Coordination Compounds

To determine the number of complexes with an even number of electrons in the $\mathrm{t_{2g}}$ orbitals, we first need to determine the electronic configuration of the metal ions in each complex and then find the distribution of electrons among the $\mathrm{t_{2g}}$ and $\mathrm{e_{g}}$ orbitals.

Let's consider each complex one by one:

1. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

The oxidation state of Fe is +2. The electronic configuration of $\mathrm{Fe}^{2+}$ is $\left[\mathrm{Ar}\right]3d^6$.

In an octahedral field, the splitting pattern for the 3d orbitals is such that: $t_{2g}$ (lower energy) and $e_g$ (higher energy).

So, the electronic configuration in octahedral field will be: $t_{2g}^4 e_g^2$. Therefore, there are 4 electrons in the $t_{2g}$ orbitals (even).

2. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

The oxidation state of Co is +2. The electronic configuration of $\mathrm{Co}^{2+}$ is $\left[\mathrm{Ar}\right]3d^7$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $t_{2g}^5 e_g^2$. Therefore, there are 5 electrons in the $t_{2g}$ orbitals (odd).

3. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

The oxidation state of Co is +3. The electronic configuration of $\mathrm{Co}^{3+}$ is $\left[\mathrm{Ar}\right]3d^6$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $t_{2g}^4 e_g^2$. Therefore, there are 4 electrons in the $t_{2g}$ orbitals (even).

4. $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

The oxidation state of Cu is +2. The electronic configuration of $\mathrm{Cu}^{2+}$ is $\left[\mathrm{Ar}\right]3d^9$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $t_{2g}^6 e_g^3$. Therefore, there are 6 electrons in the $t_{2g}$ orbitals (even).

5. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

The oxidation state of Cr is +2. The electronic configuration of $\mathrm{Cr}^{2+}$ is $\left[\mathrm{Ar}\right]3d^4$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $t_{2g}^3 e_g^1$. Therefore, there are 3 electrons in the $t_{2g}$ orbitals (odd).

So, the complexes with an even number of electrons in the $\mathrm{t_{2g}}$ orbitals are:

1. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (4 electrons)

1. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ (4 electrons)
   


2. $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (6 electrons)

Hence, there are 3 complexes with an even number of electrons in the $\mathrm{t_{2g}}$ orbitals. Therefore, the correct answer is option A.

To solve this problem, we need to determine the oxidation state of cobalt ($x$) and the number of ammonia molecules ($n$) in the complex $\mathrm{CoCl}_3 \cdot \mathrm{nNH}_3$, given that it produces 2 moles of $\mathrm{AgCl}$ upon reaction with excess $\mathrm{AgNO}_3$.

First, let's write the reaction between the complex and $\mathrm{AgNO}_3$.

The precipitate formation indicates that some chloride ions are free (not coordinated to the metal). Since 2 moles of $\mathrm{AgCl}$ are formed, it indicates that there are 2 chloride ions that are free to react with $\mathrm{AgNO}_3$.

Therefore, we can conclude that in the complex, 1 chloride ion is coordinated to the cobalt, and 2 chloride ions are free. This can be represented as:

$\mathrm{[Co(NH_3)_{n}Cl]Cl_2}$

Now, let's determine the oxidation state of cobalt (Co). The charge on the entire complex should be zero, so we can write the charge balance equation as follows:

The sum of the charges is:

$x + (0 \cdot n) + (-1 \cdot 1) + (-1 \cdot 2) = 0$

Solving this, we get:

$x - 1 - 2 = 0$

$x - 3 = 0$

$x = 3$

So, the oxidation state of Co is 3.

Now, we need to find the value of $n$. Since the coordination number of Co in an octahedral complex is typically 6 and we have 1 chloride ion coordinated to Co, the number of ammonia molecules coordinated to Co would be:

$n = 6 - 1 = 5$

Finally, we calculate $x + n$:

$x + n = 3 + 5 = 8$

So, the value of "$x + n$" is 8.

Therefore, the correct option is:

Option D: 8

Answer: Option D - Statement I is correct but Statement II is incorrect.

Explanation:

Statement I: $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ (trimethylamine) and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ (trimethylphosphine) can indeed act as ligands to form transition metal complexes. Ligands are molecules or ions that can donate a pair of electrons to a metal atom to form a coordinate bond. Both nitrogen and phosphorus have lone pairs of electrons that can be donated to transition metals, making $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ suitable ligands. Therefore, Statement I is correct.

Statement II: While N (Nitrogen) and P (Phosphorus) are both from group 15 of the periodic table and share some similar properties, the nature of their bonding with transition metals is not always the same. Nitrogen compounds typically form stronger bonds with metals compared to phosphorus compounds. This difference is due to several factors, including the difference in atomic sizes, electronegativity, and the extent of orbital overlap. Nitrogen, being smaller and more electronegative, can form more effective overlap and stronger bonds with metals compared to phosphorus. Therefore, Statement II is incorrect.

