Coordination Compounds

The first three electrons fill the

Sv6Kpe[] orbitals singly.Sv6Kpe[]Sv6Kpe[]

The next three electrons pair up with the electrons already in the

Sv6Kpe[] orbitals.
This results in all six electrons occupying the

Sv6Kpe[] orbitals, and the

Sv6Kpe[] orbitals remain empty.Sv6Kpe[]

${\mathrm{CN}}^{-}$ is the strongest ligand among all, hence pairing occurs in this case and for the complex ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right] , \mathrm{\mu }=0 \mathrm{BM}$

The complexes ${\left[Fe(CN{)}_6\right]}^{3-}$ and ${\left[CoC{l}_4\right]}^{2-}$ have odd-numbered d-electrons .

ΔTf = Kf ⋅ m ⋅ i and i = n ( 100% dissociation)

Where:

•ΔTf = Freezing point depression

•Kf = Freezing point depression constant

•m = Molality of the solution

•i = Van’t Hoff factor

[Co(NH₃)₆]³⁺: Co³⁺ = [Ar],3d⁶

NH₃ is strong field ligand in Co³⁺.

[Fe(CN)₆]^4-: Fe²⁺ = [Ar],3d⁶

CN⁻ is strong field ligand.

(I) ${\left[Mn(CN{)}_6\right]}^{3-}=-1.6{\Delta }_0$
(II) ${\left[Co(CN{)}_6\right]}^{4-}=-1.8{\Delta }_0$
(III) ${\left[Fe(CN{)}_6\right]}^{4-}=-2.4{\Delta }_0$
(IV) ${\left[Fe(CN{)}_6\right]}^{3-}=-2.0{\Delta }_0$

The stability of a complex can be correlated with its Crystal Field Stabilization Energy (CFSE). CFSE is the energy difference between the metal ion in the complex and its energy in the isolated state due to the splitting of d-orbitals in the presence of a ligand field. Higher CFSE values correspond to more stable complexes.

• [Ni(CO)₄] is tetrahedral → no geometrical isomerism.

• [PdCl₂(PPh₃)₂] (square planar, MA₂B₂) → cis/trans isomerism.
  So option A fails.

• [Co(en)(NH₃)₂Cl₂], [Co(NH₃)₃(NO₂)₃], [Cr(NH₃)₄Cl₂]Cl₂ are all octahedral and can show geometrical isomerism.

Set-II

• [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅(SO₄)]Cl differ in which anion is inside the coordination sphere → ionization isomers.

• [Co(NH₃)₅][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are coordination isomers, not ionization isomers.

• [Cr(H₂O)₆]Cl₃ vs [Cr(H₂O)₅Cl]Cl₂·H₂O are hydrate/coordination isomers, not ionization isomers.

Hence only option C has both complexes in Set-I showing geometrical isomerism and Set-II giving ionization isomerism.

In the case of weak field ligands, pairing will not take place until all the orbitals are singly occupied. Also, weak field ligands form high spin complexes. For weak field ligand, configuration is ${{\mathrm{t}}_{2\mathrm{g}}}^3{\mathrm{e}}_{\mathrm{g}}^1$

$$\begin{gathered} \text{ Reaction and colour changes } \\ {\left[\underset{\text{ Gren }}{Ni{\left(H_{2}O\right)}_6}\right]}^{2+}\left(aq\right)+en\left(aq\right)\to {\left[Ni{\left(H_{2}O\right)}_4\left(en\right)\right]}^{2+}\left(aq\right)+2H_{2}O \\ {\left[Ni{\left(H_{2}O\right)}_4\left(en\right)\right]}^{2+}\left(aq\right)+en\left(aq\right)\to \left[\underset{\text{ Bluc/Purple }}{{\left.Ni{\left(H_{2}O\right)}_2(en{)}_2\right]}^{2+}\left(aq\right)+2H_{2}O}\right. \\ \left[\underset{\text{ Bluc/Purple }}{{\left.Ni{\left(H_{2}O\right)}_2(en{)}_2\right]}^{2+}\left(aq\right)+en\left(aq\right)\to }\right.\underset{\text{ violet }}{{\left.Ni(en{)}_3\right]}^{2+}}\left(aq\right)+2H_{2}O \end{gathered}$$

\begin{aligned}
& \mathrm{Ni}^{2+}+6 \mathrm{H}_2 \dot{\mathrm{O}} \longrightarrow\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \\
& \text { Green coloured } \\
& \text { [A] } \\
& {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+e n \longrightarrow\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4 e n\right]^{2+}+2 \mathrm{H}_2 \mathrm{O}} \\
& \text { Pale-yellow } \\
& \text { [B] } \\
& \left.\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4 e n\right]^{2+}+e n \longrightarrow \mid \mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(e n)_2\right]^{2+}+2 \mathrm{H}_2 \mathrm{O} \\
& \text { Blue/purple } \\
& \text { [C] } \\
& {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(e n)_2\right]^{2+}+e n \longrightarrow\left[\mathrm{Ni}(e n)_3\right]^{2+}+2 \mathrm{H}_2 \mathrm{O}} \\
& \text { Violet } \\
& \text { [D] }
\end{aligned}

To determine hybridization, we consider oxidation state, ligand strength, and geometry.

