Coordination Compounds

Meridional (mer) isomerism is a type of geometrical isomerism found specifically in octahedral complexes, where three identical ligands occupy a meridian plane.

Thus, it requires at least three identical bidentate or monodentate ligands.

Given the options, the complex that could potentially exist as a meridional isomer is : Option D : $[\mathrm{Co}(\mathrm{NH}_{3})_{3}(\mathrm{NO}_{2})_{3}]$ In this complex, there are three $\mathrm{NH_3}$ and three $\mathrm{NO_2}$ ligands, allowing it to form a meridional configuration where either all three $\mathrm{NH_3}$ or all three $\mathrm{NO_2}$ ligands lie along a meridian plane.

To determine which complex exhibits maximum attraction to an applied magnetic field, we need to look at the number of unpaired electrons in each complex.

The more unpaired electrons a complex has, the greater the attraction to the magnetic field.

1.

Option A: [Ni(H₂O)₆]²⁺ Ni²⁺ has a 3d⁸ electron configuration.

In an octahedral crystal field, two of the electrons will occupy the lower energy t₂g orbitals, while the remaining six electrons will fill the higher energy e_g orbitals.

There are 2 unpaired electrons in this complex.

2.

Option B: [Co(H₂O)₆]²⁺ Co²⁺ has a 3d⁷ electron configuration.

In an octahedral crystal field, two of the electrons will occupy the lower energy t₂g orbitals, while the remaining five electrons will fill the higher energy e_g orbitals.

There are 3 unpaired electrons in this complex.

3.

Option C: [Co(en)₃]³⁺ Co³⁺ has a 3d⁶ electron configuration.

In an octahedral crystal field, all six electrons will occupy the lower energy t₂g orbitals.

There are no unpaired electrons in this complex.

4.

Option D: [Zn(H₂O)₆]²⁺ Zn²⁺ has a 3d¹⁰ electron configuration, with all orbitals being completely filled.

There are no unpaired electrons in this complex.

Based on the number of unpaired electrons, the complex with the maximum attraction to an applied magnetic field is [Co(H₂O)₆]²⁺ (Option B), as it has 3 unpaired electrons.

Let's analyze each combination : A.

Chlorophyll - Co : Mismatched.

Chlorophyll has a magnesium (Mg) ion at its center, not cobalt (Co).

B.

Water hardness - EDTA : Correct match.

EDTA (ethylenediaminetetraacetic acid) is used to treat water hardness by chelating metal ions like Ca²⁺ and Mg²⁺.

C.

Photography - [Ag(CN)₂]⁻ : Mismatched.

Silver halides (AgX, where X = Cl, Br, I) are used in photography, not [Ag(CN)₂]⁻.

D.

Wilkinson catalyst - [(Ph₃P)₃RhCl] : Correct match.

Wilkinson's catalyst is a homogeneous hydrogenation catalyst with the formula [(Ph₃P)₃RhCl].

E.

Chelating ligand - D-Penicillamine : Correct match.

D-Penicillamine is a chelating agent that can bind to metal ions through multiple coordination sites.

The mismatched combinations are A and C.

So, the correct answer is : A and C Only.

Assertion A :

absorbs at a lower wavelength of light with respect to

.

Given that H2O is a stronger field ligand than Cl-, it would result in a larger splitting of the d-orbitals and hence a larger energy difference.

This would lead to absorption at a lower wavelength.

Therefore, the Assertion A is false.

Reason R : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.

While it's true that the oxidation state can influence the absorption wavelength by affecting the energy levels of the d-orbitals, the difference in the absorption wavelength in this specific case is not primarily due to the oxidation state (the oxidation state of Co in both complexes is the same (+3)), but rather due to the difference in ligand field strength (H2O vs Cl-).

Therefore, the Reason R is not providing the correct explanation for Assertion A.

So, the correct answer would be: Option A : A is false but R is true.

Among the given options, the compounds that show color due to d-d transitions are those that have partially filled d-orbitals when in the form of complex ions or compounds.

