${\rm{CuS}}{{\rm{O}}_4}$ decolourises on addition of KCN, the product is:

${\left[ {{\rm{Cu}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{3 - }}$

${\rm{Cu}}\left( {{\rm{CN}}} \right)_4^{2 - }$

${\rm{Cu}}{\left( {{\rm{CN}}} \right)_2}$

The IUPAC name of ${{\rm{K}}_2}\left[ {{\rm{Ni}}{{\left( {{\rm{CN}}} \right)}_4}} \right]$ is

Potassium tetracyanatonickelate (III)

Potassium tetracyanatonickel (II)

Potassium tetracyanonickel (III)

An aqueous solution of ${\rm{CoC}}{{\rm{l}}_2}$ on addition of excess of concentrated ${\rm{HCl}}$ turns blue to formation of

${\left[ {{\rm{CoC}}{{\rm{l}}_4}} \right]^{2 - }}$

${\left[ {{\rm{Co}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_2}{\rm{C}}{{\rm{l}}_4}} \right]^{2 - }}$

${\left[ {{\rm{Co}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_2}2{\rm{C}}{{\rm{l}}_4}} \right]^{2 - }}$

$\left[ {{\rm{Co}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_4}{\rm{C}}{{\rm{l}}_2}} \right]$

The IUPAC name of the compound $\left[ {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}{{\left( {{\rm{N}}{{\rm{O}}_3}} \right)}_2}} \right]$ is:

Tetraamminecopper(II) dinitrite

Tetraamminecopper(III) dinitrite

Which of the following organometallic compound is $\sigma \,\,{\rm{and}}\,\,\pi $ bonded?

$\left[ {{\rm{Fe}}{{\left( {{{\rm{\eta }}^5} - {{\rm{C}}_5}{{\rm{H}}_5}} \right)}_2}} \right]$

${\rm{Fe}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}$

${\left[ {{\rm{Co}}{{\left( {{\rm{CO}}} \right)}_5}{\rm{N}}{{\rm{H}}_3}} \right]^{2 + }}$

${\rm{K}}\left[ {{\rm{PtC}}{{\rm{l}}_3}\left( {{{\rm{\eta }}^2} - {{\rm{C}}_2}{{\rm{H}}_4}} \right)} \right]$

The IUPAC name of $\left[ {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right]\left[ {{\rm{NiC}}{{\rm{l}}_4}} \right]{\rm{\;}}$ is

Tetrachloro nickel (II) – tetraammine nickel (II)

Tetraammine nickel (II) –tetrachloro nickel(II)

Tetrachloro nickel (II) –tetraammine nickelate(0)

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Medium

1 of 310Question permalink

Understanding +4 | -1

The complexes

\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}

are example for isomerism

View answer and explanation

Correct option: C

Ionization

Option C (Ionization) is correct because the complexes differ in the ions present outside the coordination sphere.

Solution approach

Correct Answer + Core Reason: Ionization. The complexes exchange ions between the coordination sphere and the counter ions.

Step-by-Step Solution:

Step 1: Identify the complexes: $\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}$

Step 2: Determine the ions present in solution when each complex is dissolved in water.\nComplex 1: $\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2+} + 2\mathrm{B}{\mathrm{r}^ - }$ \nComplex 2: $\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2+} + 2\mathrm{C}{\mathrm{l}^ - }$

Step 3: Observe that the complexes have exchanged ions ( $\mathrm{Br}^-$ and $\mathrm{Cl}^-$ ) between the coordination sphere and the counter ions.

Step 4: Conclude that this is ionization isomerism.

Why other options are wrong:
- Option A (Geometrical): Geometrical isomers have the same ligands but different spatial arrangements. These complexes differ in the ions inside and outside the coordination sphere, not just their arrangement.
- Option B (Optical): Optical isomers are non-superimposable mirror images. These complexes are not chiral.
- Option D (Linkage): Linkage isomers involve ambidentate ligands that can bind through different atoms. These complexes do not have ambidentate ligands.

Key Concepts:
- Isomerism in coordination compounds
- Ionization isomerism

Memory Aids:
- Ionization isomers: Exchange of ions between the coordination sphere and the counter ions.

Quick Tips:
- Identify the ions inside and outside the coordination sphere to determine if ionization isomerism is present.

2 Hard

2 of 310Question permalink

Analyzing +4 | -1

The total number of stereoisomers for the complex

\left[\mathrm{Cr}(\mathrm{Ox})_{2} \mathrm{ClBr}\right]^{3-}

(where

\text{ox =}

oxalate) is

View answer and explanation

Correct option: A

3

Option A (3) is correct because the complex can exhibit cis and trans isomers, and the cis isomer is chiral, leading to a total of 3 stereoisomers.

Solution approach

Correct Answer + Core Reason: 3, due to cis, trans isomers and optical activity of cis isomer.

Step-by-Step Solution:

Step 1: Draw the possible isomers of the complex $\left[\mathrm{Cr}(\mathrm{Ox})_{2} \mathrm{ClBr}\right]^{3-}$ .

Step 2: The complex can exist in cis and trans forms with respect to Cl and Br.

Step 3: The trans isomer is optically inactive.

Step 4: The cis isomer is chiral and exists as a pair of enantiomers (d and l forms). So, cis isomer contributes 2 stereoisomers.

Step 5: The trans isomer contributes 1 stereoisomer.

Step 6: Total number of stereoisomers = 2 (from cis) + 1 (from trans) = 3.

Why other options are wrong:
- Option B: 2 is incorrect as it only considers either cis and trans or the enantiomers of cis.
- Option C: 4 is incorrect as it overcounts the isomers.
- Option D: 1 is incorrect as it only considers the trans isomer or fails to recognize the cis isomer's chirality.

Key Concepts:
- Stereoisomerism
- Cis-trans isomerism
- Optical isomerism
- Chirality

Memory Aids:
- ox = oxalate (bidentate ligand)

Quick Tips:
- Draw all possible isomers to avoid missing any.

3 Medium

3 of 310Question permalink

Application +4 | -1

The value of

x

on the

{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^x}

is:

View answer and explanation

Correct option: B

-2

The correct answer is -2 because the overall charge of the complex ion

{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^x}

is -2.

Solution approach

Correct Answer + Core Reason: -2. The overall charge of the complex ion is -2.

Step-by-Step Solution:

Step 1: Identify the complex: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^x}$

Step 2: Determine the oxidation state of Ni. Ni is in group 10, and its common oxidation state is +2.

Step 3: Determine the charge of the ligand CN: -1

Step 4: Set up the equation: $+2 + 4(-1) = x$

Step 5: Solve for $x$ : $2 - 4 = x \Rightarrow x = -2$

Why other options are wrong:
- Option A: Incorrect charge balance.
- Option C: Incorrect charge balance.
- Option D: Incorrect charge balance.

Key Concepts:
- Oxidation state
- Coordination complexes

Memory Aids:
- Remember the common oxidation states of elements and ligands.

Quick Tips:
- Always balance the charges in the complex to find the overall charge of the complex ion.

4 Medium

4 of 310Question permalink

Application +4 | -1

Excited state configuration of

\mathrm{M}{\mathrm{n}^{2 + }}

is

View answer and explanation

Correct option: B

t_{2g}^3e_g^2

Option B is correct. The excited state configuration of

\mathrm{M}{\mathrm{n}^{2 + }}

is

t_{2g}^3e_g^2

because one or more electrons from the

t_{2g}

orbitals get excited to the

e_g

orbitals.

Solution approach

Correct Answer + Core Reason: $t_{2g}^3e_g^2$ . In the excited state, electrons are promoted from lower energy orbitals to higher energy orbitals.

Step-by-Step Solution:
Step 1: Determine the ground state electronic configuration of $\mathrm{M}{\mathrm{n}^{2 + }}$ .
$\mathrm{M}{\mathrm{n}^{2 + }}$ has a $3{\mathrm{d}^5}$ configuration.\nIn an octahedral field, the $d$ orbitals split into $t_{2g}$ and $e_g$ orbitals.\nThe ground state configuration is $t_{2g}^3e_g^2$ (high spin) or $t_{2g}^5e_g^0$ (low spin, but Mn2+ usually forms high spin complexes).

