symmetry hence it is also optically inactive.

Hence the formula of the complex is $\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$

EDTA (ethylenediaminetetracetic acid) is a hexadentate ligand. It is also called the amino poly-carboxylic acid.

It is structure consists of the ethane group which is attached to tertiary amine groups at the terminal positions forming an ethylene diamine. Tertiary amine group is situated by the acetic acid groups $\left(-\mathrm{CH}_2 \mathrm{COOH}\right)$.

The ionisation isomers give different ions in solution. In the complex given in option (B), Cl$^-$ is replaced by NO$_2^ - $ in ionisation sphere.

$\mathrm{[Cr(H_2O)_4Cl(NO_2)]Cl}$ and $[Cr{({H_2}O)_4}C{l_2}](N{O_2})$ have different ions inside and outside the coordinate sphere and they are isomers. Therefore, they are ionisation isomers.

$\begin{aligned} & {\left[\mathrm{NiCl}_4\right]^{2-}} \\\\ & \text { Oxidation state of }\left[\mathrm{NiCl}_4\right]^{2-}=+2 \\\\ & \mathrm{Ni}(28)=[\mathrm{Ar}] 3 d^8 4 s^2 \\\\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8\end{aligned}$

Among the given complexes of Ni , only $\mathrm{Cl}^{-}$is a weak ligand and does not cause pairing of electrons to take place.

Thus, the spin only magnetic moment, with two unpaired electrons $(n=2)$ configuration is

$ \sqrt{n(n+2)}=\sqrt{2(2+2)}=2.82 \mathrm{BM}=2.82 \mathrm{BM} $

The $SC{N^ - }$ ion can coordinate through $S$ or $N$ atom giving rise to linkage isomerism

$M \leftarrow SCN\,\,\,$ thiocyanto

$M \leftarrow NCS\,\,\,$ isothiocyanato.

In [Cr(CO)$_6$]: Cr(24) = [Ar] 3$d^5$ 4$s^1$

Since CO is a strong field ligand, so pairing of electrons will take place and the configuration will become $3d^6$. There will be no unpaired electrons, so, the spin only magnetic moment is zero.

Among the given compounds, the compounds of the type [MA$_2$X$_2$] and [M(AA)$_2$X$_2$] will exhibit geometrical isomerism.

In octahedral complex the magnitude of ${\Delta _o}$ will be highest in a complex having strongest ligand. Of the given ligands $C{N^ - }$ is strongest so ${\Delta _o}$ will be highest for ${\left( {Co{{\left( {CN} \right)}_6}} \right)^{3 - }}.$ Thus option $(a)$ is correct.

In the given complex we have two bidentate ligands (i.e. $en$ and ${C_2}{O_4}$ ), so coordination number of $E$ is $6$ $\left( {2 \times 2 + 1 \times 2 = 6} \right)$

Let the oxidation state of $E$ in complex be $x,$

then $\left[ {x + \left( { - 2} \right) = 1} \right]$ or $x - 2 = 1$

or $\,\,\,\,$ $x=+3,$ so its oxidation state is $+3$

Thus option $(d)$ is correct.

The electronic configuration of Cu$^{2+}$ is [Ar]3d$^9$. It contains one unpaired electron. Hence, CuF$_2$ is coloured compound.

In all other compounds, Cu$^+$ ion is present. It has electronic configuration of Cu$^{2+}$ is [Ar]3d$^{10}$. All electrons are paired.

Hence, they are colourless.

In [Ni(CO)$_4$], Ni is in 0 oxidation state, so there are 8 electrons in 3d subshell and 2 electrons in 4s. CO is the strong ligand, so it cause pairing of electrons in 3d subshell leaving one d subshell vacant. Now, the electron from s shell shifts to d creating vacant space in s subshell and thus the hybridisation is sp$^3$.

Ni(28) = [Ar] 4s$^2$3d$^8$

CO is a strong ligand, causes coupling.

In [Ni(CN)$_4$]$^{2-}$, Ni is in +2 oxidation state so it has 8 electrons in d subshell and CN is also a strong ligand which pairs with d subshell atom and leaves one d orbital empty and thus its hybridisation is dsp$^2$.

In [Ni(CN)$_4$]$^{2-}$, the oxidation state of Ni is +2

Ni$^{2+}$ = [Ar] 3d$^8$4s$^0$

CN$^-$ is strong ligand, causes coupling.