Coordination Compounds

C. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right] \mathrm{Cl}_2$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5(\mathrm{ONO})\right] \mathrm{Cl}_2$ are linkage isomers

D. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ and $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$ are solvate isomers

An ambidentate ligand is a ligand which has two different donor atoms and either of the two ligates in complex.

(1) Ethane-1,2-diamine → didentate

(2) Ethylenediamine tetraacetate ion → hexadentate

(3) Thiocyanate $\left(\mathrm{SCN}^{-}\right) \rightarrow$ Ambidentate ligand.

(4) Oxalate → didentate

A. $[\text{Pt}(\text{Cl}_2)(\text{NH}_3)_2]$

The formula is most likely intended to be $[\text{PtCl}_2(\text{NH}_3)_2]$, which has two chloride ligands and two ammonia ligands.

1. Find the oxidation state of Platinum (Pt):

   Let the oxidation state of Pt be $x$. The charge on each chloride ligand ($\text{Cl}^-$) is -1, and ammonia ($\text{NH}_3$) is a neutral ligand (charge 0). The overall complex is neutral.

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

   $ x - 2 = 0 $

   $ x = +2 $

   So, we have Platinum in the +2 oxidation state, which is $\text{Pt}^{2+}$.

2. Determine the electronic configuration:

   The atomic number of Pt is 78. Its ground-state electronic configuration is $[\text{Xe}] 4f^{14} 5d^9 6s^1$.

   For $\text{Pt}^{2+}$, we remove two electrons (one from $6s$ and one from $5d$).

   The electronic configuration of $\text{Pt}^{2+}$ is $[\text{Xe}] 4f^{14} 5d^8$.

3. Determine the hybridization and shape:

   The coordination number is 4. For a $d^8$ configuration, particularly for metals in the 4d and 5d series like Pt, the ligands always cause the electrons to pair up, leaving one $d$-orbital empty. This leads to $dsp^2$ hybridization.

   The orbital diagram for $\text{Pt}^{2+}$ in the complex will be:

   $ \begin{array}{cc} \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\quad} & \underline{\quad} & \underline{\quad} \ \underline{\quad} \ \underline{\quad} \\ 5d & 6s & 6p \end{array} $

   The empty $5d$, $6s$, and two $6p$ orbitals hybridize to form four $dsp^2$ hybrid orbitals.

   The geometry corresponding to $dsp^2$ hybridization is Square planar.

   Therefore, A matches with III.

B. $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$

The complex ion here is $[\text{Co}(\text{NH}_3)_6]^{3+}$.

1. Find the oxidation state of Cobalt (Co):

   Let the oxidation state of Co be $x$. Ammonia ($\text{NH}_3$) is a neutral ligand. The overall charge on the complex ion is +3.

   $ x + 6(0) = +3 $

   $ x = +3 $

   So, we have Cobalt in the +3 oxidation state, which is $\text{Co}^{3+}$.

2. Determine the electronic configuration:

   The atomic number of Co is 27. Its ground-state electronic configuration is $[\text{Ar}] 3d^7 4s^2$.

   For $\text{Co}^{3+}$, we remove three electrons (two from $4s$ and one from $3d$).

   The electronic configuration of $\text{Co}^{3+}$ is $[\text{Ar}] 3d^6$.

3. Determine the hybridization and shape:

   The coordination number is 6. The ligand is $\text{NH}_3$, which is a strong-field ligand. It will cause the pairing of the $3d$ electrons.

   The electrons in the $3d$ orbitals of $\text{Co}^{3+}$ will rearrange from $[\uparrow\downarrow][\uparrow][\uparrow][\uparrow][\uparrow]$ to $[\uparrow\downarrow][\uparrow\downarrow][\uparrow\downarrow][\quad][\quad]$.

   This leaves two $3d$ orbitals vacant.

   $ \begin{array}{cc} \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\quad} \ \underline{\quad} & \underline{\quad} & \underline{\quad} \ \underline{\quad} \ \underline{\quad} \\ 3d & 4s & 4p \end{array} $

   Hybridization involves two $3d$, one $4s$, and three $4p$ orbitals, leading to $d^2sp^3$ hybridization. This is an inner orbital complex.

