Coordination Compounds

$ \text { Field strength order : } \mathrm{CO}>\mathrm{H}_2 \mathrm{O}>\mathrm{F}^{-}>\mathrm{S}^{2-} $

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

$\mathrm{Co}^{3+}($ strong field ligand $) \Rightarrow 3 \mathrm{~d}^6\left(\mathrm{t}_{2 \mathrm{~g}}^6, \mathrm{e}_{\mathrm{g}}^0\right)$,

Hybridisation : $\mathrm{d}^2 \mathrm{sp}^3$

Inner obital complex(spin paired complex)

Pairing will take place.

$ \left[\mathrm{CoF}_6\right]^{3-} $

$\mathrm{Co}^{3+}($ weak field ligand $) \Rightarrow 3 \mathrm{~d}^6\left(\mathrm{t}_{2 \mathrm{~g}}^4, \mathrm{e}_{\mathrm{g}}^2\right)$

Hybridisation : $\mathrm{sp}^3 \mathrm{~d}^2$

Outer orbital complex (spin free complex)

No pairing will take place

Let's analyze each statement separately:

Statement (I):

So, Statement I is false.

Statement (II): Prussian blue, also known as ferric ferrocyanide, is a famous pigment that contains iron in both +2 and +3 oxidation states. Its chemical formula can be represented as $\mathrm{Fe}_4\mathrm{[Fe(CN)_6]_3}$. In this formula, the iron within the square brackets, $\mathrm{[Fe(CN)_6]^{4-}}$, is in the +2 oxidation state (ferrocyanide ion), while the iron outside the square brackets is in the +3 oxidation state (ferric ion). The compound is a complex salt that arises from the reaction of ferric and ferrous ions with cyanide ions. Therefore, Statement II is also true.

With both statements being true, the correct answer is:

Option A : Statement I is false but Statement II is true.

Among the given options, the compounds that show color due to d-d transitions are those that have partially filled d-orbitals when in the form of complex ions or compounds.

Let's examine each option:

Option A: $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$

Chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ exists in the +6 oxidation state as the dichromate ion ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$). In this oxidation state, chromium does not have any electrons in the d-orbitals (it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ configuration) so there is no possibility for d-d transitions. Instead, the color is due to charge transfer transitions.

Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$

Copper(II) sulfate pentahydrate, $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$, contains copper in a +2 oxidation state. In this state, copper has a $\mathrm{[Ar]}\;3\mathrm{d}^9$ electron configuration, which means there is one unpaired electron in the d-orbitals. Due to the presence of this unpaired electron, d-d transitions can occur, and this is why $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in color.

Option C: $\mathrm{KMnO}_4$

In potassium permanganate ($\mathrm{KMnO}_4$), manganese is in the +7 oxidation state, and it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ electron configuration. Like chromium in the +6 state, manganese in the +7 state does not have any d-electrons available for d-d transitions. The deep purple color of $\mathrm{KMnO}_4$ is also due to charge transfer transitions.

Option D: $\mathrm{K}_2 \mathrm{CrO}_4$

Potassium chromate ($\mathrm{K}_2 \mathrm{CrO}_4$) contains chromium in the +6 oxidation state as the chromate ion ($\mathrm{CrO}_4^{2-}$). Like dichromate, chromium does not have d-electrons and therefore cannot undergo d-d transitions in this compound. The color is due to charge transfer transitions.

So the only compound from the options listed that shows color due to d-d transitions is Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$.

A homoleptic complex is one in which a central metal atom or ion is surrounded by only one kind of donor groups, ligands or ions. On the other hand, a heteroleptic complex contains a central metal surrounded by more than one kind of donor groups, ligands or ions.

Let's examine the given options :

Option A: $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$

This complex contains two different ligands, ammonia ($\mathrm{NH}_3$) and chloride (Cl). Therefore, it is a heteroleptic complex.

Option B: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$

Just like Option A, this complex has two types of ligands, ammonia ($\mathrm{NH}_3$) and chloride (Cl), making it a heteroleptic complex.

Option C: $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$

Again, this complex has ammonia ($\mathrm{NH}_3$) and chloride (Cl) ligands, so it is a heteroleptic complex.

Option D: $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

This complex has only cyanide (CN) ligands surrounding the nickel ion, with no other types of ligands present. Thus, this is a homoleptic complex because it is coordinated by a single type of ligand.

Therefore, Option D is the correct answer as it represents a homoleptic complex.

In order to determine the correctness of the given statements, we need to analyze the nature of the mentioned complexes.

