Coordination Compounds

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)

Among the options listed, Option B, cis-PtCl₂(NH₃)₂], is the one used as an anticancer agent. This compound is commonly known as Cisplatin. It's a platinum-based drug used in chemotherapy and is effective in treating various types of cancers, including testicular, ovarian, and bladder cancer.

The effectiveness of Cisplatin in cancer treatment lies in its ability to form platinum-DNA adducts, which interfere with DNA repair mechanisms, leading to cell apoptosis (programmed cell death) in rapidly dividing cells, such as cancer cells.