Coordination Compounds

First, let us determine the $d$-electron count of the cobalt center in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$.

1. Oxidation state and $d$-electron count

Neutral cobalt (Co) has an atomic number of 27 and an electronic configuration:

$ \mathrm{Co} \; \bigl[\mathrm{Ar}\bigr]\, 3d^7\, 4s^2. $

In the +3 oxidation state, cobalt has lost a total of 3 electrons:

First two electrons are removed from the $4s$ orbital.

The third electron is removed from the $3d$ orbitals.

Thus, $\mathrm{Co}^{3+}$ has

$ 3d^{7-1} = 3d^6. $

So in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$, the cobalt center is $d^6$.

2. High-spin vs Low-spin for $\mathrm{Co}^{3+}$

Even though $\mathrm{H}_2\mathrm{O}$ is generally considered a weak field ligand, the $\mathrm{Co}^{3+}$ ion has a relatively high charge (+3). A higher charge on the metal center usually increases the ligand-field splitting ($\Delta_\mathrm{o}$) significantly compared to lower oxidation states of the same metal.

In most typical octahedral complexes of $\mathrm{Co}^{3+}$, the splitting $\Delta_\mathrm{o}$ is large enough that the complex ends up being low spin.

Electronic configuration in an octahedral field

For a $d^6$ octahedral low-spin complex, all six electrons pair up in the lower-energy $\mathrm{t_{2g}}$ orbitals:

$ \mathrm{t_{2g}^6}\, \mathrm{e_g^0} $

That leaves 0 unpaired electrons.

3. Final answer

Therefore, the number of unpaired $d$-electrons in

$\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$ is $ \boxed{0}. $

Correct Option: A (0)

To determine the atomic number of the metal based on the provided spin-only magnetic moment value, we use the formula for calculating the spin-only magnetic moment:

$ \mu = \sqrt{n(n+2)} \mu_{B} $

where $ \mu $ is the magnetic moment in Bohr magnetons ($ \mu_{B} $), and $ n $ is the number of unpaired electrons.

Given that the spin-only magnetic moment value is $ 3.86 \, \mathrm{BM} $, we can set this value equal to the formula to solve for $ n $:

$ 3.86 = \sqrt{n(n+2)} $

Squaring both sides gives:

$ 14.8996 = n(n+2) $

Since $ n $ must be an integer and this equation doesn't solve neatly for an integer $ n $, we look for a value of $ n $ that would give a product close to $ 14.8996 $ when plugged into $ \sqrt{n(n+2)} $.

Practically, we can test the square root values near $ 3.86 $ for different $ n $ to see which one gives a close match. Considering the usual spin-only magnetic moments for common numbers of unpaired electrons:

• $ n = 1 $, $ \mu = \sqrt{3} \approx 1.73 \mu_B $


• $ n = 2 $, $ \mu = \sqrt{8} \approx 2.83 \mu_B $


• $ n = 3 $, $ \mu = \sqrt{15} \approx 3.87 \mu_B $


• $ n = 4 $, $ \mu = \sqrt{24} \approx 4.90 \mu_B $

The value $ n = 3 $ gives a magnetic moment of approximately $ 3.87 \mu_B $, which is very close to the given value, $ 3.86 \mu_B $. Therefore, the metal has 3 unpaired electrons in its d-orbital configuration.

In the +2 oxidation state, metals lose electrons from the s-orbital before the d-orbital. Given that the element is a first row transition metal, starting with scandium (Sc) at atomic number 21 which has an electronic configuration of $[Ar] 3d^1 4s^2$ in its neutral state, we can deduce the configurations:

• $ Z = 22 $ for Titanium (Ti), which has a neutral configuration of $[Ar] 3d^2 4s^2$. In the +2 oxidation state, it would have an electronic configuration of $[Ar] 3d^2$, resulting in 2 unpaired electrons.


• $ Z = 23 $ for Vanadium (V), with a neutral configuration of $[Ar] 3d^3 4s^2$. In the +2 state, its configuration would be $[Ar] 3d^3$, presenting 3 unpaired electrons, which matches our calculation.


• $ Z = 26 $ for Iron (Fe), indicating a neutral configuration of $[Ar] 3d^6 4s^2$. In the +2 state, $[Ar] 3d^6$, which would have 4 unpaired electrons, not matching the calculation.


• $ Z = 25 $ for Manganese (Mn), with a neutral configuration of $[Ar] 3d^5 4s^2$. In the +2 state, $[Ar] 3d^5$, indicating 5 unpaired electrons, which also does not match our calculation.