Statement (I) asserts that a solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green in color. This complex involves a nickel(II) cation coordinated with six water molecules. Nickel(II) has the electronic configuration $[Ar]3d^8$. In an octahedral field created by water ligands, the d-orbitals split into two sets, $t_{2g}$ and $e_g$, due to crystal field splitting. The presence of unpaired electrons allows for d-d transitions when light is absorbed, which is responsible for the characteristic color of the complex. Since typical $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ complexes are indeed green in color due to such transitions, Statement (I) is correct.

Statement (II) states that a solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colorless. In this complex, nickel(II) is surrounded by four cyanide ligands. Cyanide is a strong field ligand, which leads to a large crystal field splitting that exceeds the pairing energy of the electrons. Hence, all the electrons in the nickel(II) ion are paired, and there are no unpaired electrons available for d-d transitions. As there are no d-d transitions to absorb light in the visible region, the complex appears colorless. Therefore, Statement (II) is also correct.

Given that both statements are correct, the most appropriate answer would be :

Option B : Both Statement I and Statement II are correct.

$\left[\mathrm{Ni}(\mathrm{CO})_4\right] \rightarrow$ diamagnetic, $\mathrm{sp}^3$ hybridisation, number of unpaired electrons $=0$

$\left[\mathrm{NiCl}_4\right]^{2-}, \rightarrow$ paramagnetic, $\mathrm{sp}^3$ hybridisation, number of unpaired electrons $=2$

$\begin{aligned} & {\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { Contains } \mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^3: \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^{\mathrm{o}}} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { Contains } \mathrm{Fe}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^5: \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^2} \\ & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \text { Contains } \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^8: \mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{e}_{\mathrm{g}}^2} \\ & {\left[\mathrm{~V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { Contains } \mathrm{V}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^2: \mathrm{t}_{2 \mathrm{~g}}^2 \mathrm{eg}^{\mathrm{o}}} \end{aligned}$

(A) Strength of anionic ligands cannot be explained by CFT instead LFT i.e., ligand field theory explains the strength of ligands.

(B) VBT does not give a quantitative interpretation of kinetic stability of coordination compounds.

(C) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \rightarrow \mathrm{Ni}^{2+}$ in presence of $\mathrm{CN}^{-}$ligand.

Square planar geometry since $\rightarrow 4, d s p^2$ hybridised orbitals created.

(D) cis- $\left[\mathrm{PtCl}_2(\mathrm{en})_2\right] \Rightarrow$ It has two possible isomers

$\mathrm{Mn_2(CO)_{10}}$

Let's examine each statement for correctness:

(A) Ethane-1, 2-diamine is a chelating ligand.

Ethane-1,2-diamine, also known as ethylenediamine (en), has two nitrogen atoms that can coordinate to a metal ion, forming a ring structure in the process. Because it can form these two bonds, it can "chelate" a metal ion, thus it is correctly identified as a chelating ligand.

(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.

Aluminium is indeed produced industrially by the Hall-Héroult process, which involves the electrolysis of aluminium oxide ($\mathrm{Al_2O_3}$) dissolved in molten cryolite ($\mathrm{Na_3AlF_6}$). Cryolite acts as a solvent for the aluminium oxide and reduces the melting point of the mixture, thus decreasing energy consumption during electrolysis. This statement is correct.

(C) Cyanide ion is used as a ligand for leaching of silver.

Gold and silver are often extracted from their ores via a leaching process using a cyanide solution. The cyanide ion ($\mathrm{CN^-}$) complexes with the metal ions to form soluble complexes like [Ag(CN)$_2$]$^-$, enabling the separation of silver from the ore. So, this statement is correct as well.

(D) Phosphine act as a ligand in Wilkinson's catalyst.

Wilkinson's catalyst is $\mathrm{RhCl(PPh_3)_3}$, where PPh$_3$ stands for triphenylphosphine, a type of phosphine ligand. Phosphines are indeed ligands in Wilkinson's catalyst, and they play an important role in its catalytic activity, particularly in hydrogenation reactions. Therefore, this statement is correct.

(E) The stability constants of $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ are similar with EDTA complexes.

EDTA (ethylenediaminetetraacetic acid) forms strong complexes with many metal ions including $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$. However, the stability constants of their complexes with EDTA are not similar; the stability constant for the calcium complex is notably higher than that for the magnesium complex. Thus, this statement is incorrect.

With all the information above, we can conclude:

(A) is correct, (B) is correct, (C) is correct, (D) is correct, and (E) is incorrect.

Therefore, the correct statements are (A), (B), (C), and (D), making Option D—(A), (B), (C) only—the correct choice.