   The geometry corresponding to $d^2sp^3$ hybridization is Octahedral.

   Therefore, B matches with I.

C. $[\text{NiCl}_4]^{2-}$

1. Find the oxidation state of Nickel (Ni):

   Let the oxidation state of Ni be $x$. The charge on each chloride ligand ($\text{Cl}^-$) is -1. The overall charge on the complex ion is -2.

   $ x + 4(-1) = -2 $

   $ x - 4 = -2 $

   $ x = +2 $

   So, we have Nickel in the +2 oxidation state, which is $\text{Ni}^{2+}$.

2. Determine the electronic configuration:

   The atomic number of Ni is 28. Its ground-state electronic configuration is $[\text{Ar}] 3d^8 4s^2$.

   For $\text{Ni}^{2+}$, we remove two electrons from the $4s$ orbital.

   The electronic configuration of $\text{Ni}^{2+}$ is $[\text{Ar}] 3d^8$.

3. Determine the hybridization and shape:

   The coordination number is 4. The ligand $\text{Cl}^-$ is a weak-field ligand. It does not cause the pairing of $3d$ electrons.

   The orbital diagram for $\text{Ni}^{2+}$ is:

   $ \begin{array}{cc} \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow} \ \underline{\uparrow} & \underline{\quad} & \underline{\quad} \ \underline{\quad} \ \underline{\quad} \\ 3d & 4s & 4p \end{array} $

   Since no inner $d$-orbitals are empty, the hybridization will involve the outer orbitals: one $4s$ and three $4p$ orbitals. This gives $sp^3$ hybridization.

   The geometry corresponding to $sp^3$ hybridization is Tetrahedral.

   Therefore, C matches with IV.

D. $[\text{Fe}(\text{CO})_5]$

1. Find the oxidation state of Iron (Fe):

   The ligand is carbonyl ($\text{CO}$), which is a neutral ligand. The overall complex is neutral.

   Therefore, the oxidation state of Fe is 0.

2. Determine the electronic configuration:

   The atomic number of Fe is 26. Its ground-state electronic configuration is $[\text{Ar}] 3d^6 4s^2$. It has 8 valence electrons.

3. Determine the hybridization and shape:

   The coordination number is 5. The ligand $\text{CO}$ is a strong-field ligand. It causes the pairing of all valence electrons (from both $3d$ and $4s$) into the $3d$ orbitals.

   The 8 valence electrons will pair up in the $3d$ orbitals, resulting in a $3d^8$ configuration within the complex.

   $ \begin{array}{cc} \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\uparrow\downarrow} \ \underline{\quad} & \underline{\quad} & \underline{\quad} \ \underline{\quad} \ \underline{\quad} \\ 3d & 4s & 4p \end{array} $

   For bonding with five $\text{CO}$ ligands, one empty $3d$, one $4s$, and three $4p$ orbitals are used for hybridization. This gives $dsp^3$ hybridization.

   The geometry corresponding to $dsp^3$ hybridization is Trigonal bipyramidal.

   Therefore, D matches with II.

Conclusion:

Let's summarise our findings:

• A. $[\text{Pt}(\text{Cl}_2)(\text{NH}_3)_2]$ is Square planar (III).

• B. $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$ is Octahedral (I).

• C. $[\text{NiCl}_4]^{2-}$ is Tetrahedral (IV).

• D. $[\text{Fe}(\text{CO})_5]$ is Trigonal bipyramidal (II).

Matching these results with the given options, we get:

A-III, B-I, C-IV, D-II.

This corresponds to Option B.

The correct answer is Option B.

Conductance of any complex depends on the following factor.

(1) Number of ions produced by complex.

(2) If number of ions are same then we will check charge on complex unit.

(1) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3 \rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{+3}+3 \mathrm{Cl}^{-}$

(2) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl} \rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{+}+\mathrm{Cl}^{-}$

$\left.\begin{array}{l}\text { (3) }\left[\mathrm{Co}^{+3}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right] \\ \text { (4) }\left[\mathrm{Co}^{+2}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\end{array}\right\}$ Both complex units have no charge. Therefore both complex units have same conductance.