Given this analysis, the most appropriate answer is Option D: Statement I is correct but Statement II is incorrect.

To match List I with List II correctly, we need to identify the colors associated with each compound.

Let's analyze each compound:

1. $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2O}$ - This is known as Prussian Blue.

2. $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}$ - This complex is known to have a violet color.

3. $[\mathrm{Fe}(\mathrm{SCN})]^{2+}$ - This ion is known for its blood red color.

4. $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\cdot12 \mathrm{MoO}_3$ - This is commonly known as Ammonium phosphomolybdate, which is yellow.

Matching each item, we get:

Based on the analysis, the correct matching is:

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

Hybridisation of $\mathrm{P}$ in $\mathrm{PF}_5$ is $s p^3 d$ but the hybridisation of $\mathrm{Br}$ in $\mathrm{BrF}_5$ is $s p^3 d^2$. So, statement-I is false.

Hybridisation of $\mathrm{S}$ in $\mathrm{SF}_6$ is $s p^3 d^2$ but the hybridisation of $\mathrm{Co}^{3+}$ in $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is $d^2 s p^3$ as $\mathrm{NH}_3$ is a strong field ligand forcing the unpaired electrons to pair up. So, statement-II is false.

(A) $\mathrm{TiCl}_4: \mathrm{Ti}^{4+}: 3 d^0 4 s^0$ or $\mathrm{e}^0 \mathrm{t}_2^0$

(B) $\left[\mathrm{FeO}_4\right]^{2-}: \mathrm{Fe}^{6+}: 3 d^2 4 s^0$ or $\mathrm{e}^2 \mathrm{t}_2^0$

(C) $\left[\mathrm{FeCl}_4\right]^{-}: \mathrm{Fe}^{3+}: 3 d^5 4 s^0$ or $\mathrm{e}^2 \mathrm{t}_2^3$

(D) $\left[\mathrm{CoCl}_4\right]^{2-}: \mathrm{Co}^{2+}: 3 d^7 4 s^0$ or $\mathrm{e}^4 \mathrm{t}_2^3$

$\therefore$ Correct matching of tetrahedral complexes given in List-I with their electronic configuration given in List-II is

(A)-(III); (B)-(I); (C)-(IV); (D)-(II)

The correct IUPAC name of $[\mathrm{PtBr}_2(\mathrm{PMe}_3)_2]$ is dibromobis(trimethylphosphine)platinum(II).

Wavenumber $=\frac{1}{\lambda} \propto$ Frequency $\propto \Delta_0$

Wavenumber order : $\mathrm{D}<\mathrm{A}<\mathrm{C}<\mathrm{B}$

A $\rightarrow$ Tetrahedral (III)

B $\rightarrow$ Square planar (IV)

C $\rightarrow$ Trigonal bipyramidal (I)

D $\rightarrow$ Octahedral (II)

Identifying the electronic configuration and geometry :

The condition "no electrons in the $ t_2 $ orbital" typically applies to a tetrahedral crystal field splitting pattern. In a tetrahedral field:

The $ d $-orbitals split into two sets: $ e $ (lower energy, 2 orbitals) and $ t_2 $ (higher energy, 3 orbitals).

Electrons fill the lower-energy $ e $ set before occupying the $ t_2 $ set.

Thus, to have no electrons in $ t_2 $, either:

The metal has a $ d^0 $ configuration (no $ d $-electrons at all), or

The $ d $-electron count is so low that all electrons can fit into the $ e $ orbitals without needing to occupy $ t_2 $.

Analyzing each complex :

(i) $\mathrm{TiCl}_4$ :

Ti in TiCl₄ is in the +4 oxidation state (since TiCl₄ is neutral and each Cl is -1).

Ti (Z = 22) neutral : $ 3d^2 4s^2 $. As Ti⁴⁺, it loses all 4 valence electrons (2 from 4s and 2 from 3d), resulting in $ d^0 $.

With $ d^0 $, there are no electrons to occupy $ t_2 $ orbitals.

No electrons in $ t_2 $.

(ii) $[\mathrm{MnO}_4]^{-}$ (permanganate) :

Mn in permanganate ($\mathrm{MnO_4}^-$) is in the +7 oxidation state.

Mn (Z = 25) neutral is $ 3d^5 4s^2 $. Mn⁷⁺ means removing all 7 valence electrons, leaving $ d^0 $.

With $ d^0 $, no electrons in $ t_2 $.

No electrons in $ t_2 $.

(iii) $[\mathrm{FeO}_4]^{2-}$ (ferrate) :

Fe in $\mathrm{FeO_4}^{2-}$: Oxygen contributes -8 total. The ion is -2. Thus, Fe + (-8) = -2 → Fe = +6 oxidation state.