So, the atomic number of the metal with a +2 oxidation state and a spin-only magnetic moment value of $3.86 \mathrm{~BM}$ (corresponding to 3 unpaired electrons) is 23, which identifies the metal as Vanadium (V). Therefore,

Option B (23) is the correct answer.

$\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow 2$ unpaired electrons

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \rightarrow 4$ unpaired electrons

Above 2 complex have even number of unpaired electrons.

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

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.

$ \left[\mathrm{Co}\left(\mathrm{Cl}_{2}\right)(e n)_{2}\right]^{+} $ion has geometrical isomers.

The correct order is IV $ < $ I $ < $ II $ < $ III CFSE is higher when the complex contains strong ligand. According to spectrochemical series, the order of field strength is

$ \mathrm{F}^{-} < \mathrm{H}_{2} \mathrm{O} < \mathrm{NH}_{3} < \mathrm{CN}^{-} $

Hence, the correct order is

$ \left[\mathrm{CoF}_{6}\right]^{3-} < \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} < \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} $ $ < \left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-} $

Among the given ions the spin only magnetic moment is least for $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$.

The spin only magnetic moment is given by

$\mu_s=\sqrt{n(n+2)}$ (where, $n$ is number of unpaired electrons.)

For $\mathrm{Ti}^{3+}$ the number of unpaired electron is 1 which is

$ \mu_s=\sqrt{1(1+2)}=\sqrt{3} \mathrm{BM} $

While for $\mathrm{Mn}^{2+}$ number of unpaired electron is 5 .

$ \begin{aligned} \mu_s & =\sqrt{5(5+2)}=\sqrt{35} \mathrm{BM} \\\\ \text { For } \mathrm{Co}^{2+}, & n=3 \\\\ \mu_s & =\sqrt{3(3+2)}=\sqrt{15} \mathrm{BM} \end{aligned} $

For $\mathrm{Ni}^{2+}, n=2$

$ \mu_s=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM} $

As the ligand is same in all the given compounds, thus, the species with 5 unpaired electrons will have $t_{2 g}^3 e_g^2$ electronic configuration. Among the given options $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ has five unpaired electrons as Fe is present in +3 oxidation state.

When platinum ($\mathrm{Pt}$) is treated with aqua regia (a mixture of concentrated hydrochloric acid and concentrated nitric acid in a 3:1 ratio), it forms a chloro complex ion. The reaction of platinum with this mixture typically produces $[\mathrm{PtCl}_6]^{2-}$.

The oxidation state of platinum in the $[\mathrm{PtCl}_6]^{2-}$ complex ion can be determined as follows:

Let's assume the oxidation state of Pt is $x$. Each chloride ion ($\mathrm{Cl}^-$) has a charge of $-1$. Since there are six chloride ions, their total charge is $6 \times (-1) = -6$.

The charge balance equation for the complex ion $[\mathrm{PtCl}_6]^{2-}$ is:

$ x + 6(-1) = -2 $

Simplifying, we get:

$ x - 6 = -2 $

Adding 6 to both sides:

$ x = 4 $

Therefore, the oxidation state of platinum in $[\mathrm{PtCl}_6]^{2-}$ is $+4$.

Option C: +4

For the compound $\left[\mathrm{NiCl}_4\right]^{2-}$

$\mathrm{Cl}^{-}$is a weak ligand.

Central metal atom has +2 oxidation state.

$ \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 $

Magnetic moment is given by

$ \begin{aligned} \mu_s & =\sqrt{n(n+2)} \mathrm{BM} \\\\ \Rightarrow \quad \mu_S & =\sqrt{2(2+2)} \mathrm{BM} \\\\ & \approx 2.86 \mathrm{BM} \end{aligned} $

The complete reaction is as follows,

$ \begin{aligned} & {\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3 \longrightarrow\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}+3 \mathrm{Cl}^{-}} \\ & 3 \mathrm{Cl}^{-}+3 \mathrm{AgNO}_3 \longrightarrow 3 \mathrm{AgCl} \downarrow+3 \mathrm{NO}_3^{-} \end{aligned} $

When one mole of $\mathrm{CoH}_{12} \mathrm{O}_6 \mathrm{Cl}_3$ is treated with excess of $\mathrm{AgNO}_3, 3$ mole of AgCl are obtained. It means that $\mathrm{CoCl}_3 \mathrm{H}_{12} \mathrm{O}_6$ on dissociation in aqueous solution, all three chloride ions come in solution.