Step 2: Consider the excited state. An electron is excited from $t_{2g}$ to $e_g$ .\nIf the ground state is $t_{2g}^3e_g^2$ , an excited state could be $t_{2g}^2e_g^3$ . However, this is not one of the options.\nIf the ground state is $t_{2g}^5e_g^0$ , an excited state would be $t_{2g}^3e_g^2$ (since the question asks for *an* excited state configuration, not necessarily the *first* excited state). This matches option B.

Why other options are wrong:
- Option A: $t_{2g}^4$ is not a possible excited state configuration for $\mathrm{M}{\mathrm{n}^{2 + }}$ .
- Option C: $t_{2g}^4e_g^2$ is not a possible excited state configuration for $\mathrm{M}{\mathrm{n}^{2 + }}$ .
- Option D: $t_{2g}^5e_g^0$ is the ground state configuration, not an excited state.

Key Concepts:
- Electronic configuration
- Crystal field theory
- Excited states

Memory Aids:
- Electrons can be excited from lower energy orbitals to higher energy orbitals.

Quick Tips:
- Determine the ground state configuration first.
- Consider the possible excited states by promoting electrons.

5 Medium

5 of 310Question permalink

Analyzing +4 | -1

\mathrm{CuS}{\mathrm{O}_4}

decolourises on addition of KCN, the product is:

View answer and explanation

Correct option: B

{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{3 - }}

Option B (

{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{3 - }}

) is correct because Cu2+ is reduced to Cu+ by cyanide ions, and then forms a stable complex with excess cyanide.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{3 - }}$ . Cupric ions are reduced to cuprous ions and form a complex with cyanide.

Step-by-Step Solution:

Step 1: Write the reaction between CuSO₄ and KCN.
$2CuSO_4 + 10KCN \rightarrow 2K_3[Cu(CN)_4] + 2K_2SO_4 + (CN)_2$

Step 2: Explain the reaction.\nCuSO₄ reacts with KCN. Initially, CuCN is formed, which is unstable and decomposes to Cu₂CN₂. However, in the presence of excess KCN, Cu⁺ forms a stable complex with cyanide ions.
$Cu^{2+} + CN^- \rightarrow Cu^+ + (CN)_2$
$Cu^+ + 4CN^- \rightarrow [Cu(CN)_4]^{3-}$

Step 3: Identify the product.\nThe product is ${\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{3 - }}$ .

Why other options are wrong:
- Option A: $\mathrm{Cu}\left( {\mathrm{CN}} \right)_4^{2 - }$ is incorrect because copper is in the +1 oxidation state in the complex, not +2.
- Option C: $\mathrm{Cu}{\left( {\mathrm{CN}} \right)_2}$ is incorrect because it is an intermediate product that further reacts with KCN.
- Option D: $\mathrm{CuCN}$ is incorrect because it is an intermediate product that further reacts with KCN.

Key Concepts:
- Complex formation
- Redox reactions
- Stability of complexes

Memory Aids:
- Cyanide is a strong field ligand and can reduce Cu²⁺ to Cu⁺.

Quick Tips:
- Remember that cyanide can act as both a ligand and a reducing agent.

6 Easy

6 of 310Question permalink

Application +4 | -1

The IUPAC name of

{\mathrm{K}_2}\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]

is

View answer and explanation

Correct option: A

Potassium tetracyanonickelate (II)

Option A (Potassium tetracyanonickelate (II)) is correct because it correctly names the compound, indicating the potassium cation, the tetracyano ligands, the nickelate anion, and the oxidation state of nickel.

Solution approach

Correct Answer + Core Reason: Potassium tetracyanonickelate (II). The IUPAC naming convention requires naming the cation first, followed by the ligand and then the metal with its oxidation state.

Step-by-Step Solution:

Step 1: Identify the cation: Potassium (K).

Step 2: Identify the ligands: Cyano (CN) with a charge of -1. There are four cyano ligands.

Step 3: Identify the central metal: Nickel (Ni).

Step 4: Determine the oxidation state of nickel: The complex ion has a charge of -2. Since each cyanide ligand has a -1 charge, the total charge from the ligands is -4. Therefore, the oxidation state of nickel must be +2 to result in a -2 overall charge for the complex ion.

Step 5: Name the complex: Potassium tetracyanonickelate (II).

Why other options are wrong:
- Option B: Potassium tetracyanatonickelate (III): Incorrect oxidation state of nickel (should be II).
- Option C: Potassium tetracyanatonickel (II): Incorrect name for the complex ion (should be nickelate).
- Option D: Potassium tetracyanonickel (III): Incorrect oxidation state of nickel and incorrect name for the complex ion.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Ligand naming
- Oxidation state determination

Memory Aids:
- Cation first, then ligands, then metal.
- Use "ate" suffix for anionic complexes.

Quick Tips:
- Always specify the oxidation state of the metal using Roman numerals.

7 Medium

7 of 310Question permalink

Analyzing +4 | -1

Which of the following statements is not correct?

View answer and explanation

Correct option: A

In oxyhaemoglobin

\mathrm{F}{\mathrm{e}^{2 + }}

is paramagnetic

Option A (In oxyhaemoglobin

\mathrm{F}{\mathrm{e}^{2 + }}

is paramagnetic) is correct because in oxyhemoglobin, Fe2+ is actually diamagnetic due to strong field ligand nature of oxygen.

Solution approach

Correct Answer + Core Reason: In oxyhaemoglobin $\mathrm{F}{\mathrm{e}^{2 + }}$ is paramagnetic. In oxyhemoglobin, Fe²⁺ is diamagnetic due to strong field ligand nature of oxygen.

Step-by-Step Solution:

Step 1: Analyze the oxidation state of iron in haemoglobin and oxyhaemoglobin.\nIn both haemoglobin and oxyhaemoglobin, iron is in the +2 oxidation state (Fe²⁺).

Step 2: Determine the magnetic properties of Fe²⁺ in haemoglobin.\nIn deoxyhaemoglobin, Fe²⁺ is in a high-spin state and is paramagnetic.

Step 3: Determine the magnetic properties of Fe²⁺ in oxyhaemoglobin.\nIn oxyhaemoglobin, oxygen acts as a strong field ligand, causing the electrons in Fe²⁺ to pair up. This results in a low-spin state, making oxyhaemoglobin diamagnetic.

Step 4: Evaluate the given statements.
- Statement A: In oxyhaemoglobin $\mathrm{F}{\mathrm{e}^{2 + }}$ is paramagnetic. This is incorrect because oxyhaemoglobin is diamagnetic.
- Statement B: During respiration the size of $\mathrm{F}{\mathrm{e}^{2 + }}$ increases when it changes from diamagnetic to paramagnetic state. This is correct. When Fe²⁺ changes from diamagnetic (low spin) to paramagnetic (high spin), the ionic radius increases.
- Statement C: Four haeme groups are present in haemoglobin. This is correct.
- Statement D: Haeme is the prosthetic group and it is non-protein part. This is correct.

Why other options are wrong:
- Option B: Statement B is correct, so it is not the answer.
- Option C: Statement C is correct, so it is not the answer.
- Option D: Statement D is correct, so it is not the answer.

Key Concepts:
- Haemoglobin and oxyhaemoglobin
- Magnetic properties of transition metal complexes
- High-spin and low-spin states
- Strong field and weak field ligands

Memory Aids:
- Oxygen is a strong field ligand in oxyhaemoglobin, leading to diamagnetism.

Quick Tips:
- Remember that the magnetic properties of transition metal complexes depend on the ligand field strength.

