Coordination Compounds

$\mathrm{Ma}_3 \mathrm{~b}_3$ type complexes show Facial - Meridional isomerism

(i) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right] \Rightarrow \mathrm{Ma}_3 \mathrm{~b}_3$

(ii) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+} \Rightarrow \mathrm{Ma}_4 \mathrm{~b}_2$

(iii) $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+} \quad \Rightarrow \mathrm{M}(\mathrm{AA})_3$

(iv) $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+} \Rightarrow \mathrm{M}(\mathrm{AA})_2 \mathrm{~b}_2$

$\mathrm{a,b},=\mathrm{NH}_3, \mathrm{Cl}^{-}$

$\mathrm{AA}=\mathrm{en}$

The crystal field stabilization energy (CFSE) of a complex depends on two main factors:

Charge/Oxidation State of the Central Metal Atom: Higher oxidation states generally lead to greater crystal field splitting energy (Δ).

Field Strength of the Ligand: Stronger field ligands also result in greater splitting. Additionally, chelating ligands typically increase the field strength.

For octahedral complexes, the CFSE is calculated as:

$ \text{CFSE} = [-0.4 \, t_{2g} + 0.6 \, e_g ] \, \Delta_\circ $

Considering the complexes:

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

Co exists as Co$^{3+}$ with electronic configuration $t_{2g}^6 e_g^0$.

CFSE = $-2.4 \, (\Delta_0)_1$.

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

Co also as Co$^{3+}$ with $t_{2g}^6 e_g^0$.

CFSE = $-2.4 \, (\Delta_0)_2$.

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

Co with Co$^{2+}$ configuration $t_{2g}^5 e_g^2$.

CFSE = $-0.8 \, (\Delta_0)_3$.

$[\text{Co(NH}_3)_4]^{2+}$:

Co as Co$^{2+}$ with different configuration $e^4 t_2^3$.

CFSE = $-1.2 \, \Delta_t$, where $\Delta_t = \frac{4}{9} (\Delta_0)_3$.

The order of crystal field splitting energies, based on our criteria, is:

$ \Delta_t < (\Delta_0)_3 < (\Delta_0)_2 < (\Delta_0)_1 $

Therefore, the CFSEs of these complexes compare as follows:

$[\text{Co(en)}_3]^{3+}$ has the highest CFSE due to the $\Delta_0$ of the strongest field ligand and oxidation state.

$[\text{Co(NH}_3)_6]^{3+}$ follows with similarly high CFSE due to Co$^{3+}$.

$[\text{Co(NH}_3)_6]^{2+}$ has a moderate CFSE.

$[\text{Co(NH}_3)_4]^{2+}$ has the lowest CFSE with $\Delta_t$.

This hierarchy of CFSE is determined by both the oxidation state of the cobalt metal and the field strength of the ligands involved.

(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} $

The complexes are $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}$ It is based on the ligand strength. In spectrochemical series, ligands are weak or strong stronger ligands will split $d$-orbital energies in transition metal complexes. So, strong ligands cause larger splitting. and result in shorter wavelengths of absorbed light. For weak ligands, the splitting is lesser and wavelength is maximum. For the given ligands, $\mathrm{CN}^{-}, \mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{Cl}^{-}$, the older of increasing field (ligand) strength is $Cl^-<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{CN}^{-}$(splitting is inversely. proportional to wavelength): $C N^{-}$is a strong-field ligand. So, it will cause a larger splitting and absorb light at a shorter wavelength (higher energy). So, the complex $\left[C_0(\mathrm{CN})_6\right]^{3^-}$ has shorter wavelength$\mathrm{NH}_3$ ligand strength is lower than $C{N}^{-}$ but higher than $\mathrm{Cl}^{-}$and $\mathrm{H}_2 \mathrm{O}$. The splitting caused by $\mathrm{NH}_3$ is lesser than $\mathrm{CN}^{-}$. So, the wavelength is higher than $\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-}$ complex.$ \mathrm{Co}\left(\mathrm{NH}_3\right)_6^{3+} \text { has wavelength higher than }\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-} \text { - } $$\mathrm{H}_2 \mathrm{O}$ is a weaker ligand than $\mathrm{NH}_3$ so, the complex $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2O\right]^{3+}$ has higher wavelength than $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$Cl is the weak field ligand than $^{}$ $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ So, the wavelength is maximum for the complex containing $C^{-}$.$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} \text { has highest wavelength. } $ The correct order of wavelength maxima of the absorption band is. $ \left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+} \leq\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} $$ \text { Answer: Option (A) } $

The number of unpaired electron in the given complex ion are, $[\mathrm{Mn}(\mathrm{CN})]^{3-}$ has two unpaired electrons, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ has a single unpaired electron, $\left(\mathrm{COF}_{\mathrm{6}}\right)^{3-}$ has four electrons while $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3^-}$ in diamagnetic i.e. has zero unpaired electrons. Thus, the correct order is D, $B, A, C$.