The increasing order of magnetic moments is:

$ \mathrm{III} < \mathrm{I} < \mathrm{II} < \mathrm{IV} $

The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. The magnetic moment $\mu$ is directly proportional to $n$.

I. $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$:

Oxidation state of Mn: $+3$

Electron configuration: $[\mathrm{Ar}] 3d^4$

CN⁻ is a strong ligand causing electron pairing.

$\mathrm{Mn}^{3+}$ has 2 unpaired electrons, $n = 2$.

$\mu = \sqrt{2(2+2)} = \sqrt{8} \, \mathrm{BM}$

II. $[\mathrm{MnCl}_6]^{3-}$:

Oxidation state of Mn: $+3$

Cl⁻ is a weak ligand, leading to 4 unpaired electrons.

$n = 4$

$\mu = \sqrt{4(4+2)} = \sqrt{24} \, \mathrm{BM}$

III. $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$:

Oxidation state of Fe: $+3$

CN⁻ is a strong ligand causing electron pairing.

$\mathrm{Fe}^{3+}$ has 1 unpaired electron, $n = 1$.

$\mu = \sqrt{1(1+2)} = \sqrt{3} \, \mathrm{BM}$

IV. $[\mathrm{FeF}_6]^{3-}$:

Oxidation state of Fe: $+3$

F⁻ is a weak ligand, so no electron pairing happens.

$\mathrm{Fe}^{3+}$ has 5 unpaired electrons, $n = 5$.

$\mu = \sqrt{5(5+2)} = \sqrt{35} \, \mathrm{BM}$

Thus, the correct order based on increasing magnetic moments is $\mathrm{III} < \mathrm{I} < \mathrm{II} < \mathrm{IV}$.

Paramagnetism is influenced by the magnetic moment, which can be calculated using the formula:

$ \mu = \sqrt{n(n+2)} \, \text{BM} $

Here, $n$ represents the number of unpaired electrons in the complex. The larger the value of $n$, the greater the magnetic moment, and therefore, the stronger the paramagnetic behavior.

Let's evaluate each complex:

$[\text{Co}(\text{NH}_3)_6]^{2+}$:

Cobalt is in the +2 oxidation state. With $\text{NH}_3$ as a strong ligand, it leaves one unpaired electron.

$[\text{Co}(\text{NH}_3)_6]^{3+}$:

Cobalt is in the +3 oxidation state. Since $\text{NH}_3$ is a strong ligand, all electrons are paired, making it diamagnetic.

$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$:

Cobalt is in the +2 oxidation state. With $\text{H}_2\text{O}$ as a weak ligand, electron pairing does not occur, resulting in 3 unpaired electrons.

$[\text{Co}(\text{H}_2\text{O})_6]^{3+}$:

Cobalt is in the +3 oxidation state. It has 4 unpaired electrons.

Thus, $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ exhibits the highest degree of paramagnetism due to having the most unpaired electrons.

The correct match is I-D, II-C, III-B, IV-A.

I. $\left[\mathrm{Ni}(\mathrm{CO})_4\right] \rightarrow$ Oxidation state of Ni $=0$

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

As CO is strong ligand, pairing wiII occur

$ \begin{aligned} & II.{\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \longrightarrow \text { Oxidation state }} \\ & \text { of } \mathrm{Ni}=+2 \\ & \,\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 4 s^0 \end{aligned} $

As CN is strong ligand, pairing will occur, leaving one $d$-orbital empty Hvbridisation $=d s p^2$

$ \begin{aligned} & \text { $III.$ }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \longrightarrow \text { Oxidation state } \\ & \text { of } \mathrm{Co}=+3 \\ & \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 \end{aligned} $

As $\mathrm{NH}_3$ is a strong ligand, $3 d$ electron pair up leaving $2 d$-orbital empty.

Hybridisation $=d^2 s p^3$

IV. $\left[\mathrm{COF}_6\right]^{3-} \longrightarrow$ Oxidation state of $\mathrm{Co}=+3$

As $F$ is weak ligand, thus pairing of $d$-orbital electron is not happened.

$ \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 $

Hybridisation $=s p^3 d^2$

The coordination compound with the empirical formula $\mathrm{CoCl}_3 \cdot 4 \mathrm{NH}_3$ forms a green isomeric complex with cobalt (III) chloride and ammonia, in a 1:1 ratio (acting as an electrolyte). The molecular formula of this complex is $\left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right]^{+} \mathrm{Cl}^{-}$, denoted as $[X]$.