Let's examine each option: Option A: $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ Chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ exists in the +6 oxidation state as the dichromate ion ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$).

In this oxidation state, chromium does not have any electrons in the d-orbitals (it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ configuration) so there is no possibility for d-d transitions.

Instead, the color is due to charge transfer transitions.

Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ Copper(II) sulfate pentahydrate, $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$, contains copper in a +2 oxidation state.

In this state, copper has a $\mathrm{[Ar]}\;3\mathrm{d}^9$ electron configuration, which means there is one unpaired electron in the d-orbitals.

Due to the presence of this unpaired electron, d-d transitions can occur, and this is why $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in color.

Option C: $\mathrm{KMnO}_4$ In potassium permanganate ($\mathrm{KMnO}_4$), manganese is in the +7 oxidation state, and it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ electron configuration.

Like chromium in the +6 state, manganese in the +7 state does not have any d-electrons available for d-d transitions.

The deep purple color of $\mathrm{KMnO}_4$ is also due to charge transfer transitions.

Option D: $\mathrm{K}_2 \mathrm{CrO}_4$ Potassium chromate ($\mathrm{K}_2 \mathrm{CrO}_4$) contains chromium in the +6 oxidation state as the chromate ion ($\mathrm{CrO}_4^{2-}$).

Like dichromate, chromium does not have d-electrons and therefore cannot undergo d-d transitions in this compound.

The color is due to charge transfer transitions.

So the only compound from the options listed that shows color due to d-d transitions is Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$.

The complex $\left[\mathrm{NiCl}_{2} \mathrm{Br}_{2}\right]^{2-}$ is a tetrahedral complex due to the fact that Nickel (Ni) in this complex is in its +2 oxidation state, and thus adopts a dsp² hybridization, resulting in a tetrahedral geometry.

If $\mathrm{Ni}^{2+}$ is replaced by $\mathrm{Pt}^{2+}$, the new complex would be $\left[\mathrm{PtCl}_{2} \mathrm{Br}_{2}\right]^{2-}$.

Platinum (Pt) in its +2 oxidation state typically forms square planar complexes due to its preference for dsp² hybridization.

Geometry : $\left[\mathrm{NiBr}_2 \mathrm{Cl}_2\right]^{2-}$ is tetrahedral. $\left[\mathrm{PtBr}_2 \mathrm{Cl}_2\right]^{2-}$ is square planar.

Thus, replacing $\mathrm{Ni}^{2+}$ with $\mathrm{Pt}^{2+}$ changes the geometry from tetrahedral to square planar.

Geometrical Isomerism : Tetrahedral complexes, like $\left[\mathrm{NiBr}_2 \mathrm{Cl}_2\right]^{2-}$, do not exhibit geometrical isomerism.

Square planar complexes, like $\left[\mathrm{PtBr}_2 \mathrm{Cl}_2\right]^{2-}$, can exhibit geometrical isomerism.

Therefore, geometrical isomerism would be introduced by replacing $\mathrm{Ni}^{2+}$ with $\mathrm{Pt}^{2+}$.

Optical Isomerism : Both the $\mathrm{Ni}$-based tetrahedral complex and the $\mathrm{Pt}$-based square planar complex are optically inactive.

Optical isomerism is not observed in either case.

Magnetic Properties : The magnetic properties depend on the electronic configuration of the central metal ion.

There will be a change in magnetic properties due to the different electronic configurations of $\mathrm{Ni}^{2+}$ and $\mathrm{Pt}^{2+}$.

Based on this information, the properties expected to change when $\mathrm{Ni}^{2+}$ is replaced by $\mathrm{Pt}^{2+}$ are: Geometry (from tetrahedral to square planar) Geometrical Isomerism (introduced in the $\mathrm{Pt}$-based complex) Magnetic Properties (due to different electronic configurations) Optical isomerism remains unchanged (inactive in both cases).

Therefore, the correct answer is : Option B: A, B, and D (Geometry, Geometrical Isomerism, and Magnetic Properties).