The atomic number of $\mathrm{C}_0$ is 27 . Its electronic configuration is $[\mathrm{Ar}] 3 d^2 4 s^2$.

In $\left[\mathrm{Co}(\mathrm{ox})_3\right]^{3-}$, the oxidation state of Co is +3 .

In the presence of $\mathrm{C}_2 \mathrm{O}_4^{2-}$ (strong field ligand in case of $\mathrm{Co}^{3+}$ ) pairing of electrons takes place.

$ \left[\mathrm{Co}(\mathrm{ox})_3\right]^{3-}= $

It has no unpaired electron, so it is diamagnetic.

The total number of geometrical isomers for the given complexes are

• $\left[\mathrm{NiCl}_4\right]^{2-}$ - No geometrical isomer

• $\left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right]^{+}$- octahedral complex of the type $M A_4 B_2$. So can exist as cis and trans isomers. Thus 2 geometrical isomers are possible.

• [ $\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3$ ] - Octahedral complex of the type $M A_3 B_3$. So can exist as facial and meridional isomer. So, 2 geometrical isomers are possible.

• $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}$ - Octahedral complex of the type $M A_5 B$. All positions are equivalent relative to the single unique ligand. So no geometrical isomer is possible.

Thus, total 4 geometrical isomers are possible.

The magnetic moment of a complex depends on the number of unpaired electrons present in it.

Formula: The spin-only magnetic moment is given by $ \mu = \sqrt{n(n+2)} $ where $ n $ is the number of unpaired electrons.

Counting Unpaired Electrons for Each Complex:

I. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: Here, $ n=0 $. This means there are no unpaired electrons.

II. $\left[\mathrm{MnCl}_4\right]^{2-}$: Here, $ n=5 $. There are five unpaired electrons.

III. $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$: Here, $ n=1 $. There is one unpaired electron.

IV. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$: Here, $ n=3 $. There are three unpaired electrons.

The complexes with more unpaired electrons will have higher magnetic moments.

So, the order from least to greatest magnetic moment is:

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

Moles of ions precipitated.

$ \begin{aligned} n_{\text {ions }} & =\frac{\text { Number of ions }}{\text { Avogadro's number }}=\frac{3.6 \times 10^{22}}{6 \times 10^{23}} \\ n_{\text {ions }} & =0.06 \mathrm{~mol} \\ n_{\text {complex }} & =\text { Molarity } \times \text { Volume }=0.2 \times 0.1=0.02 \mathrm{~mol} \end{aligned} $

Number of $\mathrm{Cl}^{-}$ions per complex units,

$ Y=\frac{\text { Moles of precipitated ions }}{\text { Moles of complex }}=3 $

The complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_x \mathrm{Cl}_{3-y}\right] \mathrm{Cl}_y$.

Coordination number of $\mathrm{Co}=6$.

$ x=6 $

An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms, but not at the same time. $\mathrm{SO}_4^{2-}$ is a bidentate ligand.

II and III i.e. $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right]\left(\mathrm{NO}_3\right)_2$ and $\left[\mathrm{Pt}(\mathrm{en})\left(\mathrm{NH}_3\right) \mathrm{Cl}\right] \mathrm{NO}_3$ shows ionisation isomerism.

Ionisation isomerism occur when coordination compound with same empirical formula but different ions in solution are produced.

$\mathrm{C}_2 \mathrm{O}_4^{2-}=$ bidentate ligand

$\mathrm{H}_2 \mathrm{O}=$ monodentate ligand

$\mathrm{SO}_4^{2-}=$ bidentate ligand.

Hence, there is no ambidentate ligand.

An ambidentate ligand is a ligand that

can coordinate to central metal atom

through two different donor atom.

The correct IUPAC name of the complex $\mathrm{K}_3\left[\mathrm{Co}(\mathrm{ox})_3\right]$ is, potassium trioxalatocobaltate (III)

To determine the coordination number of chromium (Cr) in the complex $\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right]$, we need to identify the ligands and their denticity.