[CoF₆]³⁻ has Co³⁺ (d⁶) with fluoride as a weak field ligand, leading to a high-spin octahedral complex with sp³d² hybridization.

[Co(NH₃)₆]³⁺ has Co³⁺ (d⁶) with ammonia as a strong field ligand, forming a low-spin octahedral complex with d²sp³ hybridization.

[NiCl₄]²⁻ has Ni²⁺ (d⁸) with chloride as a weak field ligand, resulting in a tetrahedral complex with sp³ hybridization.

[Ni(CN)₄]²⁻ has Ni²⁺ (d⁸) with cyanide as a strong field ligand, forming a square planar complex with dsp² hybridization.

The correct answer is option (1).

The chemical formula of nitropenta ammine chromium(III) chloride is: $\left[Cr{\left(N{H}_3\right)}_5\right(N{O}_2\left)\right]C{l}_2$ It can exist in following two structures $\left[Cr{\left(N{H}_3\right)}_5\right(N{O}_2\left)\right] C{l}_2$ and Pentaamminenitrochromium (III) chloride $\left[Cr{\left(N{H}_3\right)}_{5}ONO\right]C{l}_2$ Pentaamminenitritochromium (III) chloride Therefore, the type of isomerism found in this compound is linkage isomerism as $-N{O}_2$ group is linked through N as −N$O_2$ or through O as ONO.

The electronic configuration of an octahedral Co(II) complex is t₂g⁶ e₁. Co(II) has a d^7 configuration, with 3 unpaired electrons. The magnetic moment is given by square root [n(n+2)], where n = 3 unpaired electrons, giving = 3.95 BM.

For the given coordination compounds:

1. K₃[Fe(OH)₆]:

   • Fe³⁺ (3d⁵, high spin in weak field ligand OH⁻)
   • Unpaired electrons = 5
   • Magnetic moment = √[n(n+2)] = √[5(5+2)] = √35 ≈ 5.91 BM

2. K₄[Fe(OH)₆]:

   • Fe²⁺ (3d⁶, high spin in weak field ligand OH⁻)
   • Unpaired electrons = 4
   • Magnetic moment = √[4(4+2)] = √24 ≈ 4.90 BM

When the complex $\left[Co{\left(N{H}_3\right)}_{5}Cl]C{l}_2\right.$ dissolves in water, it dissociates into three ions:

$\left[Co{\left(N{H}_3\right)}_{5}Cl\right]C{l}_2\to {\left[Co{\left(N{H}_3\right)}_{5}Cl\right]}^{2+}+2C{l}^{-}$

When silver nitrate $\left(AgN{O}_3\right)$ is added to the solution, the free $C{l}^{-}$ ions react with $A{g}^{+}$to form 2 moles of silver chloride ( AgCl ) precipitate.

$\left[Co{\left(N{H}_3\right)}_{5}Cl\right]C{l}_2+2AgN{O}_3\to \left[Co{\left(N{H}_3\right)}_{5}Cl\right] {\left(N{O}_3\right)}_2+2AgCl\left(s\right)$

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

• An ambidentate ligand can coordinate through two different atoms (e.g., SCN⁻ via S or N, NO₂⁻ via N or O, CN⁻ via C or N).

• $\mathrm{C_2O_4^{2-}}$ (oxalate) is bidentate (binds through two oxygens) but not ambidentate (it does not offer distinct donor atoms for alternative attachment).

EDTA (ethylenediaminetetraacetic acid) is a hexadentate ligand, meaning it can donate six pairs of electrons to a metal ion — four from carboxylate oxygens and two from amine nitrogens.

When Ba²⁺ (or any alkaline earth metal like Ca²⁺, Mg²⁺) forms a complex with EDTA, the ligand wraps around the metal ion, coordinating through all six donor sites. This results in a coordination number of 6, which corresponds to an octahedral geometry.

${\left[{\mathrm{MnF}}_6\right]}^{3-} \to {\mathrm{Mn}}^{3+ }=3{\mathrm{d}}^4 \mathrm{configuration}$.

For the given coordination compounds:

1. K₃[Fe(OH)₆]:

   • Fe³⁺ (3d⁵, high spin in weak field ligand OH⁻)
   • Unpaired electrons = 5
   • Magnetic moment = √[n(n+2)] = √[5(5+2)] = √35 ≈ 5.91 BM

2. K₄[Fe(OH)₆]:

   • Fe²⁺ (3d⁶, high spin in weak field ligand OH⁻)
   • Unpaired electrons = 4
   • Magnetic moment = √[4(4+2)] = √24 ≈ 4.90 BM

To determine the d⁴ configuration, we calculate the oxidation state of the metal in each complex and check its electronic configuration.