$\mathrm{Ni}^{2+}+2 \mathrm{dmg}^{-} \rightarrow\left[\mathrm{Ni}(\mathrm{dmg})_2\right]$

Rosy red/Bright Red precipitate

Ziegler catalyst $\rightarrow$ Titanium

Blood pigment $\rightarrow$ Iron

Wilkinson catalyst $\rightarrow$ Rhodium

Vitamin $\mathrm{B}_{12} \rightarrow$ Cobalt

The catalyst used in Wacker's process is $\mathrm{PdCl_2}$

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{+3}-\mathrm{d}^2 \mathrm{sp}^3$ hybridization

$\mathrm{BrF}_5-\mathrm{sp}^3 \mathrm{d}^2$ hybridization

$\left[\mathrm{PtCl}_4\right]^{-2}-\mathrm{dsp}^2$ hybridization

$\mathrm{SF}_6-\mathrm{sp}^3 \mathrm{d}^2$ hybridization

$\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^5$

F : Weak field Ligand

No. of unpaired electron's $=5$

$\begin{aligned} & \mu=\sqrt{5(5+2)} \\ & \mu=\sqrt{35} \mathrm{~BM} \\ & {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{V}^{+2}: 3 \mathrm{~d}^3} \end{aligned}$

No. of unpaired electron's $=3$

$\begin{aligned} & \mu=\sqrt{3(3+2)} \\ & \mu=\sqrt{15} \mathrm{~BM} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6} \end{aligned}$

H$_2$O : Weak field Ligand

No. of unpaired electron's $=4$

$\begin{aligned} & \mu=\sqrt{4(4+2)} \\ & \mu=\sqrt{24} \mathrm{~BM} \end{aligned}$

Lead chromate ( $\mathrm{PbCrO}_4$ ) dissolves in hot NaOH solution to form products due to a chemical reaction. The reaction involves the formation of a compound where lead ( $\mathrm{Pb}$ ) is coordinated by hydroxide ions ( $\mathrm{OH}^-$ ).

The balanced chemical reaction is:

$\mathrm{PbCrO}_4 + 4\mathrm{NaOH} (hot, excess) \rightarrow \left[\mathrm{Pb}(\mathrm{OH})_4\right]^{2-} + \mathrm{Na}_2\mathrm{CrO}_4$

In this reaction, lead forms a dianionic complex ( $\left[\mathrm{Pb}(\mathrm{OH})_4\right]^{2-}$ ) with hydroxide ions. This indicates that the lead is in the center of a complex with a coordination number of four, meaning it is directly bound to four hydroxide ions. Therefore, the correct description of the product formed with lead is a dianionic complex with a coordination number of four.

The crystal field splitting energy ($\Delta_0$) is proportional to the electrostatic interaction between the metal ion and the ligands. This interaction is affected by the size of the metal ion, as well as the size of the ligands.

The size of the metal ion can be estimated by its ionic radius, which is the distance from the nucleus to the outermost electron shell. A smaller ionic radius corresponds to a greater electrostatic interaction with the ligands, resulting in a larger $\Delta_0$ value.

Out of the given options, the ionic radii of the metal ions are:

- $\mathrm{Ti}^{3+}$: 67 pm

- $\mathrm{Cr}^{3+}$: 62 pm

- $\mathrm{Mn}^{3+}$: 65 pm

- $\mathrm{Fe}^{3+}$: 65 pm

The smaller ionic radius of $\mathrm{Cr}^{3+}$ compared to the other metal ions indicates a stronger electrostatic interaction with the ligands, resulting in a higher crystal field splitting energy ($\Delta_0$).

Therefore, the correct answer is option (C) $\left[\mathrm{Cr}\left(\mathrm{OH}_2\right)_6\right]^{3+}$, as it has the highest tendency to attract ligands due to its smaller ionic radius.

$\left[\mathrm{Cr}(\mathrm{Ox})_2 \mathrm{ClBr}\right]^{-3}$

- No. of isomers -


- This structure has plane of symmetry, So no optical isomerism will be shown.


- This structure does not contain plane of symmetry, So two forms $d$ as well as 1 will be shown.

To determine which complex exhibits maximum attraction to an applied magnetic field, we need to look at the number of unpaired electrons in each complex. The more unpaired electrons a complex has, the greater the attraction to the magnetic field.

1. Option A: [Ni(H₂O)₆]²⁺ Ni²⁺ has a 3d⁸ electron configuration. In an octahedral crystal field, two of the electrons will occupy the lower energy t₂g orbitals, while the remaining six electrons will fill the higher energy e_g orbitals. There are 2 unpaired electrons in this complex.