Fe (Z = 26) neutral is $ 3d^6 4s^2 $. Fe⁶⁺ means removing 6 electrons from the valence shell. After losing 2 from 4s and 4 from 3d, we get $ d^2 $.

In a tetrahedral field, $ d^2 $ fills the lower $ e $ orbitals (2 electrons into e orbitals).

No need to occupy $ t_2 $ orbitals since both electrons fit into e.

No electrons in $ t_2 $.

(iv) $[\mathrm{FeCl}_4]^{-}$ :

Fe in $\mathrm{FeCl_4}^- $: Cl total charge = -4, complex = -1, so Fe = +3.

Fe(III) is $ d^5 $.

In a tetrahedral field, with 5 $ d $-electrons, after filling the 2 $ e $ orbitals, we have 3 more electrons that must go into $ t_2 $.

Has electrons in $ t_2 $.

(v) $[\mathrm{CoCl}_4]^{2-}$:

Co in $\mathrm{CoCl_4}^{2-}$ : Cl total = -4, complex = -2, so Co = +2.

Co(II) is $ d^7 $.

For $ d^7 $, even after filling the $ e $ set (2 orbitals), we have 5 more electrons left, which must occupy the $ t_2 $ orbitals.

Has electrons in $ t_2 $.

Counting the complexes with no electrons in $ t_2 $ :

TiCl₄: No electrons in $ t_2 $.

[MnO₄]⁻: No electrons in $ t_2 $.

[FeO₄]²⁻: No electrons in $ t_2 $.

[FeCl₄]⁻: Has electrons in $ t_2 $.

[CoCl₄]²⁻: Has electrons in $ t_2 $.

Total with no electrons in $ t_2 $ = 3.

Answer :

3

To determine the number of geometrical isomers for the given complex $\mathrm{MABXL}$, where $\mathrm{M}$ is a metal and $\mathrm{A}$, $\mathrm{B}$, $\mathrm{X}$, and $\mathrm{L}$ are unidentate ligands, we need to consider the geometry implied by the $\mathrm{sp}^3$ hybridization of the metal. The $\mathrm{sp}^3$ hybridization suggests a tetrahedral geometry for the metal complex.

In a tetrahedral complex, geometrical isomerism does not arise. This is because geometrical isomerism (also known as cis-trans isomerism) typically occurs in square planar or octahedral complexes where ligands can be positioned differently around the central metal atom to give distinct isomers. In a tetrahedral complex, every position is equivalent due to the symmetric arrangement of the ligands around the metal center, making it impossible to distinguish between different sides of the molecule as you would with a square planar or octahedral complex.

Given that the complex follows tetrahedral geometry because of $\mathrm{sp}^3$ hybridization and that tetrahedral complexes do not exhibit geometrical isomerism, the number of geometrical isomers exhibited by the complex is:

Option A: 0

This is because in a tetrahedral arrangement of ligands around a metal center, no geometrical isomers can be formed due to the lack of cis or trans arrangements possible within such a symmetric geometry.

The correct answer is Option C:

$ \mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3 $.

Here's why:

The spectrochemical series is a list of ligands arranged in order of increasing field strength. This means that ligands higher on the series cause a larger splitting of the d-orbitals in a transition metal complex, resulting in a larger crystal field stabilization energy (CFSE).

Here's a breakdown of the factors influencing the spectrochemical series:

• Charge Density: Ligands with a higher charge density (more negative charge concentrated in a smaller space) will interact more strongly with the metal ion's d-orbitals. For example, $ \mathrm{F}^{-} $ has a higher charge density than $ \mathrm{Br}^{-} $, making it a stronger field ligand.
• Electronegativity: More electronegative atoms within a ligand draw electron density away from the metal center, weakening the metal-ligand bond and reducing field strength.
• π-Backbonding: Ligands that can participate in π-backbonding (donation of electrons from the metal d-orbitals to empty ligand orbitals) can strengthen the metal-ligand bond and increase field strength. For example, $ \mathrm{CN}^{-} $ is a strong field ligand due to its ability to engage in π-backbonding with transition metals.

Let's analyze why the other options are incorrect:

• Option A: This order is incorrect because $ \mathrm{NH}_3 $ is a stronger field ligand than halides, which is not reflected in this arrangement.
• Option B: This order is incorrect because $ \mathrm{CN}^{-} $ is a much stronger field ligand than $ \mathrm{NH}_3 $.
• Option D: This order is incorrect because $ \mathrm{CN}^{-} $ is a much stronger field ligand than $ \mathrm{Br}^{-} $.

Therefore, the correct order in increasing field strength is: $ \mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3 $.

So, the complex with minimum number of unpaired $\mathrm{e}^{\ominus}$ is option (2) $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$.

Balancing charges,

$\begin{aligned} & 3-y=-1 \\ & \Rightarrow y=4 \\ & x=2 \\ & x+y=6 \\ & \end{aligned}$

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.

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.

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.

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 :

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

E < A < D < C < B