8 Medium

8 of 310Question permalink

Analyzing +4 | -1

Which one of the following is an inner orbital complex as well as diamagnetic in nature?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because Co3+ forms a low-spin, inner orbital complex with NH3, making it diamagnetic.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ . Co³⁺ forms a low-spin, inner orbital complex with NH₃, making it diamagnetic.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the metal ions.
- Cr³⁺: [Ar] 3d³
- Co³⁺: [Ar] 3d⁶
- Ni²⁺: [Ar] 3d⁸
- Zn²⁺: [Ar] 3d¹⁰

Step 2: Analyze the ligand field strength of NH₃.\nNH₃ is a strong field ligand, which can cause pairing of electrons in the d orbitals.

Step 3: Determine the hybridization and magnetic properties of each complex.
- ${\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ : Cr³⁺ (d³) has 3 unpaired electrons, so it is paramagnetic. It uses outer d-orbitals in hybridization.
- ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ : Co³⁺ (d⁶) with strong field NH₃ forms a low-spin complex with no unpaired electrons, so it is diamagnetic. It undergoes $d^2sp^3$ hybridization (inner orbital complex).
- ${\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ : Ni²⁺ (d⁸) has 2 unpaired electrons, so it is paramagnetic. It uses outer d-orbitals in hybridization.
- ${\left[ {\mathrm{Zn}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ : Zn²⁺ (d¹⁰) has all paired electrons, so it is diamagnetic. However, it is an outer orbital complex.

Step 4: Identify the complex that is both inner orbital and diamagnetic.
${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ is the only complex that is both inner orbital ( $d^2sp^3$ hybridization) and diamagnetic (no unpaired electrons).

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ is paramagnetic.
- Option C: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ is paramagnetic.
- Option D: ${\left[ {\mathrm{Zn}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ is an outer orbital complex.

Key Concepts:
- Inner and outer orbital complexes
- Diamagnetism and paramagnetism
- Crystal field theory
- Hybridization

Memory Aids:
- Strong field ligands favor inner orbital complexes and can lead to diamagnetism.

Quick Tips:
- Determine the electronic configuration and ligand field strength to predict the magnetic properties and hybridization of a complex.

9 Hard

9 of 310Question permalink

Application +4 | -1

A solution made by dissolving

0.875 \mathrm{~g}

of compound

\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_3

in

25 \mathrm{~g}

of water freezes at

-0.56^{\circ} \mathrm{C}

.

The structural formula of the complex, if cryoscopic constant of water is

1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

, is

View answer and explanation

Correct option: C

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}

Option C (

\left[\mathrm{Co}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Cl}

) is correct because the calculated Van't Hoff factor matches the dissociation of the complex into two ions.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Co}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$ . The Van't Hoff factor indicates the complex dissociates into two ions in solution.

Step-by-Step Solution:

Step 1: Calculate the molality (m) of the solution.
$m = \frac{{\text{{moles of solute}}}}{{\text{{mass of solvent in kg}}}}$

Step 2: Calculate the molar mass of $\mathrm{Co}(\mathrm{NH}_3)_4 \mathrm{Cl}_3$ .\nMolar mass = 58.93 + 4(14.01 + 3(1.01)) + 3(35.45) = 58.93 + 4(17.04) + 106.35 = 58.93 + 68.16 + 106.35 = 233.44 g/mol

Step 3: Calculate the number of moles of the solute.\nMoles = $\frac{{0.875 \text{{ g}}}}{{233.44 \text{{ g/mol}}}} = 0.00375 \text{{ mol}}$

Step 4: Calculate the molality of the solution.
$m = \frac{{0.00375 \text{{ mol}}}}{{0.025 \text{{ kg}}}} = 0.15 \text{{ mol/kg}}$

Step 5: Calculate the Van't Hoff factor (i) using the freezing point depression formula.
$\Delta T_f = i K_f m$
$0.56 = i \times 1.86 \times 0.15$
$i = \frac{{0.56}}{{1.86 \times 0.15}} = \frac{{0.56}}{{0.279}} = 2.007 \approx 2$

Step 6: Determine the structural formula based on the Van't Hoff factor.\nSince i ≈ 2, the complex dissociates into two ions in solution. This means one chloride ion is outside the coordination sphere.\nTherefore, the structural formula is $\left[\mathrm{Co}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$ .

Why other options are wrong:
- Option A: $\left[\mathrm{Co}(\mathrm{NH}_3)_4\right] \mathrm{Cl}_3$ : This complex would dissociate into 4 ions, giving i ≈ 4, which is not the case.
- Option B: $\left[\mathrm{Co}(\mathrm{NH}_3)_4 \mathrm{Cl}\right] \mathrm{Cl}_2$ : This complex would dissociate into 3 ions, giving i ≈ 3, which is not the case.
- Option D: $\left[\mathrm{Co}(\mathrm{NH}_3)_3 \mathrm{Cl}_3\right] \mathrm{NH}_3$ : This complex would not dissociate into ions, giving i ≈ 1, which is not the case.

Key Concepts:
- Freezing point depression
- Van't Hoff factor
- Coordination complexes
- Molality

Memory Aids:
- $\Delta T_f = i K_f m$ (Freezing point depression formula)
- Van't Hoff factor (i) indicates the number of ions formed in solution.

Quick Tips:
- Always calculate the Van't Hoff factor first to determine the number of ions formed.
- Remember to consider the molar mass of the complex when calculating molality.
- Pay attention to the number of ions outside the coordination sphere.

10 Hard

10 of 310Question permalink

Understanding +4 | -1

Assertion : Square planar complex

{\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}

has two optical isomers.

Reason : Mirror image of

{\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}

is non-super imposable.

View answer and explanation

Correct option: D

If both Assertion & Reason are false

Option D is correct because square planar complexes of the type

{\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}

do not have optical isomers, and their mirror images are superimposable.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are false. Square planar complexes of the type ${\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ do not have optical isomers.

Step-by-Step Solution:

Step 1: Define optical isomerism: Optical isomers are stereoisomers that are non-superimposable mirror images of each other.

Step 2: Consider square planar complexes: Square planar complexes are planar, and their mirror images are superimposable. Therefore, they do not exhibit optical isomerism.

Step 3: Analyze the assertion: The assertion states that square planar complex ${\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ has two optical isomers, which is false.

Step 4: Analyze the reason: The reason states that the mirror image of ${\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ is non-superimposable, which is also false.

Step 5: Conclusion: Both the assertion and the reason are false.

Why other options are wrong:
- Option A: Both the assertion and the reason are false.
- Option B: Both the assertion and the reason are false.
- Option C: The reason is false.

Key Concepts:
- Optical isomerism
- Square planar complexes
- Coordination complexes

Memory Aids:
- Square planar = no optical isomers.

Quick Tips:
- Remember that square planar complexes do not exhibit optical isomerism because they are planar and their mirror images are superimposable.

11 Easy

11 of 310Question permalink

Application +4 | -1

The pair of compounds having metals in their highest oxidation state is

View answer and explanation

Correct option: B

{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - },\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

Option B (

{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - },\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

) is correct because Mn in

{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }

and Cr in

\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

are in their highest oxidation states, +7 and +6 respectively.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - },\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ . The oxidation states of Mn and Cr are +7 and +6, respectively, which are their highest possible oxidation states.

Step-by-Step Solution:

Step 1: Determine the oxidation state of each metal in each compound.

Step 2: For $\mathrm{Mn}{\mathrm{O}_2}$ , let the oxidation state of Mn be x. Then, $x + 2(-2) = 0$ , so $x = +4$ .

Step 3: For $\mathrm{FeC}{\mathrm{l}_3}$ , the oxidation state of Fe is +3.

Step 4: For ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$ , let the oxidation state of Mn be x. Then, $x + 4(-2) = -1$ , so $x = +7$ .

Step 5: For $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ , let the oxidation state of Cr be x. Then, $x + 2(-2) + 2(-1) = 0$ , so $x = +6$ .

Step 6: For ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , let the oxidation state of Fe be x. Then, $x + 6(-1) = -3$ , so $x = +3$ .