2. Option B: [Co(H₂O)₆]²⁺ Co²⁺ has a 3d⁷ electron configuration. In an octahedral crystal field, two of the electrons will occupy the lower energy t₂g orbitals, while the remaining five electrons will fill the higher energy e_g orbitals. There are 3 unpaired electrons in this complex.

3. Option C: [Co(en)₃]³⁺ Co³⁺ has a 3d⁶ electron configuration. In an octahedral crystal field, all six electrons will occupy the lower energy t₂g orbitals. There are no unpaired electrons in this complex.

4. Option D: [Zn(H₂O)₆]²⁺ Zn²⁺ has a 3d¹⁰ electron configuration, with all orbitals being completely filled. There are no unpaired electrons in this complex.

Based on the number of unpaired electrons, the complex with the maximum attraction to an applied magnetic field is [Co(H₂O)₆]²⁺ (Option B), as it has 3 unpaired electrons.

Let's analyze each combination :

A. Chlorophyll - Co : Mismatched. Chlorophyll has a magnesium (Mg) ion at its center, not cobalt (Co).

B. Water hardness - EDTA : Correct match. EDTA (ethylenediaminetetraacetic acid) is used to treat water hardness by chelating metal ions like Ca²⁺ and Mg²⁺.

C. Photography - [Ag(CN)₂]⁻ : Mismatched. Silver halides (AgX, where X = Cl, Br, I) are used in photography, not [Ag(CN)₂]⁻.

D. Wilkinson catalyst - [(Ph₃P)₃RhCl] : Correct match. Wilkinson's catalyst is a homogeneous hydrogenation catalyst with the formula [(Ph₃P)₃RhCl].

E. Chelating ligand - D-Penicillamine : Correct match. D-Penicillamine is a chelating agent that can bind to metal ions through multiple coordination sites.

The mismatched combinations are A and C. So, the correct answer is :

A and C Only.

In order to match these complexes with their respective crystal field splitting energies (CFSE), we need to understand the coordination number, the geometry of the complex, and the nature of the ligand.

The crystal field splitting energy, often denoted as Δ₀ or Δ, is the energy difference between the higher-energy and lower-energy sets of d-orbitals in a transition metal ion when in a crystal field created by a set of ligands.

The sign of CFSE is often taken as negative because it represents a lowering in energy due to the field of the ligands. The magnitude of CFSE depends on the nature of the ligand, with ligands causing larger splitting being known as strong-field ligands and those causing smaller splitting being known as weak-field ligands.

Based on spectrochemical series, we know the strength of ligands as follows: F⁻ < H₂O < NH₃ < CN⁻

The crystal field splitting energy also depends on the configuration of the d electrons in the transition metal ion. For octahedral complexes, the d orbitals split into two sets with an energy difference Δ₀: a lower-energy set (dxy, dyz, dxz) called t₂g, and a higher-energy set (dz², dx²-y²) called eg.

Now let's analyze each complex:

1. [Cu(NH₃)₆]²⁺: Cu²⁺ ion has a d⁹ configuration. It is an octahedral complex with a strong field ligand (NH₃), hence the splitting will be large.

2. [Ti(H₂O)₆]³⁺: Ti³⁺ ion has a d¹ configuration. It is an octahedral complex with a weak field ligand (H₂O), hence the splitting will be small.

3. [Fe(CN)₆]³⁻: Fe³⁺ ion has a d⁵ configuration. It is an octahedral complex with a very strong field ligand (CN⁻), hence the splitting will be very large.

4. [NiF₆]⁴⁻: Ni²⁺ ion has a d⁸ configuration. It is an octahedral complex with a very weak field ligand (F⁻), hence the splitting will be very small.

So, the correct order of CFSE values for these complexes should be:

[NiF₆]⁴⁻ < [Ti(H₂O)₆]³⁺ < [Cu(NH₃)₆]²⁺ < [Fe(CN)₆]³⁻

Therefore, the correct answer is:

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

Assertion A : $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ absorbs at a lower wavelength of light with respect to $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$.

Given that H₂O is a stronger field ligand than Cl^-, it would result in a larger splitting of the d-orbitals and hence a larger energy difference. This would lead to absorption at a lower wavelength. Therefore, the Assertion A is false.

Reason R : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.