When this complex reacts with silver nitrate ($\mathrm{AgNO}_3$), the following equation represents the reaction:

$ \begin{aligned} &{\left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right]^{+} \mathrm{Cl}^{-}(\mathrm{aq}) + \mathrm{Ag}^{+} \mathrm{NO}_3^{-}(\mathrm{aq})} \\ & \quad \rightarrow \mathrm{AgCl}(\mathrm{s}) + \left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{NO}_3(\mathrm{aq}) \end{aligned} $

Initially, 0.1 moles of the complex $[X]$ are present in a 100 mL solution, calculated as follows:

Given: $\frac{100 \times 1}{1000} \mathrm{~mol} = 0.1 \mathrm{~mol}$

Therefore, for the reaction:

0.1 mol of $[X]$ is consumed,

0.1 mol of $\mathrm{AgNO}_3$ is used,

0.1 mol of $\mathrm{AgCl}$ is formed.

This reflects a stoichiometric conversion of reactants to products, where 0.1 mol of $\mathrm{AgCl}$ is precipitated as solid.

$ \text { The complete reaction is a follows } $

$ \begin{aligned} &\text { Coordination number }\\ &\begin{aligned} & {\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}=2} \\ & {\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}=4} \end{aligned} \end{aligned} $

Strong ligands are those ligands that produce large spliting between the $d$-orbitals and form low spin complex. Order of ligands is,

$ \begin{aligned} & \mathrm{Br}^{-}<\mathrm{S}^{2-}<\mathrm{OH}^{-}<\mathrm{C}_2 \mathrm{O}_4^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3 \\ & <\mathrm{en}<\mathrm{CN}^{-}<\mathrm{CO} . \end{aligned} $

Hence, total of 4 ligands are stronger than $\mathrm{H}_2 \mathrm{O}$.

The correct order of number of unpaired electron is

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{MnCl}_6\right]^{3-}$ $<\left[\mathrm{FeF}_6\right]^{3-}$

i.e., IV < III < I < II

IV. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \rightarrow$ The central atom has +3 oxidation state. $\mathrm{Fe}^{3+}=[A r] 3 d^5$

As $\mathrm{CN}^{-}$is strong ligand, pairing of electron will occur it has only 1 unpaired electron. $\left(t_{2 g}^{221} e_g^0\right)$

III. $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-} \rightarrow$ The central atom has 3 oxidation state.

$ \mathrm{Mn}^{3+}=[A r] 3 d^4 $

2 unpaired electrons are present.

I. $\left[\mathrm{MnCl}_6\right]^{3-} \rightarrow$ Oxidation state of central atom is +3 .

$ \mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 d^4 $

As $\mathrm{Cl}^{-}$is weak ligand, pairing of electron will not happen and therefore number of unpaired electrons $=4$

II. $\left[\mathrm{FeF}_6\right]^{3-} \rightarrow$ Central atom has +3 oxidation state.

$\mathrm{F}^{-}$is a weak ligand, hence no pairing of electron happens. Hence, number of unpaired electron is 5 .

In the complex ion $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$, the central nickel atom is in the +2 oxidation state, giving it an electronic configuration of $3d^8$. Ammonia, being a strong ligand, causes the electrons to pair up in the $t_{2g}$ orbitals rather than occupying the $e_g$ level.

Therefore, the outer electronic configuration becomes $t_{2g}^6 e_g^2$. Since all electrons in the $t_{2g}$ orbitals are paired, there are no unpaired electrons in these orbitals.

Complex $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$, oxidation state of $\mathrm{Mn}=+3$

$ \mathrm{Mn}^{3+}:[\mathrm{Ar}] 3 d^4 4 s^0 $

It has 2 unpaired electrons.

Spin only magnetic moment is given by

$ \begin{aligned} \mu_s & =\sqrt{n(n+2)} \\ \Rightarrow \mu_s & =\sqrt{2(2+2)}=\sqrt{8} \text { or } \mu_s=2.84 \mathrm{BM} \end{aligned} $

For complex, $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$, oxidation state of $\mathrm{Co}=+3$

$ \mathrm{Co}^{3+}=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6 $

Since oxalate is a strong field ligand, pairing will occur. There is no unpaired electrons ( $n=0$ ).

$ \text { Hence, } \mu_s=0 $