Step 7: For ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_3}} \right]}$ , let the oxidation state of Co be x. Then, $x + 3(-1) = 0$ , so $x = +3$ .

Step 8: For ${\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }}$ , let the oxidation state of Ni be x. Then, $x + 4(-1) = -2$ , so $x = +2$ .

Step 9: For ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^ - }$ , let the oxidation state of Co be x. Then, $x + 4(-1) = -1$ , so $x = +3$ .

Step 10: Identify the pair with metals in their highest oxidation states. Mn can have a maximum oxidation state of +7, and Cr can have a maximum oxidation state of +6. Fe can have a maximum oxidation state of +6, Ni can have a maximum oxidation state of +4, and Co can have a maximum oxidation state of +5.

Step 11: Compare the oxidation states with the maximum possible oxidation states. In option B, Mn is in +7 state and Cr is in +6 state, which are their highest oxidation states.

Why other options are wrong:
- Option A: In $\mathrm{Mn}{\mathrm{O}_2}$ , Mn is in +4 oxidation state, which is not its highest (+7). In $\mathrm{FeC}{\mathrm{l}_3}$ , Fe is in +3 oxidation state, which is not its highest (+6).
- Option C: In ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , Fe is in +3 oxidation state, which is not its highest (+6). In ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_3}} \right]}$ , Co is in +3 oxidation state, which is not its highest (+5).
- Option D: In ${\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }}$ , Ni is in +2 oxidation state, which is not its highest (+4). In ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^ - }$ , Co is in +3 oxidation state, which is not its highest (+5).

Key Concepts:
- Oxidation state calculation
- Maximum oxidation states of transition metals

Memory Aids:
- Remember the common oxidation states of oxygen (-2) and chlorine (-1).
- The sum of oxidation states in a neutral compound is zero, and in an ion, it equals the charge of the ion.

Quick Tips:
- Know the maximum oxidation states of common transition metals.
- Practice oxidation state calculations frequently.

12 Hard

12 of 310Question permalink

Understanding +4 | -1

Which of the following compounds is not coloured yellow?

View answer and explanation

Correct option: A

Z{n_2}[Fe{(CN)_6}]

Option A (

Z{n_2}[Fe{(CN)_6}]

) is correct because it is white or bluish-white, while the other options are yellow.

Solution approach

Correct Answer + Core Reason: $Z{n_2}[Fe{(CN)_6}]$ is not yellow; it is white or bluish-white.

Step-by-Step Solution:
Step 1: Recall the colors of the given compounds.
Step 2: Identify the compound that is not yellow.

$Z{n_2}[Fe{(CN)_6}]$ is Zinc ferrocyanide, which is a white or bluish-white precipitate.
${K_3}[Co{(N{O_2})_6}]$ is Potassium hexanitrocobaltate(III), which is yellow.
${(N{H_4})_3}[As{(M{o_3}{O_{10}})_4}]$ is Ammonium molybdophosphate, which is yellow.
$BaCr{O_4}$ is Barium chromate, which is yellow.

Why other options are wrong:
- Option B: ${K_3}[Co{(N{O_2})_6}]$ is yellow.
- Option C: ${(N{H_4})_3}[As{(M{o_3}{O_{10}})_4}]$ is yellow.
- Option D: $BaCr{O_4}$ is yellow.

Key Concepts:
- Colors of inorganic compounds
- Precipitation reactions

Memory Aids:
- Remember the common colors of some important inorganic compounds.

Quick Tips:
- Memorize the colors of common precipitates and complexes.

13 Easy

13 of 310Question permalink

Remembering +4 | -1

An imperfect complex of a complex compound is 100% ionized; the compound is called:

View answer and explanation

Correct option: A

Double salt

Option A (Double salt) is correct because double salts dissociate into their constituent ions when dissolved in water, exhibiting 100% ionization.

Solution approach

Correct Answer + Core Reason: Double salt. Double salts completely dissociate into ions in solution.

Step-by-Step Solution:

Step 1: Understand the definition of a double salt.

Step 2: Double salts are ionic compounds formed from the combination of two different salts. When dissolved in water, they dissociate completely into their constituent ions.

Step 3: This complete dissociation leads to 100% ionization of the complex.

Step 4: Complex salts, acid salts, and normal salts do not necessarily undergo 100% ionization in solution.

Why other options are wrong:
- Option B: Complex salts contain complex ions that do not fully dissociate in solution. Therefore, they are not 100% ionized.
- Option C: Acid salts are salts that contain an ionizable hydrogen atom. Their ionization depends on the specific salt and is not necessarily 100%.
- Option D: Normal salts are salts that do not contain ionizable hydrogen or hydroxide ions. Their ionization is also not necessarily 100% and depends on the salt's solubility and dissociation constant.

Key Concepts:
- Double salts
- Complex salts
- Acid salts
- Normal salts
- Ionization

Memory Aids:
- Double salts = Double the ions (complete dissociation)

Quick Tips:
- Remember that double salts are mixtures of two salts and dissociate completely.

14 Hard

14 of 310Question permalink

Understanding +4 | -1

Assertion : In the complex

{{\text{K}}_{\text{2}}}\left[ {{\text{PtC}}{{\text{l}}_{\text{6}}}} \right]

, coordination number of Pt is 6.

Reason : In the complex six coordinate bonds are formed between Pt and chloro ligands.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.

Option A is correct because the coordination number of Pt is indeed 6, and this is due to the formation of six coordinate bonds with the chloro ligands.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. The coordination number is determined by the number of ligands directly bonded to the central metal atom.

Step-by-Step Solution:

Step 1: Identify the complex: ${{\text{K}}_{\text{2}}}\left[ {{\text{PtC}}{{\text{l}}_{\text{6}}}} \right]$

Step 2: Determine the central metal atom: Platinum (Pt).

Step 3: Identify the ligands directly bonded to the central metal atom: Six chloride ions (Cl $^-$ ).

Step 4: The coordination number is the number of ligands directly attached to the central metal atom. In this case, there are six chloride ligands attached to Pt.

Step 5: Therefore, the coordination number of Pt is 6.

Step 6: The reason states that six coordinate bonds are formed between Pt and chloro ligands, which is the explanation for the coordination number being 6.

Why other options are wrong:
- Option B: The reason is a correct explanation of the assertion, so this option is incorrect.
- Option C: Both the assertion and the reason are true, so this option is incorrect.
- Option D: Both the assertion and the reason are true, so this option is incorrect.

Key Concepts:
- Coordination number
- Coordination complexes
- Ligands

Memory Aids:
- Coordination number is the number of ligands directly bonded to the central metal atom.

Quick Tips:
- Always identify the central metal atom and the ligands attached to it to determine the coordination number.

15 Medium

15 of 310Question permalink

Analyzing +4 | -1

Both geometrical and optical isomerisms are shown by

View answer and explanation

Correct option: A

{\left[ {CO{{\left( {en} \right)}_2}C{l_2}} \right]^ + }

Option A (

{\left[ {CO{{\left( {en} \right)}_2}C{l_2}} \right]^ + }

) is correct because it can exhibit both geometrical (cis and trans) and optical isomerism (for the cis isomer).

Solution approach

Correct Answer + Core Reason: ${\left[ {CO{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$ . This complex exhibits both geometrical and optical isomerism.

Step-by-Step Solution:

Step 1: Analyze each complex to determine the possible types of isomerism.

Step 2: Option A: ${\left[ {CO{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$ : This complex can exist as cis and trans isomers (geometrical isomerism). The cis isomer is chiral and exhibits optical isomerism.

Step 3: Option B: ${\left[ {CO{{\left( {N{H_3}} \right)}_5}Cl} \right]^{2 + }}$ : This complex does not exhibit geometrical or optical isomerism.

Step 4: Option C: ${\left[ {CO{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }$ : This complex can exhibit geometrical isomerism (cis and trans), but it does not exhibit optical isomerism.