While it's true that the oxidation state can influence the absorption wavelength by affecting the energy levels of the d-orbitals, the difference in the absorption wavelength in this specific case is not primarily due to the oxidation state (the oxidation state of Co in both complexes is the same (+3)), but rather due to the difference in ligand field strength (H₂O vs Cl^-). Therefore, the Reason R is not providing the correct explanation for Assertion A.

So, the correct answer would be:

Option A : A is false but R is true.

The complex $\left[\mathrm{NiCl}_{2} \mathrm{Br}_{2}\right]^{2-}$ is a tetrahedral complex due to the fact that Nickel (Ni) in this complex is in its +2 oxidation state, and thus adopts a dsp² hybridization, resulting in a tetrahedral geometry.

If $\mathrm{Ni}^{2+}$ is replaced by $\mathrm{Pt}^{2+}$, the new complex would be $\left[\mathrm{PtCl}_{2} \mathrm{Br}_{2}\right]^{2-}$. Platinum (Pt) in its +2 oxidation state typically forms square planar complexes due to its preference for dsp² hybridization.

1. Geometry :

   
   
   • $\left[\mathrm{NiBr}_2 \mathrm{Cl}_2\right]^{2-}$ is tetrahedral.
   • $\left[\mathrm{PtBr}_2 \mathrm{Cl}_2\right]^{2-}$ is square planar.
   • Thus, replacing $\mathrm{Ni}^{2+}$ with $\mathrm{Pt}^{2+}$ changes the geometry from tetrahedral to square planar.



2. Geometrical Isomerism :

   
   
   • Tetrahedral complexes, like $\left[\mathrm{NiBr}_2 \mathrm{Cl}_2\right]^{2-}$, do not exhibit geometrical isomerism.
   • Square planar complexes, like $\left[\mathrm{PtBr}_2 \mathrm{Cl}_2\right]^{2-}$, can exhibit geometrical isomerism.
   • Therefore, geometrical isomerism would be introduced by replacing $\mathrm{Ni}^{2+}$ with $\mathrm{Pt}^{2+}$.



3. Optical Isomerism :

   
   
   • Both the $\mathrm{Ni}$-based tetrahedral complex and the $\mathrm{Pt}$-based square planar complex are optically inactive.
   • Optical isomerism is not observed in either case.



4. Magnetic Properties :

   
   
   • The magnetic properties depend on the electronic configuration of the central metal ion.
   • There will be a change in magnetic properties due to the different electronic configurations of $\mathrm{Ni}^{2+}$ and $\mathrm{Pt}^{2+}$.

Based on this information, the properties expected to change when $\mathrm{Ni}^{2+}$ is replaced by $\mathrm{Pt}^{2+}$ are:

• Geometry (from tetrahedral to square planar)
• Geometrical Isomerism (introduced in the $\mathrm{Pt}$-based complex)
• Magnetic Properties (due to different electronic configurations)

Optical isomerism remains unchanged (inactive in both cases).

Therefore, the correct answer is :

Option B: A, B, and D (Geometry, Geometrical Isomerism, and Magnetic Properties).

In this problem, we are matching chemical compounds with their corresponding color. Let's go over each one :

A. $Mg(NH_4)PO_4$ - Magnesium ammonium phosphate is white in color, so A matches with II.

B. $K_3[Co(NO_2)_6]$ - Potassium hexanitritocobaltate(III) is a yellow complex, so B matches with III.

C. $MnO(OH)_2$ - Manganese hydroxide is brown in color, so C matches with I.

D. $Fe_4[Fe(CN)_6]_3$ - Prussian blue or Ferric ferrocyanide is deep blue in color, so D matches with IV.

So, the correct option is Option B: A-II, B-III, C-I, D-IV.

For spin-only magnetic moment, we use the formula:

$\mathrm{Magnetic ~moment} = \sqrt{n(n+2)} \cdot \mathrm{BM}$

where $n$ is the total number of unpaired electrons.

For $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$, the electronic configuration of $\mathrm{Fe^{3+}}$ is $\mathrm{d^5}$. Here, all the five electrons will be unpaired due to the high spin nature of Fe$^{3+}$. Hence, $n=5$ and magnetic moment

$= \sqrt{5(5+2)} \cdot \mathrm{BM} \approx 5.92 \mathrm{BM}$

For $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, the $\mathrm{Fe^{3+}}$ ion has electronic configuration $\mathrm{d^5}$. In this case, the strong field ligand cyanide ($\mathrm{CN}^{-}$) will pair up the electrons in the $\mathrm{d}$ orbital. So, there will be only one unpaired electron. Hence, $n=1$ and magnetic moment

$= \sqrt{1(1+2)} \cdot \mathrm{BM} \approx 1.732 \mathrm{BM}$

Therefore, the answer is 5.92 B.M. and 1.732 B.M.