Explanation

A. $[ \mathrm{NiCl}_4 ]^{2-}$

Nickel Oxidation State: $ \text{Ni}^{2+} $

Electron Configuration: $ 3d^8 $

Hybridization: $ sp^3 $

Unpaired Electrons: 2

Magnetic Property: Paramagnetic

B. $\mathrm{Ni}(\mathrm{CO})_4$

Nickel Oxidation State: $\text{Ni}$ (neutral)

Electron Configuration: $ 3d^8 4s^2 $

Hybridization: $ sp^3 $

Unpaired Electrons: 0

Magnetic Property: Diamagnetic

C. $[ \mathrm{Ni}(\mathrm{CN})_4 ]^{2-}$

Nickel Oxidation State: $ \text{Ni}^{2+} $

Electron Configuration: $ 3d^8 $

Hybridization: $ dsp^2 $

Unpaired Electrons: 0

Magnetic Property: Diamagnetic

D. $[ \mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6 ]^{2+}$

Nickel Oxidation State: $ \text{Ni}^{2+} $

Electron Configuration: $ 3d^8 $

Hybridization: $ sp^3d^2 $

Unpaired Electrons: 2

Magnetic Property: Paramagnetic

E. $\mathrm{Ni}(\mathrm{PPh}_3)_4$

Nickel Oxidation State: $\text{Ni}$ (neutral)

Electron Configuration: $ 3d^8 4s^2 $

Hybridization: $ sp^3 $

Unpaired Electrons: 0

Magnetic Property: Diamagnetic

Overall, complexes A and D are paramagnetic as they have unpaired electrons.

$\begin{aligned} & \lambda \propto \frac{1}{\text { strength of ligand }} \\ & \lambda \propto \frac{1}{\text { splitting }} \end{aligned}$

A. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\quad$; 475 nm

B. $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \quad ; \quad 310 \mathrm{~nm}$

C. $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{2+} \quad ; \quad 600 \mathrm{~nm}$

D. $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \quad ; \quad 498 \mathrm{~nm}$

Order of $\lambda=\mathrm{C}>\mathrm{D}>\mathrm{A}>\mathrm{B}$

Ethylene diaminetetraacetate

It is hexadentate as it can bind through two nitrogen and four oxygen atoms to a central metal ion.

Ambidentate ligands are those ligands which have two different donor atoms and either of two donor atom is attached to the metal during complex formation.

A. Metal-Metal linkage present.

B. Terminal CO groups are present

C. In $[\mathrm{Mn}_2(\mathrm{CO})_{10}]$ bridging $\mathrm{CO}$ groups are not present.

D. Metal is in zero oxidation state.

$ \text { In }\left[\mathrm{Co}\left(\mathrm{NH}_3\right) 6\right]^{3+}, \mathrm{Co}^{3+} \text { ion is having } 3 \mathrm{~d}^6 \text { configuration. } $

$ \text { In presence of } \mathrm{NH}_3 \text { ligand, pairing of electrons takes place and it becomes diamagnetic complex ion. } $

$\therefore \quad\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is octahedral with $\mathrm{d}^2 \mathrm{sp}^3$ hybridisation and it is diamagnetic in nature. In case of [CoF6]3-, Co is in +3 oxidation state and it is having 3d6 configuration.

In presence of weak field $\mathrm{F}$ - ligand, pairing does not take place.

$ \therefore \quad \ln \left[\mathrm{CoF}_6\right]^{3-}, \mathrm{Co}^{3+} \text { is } \mathrm{sp}^3 \mathrm{~d}^2 \text { hybridised with four unpaired electrons, so it is paramagnetic in nature. } $

To determine the correctness of these statements and choose the correct option, we need to understand the concepts of homoleptic and heteroleptic complexes:

Homoleptic Complex: A complex in which a central atom is surrounded by only one kind of donor atom or identical ligands is known as a homoleptic complex. An example would be $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, where all the ligands are ammonia ($\mathrm{NH}_3$).