Step 5: Option D: ${\left[ {Cr{{\left( {OX} \right)}_3}} \right]^3}$ (where OX is oxalate): This complex exhibits optical isomerism but not geometrical isomerism.

Step 6: Final Answer: ${\left[ {CO{{\left( {en} \right)}_2}C{l_2}} \right]$

16 Hard

16 of 310Question permalink

Understanding +4 | -1

Assertion : All tetrahedral complexes are mainly high spin and the low spin configurations are rarely observed.
Reason:

\Delta {\text{t}}

is always much smaller even with stronger field ligands and it is never energetically favourable to pair up the electrons.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.

Option A is correct because tetrahedral complexes usually have high spin configurations due to the smaller crystal field splitting energy (

\Delta t

), and this is because it's rarely energetically favorable to pair up electrons.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. Tetrahedral complexes are generally high spin due to the smaller crystal field splitting energy.

Step-by-Step Solution:

Step 1: Understand the concept of crystal field splitting energy (CFSE).

Step 2: Recall that tetrahedral complexes have a smaller CFSE ( $\Delta t$ ) compared to octahedral complexes ( $\Delta o$ ).

Step 3: The smaller $\Delta t$ means that the energy required to pair up electrons is often greater than the energy difference between the $e$ and $t_2$ sets of orbitals.

Step 4: Therefore, electrons tend to occupy the higher energy $t_2$ orbitals before pairing up in the lower energy $e$ orbitals, resulting in high spin configurations.

Step 5: The reason correctly explains why tetrahedral complexes are mainly high spin: $\Delta t$ is always much smaller, even with strong field ligands, making it energetically unfavorable to pair up electrons.

Why other options are wrong:
- Option B: The reason is a correct explanation of the assertion, so this option is incorrect.
- Option C: Both the assertion and the reason are true, so this option is incorrect.
- Option D: Both the assertion and the reason are true, so this option is incorrect.

Key Concepts:
- Crystal field theory
- Crystal field splitting energy (CFSE)
- Tetrahedral complexes
- High spin vs. low spin configurations

Memory Aids:
- Tetrahedral = Tiny splitting, so high spin.

Quick Tips:
- Remember that $\Delta t \approx \frac{4}{9} \Delta o$ . Tetrahedral splitting is always smaller than octahedral splitting.

17 Hard

17 of 310Question permalink

Understanding +4 | -1

An aqueous solution of

\mathrm{CoC}{\mathrm{l}_2}

on addition of excess of concentrated

\mathrm{HCl}

turns blue to formation of

View answer and explanation

Correct option: A

{\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }}

Option A (

{\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }}

) is correct because the addition of excess concentrated HCl to an aqueous solution of

\mathrm{CoC}{\mathrm{l}_2}

leads to the formation of the tetrahedral complex

{\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }}

which is blue in color.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }}$ is formed, which is blue.

Step-by-Step Solution:
Step 1: Write the reaction of $\mathrm{CoC}{\mathrm{l}_2}$ with excess $\mathrm{HCl}$ .
Step 2: Identify the product formed.
Step 3: Relate the color change to the formation of the product.
\nIn aqueous solution, $\mathrm{CoC}{\mathrm{l}_2}$ exists as ${\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ , which is pink. Upon addition of excess $\mathrm{HCl}$ , the following reaction occurs:
${\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }} + 4\mathrm{C}{\mathrm{l}^ - } \rightleftharpoons {\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }} + 6{\mathrm{H}_2}\mathrm{O}$ \nThe tetrahedral complex ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }}$ is blue.

Why other options are wrong:
- Option B: ${\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}\mathrm{C}{\mathrm{l}_4}} \right]^{2 - }}$ is not the primary species formed in excess $\mathrm{HCl}$ .
- Option C: ${\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}2\mathrm{C}{\mathrm{l}_4}} \right]^{2 - }}$ is an incorrect formula.
- Option D: $\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]$ is not the primary species formed in excess $\mathrm{HCl}$ .

Key Concepts:
- Coordination complexes
- Ligand exchange reactions
- Color of coordination complexes

Memory Aids:
- Remember that chloride ions are strong ligands that can replace water ligands in the coordination sphere.

Quick Tips:
- Know the common colors of different coordination complexes.

18 Hard

18 of 310Question permalink

Understanding +4 | -1

Homoleptic octahedral complexes of a metal ion '

{M^{3 + }}

' with three monodentate ligands

{L_1},{L_2}

and

{L_3}

absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is

View answer and explanation

Correct option: C

{L_3} < {L_1} < {L_2}

Option C (

{L_3} < {L_1} < {L_2}

) is correct because the energy absorbed is inversely proportional to the wavelength, and the ligand strength is directly proportional to the energy absorbed. Red < Green < Blue implies

{L_3} < {L_1} < {L_2}

.

Solution approach

Correct Answer + Core Reason: ${L_3} < {L_1} < {L_2}$ because ligand strength is proportional to energy absorbed, which is inversely proportional to wavelength.

Step-by-Step Solution:
Step 1: Relate the wavelength of absorbed light to the energy of the transition.
Step 2: Relate the energy of the transition to the ligand field strength.
Step 3: Arrange the ligands in increasing order of field strength.
\nThe energy of absorbed light is inversely proportional to the wavelength: $E = \frac{{hc}}{\lambda}$ .\nSince the complex with ${L_1}$ absorbs green light, ${L_2}$ absorbs blue light, and ${L_3}$ absorbs red light, the order of wavelength is: red > green > blue.\nTherefore, the order of energy is: red < green < blue.\nSince stronger field ligands cause a larger splitting of d-orbitals, the order of ligand strength is: ${L_3} < {L_1} < {L_2}$

Why other options are wrong:
- Option A: ${L_1} < {L_2} < {L_3}$ is incorrect because red light has the longest wavelength and thus the lowest energy, implying the weakest ligand field.
- Option B: ${L_2} < {L_1} < {L_3}$ is incorrect because blue light has the shortest wavelength and thus the highest energy, implying the strongest ligand field.
- Option D: ${L_3} < {L_2} < {L_1}$ is incorrect because the order of absorbed wavelengths is red < green < blue.

Key Concepts:
- Ligand field strength
- Spectrochemical series
- Relationship between wavelength and energy

Memory Aids:
- Remember that shorter wavelengths correspond to higher energy and stronger ligand fields.

Quick Tips:
- Know the relationship between color, wavelength, energy, and ligand field strength.

19 Medium

19 of 310Question permalink

Application +4 | -1

Which complex is diamagnetic?

View answer and explanation

Correct option: A

{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4 - }}

Option A ([Fe(CN)6]^4-) is correct because Fe in this complex has a d⁶ configuration, and CN- is a strong field ligand, causing pairing of all electrons, resulting in a diamagnetic complex.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4 - }}$ . This complex is diamagnetic because all electrons are paired.

Step-by-Step Solution:

Step 1: Determine the oxidation state and electronic configuration of the metal ion in each complex.

Step 2: Option A: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4 - }}$ . Fe is in +2 oxidation state. It has $d^6$ configuration. CN- is a strong field ligand, so it pairs up the electrons, resulting in a diamagnetic complex.

Step 3: Option B: ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{3 + }}$ . Cu is in +3 oxidation state. It has $d^8$ configuration. It is likely to be paramagnetic.

Step 4: Option C: ${\left[ {\mathrm{Ti}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ . Ti is in +3 oxidation state. It has $d^1$ configuration. It is paramagnetic.

Why other options are wrong:
- Option B: Copper(III) complex is likely paramagnetic.
- Option C: Titanium(III) complex is paramagnetic.

Key Concepts:
- Diamagnetism and paramagnetism
- Electronic configuration
- Coordination complexes
- Strong and weak field ligands

Memory Aids:
- Strong field ligands cause pairing of electrons.

Quick Tips:
- Determine the number of unpaired electrons to determine if a complex is paramagnetic or diamagnetic.