(A) [FeO₄]²⁻ → Fe in +6 state → d² (not d⁴)

(B) [Mn(CN)₆]³⁻ → Mn in +3 state → d⁴

(C) [Fe(CN)₆]³⁻ → Fe in +3 state → d⁵ (not d⁴)

(D) Cr₂(OCOCH₃)₄(H₂O)₂ → Cr in +2 state → d⁴

(E) [NiF₆]²⁻ → Ni in +4 state → d⁶ (not d⁴)

${\mathrm{CN}}^{-}$ is the strongest ligand among all, hence pairing occurs in this case and for the complex ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right] , \mathrm{\mu }=0 \mathrm{BM}$

$\mathrm{K}\to {\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$

$x+6(-1)=3$

$x=+3$

${\mathrm{Fe}}^{3+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^5$ (Paramagnetic)

$\mathrm{N}\to \left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2$

${\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$ (Paramagnetic)

To determine paramagnetism, we analyze ligand field effects and unpaired electrons.

[NiCl₄]²⁻: Ni²⁺ (3d⁸), Cl⁻ is a weak field ligand, resulting in a high-spin tetrahedral geometry with two unpaired electrons, making it paramagnetic.

[Ni(CO)₄]: Ni⁰ (3d⁸ 4s²), CO is a strong field ligand, forming a low-spin tetrahedral complex with paired electrons, making it diamagnetic.

[Ni(CN)₄]²⁻: Ni²⁺, CN⁻ forms a square planar complex with paired electrons, making it diamagnetic.

[Fe(CO)₅]: Fe⁰ (3d⁶ 4s²), CO forms a low-spin d⁶ complex with paired electrons, making it diamagnetic.

Correct answer: [NiCl₄]²⁻.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}} =0.{558}^0\mathrm{C}, {\mathrm{K}}_{\mathrm{f}} =1.86 \mathrm{amd} \mathrm{m}=0.1 \\ ∆{\mathrm{T}}_{\mathrm{f}} =\mathrm{i}\times \mathrm{m}\times {\mathrm{K}}_{\mathrm{f} } \\ 0.558 =\mathrm{i} \times 0.1\times 1.86 \Rightarrow \mathrm{i}=3 \\ \mathrm{Hence} , \mathrm{the} \mathrm{complex} \mathrm{gives} 3 \mathrm{ions} \mathrm{after} \mathrm{dissociation}. \\ \mathrm{The} \mathrm{Formula} \mathrm{of} \mathrm{complex} \mathrm{must} \mathrm{be} \left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2 \end{gathered}$$

For same metal ion (Co³⁺, octahedral), λmax ∝ 1/Δ₀.
Ligand field strength: Cl⁻ < H₂O < NH₃ < CN⁻.
So Δ₀: [Co(CN)₆]³⁻ > [Co(NH₃)₆]³⁺ > [Co(NH₃)₅(H₂O)]³⁺ > [Co(NH₃)₅Cl]²⁺.
Therefore λmax order is the reverse:

[Co(CN)₆]³⁻ < [Co(NH₃)₆]³⁺ < [Co(NH₃)₅(H₂O)]³⁺ < [Co(NH₃)₅Cl]²⁺.

For the complex \( [\text{Co}(\text{NH}_3)_6]^{3+} \), \( \text{Co} \) is in the \( +3 \) oxidation state and has a \( 3d^6 \) configuration. Ammonia \( (\text{NH}_3) \) is a strong-field ligand, which causes electron pairing in the \( 3d \) orbitals. This results in an inner orbital complex with no unpaired electrons, making it a spin-paired complex.

For the complex \( [\text{CoF}_6]^{3-} \), \( \text{Co} \) is also in the \( +3 \) oxidation state with a \( 3d^6 \) configuration. Fluoride \( (\text{F}^-) \) is a weak-field ligand, which does not cause electron pairing. This results in an outer orbital complex with four unpaired electrons, making it a spin-free complex.

• The final name is constructed as follows: dibromidobis(triphenylphosphine)platinum(II).

From the NCERT spectrochemical series (weak → strong): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < … < CN⁻; hence among the given, $({\mathrm{Br}}^{-}<{\mathrm{F}}^{-}<{\mathrm{H}}_2\mathrm{O}<{\mathrm{NH}}_3).$

${\mathrm{CoCl}}_3.5{\mathrm{NH}}_3$ give 2 moles of AgCl ppt in the presence of excess amount of ${\mathrm{AgNO}}_3$.

${\mathrm{CoCl}}_2.5{\mathrm{NH}}_3 +2{\mathrm{AgNO}}_3\to 2\mathrm{AgCl} +\left[\mathrm{CoCl}{\left({\mathrm{NH}}_3\right)}_5\right]{\left({\mathrm{NO}}_3\right)}_2$

Co form the co-ordination complex having coordination number 6.

WFL will not cause pairing & high spin arrangement due to greater no. unpaired electrons.

$EDT{A}^{4-}$Hexadentate ligand

${\left[Ca\right(EDTA\left)\right]}^{2-}$

So Coordination environment is octahedral

The oxidation states are $F{e}^{3+}$ in $K_3\left[Fe(OH{)}_6\right]$ (high-spin $d^5,5.92$ B.M.) and $F{e}^{2+}$ in $K_4\left[Fe(OH{)}_6\right]$ (high-spin $d^6$, 4.90 B.M.).