Option A : $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$ (Prussian Blue)

Prussian blue is a very stable and intensely colored compound, but it is not soluble in water. The individual cyanide ligands and the iron ions are bonded together strongly, which results in a crystalline structure that does not easily dissociate into its components in water.

Option B : $\left(\mathrm{NH}_{4}\right)_{3}\left[\mathrm{As}\left(\mathrm{Mo}_{3} \mathrm{O}_{10}\right)_{4}\right]$ (Ammonium Arseno Molybdate)

In this compound, the arsenic and molybdenum atoms are coordinated in a complex polyatomic anion. Despite the presence of ammonium ions, which are generally soluble, the overall complex has a low solubility due to the strong interactions between arsenic, molybdenum, and oxygen within the anion.

Option C : $\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right]$

This is a potassium salt of an anionic complex. Although potassium salts are generally soluble in water, the nitrite ligands attached to the cobalt in the anion form a tightly bonded structure, resulting in very poor solubility in water.

Option D : $\left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}(\mathrm{OAc})_{6}\right] \mathrm{Cl}$

This complex is composed of iron atoms coordinated with hydroxide and acetate ligands, and a chloride counterion. Hydroxide and acetate ions are both polar and capable of forming hydrogen bonds with water molecules, which is a key factor in solubility. The chloride ion is also soluble in water. The presence of these ions, combined with the likely overall charge of the complex (since it's paired with a counterion), makes it soluble in water.

Option A :

- $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$ (Oxalate) is a bidentate ligand. It always binds through its two oxygen atoms.

- $\mathrm{NO}_{2}{ }^{-}$ (Nitrite) is an ambidentate ligand. It can coordinate either through nitrogen or oxygen.

- $\mathrm{NCS}^{-}$ (Thiocyanate) can also bind through either nitrogen or sulfur, making it an ambidentate ligand.

Option B :

- $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$ (Oxalate) as mentioned earlier, is a bidentate ligand.

- Ethylene diamine (en) is also a bidentate ligand, as it always binds through its two nitrogen atoms.

- $\mathrm{H}_{2} \mathrm{O}$ (Water) is a monodentate ligand, binding only through one oxygen atom.

Option C :

- $\mathrm{NO}_{2}^{-}$ as mentioned, is an ambidentate ligand.

- $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$ as mentioned, is a bidentate ligand.

- $\mathrm{EDTA}^{4-}$ is a hexadentate ligand. It always binds through six donor atoms and is not ambidentate.

Option D :

- $\mathrm{EDTA}^{4-}$ as mentioned, is a hexadentate ligand.

- $\mathrm{NCS}^{-}$ as mentioned, is an ambidentate ligand.

- $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$ as mentioned, is a bidentate ligand.

From the above analysis, it's clear that Option B is the one that does not contain any ambidentate ligands. So, the correct answer is Option B.

Meridional (mer) isomerism is a type of geometrical isomerism found specifically in octahedral complexes, where three identical ligands occupy a meridian plane. Thus, it requires at least three identical bidentate or monodentate ligands.

Given the options, the complex that could potentially exist as a meridional isomer is :

Option D : $[\mathrm{Co}(\mathrm{NH}_{3})_{3}(\mathrm{NO}_{2})_{3}]$

In this complex, there are three $\mathrm{NH_3}$ and three $\mathrm{NO_2}$ ligands, allowing it to form a meridional configuration where either all three $\mathrm{NH_3}$ or all three $\mathrm{NO_2}$ ligands lie along a meridian plane.

The number of unpaired electrons in a complex can be determined by the oxidation state of the central metal atom and the ligand field splitting energy. In general, a complex with a high-spin configuration will have more unpaired electrons than a complex with a low-spin configuration.

In the given complexes, the oxidation states of the central metal atoms are:

• Fe(II) in [Fe(CN)₆]^3-


• Fe(III) in [FeF₆]^3-


• Co(III) in [CoF₆]^3-


• Cr(III) in [Cr(ox)₃]3-


• Ni(0) in [Ni(CO)₄]

The ligand field splitting energy of a complex depends on the type of ligand. In general, ligands that are strong field ligands will split the d orbitals more than ligands that are weak field ligands.

