In the silver plating of ${\rm{Cu}},{\rm{\;K}}\left[ {{\rm{Ag}}{{\left( {{\rm{CN}}} \right)}_2}} \right]$ is used instead of ${\rm{AgN}}{{\rm{O}}_3}.$ The reason is:

Less availability of ${\rm{A}}{{\rm{g}}^ + }$ ion as ${\rm{Cu}}$ cannot displace ${\rm{Ag}}$ from ${\rm{Ag}}{\left( {{\rm{CN}}} \right)_2}$

A thin layer of ${\rm{Ag}}$ is formed on ${\rm{Cu}}$

$A{{\rm{g}}^ + }$ ions are completely removed from solution

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Hard, Page 2

31 Hard

31 of 79Question permalink

Application +4 | -1

On treatment of

100 \mathrm{~mL}

of

0.1 \mathrm{M}

solution of the complex

\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3

with excess of

\mathrm{AgNO}_3, 4.305 \mathrm{~g}

of

\mathrm{AgCl}

was obtained. What is the positive charge present on the complex ion?

View answer and explanation

Correct option: A

3

Option A (3) is correct because the amount of AgCl precipitated corresponds to 3 chloride ions being outside the coordination sphere, resulting in a +3 charge on the complex ion.

Solution approach

Correct Answer + Core Reason: 3. The positive charge on the complex ion is determined by the number of chloride ions outside the coordination sphere.

Step-by-Step Solution:

Step 1: Calculate the number of moles of AgCl precipitated.\nMolar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol\nMoles of AgCl = (Mass of AgCl) / (Molar mass of AgCl) = 4.305 g / 143.32 g/mol = 0.03 moles

Step 2: 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 3: Determine the number of Cl- ions precipitated per complex molecule.\nMoles of AgCl / Moles of complex = 0.03 moles / 0.01 moles = 3\nThis means 3 Cl- ions are precipitated per complex molecule.

Step 4: Determine the charge on the complex ion.\nSince 3 Cl- ions are precipitated, they must be outside the coordination sphere. Therefore, the complex must be of the form $\left[Co(NH_3)_6\right]^{3+}3Cl^-$ \nThe positive charge on the complex ion is +3.

Why other options are wrong:
- Option B (0): This is incorrect because it implies that no chloride ions are outside the coordination sphere, which contradicts the precipitation of AgCl.
- Option C (0): This is incorrect because it implies that no chloride ions are outside the coordination sphere, which contradicts the precipitation of AgCl.
- Option D (0): This is incorrect because it implies that no chloride ions are outside the coordination sphere, which contradicts the precipitation of AgCl.

Key Concepts:
- Coordination Complexes
- Precipitation Reactions
- Moles Calculation
- Charge on Complex Ions

Memory Aids:
- The number of moles of AgCl precipitated corresponds to the number of Cl- ions outside the coordination sphere.

Quick Tips:
- Calculate the moles of complex and the moles of precipitate to find the ratio of ions precipitated per complex molecule. This ratio gives the number of Cl- ions outside the coordination sphere.

32 Hard

32 of 79Question permalink

Understanding +4 | -1

Which of the following sets is/are example(s) of co-ordination isomerism is complex?

View answer and explanation

Correct option: A

\left[{Co}\left({NH}_3\right)_6\right][{Cr}({CN})_6]

and

\left[{Co}\left({CN}_6\right)\right]\left[{Cr}\left({NH}_3\right)_6\right]

Option A is correct because

\left[{Co}\left({NH}_3\right)_6\right][{Cr}({CN})_6]

and

\left[{Co}\left({CN}_6\right)\right]\left[{Cr}\left({NH}_3\right)_6\right]

involve the exchange of ligands between the cationic and anionic coordination entities.

Solution approach

Correct Answer + Core Reason: $\left[{Co}\left({NH}_3\right)_6\right][{Cr}({CN})_6]$ and $\left[{Co}\left({CN}_6\right)\right]\left[{Cr}\left({NH}_3\right)_6\right]$ . This pair exhibits coordination isomerism due to the exchange of ligands between the complex ions.

Step-by-Step Solution:

Step 1: Define coordination isomerism: Coordination isomers are coordination compounds in which the ligands are exchanged between the cationic and anionic complex ions.