For spin-only magnetic moment, we use the formula:

where $n$ is the total number of unpaired electrons. For

, the electronic configuration of $\mathrm{Fe^{3+}}$ is $\mathrm{d^5}$.

Here, all the five electrons will be unpaired due to the high spin nature of Fe$^{3+}$.

Hence, $n=5$ and magnetic moment

For

, the $\mathrm{Fe^{3+}}$ ion has electronic configuration $\mathrm{d^5}$.

In this case, the strong field ligand cyanide ($\mathrm{CN}^{-}$) will pair up the electrons in the $\mathrm{d}$ orbital.

So, there will be only one unpaired electron.

Hence, $n=1$ and magnetic moment

Therefore, the answer is 5.92 B.M. and 1.732 B.M.

Option A : -

(Oxalate) is a bidentate ligand. It always binds through its two oxygen atoms. -

(Nitrite) is an ambidentate ligand. It can coordinate either through nitrogen or oxygen. -

(Thiocyanate) can also bind through either nitrogen or sulfur, making it an ambidentate ligand. Option B : -

(Oxalate) as mentioned earlier, is a bidentate ligand. - Ethylene diamine (en) is also a bidentate ligand, as it always binds through its two nitrogen atoms. -

(Water) is a monodentate ligand, binding only through one oxygen atom. Option C : -

as mentioned, is an ambidentate ligand. -

as mentioned, is a bidentate ligand. -

is a hexadentate ligand. It always binds through six donor atoms and is not ambidentate. Option D : -

as mentioned, is a hexadentate ligand. -

as mentioned, is an ambidentate ligand. -

as mentioned, is a bidentate ligand.

From the above analysis, it's clear that Option B is the one that does not contain any ambidentate ligands.

So, the correct answer is Option B.

In the complex [Co(NH$_3$)$_6$]$^{3+}$, Co is in the +3 oxidation state.

This means it has a configuration of $d^6$.

NH$_3$ is a strong field ligand, which causes the d-electrons to pair up.

Therefore, this complex is a low-spin (also referred to as spin-paired) complex.

The term "low-spin" means that the electrons prefer to pair up in the lower energy d-orbitals rather than occupy the higher energy orbitals.

When all the electrons are paired, the complex is diamagnetic, which means it's not attracted to an external magnetic field.

The hybridization of this complex is $d^2sp^3$, also known as inner orbital complex, because the inner d-orbitals are used in the hybridization.

The complexes [CoF$_6$]$^{3-}$ and [CoCl$_6$]$^{3-}$ are both octahedral, but they do have unpaired electrons because $F^-$ and $Cl^-$ are not strong enough field ligands to cause pairing of all the electrons in the d-orbitals.

Therefore, they are not low-spin or diamagnetic.

Lastly, the complex [NiCl$_4$]$^{2-}$ is not octahedral.

Because $Cl^-$ is a weak field ligand, the electrons in the d-orbitals do not pair up, and the $Ni^{2+}$ ion, with a $d^8$ configuration, undergoes sp^3 hybridization, resulting in a tetrahedral complex.

Therefore, the only octahedral, diamagnetic, and low-spin complex in the options provided is [Co(NH$_3$)$_6$]$^{3+}$ (Option A).

Let's analyze each statement separately: Statement (I): So, Statement I is false.

Statement (II): Prussian blue, also known as ferric ferrocyanide, is a famous pigment that contains iron in both +2 and +3 oxidation states.

Its chemical formula can be represented as $\mathrm{Fe}_4\mathrm{[Fe(CN)_6]_3}$.

In this formula, the iron within the square brackets, $\mathrm{[Fe(CN)_6]^{4-}}$, is in the +2 oxidation state (ferrocyanide ion), while the iron outside the square brackets is in the +3 oxidation state (ferric ion).

The compound is a complex salt that arises from the reaction of ferric and ferrous ions with cyanide ions.

Therefore, Statement II is also true.

With both statements being true, the correct answer is: Option A : Statement I is false but Statement II is true.