Coordination Compounds

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 .

$ \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.

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

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

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

Hence, the correct order is

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

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

The spin only magnetic moment is given by

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplifying, we get:

$ x - 6 = -2 $

Adding 6 to both sides:

$ x = 4 $

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

Option C: +4

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

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

Central metal atom has +2 oxidation state.

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

Magnetic moment is given by

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

The complete reaction is as follows,

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

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

Thus, the sum of oxidation state and coordination number of central metal atom is maximum in $\mathrm{K}_2\left[\mathrm{PtCl}_6\right]$.

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

(A) $\left[\mathrm{CoF}_6\right]^{3-}$

$ { }_{27} \mathrm{Co}=[\mathrm{Ar}] 3 d^7 4 s^2 $

Oxidation state of Co in above complex

$ =(+3) $

$ \mathrm{Co}^{3+}=\left[A_0\right] 3 d^6 4 s^0 $

Since (F) is a weak field ligand, hence electrons will not pair up.

So, number of unpaired electrons, $n=4$ Magnetic moment, $\mu=\sqrt{n(n+2)} \mathrm{BM}$

$ =\sqrt{4 \times 6}=\sqrt{24} $

(B) Similarly for $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

Since oxalate is a strong field-ligand (electrons will pair up).

$ \therefore \quad[n=0] \text { and }[\mu=0] $

(C) $\left[\mathrm{FeF}_6\right]^{3-}$

$ \mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2 $

Since (F) is a weak-field ligand, electrons will not pair up.

$ \begin{aligned} & (n=5) \\ \therefore \quad & \mu=\sqrt{5(7)}=\sqrt{35} \end{aligned} $

(D) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$

$ \begin{aligned} { }_{25} \mathrm{Mn} & =[\mathrm{Ar}] 3 d^5 4 s^2 \\ \mathrm{Mn}^{3+} & =[\mathrm{Ar}] 3 d^{4+} 4 s^0 \end{aligned} $

CN being a strong field ligand, will pair up electron.

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

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

In this complex, oxidation state of cobalt is +3 .

E.C. of $\mathrm{Co}^{3+}=[\mathrm{Ar}]^{18} 3 d^6$

( $\mathrm{NH}_3$ is strong field ligand, thus electrons get paired up.)

• It is diamagnetic because of the absence of unpaired electrons.

• As inner $d$-orbital is involved, thus it is an inner orbital complex.

According to Werner's theory, secondary valency is represented by the number of groups bonded to the central metal atom/ion in a coordination complex.

$\operatorname{In}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Br}_2$, oxidation state of Cr is +2

The electronic configuration of $\mathrm{Cr}^{2+}=[\mathrm{Ar}], 3 d^4 4 s^0 4 p^0 4 d$

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

Since, there are four unpaired electrons, thus the complex would be paramagnetic.

In $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ oxidation state of Fe is +2

$ \mathrm{Fe}^{2+}=[\mathrm{Ar}], 3 d^6, 4 s^0, 4 p^0 $

No unpaired electrons are there. Thus the complex will be diamagnetic. In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ oxidation state of Ni is 0 .

$ \mathrm{Ni}^0=[\mathrm{Ar}], 3 d^8 $

No unpaired electrons are there. Thus the complex will be diamagnetic.

In $\mathrm{Na}_2\left[\mathrm{NiCl}_4\right]$ oxidation state of Ni is +2

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

Since, there are 2 unpaired electrons. Thus, the complex is paramagnetic. Thus, the incorrect match is $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ paramagnetic.

$ \begin{aligned} &\text { }\\ &\begin{aligned} (A)\,\, \mathrm{FeCl}_3(a q)+ & \mathrm{NH}_3(a q) \longrightarrow \underset{\text { Brown ppt. }}{\mathrm{Fe}(\mathrm{OH})_3}+\mathrm{NH}_4 \mathrm{Cl} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} (B)\,\, \mathrm{AgCl}(a q)+ & \mathrm{NH}_3(a q) \longrightarrow \underset{\text { Soluble colourless }}{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Cl}} \end{aligned} \end{aligned} $

$ \text { (C) } \mathrm{Cu}^{2+}(a q)+\mathrm{NH}_3(a q) \longrightarrow \underset{\substack{\text { Deep blue } \\ \text { complex }}}{\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}}$

$ \begin{aligned} &\text { ∴ The correct option is }\\ &\mathrm{A} \rightarrow \mathrm{III}, \mathrm{~B} \rightarrow \mathrm{IV}, \mathrm{C} \rightarrow \mathrm{II} . \end{aligned} $

Secondary valences are non-ionisable valences.