The following table shows the number of unpaired electrons in each complex :

Complex	Oxidation state	Ligand field splitting energy	Number of unpaired electrons
[Fe(CN)6]3-	Fe(II)	Low-spin	1
[FeF6]3-	Fe(III)	High-spin	5
[CoF6]3-	Co(III)	High-spin	4
[Cr(ox)3]3-	Cr(III)	High-spin	3
[Ni(CO)4]	Ni(0)	None	0

Therefore, the correct order of the number of unpaired electrons in the given complexes is :

E < A < D < C < B

$ \begin{aligned} & \text{(A)} ~~ {\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{Ti}^{2+} \Rightarrow 3 \mathrm{~d}^2 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=2 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\ & \mathrm{CFSE}=[-0.4 \times 2+0.6 \times 0] \Delta_0 \\\\ & \quad=-0.8 \Delta \end{aligned} $

$ \begin{aligned} & \text{(B)} ~~ {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^3 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 0] \Delta_0 \\\\ & =-1.2 \Delta_0 \end{aligned} $

$ \begin{aligned} & \text{(C)} ~~{\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Mn}^{3+} \Rightarrow 3 \mathrm{~d}^4 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=1 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 1] \Delta_0 \\\\ & =-0.6 \Delta_0 \end{aligned} $

$ \begin{aligned} & \text{(D)} ~~{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \quad \mathrm{e}_{\mathrm{g}}=2 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 2] \Delta_0 \\\\ & \quad=0 \Delta_0 \end{aligned} $

In the complex [Co(NH$_3$)$_6$]$^{3+}$, Co is in the +3 oxidation state. This means it has a configuration of $d^6$. NH$_3$ is a strong field ligand, which causes the d-electrons to pair up. Therefore, this complex is a low-spin (also referred to as spin-paired) complex.

The term "low-spin" means that the electrons prefer to pair up in the lower energy d-orbitals rather than occupy the higher energy orbitals. When all the electrons are paired, the complex is diamagnetic, which means it's not attracted to an external magnetic field. The hybridization of this complex is $d^2sp^3$, also known as inner orbital complex, because the inner d-orbitals are used in the hybridization.

The complexes [CoF$_6$]$^{3-}$ and [CoCl$_6$]$^{3-}$ are both octahedral, but they do have unpaired electrons because $F^-$ and $Cl^-$ are not strong enough field ligands to cause pairing of all the electrons in the d-orbitals. Therefore, they are not low-spin or diamagnetic.

Lastly, the complex [NiCl$_4$]$^{2-}$ is not octahedral. Because $Cl^-$ is a weak field ligand, the electrons in the d-orbitals do not pair up, and the $Ni^{2+}$ ion, with a $d^8 $ configuration, undergoes sp^3 hybridization, resulting in a tetrahedral complex.

Therefore, the only octahedral, diamagnetic, and low-spin complex in the options provided is [Co(NH$_3$)$_6$]$^{3+}$ (Option A).

For option (A)

$ \begin{aligned} & \mathrm{Cr}^{+3}: 3 \mathrm{~d}^3 \\\\ & \mathrm{CN}^{-} \rightarrow \mathrm{SFL} \end{aligned} $

No. of unpaired electrons $=3$

For option (B)

$ \begin{aligned} & \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6 \\\\ & \mathrm{H}_2 \mathrm{O}: \mathrm{WFL} \end{aligned} $

No. of unpaired electrons $=4$

For option (C)

$ \begin{aligned} & \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $

No. of unpaired electrons $=0$

For option (D)

$ \begin{aligned} & \mathrm{Ni}^{+2}: 3 \mathrm{~d}^8 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $

No. of unpaired electrons $=2$

Option A : $\mathrm{Na}_{3}\left[\mathrm{CoCl}_{6}\right]$

This is an octahedral complex, but it is not diamagnetic. Co in this complex is in +3 oxidation state, and the d-electron configuration is $t_{2g}^6 e_{g}^0$, leading to two unpaired electrons.

Option B : $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$

This is also an octahedral complex, but it is not diamagnetic. The Co ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^5 e_{g}^2$, leading to 3 unpaired electrons.

Option C : $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$

This complex is both octahedral and diamagnetic. Co in this complex is in +3 oxidation state and the d-electron configuration is $t_{2g}^6 e_{g}^0$, which means no unpaired electrons. This complex is more stable due to a strong field ligand (CN^-) which leads to a greater splitting of d-orbitals and more pairing of electrons.

Option D : $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$

This is an octahedral complex, and it is diamagnetic. The Ni ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^8 e_{g}^0$, leading to no unpaired electrons. However, NH₃ is a weaker field ligand compared to CN^-, so the complex is less stable than Option C.

Hence, the complex that is octahedral, diamagnetic, and the most stable among the given options is $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$.

