Coordination Compounds

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.

The increasing order of magnetic moments is:

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

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

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

Oxidation state of Mn: $+3$

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

CN⁻ is a strong ligand causing electron pairing.

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

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

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

Oxidation state of Mn: $+3$

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

$n = 4$

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

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

Oxidation state of Fe: $+3$

CN⁻ is a strong ligand causing electron pairing.

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

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

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

Oxidation state of Fe: $+3$

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

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

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

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

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

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

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

Let's evaluate each complex:

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

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

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

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

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

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

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

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

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

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

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

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

As CO is strong ligand, pairing wiII occur

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

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

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

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

Hybridisation $=d^2 s p^3$

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

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

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

Hybridisation $=s p^3 d^2$

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

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

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

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

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

Therefore, for the reaction:

0.1 mol of $[X]$ is consumed,

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

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

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

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

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

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

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

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

The correct order of number of unpaired electron is

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

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

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

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

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

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

2 unpaired electrons are present.

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

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

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

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

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

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

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

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

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

It has 2 unpaired electrons.

Spin only magnetic moment is given by

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

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

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

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

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

Element '$X$' is forming a complex of the type $\left[X \mathrm{Cl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}$.

Total number of ligands attached to $X$ is 6 (one Cl and five $\mathrm{H}_2 \mathrm{O}$).

$\therefore$ Covalency of $X$ is 6.

Cl is a unidentate ligand with charge $-$1.

$\mathrm{H}_2 \mathrm{O}$ is a unidentate ligand with charge 0.

$\begin{array}{rr} x+(-1)+0 & =+2 \\ x-1 & =+2 \\ x=+2+1 & =+3 \end{array}$

$\therefore$ Oxidation state of $x$ is +3.

Presence of vacant $d$-orbitals in $\mathrm{Si}, \mathrm{Ge}$ and Sn , makes the existence of species $\left[\mathrm{SiF}_6\right]^{2-},\left[\mathrm{GeCl}_6\right]^{2-}$ and $\left[\mathrm{Sn}(\mathrm{OH})_6\right]^{2-}$ feasible. But $\left[\mathrm{SiCl}_6\right]^{2-}$ cannot exist as six large $\mathrm{Cl}^{-}$ cannot be accommodated around $\mathrm{Si}^{4+}$ ion due to smaller size of cation.

Homoleptic complexes are the complexes in which all the ligands attached to central metal atom/ion are same or identical.

$\therefore \mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is a homoleptic complex whereas other complexes are not.

Crystal field theory considered ligands as point charges but does not explain it. But, it explain the formation and structure of the complexes, colour and their magnetic properties. Covalent character of metal-ligand bonding was not explained. It considered them as point charges. Thus, explaining ionic character only. Thus, CFFT explained II, III and IV statements only.

Oxidation state of Ti in $\mathrm{TiCl}_3$ is +3. Electronic configuration is [Ar] $3 d^1 4 s^{\circ}$. In absence of ligands, no splitting occurs, in $d$-orbitals of $\mathrm{Ti}^{3+}$ ions. So, no transitions are possible. Hence, $\mathrm{TiCl}_3$ is colourless. Oxidation state of Ti in $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ is + 3. In presence of a ligand, splitting in $d$-orbitals occur. Hence, $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ is coloured.

$\mathrm{TiCl}_3(X) \longrightarrow$ colourless

$\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3(Y) \longrightarrow$ coloured

$\mathrm{\mathop {Al{F_3}(s)}\limits_{Alu\min ium\,trifluoride} \buildrel {HF} \over \longrightarrow Al{F_3}(HF)\buildrel {KF} \over \longrightarrow \mathop {{K_3}(Al{F_6})}\limits_{Potassium\,hexafluorido\,alu{\mathop{\rm minate}\nolimits} \,(III)}}$

$\mathrm{AlF}_3$ is soluble in HF only in presence of KF. It is due to formation of $\mathrm{K}_3\left[\mathrm{AlF}_6\right]$.

$\mathrm{AlF}_3$ is insoluble in anhydrous $\mathrm{HF}$ because $\mathrm{F}^{-}$ ion is not available for $\mathrm{H}$-bonding.

KCN $\mathrm{\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{S_2}O_3^{2 - }}^{NaOH}}}$ $\mathrm{\mathop {KSCN}\limits_{Potassium\,thiocyanate}}$

$\mathrm{FeCl_3+3KSCN\rightarrow}$ $\mathrm{\mathop {Fe{{(SCN)}_3}}\limits_{Blood\,red\,colour}}$ + 3KCl

Reaction of FeCl$_3$ with KSCN gives a blood red solution.

(a)

$\begin{aligned} & {\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}} \\ & \mathrm{Ni}^{+2}=3 d^8 \\ & 2 \text { unpaired electrons are present. So, } \\ & \therefore \text { Nature = paramagnetic } \\ & \text { Shape }=\text { square planar } \end{aligned}$

(b)

$\begin{aligned} & {\left[\mathrm{Ni}(\mathrm{CO})_4\right]} \\ & \mathrm{Ni}=3 d^8 4 s^2 \\ & \text { Shape }=\text { tetrahedral } \end{aligned}$

As CO is strong ligand, it causes pairing of electrons, therefore nature is diamagnetic.

(c)

$\begin{aligned} & {\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}} \\ & \mathrm{Ni}^{2+}=3 d^8 \\ & \text { Shape }=\text { tetrahedral } \end{aligned}$

As $\mathrm{Cl}^{-}$ is weak ligand, no pairing occurs therefore, nature is paramagnetic.

(d)

$\begin{aligned} & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} \\ & \mathrm{Ni}^{2+}=3 d^8 4 s^0 \end{aligned}$

Shape $=$ octahedral and nature is paramagnetic.

Coordination number is defined as number of coordinate bonds of central metal atom with the ligand. Here, en is bidentate ligand and Cl$^-$ is monodentate ligand.

In the given complex, Co binds with two a bidentate ligand en (2 $\times$ 2 = 4) and two monodentate ligand (2 $\times$ 1 = 2).

$\therefore$ Its coordination number is, 4 + 2 = 6.