Coordination Compounds

(A) $[ \text{Fe(CN)}_5 \text{NO} ]^{2-}$: This is a heteroleptic complex with iron in the $\text{Fe}^{2+}$ oxidation state, resulting in a $3d^6$ electronic configuration. It forms a low spin complex with electron configuration $\text{t}_{2g}^6 \text{e}_g^0$ due to the strong field ligands (SFL).

(B) $[ \text{CoF}_6 ]^{3-}$: This homoleptic complex contains cobalt in the $\text{Co}^{3+}$ oxidation state, leading to a $3d^6$ electron configuration. It is a high spin complex with an $\text{sp}^3 \text{d}^2$ hybridization due to weak field ligands (WFL).

(C) $[ \text{Fe(CN)}_6 ]^{4-}$: This homoleptic complex has iron in the $\text{Fe}^{2+}$ state, giving a $3d^6$ configuration. It forms a low spin complex with hybridization $\text{d}^2 \text{sp}^3$ and electron configuration $\text{t}_{2g}^6 \text{e}_g^0$, due to the strong field nature of the cyanide ligands.

(D) $[ \text{Co(NH}_3 )_6 ]^{3+}$: This homoleptic complex contains cobalt in the $\text{Co}^{3+}$ state, resulting in a $3d^6$ electron configuration. It forms a low spin complex with hybridization $\text{d}^2 \text{sp}^3$ and configuration $\text{t}_{2g}^6 \text{e}_g^0$ due to the strong field ligands.

(E) $[ \text{Cr(H}_2\text{O})_6 ]^{2+}$: This homoleptic complex with chromium in the $\text{Cr}^{2+}$ state results in a $3d^4$ electronic configuration. It is a high spin complex with hybridization $\text{d}^2 \text{sp}^3$ and configuration $\text{t}_{2g}^3 \text{e}_g^1$ due to weak field ligands.

In summary, the homoleptic complexes that are low spin are:

(C) $[ \text{Fe(CN)}_6 ]^{4-}$

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

The magnetic behavior of complexes can provide insights into their geometry and oxidation states. Given two complexes, $[ \mathrm{NiCl}_4 ]^{2-}$ and $[ \mathrm{Ni}(\mathrm{CO})_4 ]$, we need to analyze the magnetic properties to deduce these characteristics.

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

Oxidation State: In $[\mathrm{NiCl}_4]^{2-}$, nickel is in the +2 oxidation state ($\mathrm{Ni}^{2+}$). The electron configuration for $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}]\, 3d^8\, 4s^0$.

Geometry and Hybridization: The complex shows paramagnetic properties, indicating the presence of unpaired electrons. The hybridization in this case is $\mathrm{sp}^3$, which corresponds to a tetrahedral geometry. Thus, there are two unpaired electrons leading to its paramagnetism.

$[\mathrm{Ni}(\mathrm{CO})_4]$

Oxidation State: In this complex, nickel is in the zero oxidation state ($\mathrm{Ni}(0)$). The electron configuration is $[\mathrm{Ar}]\, 3d^{10}\, 4s^0$ after rearrangement.

Geometry and Hybridization: The compound is diamagnetic, indicating all electrons are paired. Its hybridization is also $\mathrm{sp}^3$, meaning the geometry is tetrahedral, resulting in no unpaired electrons and confirming its diamagnetic nature.

In summary, $[\mathrm{NiCl}_4]^{2-}$ is a tetrahedral complex with nickel in a +2 oxidation state and is paramagnetic. In contrast, $[\mathrm{Ni}(\mathrm{CO})_4]$ is a tetrahedral complex with nickel in a 0 oxidation state and is diamagnetic.

For complex $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{SCN})_6\right]$

Calculation of CFSE

$ \begin{aligned} & =(-0.4 \times 3+0.6 \times 2) \Delta_0 \\\\ & =0 \Delta_0 \end{aligned} $

$\begin{aligned} & \stackrel{+2}{\text { (A) } \mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]} \\ & \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^{\circ},\left(\mathrm{CN}^{-} \text {is } \mathrm{S} . \mathrm{F} . \mathrm{L}\right) \\ & \text { Pre hybridization state of } \mathrm{Ni}^{+2} \end{aligned}$

(B) $[\mathrm{Ni}(\mathrm{CO})_4]$

$\mathrm{Ni}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$

$\mathrm{CO}$ is S.F.L , so pairing occur

Pre hybridization state of $\mathrm{Ni}$

(C) $[\mathrm{Co}(\mathrm{NH}_3)_6] \mathrm{Cl}_3$

$\mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0$

With $\mathrm{Co}^{3+}, \mathrm{NH}_3$ act as S.F.L

(D) $\mathrm{Na}_3[\mathrm{CoF}_6]$

$\mathrm{Co}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^6\left(\mathrm{~F}^{\ominus}: \text { W.F.L }\right)$

The coordination environment of the $\mathrm{Ca}^{2+}$ ion when it forms a complex with $\mathrm{EDTA}^{4-}$ (ethylenediaminetetraacetic acid) is octahedral. Ethylenediaminetetraacetic acid (EDTA) is a hexadentate ligand, which means it has six donor atoms that can bind to a central metal ion. In the case of EDTA, these donor atoms are four oxygen atoms from its four carboxyl groups and two nitrogen atoms from its two amine groups.

When $\mathrm{Ca}^{2+}$ forms a complex with $\mathrm{EDTA}^{4-}$, all six donor atoms from EDTA coordinate with the calcium ion. As a result, the coordination number of the calcium ion is 6, leading to an octahedral geometry. This is because the octahedral geometry is the most common and energetically favorable arrangement for a coordination number of 6. In such a geometry, the six ligands are placed at equal distances from the central ion and at 90° angles relative to adjacent ligands, maximizing the distance between all ligands to minimize repulsion.

The correct answer is Option B, octahedral.

To evaluate the assertion and the reason, let's first analyze the given complex ion $[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$.

1. Assertion (A): The total number of geometrical isomers shown by $[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$ complex ion is three.

2. Reason (R): $[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$ complex ion has an octahedral geometry.

First, we know that $[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$ is a coordination complex with $\mathrm{Co}$ (Cobalt) in the center coordinated to two ethylenediamine (en) ligands and two chlorides (Cl). Ethylenediamine is a bidentate ligand, meaning it binds through two donor atoms, giving the overall complex an octahedral geometry around the central cobalt ion.

To verify the assertion about the geometrical isomers:

In an octahedral complex, the two chlorides and the two $\mathrm{en}$ ligands can have different spatial arrangements relative to each other. Specifically, for $[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$, the possible geometrical isomers are:

• cis-isomer: where the two chlorides are adjacent to each other.
• trans-isomer: where the two chlorides are opposite each other.

These two configurations are the typical geometrical isomers for this type of complex. Thus, the assertion that there are three geometrical isomers appears incorrect, as the common understanding is that there are only two geometrical isomers for this type of octahedral complex.

Now, let’s consider the reason:

The reason states that $[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$ has an octahedral geometry. This statement is indeed correct because the coordination number of 6 (from the four donor atoms of the two $\mathrm{en}$ ligands and the two chloride ions) leads to an octahedral geometry.

Therefore, the most appropriate answer is:

Option B: (A) is not correct but (R) is correct

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