20 Hard

20 of 310Question permalink

Analyzing +4 | -1

Select the correct statement(s).

View answer and explanation

Correct option: B

The octahedral complex formed by joining the

\mathrm{CN}^{-}

ligands to

\mathrm{Fe}^{+3}

ion has theoretical value of spin magnetic moment equal to 1.73 B.M.

Option B is correct because the octahedral complex of

\mathrm{Fe}^{+3}

with

\mathrm{CN}^{-}

has one unpaired electron, resulting in a spin magnetic moment of approximately 1.73 B.M.

Solution approach

Correct Answer + Core Reason: The octahedral complex formed by joining the $\mathrm{{CN}}^{-}$ ligands to $\mathrm{{Fe}}^{+3}$ ion has theoretical value of spin magnetic moment equal to 1.73 B.M. This is because the complex is low spin with one unpaired electron.

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: Stability order of complexes is incorrect. Oxalate is a chelating ligand and forms more stable complexes. The correct order is $\left[\mathrm{{CrF}}_6\right]^{3-} < \left[\mathrm{{Cr}}(\mathrm{{H}}_2 \mathrm{{O}})_6\right]^{3+} < \left[\mathrm{{Cr}}(\mathrm{{ox}})_3\right]^{3-}$ .
- B: $\left[\mathrm{{Fe}}(\mathrm{{CN}})_6\right]^{3-}$ : $Fe^{3+}$ is $d^5$ . $CN^-$ is a strong field ligand, so it pairs up electrons. Electronic configuration is $t_{2g}^5 e_g^0$ . One unpaired electron. The spin-only magnetic moment is $\sqrt{{n(n+2)}} = \sqrt{{1(1+2)}} = \sqrt{{3}} \approx 1.73$ B.M. This statement is correct.
- C: In $\mathrm{{CuSO}}_4 \cdot 5 \mathrm{{H}}_2 \mathrm{{O}}$ , only $\mathrm{{Cu}}-\mathrm{{O}}$ coordinate bonds are present. The copper ion is coordinated to four water molecules, and the fifth water molecule is hydrogen-bonded. The sulfate ion is ionically bonded. This statement is incorrect.
- D: Stability of complex is incorrect. en (ethylenediamine) is a chelating ligand and forms more stable complexes than monodentate ligands like ammonia. The correct order is $\left[\mathrm{{Ni}}(\mathrm{{en}})_3\right]^{2+} > \left[\mathrm{{Ni}}(\mathrm{{NH}}_3)_6\right]^{2+}$ .

Step 2: Conclusion: Only statement B is correct.

Why other options are wrong:
- Option A: The stability order is incorrect due to the chelate effect of oxalate.
- Option C: $\mathrm{{CuSO}}_4 \cdot 5 \mathrm{{H}}_2 \mathrm{{O}}$ has only $\mathrm{{Cu}}-\mathrm{{O}}$ coordinate bonds, not $\mathrm{{Cu}}-\mathrm{{S}}$ coordinate bonds.
- Option D: The stability order is incorrect due to the chelate effect of ethylenediamine (en).

Key Concepts:
- Stability of complexes
- Chelate effect
- Spin-only magnetic moment
- Coordination compounds

Memory Aids:
- Chelate effect: Chelating ligands form more stable complexes than monodentate ligands.
- Spin-only magnetic moment formula: $\mu = \sqrt{{n(n+2)}}$ B.M., where n is the number of unpaired electrons.

Quick Tips:
- Remember the chelate effect when comparing the stability of complexes.
- Use the spin-only formula to calculate the magnetic moment.

21 Easy

21 of 310Question permalink

Understanding +4 | -1

The two compounds pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt(III) chloride represent:

View answer and explanation

Correct option: D

No isomerism

Option D is correct. The two compounds are not isomers because they differ only in the counter-ion (bromide vs. chloride), not in the arrangement of ligands around the central metal ion.

Solution approach

Correct Answer + Core Reason: D. No isomerism. They differ only in the counter-ion.

Step-by-Step Solution:

Step 1: Identify the two compounds: pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt(III) chloride.

Step 2: Write the formulas of the compounds: $\Co(NH_3)_5SO_4\Br$ and $\Co(NH_3)_5SO_4\Cl$ .

Step 3: Observe the difference between the compounds: The only difference is the counter-ion outside the coordination sphere (bromide vs. chloride).

Step 4: Since the compounds only differ in the counter-ion and not in the arrangement of ligands around the central metal ion, they are not isomers.

Why other options are wrong:
- Option A: Linkage isomerism requires an ambidentate ligand, which is not the case here.
- Option B: Ionization isomerism involves the exchange of ligands inside and outside the coordination sphere, which is not the case here.
- Option C: Coordination isomerism requires two or more complex ions, which is not the case here.

Key Concepts:
- Isomerism in coordination compounds
- Coordination sphere
- Counter-ions

Memory Aids:
- Isomers have the same formula but different arrangements.

Quick Tips:
- Check if the compounds have the same formula and different arrangements before classifying them as isomers.

22 Easy

22 of 310Question permalink

Application +4 | -1

The IUPAC name of the compound

\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}{{\left( {\mathrm{N}{\mathrm{O}_3}} \right)}_2}} \right]

is:

View answer and explanation

Correct option: B

Dinitratotetraamminecopper(II)

The correct answer is B (Dinitratotetraamminecopper(II)). The IUPAC name follows the rules for naming coordination compounds: ligands are named alphabetically, followed by the metal with its oxidation state in Roman numerals.

Solution approach

Correct Answer + Core Reason: Dinitratotetraamminecopper(II). Follows IUPAC naming conventions.

Step-by-Step Solution:

Step 1: Identify the ligands and the metal. The ligands are ammine ( $\mathrm{N}{\mathrm{H}_3}$ ) and nitrate ( $\mathrm{N}{\mathrm{O}_3}$ ). The metal is copper (Cu).

Step 2: Determine the oxidation state of the metal. The complex has a charge of 0. Nitrate has a charge of -1, and there are two nitrates, so the total negative charge is -2. Therefore, copper must have an oxidation state of +2.

Step 3: Name the ligands alphabetically. Ammine comes before nitrate.

Step 4: Write the name: Tetraammine dinitrato copper(II).

Why other options are wrong:
- Option A: Cuprammonium nitrate: This is an older, less systematic name.
- Option C: Tetraamminecopper(II) dinitrite: Nitrate is $\mathrm{N}{\mathrm{O}_3}^-$ , not nitrite ( $\mathrm{N}{\mathrm{O}_2}^-$ ).
- Option D: Tetraamminecopper(III) dinitrite: The oxidation state of copper is +2, not +3, and nitrate is $\mathrm{N}{\mathrm{O}_3}^-$ , not nitrite ( $\mathrm{N}{\mathrm{O}_2}^-$ ).

Key Concepts:
- IUPAC nomenclature
- Coordination compounds
- Oxidation state

Memory Aids:
- Ligands named alphabetically, then metal with oxidation state.

Quick Tips:
- Follow the IUPAC naming rules systematically.

23 Easy

23 of 310Question permalink

Understanding +4 | -1

Which of the following organometallic compound is

\sigma \,\,\mathrm{and}\,\,\pi

bonded?

View answer and explanation

Correct option: C

\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]

Option C (

\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]

) is correct because Ferrocene has both sigma and pi bonding between Fe and the cyclopentadienyl rings.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]$ . Ferrocene exhibits both sigma and pi bonding.

Step-by-Step Solution:

Step 1: Define $\sigma$ and $\pi$ bonding in organometallic compounds: $\sigma$ bonding involves direct overlap of orbitals, while $\pi$ bonding involves sideways overlap of p-orbitals.