1. Identify the central metal atom: The central metal atom is Chromium (Cr).

2. Identify the ligands:

   • $\mathrm{H}_2 \mathrm{O}$ (aqua)

   • $\mathrm{C}_2 \mathrm{O}_4$ (oxalato, derived from oxalic acid, $\mathrm{C}_2 \mathrm{O}_4^{2-}$)

3. Determine the denticity of each ligand:

   • $\mathrm{H}_2 \mathrm{O}$ (water): This is a monodentate ligand, meaning each water molecule forms one coordinate bond with the central metal atom.

   • $\mathrm{C}_2 \mathrm{O}_4^{2-}$ (oxalate): This is a bidentate ligand, meaning each oxalate ion forms two coordinate bonds with the central metal atom, typically through its two oxygen atoms.

4. Calculate the total number of coordinate bonds:

   • There are two $\mathrm{H}_2 \mathrm{O}$ ligands. Since each is monodentate, they contribute $2 \times 1 = 2$ coordinate bonds.

   • There are two $\mathrm{C}_2 \mathrm{O}_4^{2-}$ ligands. Since each is bidentate, they contribute $2 \times 2 = 4$ coordinate bonds.

5. Sum the coordinate bonds to find the coordination number:

   Coordination number = (bonds from $\mathrm{H}_2 \mathrm{O}$) + (bonds from $\mathrm{C}_2 \mathrm{O}_4^{2-}$)

   Coordination number = $2 + 4 = 6$

Therefore, the coordination number of chromium in the given complex is 6.

The final answer is $\boxed{6}$.

The complex ion $\left[\mathrm{Mn}(\mathrm{Cl})_6\right]^{3-}$ has a spin-only magnetic moment of 4.90 B.M.

Oxidation State and Electron Configuration:

For $\mathrm{Mn}$ in this ion, the oxidation state is +3. This means manganese has lost three electrons.

The electron configuration for $\mathrm{Mn}^{3+}$ is $[\mathrm{Ar}]\,3d^4.$ This tells us the ion has 4 electrons in the 3d orbitals.

Finding the Magnetic Moment:

The formula for the spin-only magnetic moment is:

$ \mu = \sqrt{n(n+2)} \text{ B.M.} $

Here, $n$ is the number of unpaired electrons. For $\mathrm{Mn}^{3+}$, $n = 4$ because there are 4 unpaired electrons in the $3d$ orbitals.

Plug in the value of $n$:

$ \mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90\ \text{B.M.} $

$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ has $t_{2 g}^3 e_9^2$ configuration

$\mathrm{H}_2 \mathrm{O}$ is a weak field ligand, means it does not cause pairing of $e^{-}$.

The $3 d$ electron will occupy the $d$-orbital with 5 -single electrons is each orbital. $d$ splits into $t_{2 g}$ has 3 electrons and $e_g$ has 2 electrons.

We are given the compound:

$ \mathrm{Fe}_x\left[\mathrm{Fe}_y(\mathrm{CN})_6\right]_3 $

and asked to find the values of $x$ and $y$.

Step 1. Identify familiar Prussian blue–type compounds

Prussian blue and related compounds have the general formula:

$ \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 $

This often corresponds to a mixed-valence compound containing both Fe²⁺ and Fe³⁺ ions.

Step 2. Match with the given general form

Given formula: $\mathrm{Fe}_x[\mathrm{Fe}_y(\mathrm{CN})_6]_3$

If we compare:

$ \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 $

we identify:

$ x = 4, \quad y = 1 $

✅ Answer: Option B (4,1)

Option (d) is incorrect regarding magnitude of crystal field splitting $\left.\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{2+}>\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

This is because $\mathrm{NH}_3$ is strong ligand than $\mathrm{H}_2 \mathrm{O}$ and strong ligand causes larger splitting of $d$-orbitals.

Complexes given in

I, II and IV exhibit geometrical isomerism

Geometrical isomerism, also called cis-trans isomerism, occurs in coordination complexes when ligands can occupy different position around the central metal ions.

This is found in square planar and octahedral complex.

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

(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:

	LIST I (Compound)		LIST II (Colour)
A.	$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2O}$	III.	Prussian Blue
B.	$\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}$	I.	Violet
C.	$[\mathrm{Fe}(\mathrm{SCN})]^{2+}$	II.	Blood Red
D.	$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\cdot12 \mathrm{MoO}_3$	IV.	Yellow

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,

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