Heteroleptic Complex: A complex in which a central atom is surrounded by more than one kind of donor atom or different ligands is known as a heteroleptic complex, such as $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^{+}$ which contains two different types of ligands, ammonia ($\mathrm{NH}_3$) and chloride ($\mathrm{Cl}^-$).

Statement I: $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is a homoleptic complex whereas $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^{+}$ is a heteroleptic complex.

This statement is accurate because $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ indeed is a homoleptic complex as it consists only of ammonia ligands. Conversely, $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^{+}$ is a heteroleptic complex, containing both ammonia and chloride ligands.

Statement II: Complex $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ has only one kind of ligands but $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^{+}$ has more than one kind of ligands.

This statement is also true. As explained, in $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, there is only one type of ligand (ammonia), and in $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^{+}$, there are two different types of ligands (ammonia and chloride), which is consistent with the definition provided.

Conclusion: Both Statement I and Statement II are true. Therefore, the correct choice is:

Option A: Both Statement I and Statement II are true

Maximum covalency of boron is four.

Complex salt is $\mathrm{K}_2\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$

Double salt is $\mathrm{KAl}\left(\mathrm{SO}_4\right)_2 \cdot 12 \mathrm{H}_2 \mathrm{O}$ (potash alum)

Given complex compounds exhibit solvate isomerism having co-ordination number = 6.

(1) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right]$

(2) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{CO}_{3}\right)\right] \mathrm{Cl}$

(3) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right] \mathrm{Cl}_{3}$

(4) $\mathrm{K}_{3}\left[\mathrm{Al}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$

Option 4 contain all ligands are of same type i.e. why complex will be homoleptic.

due to Chelation effect of (en).

(a) $[Co{(N{H_3})_5}N{O_2}]C{l_2}$ and $[Co{(N{H_3})_5}ONO]C{l_2}$ $\to$ linkage isomerism

(b) $[Cr{(N{H_3})_6}][Co{(CN)_6}]$ and $[Cr{(CN)_6}][Co{(N{H_3})_6}]$ $\to$ coordination isomerism

(c) $[Co{(N{H_3})_5}(S{O_4})]Br$ and $[Co{(N{H_3})_5}Br]S{O_4}$ $\to$ ionisation isomerism

(d) $[Cr{({H_2}O)_6}]C{l_3}$ and $[Cr{({H_2}O)_5}Cl]C{l_2}\,.\,{H_2}O$ $\to$ solvate isomerism

(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)

In case of metal carbonyls, the bonding has both $\sigma$ and $\pi$ nature, where ligand to metal bond is '$\sigma$' (coordinate) bond and metal to ligand bond is '$\pi$' (synergic) bond.

[Ag(H₂O)₂][Ag(CN)₂]

IUPAC name : diaquasilver(I) dicyanidoargentate(I)

Stronger the field strength of ligand, higher will be the energy absorbed by the complex.

$\Rightarrow$ 'en' has a stronger field strength than 'H₂O' according to spectrochemical series

$\therefore$ Correct order of energy absorbed will be :

[Ni(en)₃]²⁺ > [Ni(H₂O)₂(en)₂]²⁺ > [Ni(H₂O)₄(en)]²⁺

i.e. (C) > (A) > (B)

Oxalate $\left(\mathrm{C}_2 \mathrm{O}_4^{2-}\right)$ is a bidentate strong field ligand, thus it can use six orbitals of $\mathrm{Co}^{3+}$ ion having $d^2 s p^3$-hybridisation.

In $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}, \mathrm{Co}^{3+}$ has all the paired electrons, therefore it is a low spin complex.

Hence, option (a) is the correct.

The wavelength of absorption in visible region for a complex (e.g. $\mathrm{Co}^{3+}$) depends on the nature of ligand, i.e. stronger be the ligand (i.e. higher the value of $\Delta_o$) more is the wavelength.

$\therefore$ Strength of ligands in the spectrochemical series are in order: $\mathrm{CN}^{-}>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{O}>\mathrm{Cl}^{-}$.

Thus, option (a) is the correct.

Key idea Complexes with square planar geometry are optically inactive.