Step 2: Analyze the options:
- A. $\mathrm{Fe}{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}$ : This involves only sigma bonds between Fe and methyl groups.
- B. ${\left[ {\mathrm{Co}{{\left( {\mathrm{CO}} \right)}_5}\mathrm{N}{\mathrm{H}_3}} \right]^{2 + }}$ : This involves sigma bonds between Co and CO, and Co and $NH_3$ , and pi backbonding between Co and CO.
- C. $\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]$ : Ferrocene has both sigma and pi bonding between Fe and the cyclopentadienyl rings.
- D. $\mathrm{K}\left[ {\mathrm{PtC}{\mathrm{l}_3}\left( {{\mathrm{\eta }^2} - {\mathrm{C}_2}{\mathrm{H}_4}} \right)} \right]$ : Zeise's salt has sigma and pi bonding between Pt and ethylene.

Step 3: Conclusion: Ferrocene is the best answer because the cyclopentadienyl ligands form both sigma and pi bonds with the iron center.

Why other options are wrong:
- Option A: $\mathrm{Fe}{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}$ : Only sigma bonds are present.
- Option B: ${\left[ {\mathrm{Co}{{\left( {\mathrm{CO}} \right)}_5}\mathrm{N}{\mathrm{H}_3}} \right]^{2 + }}$ : While pi backbonding exists between Co and CO, the bonding with $NH_3$ is purely sigma.
- Option D: $\mathrm{K}\left[ {\mathrm{PtC}{\mathrm{l}_3}\left( {{\mathrm{\eta }^2} - {\mathrm{C}_2}{\mathrm{H}_4}} \right)} \right]$ : While Zeise's salt has sigma and pi bonding, Ferrocene is a more classic example of both.

Key Concepts:
- Sigma bonding
- Pi bonding
- Organometallic compounds
- Ferrocene
- Zeise's salt

Memory Aids:
- Remember Ferrocene as a classic example of sigma and pi bonding in organometallic chemistry.

Quick Tips:
- Understand the nature of bonding between metals and ligands in organometallic compounds.

24 Hard

24 of 310Question permalink

Understanding +4 | -1

The theory that can completely/ properly explain the nature of bonding in

[Ni{(CO)_4}]

is

View answer and explanation

Correct option: D

Molecular orbital theory

Option D (Molecular orbital theory) is correct because it provides the most complete description of bonding in coordination complexes, including back-bonding effects, which are crucial for understanding the stability of carbonyl complexes like

[Ni{(CO)_4}]

.

Solution approach

Correct Answer + Core Reason: Molecular orbital theory. It provides the most complete description of bonding, including back-bonding.

Step-by-Step Solution:

Step 1: Consider Werner's theory. Werner's theory explains the primary and secondary valencies of metal ions but doesn't fully describe the nature of bonding.

Step 2: Consider Crystal Field Theory (CFT). CFT explains the splitting of d-orbitals in the presence of ligands but doesn't account for the covalent character of the bonds.

Step 3: Consider Valence Bond Theory (VBT). VBT can explain the geometry and magnetic properties of complexes but doesn't fully explain the bonding in carbonyl complexes, particularly the back-bonding.

Step 4: Consider Molecular Orbital Theory (MOT). MOT provides a comprehensive description of bonding, including the formation of sigma and pi bonds, and accounts for the synergistic effect of sigma donation and pi back-donation in carbonyl complexes. The CO ligand donates electron density to the metal (sigma donation), and the metal donates electron density back to the CO ligand (pi back-donation), strengthening the M-C bond and weakening the C-O bond.

Why other options are wrong:
- Option A: Werner's theory is too basic and doesn't explain the nature of bonding.
- Option B: Crystal field theory focuses on d-orbital splitting and doesn't fully describe the bonding.
- Option C: Valence bond theory doesn't adequately explain back-bonding in carbonyl complexes.

Key Concepts:
- Molecular orbital theory
- Coordination complexes
- Back-bonding
- Carbonyl complexes

Memory Aids:
- MOT is the most complete theory.

Quick Tips:
- Remember that MOT provides the most accurate description of bonding in complexes, especially those with pi-acceptor ligands like CO.

25 Easy

25 of 310Question permalink

Application +4 | -1

The IUPAC name of

\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]\mathrm{\;}

is

View answer and explanation

Correct option: C

Tetraammine nickel (II) –tetrachloro nickelate(II)

Option C (Tetraammine nickel (II) –tetrachloro nickelate(II)) is correct because it follows the IUPAC naming conventions for coordination compounds, correctly identifying the cation and anion.

Solution approach

Correct Answer + Core Reason: Tetraammine nickel (II) –tetrachloro nickelate(II). The cation is named first, followed by the anion. The oxidation state of Ni is +2, and the anion is a complex.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The complex is $\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]\mathrm{\;}$ which dissociates into $\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }$ and $\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }$ .

Step 2: Name the cation. The cation is a complex ion with Ni as the central metal and four ammonia ligands. The name is tetraammine nickel(II).

Step 3: Name the anion. The anion is a complex ion with Ni as the central metal and four chloride ligands. The name is tetrachloronickelate(II).

Step 4: Combine the names. The IUPAC name is Tetraammine nickel (II) –tetrachloro nickelate(II).

Why other options are wrong:
- Option A: Tetrachloro nickel (II) – tetraammine nickel (II) - Incorrect order, anion should be named after cation.
- Option B: Tetraammine nickel (II) –tetrachloro nickel(II) - Incorrect, the anionic complex should end with -ate.
- Option D: Tetrachloro nickel (II) –tetraammine nickelate(0) - Incorrect order and incorrect oxidation state.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Naming of complex ions
- Identifying cation and anion
- Oxidation state determination

Memory Aids:
- Cation is named before anion.
- Anionic complexes end with -ate.

Quick Tips:
- Always identify the cation and anion first.
- Determine the oxidation state of the metal.
- Follow IUPAC rules for naming ligands and metal ions.

26 Medium

26 of 310Question permalink

Remembering +4 | -1

The crystal field splitting energy for octahedral 

\left(\Delta_0\right)

and tetrahedral

\left(\Delta_t\right)

complexes is related to

View answer and explanation

Correct option: C

{\Delta _t} = \frac{4}{9}{\Delta _0}

Option C is correct because the relationship between crystal field splitting energy for octahedral (Δo) and tetrahedral (Δt) complexes is given by Δt = (4/9)Δo.

Solution approach

Correct Answer + Core Reason: ${\Delta _t} = \frac{4}{9}{\Delta _0}$ . This is the standard relationship between tetrahedral and octahedral splitting energies.

Step-by-Step Solution:

Step 1: Recall the relationship between the crystal field splitting energy for tetrahedral and octahedral complexes.

Step 2: The crystal field splitting energy for tetrahedral complexes ( ${\Delta _t}$ ) is approximately 4/9 of the crystal field splitting energy for octahedral complexes ( ${\Delta _0}$ ). This is because there are fewer ligands in a tetrahedral complex (4) compared to an octahedral complex (6), and the geometry leads to smaller splitting.

Step 3: Write the formula: ${\Delta _t} = \frac{4}{9}{\Delta _0}$

Why other options are wrong:
- Option A: ${\Delta _0} = \frac{4}{9}{\Delta _t}$ is incorrect because it reverses the relationship. Octahedral splitting is larger than tetrahedral splitting.
- Option B: ${\Delta _t} = \frac{1}{2}{\Delta _0}$ is incorrect. The correct fraction is 4/9.
- Option D: ${\Delta _0} = - 2{\Delta _t}$ is incorrect. The splitting energies are related by a factor of 4/9, and the negative sign is not applicable here.

Key Concepts:
- Crystal field splitting energy
- Octahedral complexes
- Tetrahedral complexes

Memory Aids:
- Tetrahedral splitting is smaller than octahedral splitting.
- ${\Delta _t} = \frac{4}{9}{\Delta _0}$

Quick Tips:
- Remember the 4/9 relationship for quick calculations.

27 Easy

27 of 310Question permalink

Understanding +4 | -1

Complexes with

C{N^ - }

ligands are usually:

View answer and explanation

Correct option: B

Low spin complexes

The correct answer is B (Low spin complexes) because

C{N^-}

is a strong field ligand, which causes pairing of electrons and results in low spin complexes.

