Coordination Compounds

In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right]_6\right]^{3+}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6, \mathrm{NH}_3$ is S.F.L Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$ In $\mathrm{SF}_6$, Hybridisation state of sulphur is $\mathrm{sp}^3 \mathrm{~d}^2$ In $\left[\mathrm{CrF}_6\right]^{3-}, \mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^3$ Hybridisation state of $\mathrm{Cr}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$ $\left[\mathrm{CoF}_6\right]^{3-}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6 \mathrm{~F}^{-}$is W.F.L Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{sp}^3 \mathrm{~d}^2$ $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^4 \mathrm{CN}^{-}$is S.F.L Hybridisation state of $\mathrm{Mn}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$ $\left[\mathrm{MnCl}_6\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^4 \mathrm{Cl}^{-}$is W.F.L Hybridisation state of $\mathrm{Cl}^{-}$is $\mathrm{sp}^3 \mathrm{~d}^2$ Total number of $\mathrm{sp}^3 \mathrm{~d}^2$ hybridized molecules is 3

Homoleptic complex of type $\left[\mathrm{Ma}_6\right]$ (Where $\mathrm{a} \Rightarrow$ monodentate ligand) cannot show geometrical as well as optical isomerism.

Cis-platin and trans-platin has formula $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ which is a heteroleptic complex of platinum.

Determine the oxidation state of manganese in

Chloride (Cl⁻) has a charge of -1. With six chloride ions, the total charge contributed by the ligands is

Let the oxidation state of Mn be

Then:

So, manganese is in the +3 oxidation state.

Find the d-electron count for Mn(III): The neutral manganese atom has the electronic configuration

Removing three electrons (first from the 4s, then from the 3d) gives:

Therefore, Mn(III) is a

system.

Predict the spin state in an octahedral complex: Chloride (Cl⁻) is a weak-field ligand, meaning it produces a relatively small crystal field splitting (

). When

is small, the pairing energy is higher than

so electrons prefer to remain unpaired. For a

configuration in a high-spin octahedral complex, the electrons will be distributed as: Three electrons in the three

orbitals (one in each) One electron in one of the

orbitals This results in a total of 4 unpaired electrons, making the complex paramagnetic.

Determine the type of hybridization: In octahedral complexes, the two common hybridizations are:

(inner orbital complex), typically seen in low-spin complexes where inner 3d orbitals are available because the electrons are paired.

(outer orbital complex), seen in high-spin complexes where the inner

orbitals are occupied by unpaired electrons. Since

is high-spin (because of Cl⁻ being a weak-field ligand), the inner 3d orbitals are not available for hybridization.

Therefore, the complex uses the outer orbitals (the 4s, 4p, and 4d orbitals), leading to an

hybridization. Conclusion: The hybridization of

is

It is paramagnetic with four unpaired electrons. Thus, the correct answer is: Option A:

, paramagnetic with four unpaired electrons.

Identify the oxidation state and d-electron count for each complex: For

: Ethylenediamine (en) is neutral. Oxidation state of cobalt is +3. Cobalt (atomic number 27) in the +3 state gives a

configuration. With strong/moderate field ligands like en, the complex is low spin, leading to a configuration of

. For

: Each fluoride ion is

so, with six of them, Co must be in the +3 state again.

However, since fluoride is a weak field ligand, this complex adopts a high spin configuration for

, resulting in

. For

: Water is neutral. Manganese in the +2 state gives a

configuration. With weak field water, the

configuration is high spin:

. For

: Zinc in the +2 state has a

configuration. In an octahedral field, all 10 electrons fill the orbitals as

.

Calculate the crystal field stabilization energy (CFSE) for each case: The CFSE in an octahedral field can be estimated as:

For

(low-spin

in

):

For

(high-spin

in

):

For

(high-spin

in

):

For

(for

in

):

Compare the CFSE values:

gives a CFSE of

(most negative, hence highest stabilization).

The others result in less stabilization (or net zero).

Conclusion: The complex with the highest CFSE is the one with the configuration

, which corresponds to

. Thus, the answer is: Option A:

.

The energy of absorbed light by a complex can be expressed as $E = hv = \dfrac{hC}{\lambda}$, where $E$ is the energy, $h$ is Planck's constant, $v$ is the frequency, $C$ is the speed of light, and $\lambda$ is the wavelength of the absorbed light.

This formula demonstrates that energy ($E$) is inversely proportional to the wavelength ($\lambda$).

In this context, all cobalt complexes are in the +3 oxidation state.

The Crystal Field Stabilization Energy (CFSE) is influenced by the strength of the ligands surrounding the metal ion.

As the ligand field strength increases, the CFSE also increases.

The order of field strength for the ligands in this case is: $\text{CN}^- > \text{NH}_3 > \text{H}_2\text{O} > \text{Cl}^-$ This implies the following order for CFSE in the given complexes: $\text{C} > \text{B} > \text{A} > \text{D}$ Because energy is inversely proportional to wavelength, complexes absorbing at higher energies will absorb at shorter wavelengths.

Therefore, the order of the complexes in terms of the wavelength of light absorbed, from longest to shortest wavelength, is: $\text{D} > \text{A} > \text{B} > \text{C}$

.tg .tg List - I List -II (A)

(III)

(B)

(IV)

(C)

(I)

(D)

(II)

The number of unpaired electrons responsible for the paramagnetic nature of each complex species is calculated as follows: Complex: $[Fe(CN)_6]^{3-}$ Ion: $Fe^{3+}$ Electronic Configuration: $3d^5$ Orbital Population: $t_{2g}^{2,2,1}$ $e_g^{0,0}$ Unpaired Electrons: 1 Complex: $[FeF_6]^{3-}$ Ion: $Fe^{3+}$ Electronic Configuration: $3d^5$ Orbital Population: $t_{2g}^{1,1,1}$ $e_g^{1,1}$ Unpaired Electrons: 5 Complex: $[CoF_6]^{3-}$ Ion: $Co^{3+}$ Electronic Configuration: $3d^6$ Orbital Population: $t_{2g}^{2,1,1}$ $e_g^{1,0}$ Unpaired Electrons: 4 Complex: $[Mn(CN)_6]^{3-}$ Ion: $Mn^{3+}$ Electronic Configuration: $3d^4$ Orbital Population: $t_{2g}^{2,1,1}$ $e_g^{0,0}$ Unpaired Electrons: 2 Each complex's paramagnetic nature is due to the specific number of unpaired electrons as indicated above.

$=[-0.4 \times 6+0.6 \times(0)] \Delta_0=-2.4 \Delta_0$

(3) $\quad\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4} \quad \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0$

Geometrical isomer $=2(1$ cis +1 trans $)$ Optical isomer $=3$ ( 2 optically active +1 optically inactive) Stereoisomer $=3$

Wilkinson's catalyst is [RhCl(PPh3)3] Here central atom is Rh.

Oxidation state of Rh here is = + 1 $\therefore$ Electronic configuration of Rh+ = [Kr]4d8 And the complex is square planar