Among the given options, only option (a) has square planar geometry with $d s p^2$-hybridisation. Rest all other option show non-superimposable mirror images, thus their structure are optically active. Hence, [ $\mathrm{RhCl}(\mathrm{CO})(\mathrm{P}. \mathrm{Ph}_3)(\mathrm{NH}_3)]$ is optically inactive and option (a) is the correct.

$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ contains $\mathrm{Co}^{3+}$ as central metal ion.

The electronic configuration of $\mathrm{Co}^{3+}$ ion $=[\mathrm{Ar}]_{18} 3 d^6$

The splitting in octahedral field is shown below

$\mathrm{Co}^{3+}=3 d^6\left(\text { or } t_{2 g}^6 e_g\right)$

$\because$ There is no unpaired electron, so it is diamagnetic in nature.

Since, the coordination number of $\mathrm{Mn}^{2+}$ ion in the complex ion is 4 , it will have either tetrahedral ($s p^3$-hybridisation) or square planar ($d s p^2$-hybridisation) geometry. But the fact that the magnetic moment of the complex ion is 5.9 BM suggests that it should have tetrahedral geometry rather than square planar because of the presence of five unpaired electrons in the $d$-orbitals.

The number of unpaired electrons can be determined as follows:

As we know that, magnetic moment $=\sqrt{n(n+2)}$

$5.9=\sqrt{n(n+2)}$

On squaring both sides, the above equation becomes

$\begin{aligned} (5.9)^2 & =n^2+2 n \\ n^2+2 n-35 & =0 \end{aligned}$

On solving the above equation, we get $n=5$

$[\mathrm{Pt}(\mathrm{Py})_4][\mathrm{PtCl}_4]$

Let the oxidation state of $\mathrm{Pt}$ is $x$.

$x+0 \times 4+x+(-1) \times 4=0$

or, $2 x=+4 \text { or, } x=+2$

Thus, it is the complex in which the central metal atom in cationic and anionic parts have same oxidation state.

The IUPAC naming of coordination complexes follows specific rules, where ligands are named in alphabetical order, followed by the central metal and its oxidation state in parentheses.

For the complex $[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3 \mathrm{Br}\left(\mathrm{NO}_2\right) \mathrm{Cl}] \mathrm{Cl}$:

Identify the ligands in alphabetical order: The ligands within the coordination sphere include ammine ($\mathrm{NH}_3$), bromo ($\mathrm{Br}$), chloro ($\mathrm{Cl}$), and nitro ($\mathrm{NO}_2$). Arranging these in alphabetical order: ammine, bromo, chloro, nitro.

Number of each type of ligand:

Ammine: 3

Bromo: 1

Chloro: 1

Nitro: 1

Naming the ligands:

Ammine is represented as "triamine" due to three molecules.

Bromo, chloro, and nitro are prefixed with the number when there are multiple, but here there is only one of each, so no prefix number is necessary, just bromo, chloro, nitro.

Determining the oxidation state of platinum:

The overall charge of the complex is neutral.

Chloride ion outside contributes -1 charge.

Ammine is neutral, bromo is -1, chloro is -1, and nitro is -1.

Let the oxidation state of platinum be $ x $.

Solving for $ x $:

$ x + 0 \times 3 - 1 - 1 - 1 = +1 $

$ x - 3 = +1 $

$ x = +4 $

The name of the complex:

Considering the alphabetical order of ligands and the name of the metal followed by its oxidation state:

$ \text{triamine bromidochloridonitro platinum(IV) chloride} $

Thus, the IUPAC name of the complex is: Option B: triamine bromidochloridonitro platinum (IV) chloride.

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{+}$ shows geometrical isomerism because it is a $M A_4 B_2$ type coordination compound which contains two set of equivalent ligands, four $\mathrm{H}_2 \mathrm{O}$ and $2 \mathrm{Cl}$.

Hence, the possible geometrical isomers are:

Hence, correct option is (a).

Assertion and reason both are correct and reason is the correct explanation of the assertion. Toxic metal ions are removed by chelating ligands. When a solution of chelating ligand is added to solution containing toxic metal, ligands chelate the metal ions by formation of stable complex.