Solution approach

Correct Answer + Core Reason: Low spin complexes. $CN^-$ is a strong field ligand, leading to a large crystal field splitting and favoring low spin complexes.

Step-by-Step Solution:

Step 1: Identify the ligand. The ligand is $CN^-$ .

Step 2: Determine the strength of the ligand based on the spectrochemical series. $CN^-$ is a strong field ligand.

Step 3: Relate the ligand strength to the spin state of the complex. Strong field ligands cause a large crystal field splitting ( $\Delta_o$ ), which is greater than the pairing energy (P). This favors pairing of electrons and results in low spin complexes.

Step 4: Conclusion: Complexes with $CN^-$ ligands are usually low spin complexes.

Why other options are wrong:
- Option A (High spin complexes): High spin complexes are favored by weak field ligands, not strong field ligands like $CN^-$ .
- Option C (Both (1) and (2)): This is incorrect because $CN^-$ only forms low spin complexes.
- Option D (None of these): This is incorrect because $CN^-$ forms low spin complexes.

Key Concepts:
- Spectrochemical series
- Crystal field splitting energy ( $\Delta_o$ )
- Pairing energy (P)
- High spin vs. low spin complexes

Memory Aids:
- Strong field ligands = Low spin complexes
- Weak field ligands = High spin complexes

Quick Tips:
- Memorize the spectrochemical series to determine the relative strengths of ligands.

28 Hard

28 of 310Question permalink

Remembering +4 | -1

What is the overall formation equilibrium constant for the ion

{\left[ {M{L_4}} \right]^{2 - }}

ion, given that

{\mathrm{\beta }_4}

for this complex is

2.5 \times {10^{13}}?

View answer and explanation

Correct option: A

2.5 \times {10^{13}}

Option A (

2.5 \times {10^{13}}

) is correct because the overall formation equilibrium constant,

\beta_4

, is directly given as

2.5 \times 10^{13}

.

Solution approach

Correct Answer + Core Reason: $2.5 \times {10^{13}}$ . The overall formation constant is given directly.

Step-by-Step Solution:

Step 1: Understand the concept of formation constant: The overall formation constant ( $\beta_n$ ) represents the equilibrium constant for the formation of a complex from its constituent metal ion and ligands.

Step 2: Identify the given information: The overall formation constant $\beta_4$ for the complex ${\left[ {M{L_4}} \right]^{2 - }}$ is given as $2.5 \times {10^{13}}$ .

Step 3: Conclusion: Therefore, the overall formation equilibrium constant is $2.5 \times {10^{13}}$ .

Why other options are wrong:
- Option B: $5 \times {10^{ - 13}}$ : This is incorrect; it seems to be an attempt to take the inverse or perform some other incorrect calculation with the given value.
- Option C: $2.5 \times {10^{ - 14}}$ : This is incorrect; it seems to be a misunderstanding of the exponent.
- Option D: $4.0 \times {10^{ - 13}}$ : This is incorrect; there is no basis for this value based on the given information.

Key Concepts:
- Formation constant
- Equilibrium constant
- Coordination complexes

Memory Aids:
- The overall formation constant is directly related to the stability of the complex.

Quick Tips:
- Pay close attention to the definition and units of equilibrium constants.

29 Hard

29 of 310Question permalink

Application +4 | -1

On treatment of 100 mL of 0.1 M solution of

COC{l_3}.6{H_2}O

with excess of

AgN{O_3};\,1.2 \times {10^{22}}

ions are precipitated. The complex is

View answer and explanation

Correct option: D

\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O

Option D (

\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O

) is correct because it corresponds to the precipitation of 2 chloride ions per complex molecule, matching the given ion precipitation data.

Solution approach

Correct Answer + Core Reason: $\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O$ . This complex precipitates two chloride ions as AgCl.

Step-by-Step Solution:

Step 1: Calculate the number of moles of the complex.\nMolarity = 0.1 M, Volume = 100 mL = 0.1 L\nMoles of complex = Molarity * Volume = 0.1 M * 0.1 L = 0.01 moles

Step 2: Calculate the number of moles of Ag+ ions precipitated.\nNumber of Ag+ ions precipitated = $1.2 \times 10^{22}$ \nMoles of Ag+ ions = (Number of ions) / (Avogadro's number) = $(1.2 \times 10^{22}) / (6.022 \times 10^{23}) \approx 0.02$ moles

Step 3: Determine the number of Cl- ions precipitated per complex molecule.\nMoles of Ag+ ions / Moles of complex = 0.02 moles / 0.01 moles = 2\nThis means 2 Cl- ions are precipitated per complex molecule.

Step 4: Determine the formula of the complex.\nSince 2 Cl- ions are precipitated, they must be outside the coordination sphere. Therefore, the complex must be of the form $\left[Co(H_2O)_xCl_y\right]Cl_2.zH_2O$ \nGiven the overall formula $CoCl_3.6H_2O$ , we can deduce that x + z = 6 and y + 2 = 3, so y = 1. Therefore x = 5 and z = 1.\nThus, the complex is $\left[Co(H_2O)_5Cl\right]Cl_2.H_2O$

Why other options are wrong:
- Option A: $\left[ {Co{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl.2{H_2}O$ precipitates only 1 Cl- ion.
- Option B: $\left[ {Co{{\left( {{H_2}O} \right)}_3}C{l_3}} \right].3{H_2}O$ precipitates 0 Cl- ions.
- Option C: $\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$ precipitates 3 Cl- ions.

Key Concepts:
- Coordination Complexes
- Precipitation Reactions
- Moles Calculation
- Avogadro's Number

Memory Aids:
- Remember that only ions outside the coordination sphere will precipitate with Ag+.

Quick Tips:
- Calculate the moles of complex and the moles of precipitate to find the ratio of ions precipitated per complex molecule.

30 Hard

30 of 310Question permalink

Application +4 | -1

The specific rotation of a pure enantiomer is

+ 16^\circ .

Its observed rotation if it is isolated from a reaction with 25% recemisation and 75% retention is

View answer and explanation

Correct option: B

+ 12^\circ

Option B (

+ 12^\circ

) is correct because the observed rotation is proportional to the enantiomeric excess, which is reduced due to racemization.

Solution approach

Correct Answer + Core Reason: $+ 12^\circ$ . The observed rotation is calculated based on the enantiomeric excess after racemization.

Step-by-Step Solution:

Step 1: Calculate the percentage of each enantiomer after racemization.\nSince there is 25% racemization, it means 25% of the pure enantiomer has converted into a racemic mixture (50% (+) and 50% (-)).\nTherefore, the remaining pure enantiomer is 75% of the original, and 25% has become racemic.

Step 2: Determine the composition of the mixture.\nThe mixture contains 75% retention and 25% racemization. This means 75% remains as the original enantiomer, and the 25% that racemized is now 12.5% of the original enantiomer and 12.5% of its enantiomer.

Step 3: Calculate the enantiomeric excess (ee).\nee = % major enantiomer - % minor enantiomer = (75 + 12.5) - 12.5 = 75%

Step 4: Calculate the observed rotation.\nObserved rotation = (ee/100) * Specific rotation = (75/100) * (+16°) = +12°

Why other options are wrong:
- **Option A ( $- 12^\circ$ ): This is incorrect because the rotation should be positive since there is more of the + enantiomer than the - enantiomer.
- Option C ( $+ 16^\circ$ ): This is incorrect because it does not account for the racemization that has occurred, which reduces the observed rotation.
- Option D ( $- 16^\circ$ ): This is incorrect because the rotation should be positive since there is more of the + enantiomer than the - enantiomer and it also does not account for the racemization.

Key Concepts:
- Specific Rotation
- Enantiomeric Excess (ee)
- Racemization

Memory Aids:**
- Observed rotation = (ee/100) * Specific rotation

Quick Tips:
- Remember that racemization reduces the enantiomeric excess and thus the observed rotation.

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