$ \begin{array}{lll} \hline \text { Formula } & \begin{array}{l} \text { Moles of } \\ \text { AgCl ppt. } \end{array} & \begin{array}{l} \text { Number of } \\ \text { non-ionisable } \\ \text { groups } \end{array} \\ \hline \mathrm{CoCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O} & 3 & 6 \\ \hline \mathrm{NiCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} & 2 & 6 \\ \hline \mathrm{Co}\left(\mathrm{SO}_4\right) \mathrm{Br} \cdot 5 \mathrm{NH}_3 & 1 & 6 \\ \hline \end{array} $

$ \begin{aligned} &\text { ∴ The correct option is }\\ &\mathrm{I} \rightarrow 6, \mathrm{II} \rightarrow 6, \mathrm{III} \rightarrow 6 \text {. } \end{aligned} $

In $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, $\mathrm{H}_2 \mathrm{O}$ is a weak field ligand. Hence, it will not pair up the electrons in metals orbital.

Both have 4 unpaired electrons in $d$-orbital.

$ \begin{aligned} & \mu=\sqrt{n(n+2)} \mathrm{BM} \\ & \mu=\sqrt{24} \mathrm{BM} \end{aligned} $

Therefore, both have same magnetic moment (BM).

A metal complex absorb orange light. The colour in which it appears is green-blue as it is the complementary colour of orange.

$ I\left[\mathrm{CO}_2(\mathrm{CO})_8\right]: 2 \mathrm{CO} \text { ligand bridges } $

$ {II\left[\mathrm{Fe}_3(\mathrm{CO})_{12}\right]: 4 \mathrm{CO} \text { ligand bridges }}{} $

$ \text { } III\left[\mathrm{Mn}_2(\mathrm{CO})_{10}\right] \text { : Zero CO ligand bridges } $

$ \text { }IV\left[\mathrm{Fe}_2(\mathrm{CO})_9\right]: 3 \mathrm{CO} \text { ligand bridges } $

A spectrochemical series is a list of ligands ordered by ligand strength, weak field $\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{2-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}< \mathrm{NO}_3^{-}<\mathrm{F}^{-}<\mathrm{C}_2 \mathrm{O}_4^{2-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NCS}^{-}< \mathrm{CH}_3 \mathrm{CN}<\mathrm{NH}_3<$ en $<$ bipy $<$ phen $< \mathrm{NO}_2<\mathrm{PPh}_3<\mathrm{CN}^{-}<\mathrm{CO}$ strong field

$ \text { So, according to series } \mathrm{III}>\mathrm{I}>\mathrm{IV}>\mathrm{II} \text {. } $

When a ligand are arranged in order of the magnitude of crystal field splitting, the arrangement, thus obtained is called spectrochemical series, the series is as follows:

$ \mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{Cl}^{-}<\mathrm{NO}_3^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{en} $

Hence, among the given options, $\mathrm{I}^{-}<\mathrm{Cl}^{-}<\mathrm{H}_2 \mathrm{O}<$ en

Thus, option (d) is the correct answer.

$ \begin{aligned} & \text {} \left.\stackrel{\mathrm{III}}{[\mathrm{Ti}}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right.] \mathrm{Cl}_3 \xrightarrow[\mathrm{O}_2]{250^{\circ} \mathrm{C}} \stackrel{\mathrm{IV}}{\mathrm{Ti}} \\ & \left(\mathrm{Ti}^{3+}: 3 d^1\right) \quad\left(\mathrm{Ti}^{4+}: 3 d^0\right) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text { Colourless } \end{aligned} $

Violet or purple colours is due to $d-d$ transition of one unpaired $d$-electron.

$\left[\stackrel{0}{\mathrm{Ni}}(\mathrm{CO})_4\right] \Rightarrow \mathrm{CO}$ is a neutral ligand. So, in metal carbonyls metal is present in their zero valent state.

$ \text { In }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} $

In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}=$ show rearrangement and has $d^2 s p^3$ (inner) hybridisation and will be diamagnetic, i.e.

$ \text { Hence, option (b) is the correct answer. } $

$ \begin{aligned} &\text { }\\ &\begin{aligned} A_{x-2}\left[B(C)_x\right]_2 & \longrightarrow(x-2) A^{+2}+\left[B(C)_x\right]^{-(x-2)} \\ A_{x-2}\left[B(C)_x\right] \cdot & A^{+2} \text { and }\left[B(C)_2\right]^{-(x-2)} \\ i & =1-\alpha+x \alpha-2 \alpha+2 \alpha \\ (x-1) \alpha & =3 \text { or } x-1=3 / \alpha \\ \alpha & =0.75, i=4, x=5 \end{aligned} \end{aligned} $

Hydrated copper sulphate $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in colour. The ligand (water) molecules causes splitting of $d$-orbitals. This facilitated $d-d$ transition and colour.

Anhydrous copper sulphate $\mathrm{CuSO}_4$ is colourless. In the absence of ligand (water) molecules, splitting of $d$-orbitals is not possible. Hence, $d$ - $d$ transition is not possible. Hence, option (b) is correct.