The spin-only magnetic moment of a complex ion is calculated using the formula $\mu = \sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons in the complex ion. In the case of transition metal complexes, the number of unpaired electrons and hence the magnetic moment can be influenced by the field strength of the ligands (the Ligand Field Theory).

Here, CN$^-$ is a strong-field ligand (or high-spin ligand) that can cause pairing of electrons, while F$^-$ and Br$^-$ are considered weak-field ligands (or low-spin ligands) that do not promote electron pairing.

1. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$: In this complex, Fe is in +3 oxidation state and has 5 d-electrons. Due to the strong field ligand, CN$^-$, all the 5 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$.




2. $\left[\mathrm{CoF}_{6}\right]^{3-}$: Here, Co is in +3 oxidation state and has 6 d-electrons. As F$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 4 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{4(4+2)} = \sqrt{24}$.




3. $\left[\mathrm{MnBr}_{4}\right]^{2-}$: Here, Mn is in +2 oxidation state and has 5 d-electrons. Since Br$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 5 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35}$.




4. $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$: In this complex, Mn is in +3 oxidation state and has 4 d-electrons. Due to the strong field ligand, CN$^-$, all the 4 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$.

So the correct order of spin-only magnetic moments is:

$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}$

To determine the IUPAC name of the compound K₃[Co(C₂O₄)₃], let's follow the nomenclature rules for coordination compounds:

Identify the Cation and Anion: The compound consists of potassium ions (K⁺) and a complex anion [Co(C₂O₄)₃]³⁻.

Name the Cation First: The cation is named first. So, we start with "Potassium."

Determine the Ligand Name: The ligand C₂O₄²⁻ is called "oxalato."

Count the Number of Ligands: There are three oxalato ligands. Since "oxalato" does not contain any numerical prefixes (like di-, tri-), we use the prefixes "di-", "tri-", etc., for simple ligands. So, we use "trioxalato."

Name the Central Metal Atom: Since the complex ion is an anion, the metal name ends with the suffix "-ate." Therefore, "cobalt" becomes "cobaltate."

Specify the Oxidation State: The oxidation state of cobalt in this complex can be calculated:

Let the oxidation state of Co be x.

Each oxalato ligand has a charge of -2.

The overall charge of the complex ion is -3.

So, x + 3(-2) = -3 ⇒ x -6 = -3 ⇒ x = +3.

Therefore, we indicate the oxidation state as (III).

Combining all these, the IUPAC name is Potassium trioxalatocobaltate(III).

Answer: Option C

Potassium trioxalatocobaltate(III)

Unpaired electron $=1$

$ \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1 \times 3}=1.74 \text { B.M. } $

$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ No pairing because $\mathrm{H}_2 \mathrm{O}$ is WFL

Number of unpaired electrons $=5, \mu=5.92 \mathrm{BM}$

Assertion is true, Reason is true but not correct explanation.

Complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}$ will have two isomer one linked through N (Nitro) and one through O (Nitrite).

CN^– is strongest field ligand among given ligands. So maximum splitting in d-orbitals take place.

$A=\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - double salt

B. $\mathrm{CuSO}_{4} \cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ $=\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O} $ - complex salt

C. $\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$ - double salt

D. $\mathrm{Fe}(\mathrm{CN})_{2} \cdot 4 \mathrm{KCN}$ = $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ - complex salt

Option A, $\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, is a double salt because it is formed by the combination of two different salts: iron(II) sulfate ($\mathrm{FeSO}_{4}$) and ammonium sulfate ($\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$).

Option B, $\mathrm{CuSO}_{4}\cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$, is a complex salt because it contains a central copper ion coordinated to several ammonia molecules.

In this compound, the copper(II) ion (Cu²⁺) acts as the central metal ion, while the four ammonia molecules (NH₃) act as ligands, coordinating to the copper ion through their lone pairs of electrons. The sulfate ion (SO₄^2-) and water molecules (H₂O) are not involved in the coordination sphere and simply crystallize along with the complex.

Option C, $\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$, is also a double salt, formed by the combination of two different salts: potassium sulfate ($\mathrm{K}_{2} \mathrm{SO}_{4}$) and aluminum sulfate ($\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$).

The 24 water molecules in the formula are not directly involved in the salt formation but instead function to stabilize the crystal lattice by forming hydrogen bonds with the sulfate ions and other water molecules.

Option D, $\mathrm{Fe}(\mathrm{CN})_{2}\cdot4 \mathrm{KCN}$, is not an example of a double salt. It is a complex salt, formed by the coordination of iron(II) ions to cyanide ligands and potassium ions.