Step 2: Analyze option A: $\left[{Co}\left({NH}_3\right)_6\right][{Cr}({CN})_6]$ and $\left[{Co}\left({CN}_6\right)\right]\left[{Cr}\left({NH}_3\right)_6\right]$ involve the exchange of $NH_3$ and $CN$ ligands between the cobalt and chromium complex ions. This is coordination isomerism.

Step 3: Analyze option B: $\left[{Cr}\left({H}_2 {O}\right)_5 {Cl}\right] {Cl}_2 \cdot {H}_2 {O}$ and $\left[{Cr}\left({H}_2 {O}\right)_4 {Cl}_2\right] {Cl}. 2{H}_2 {O}$ are hydrate isomers, not coordination isomers.

Step 4: Analyze option C: $\left[{Co}\left({NH}_3\right)_5 {Br}\right] {SO}_3$ and $\left[{Cr}\left({NH}_3\right)_5 {SO}_4\right] {Br}$ are ionization isomers, not coordination isomers.

Step 5: Analyze option D: $\left[{Pt}\left({NH}_3\right)_2 {Cl}_2\right]$ and $\left[{Pt}\left({NH}_3\right)_4\right][{PtCl}_4]$ are not isomers.

Step 6: Conclusion: Only option A exhibits coordination isomerism.

Why other options are wrong:
- Option B: These are hydrate isomers, not coordination isomers.
- Option C: These are ionization isomers, not coordination isomers.
- Option D: These are not isomers.

Key Concepts:
- Coordination isomerism
- Coordination complexes
- Isomers

Memory Aids:
- Coordination isomers: exchange of ligands between complex ions.

Quick Tips:
- Look for complexes with both cationic and anionic complex ions. Check if the ligands are exchanged between them.

33 Hard

33 of 79Question permalink

Analyzing +4 | -1

Which of the following will have maximum stabilization due to crystal field?

View answer and explanation

Correct option: C

{[Co{(CN)_6}]^{ - 3}}

Option C (

{[Co{(CN)_6}]^{ - 3}}

) is correct because it exhibits the highest crystal field stabilization energy due to the strong field ligand CN- and the resulting low spin configuration.

Solution approach

Correct Answer + Core Reason: ${[Co{(CN)_6}]^{ - 3}}$ . Strong field ligands like CN- cause greater splitting and thus higher stabilization.

Step-by-Step Solution:

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

Step 2: Determine the electronic configuration of the metal ion.

Step 3: Consider the ligand field strength. Strong field ligands cause large splitting ( $\Delta_o$ ) and favor low-spin complexes, leading to higher CFSE.

Step 4: Calculate the Crystal Field Stabilization Energy (CFSE) for each complex.
- A. ${[Ti{({H_2}O)_6}]^{3 + }}$ : $Ti^{3+}$ is $d^1$ , CFSE = -0.4 $\Delta_o$
- B. ${[Co{({H_2}O)_6}]^{2 + }}$ : $Co^{2+}$ is $d^7$ , high spin, CFSE = -0.8 $\Delta_o$ + 2P
- C. ${[Co{(CN)_6}]^{ - 3}}$ : $Co^{3+}$ is $d^6$ , low spin, CFSE = -2.4 $\Delta_o$ + 3P
- D. ${[Cu{(N{H_3})_4}]^{2 + }}$ : $Cu^{2+}$ is $d^9$ , square planar, CFSE is complex but generally lower than C.

Step 5: Compare the CFSE values. Since CN- is a strong field ligand, it will result in a large $\Delta_o$ and a low-spin complex for $Co^{3+}$ , leading to a higher CFSE compared to the other complexes.

Why other options are wrong:
- Option A: ${[Ti{({H_2}O)_6}]^{3 + }}$ : $Ti^{3+}$ has only one d electron, leading to a relatively small CFSE.
- Option B: ${[Co{({H_2}O)_6}]^{2 + }}$ : $Co^{2+}$ with water ligands forms a high-spin complex, resulting in a smaller CFSE compared to the low-spin complex with cyanide.
- Option D: ${[Cu{(N{H_3})_4}]^{2 + }}$ : Although Copper (II) complexes can have significant CFSE, the complex is square planar, and the CFSE is generally less than the low spin octahedral complex.

Key Concepts:
- Crystal Field Theory (CFT)
- Crystal Field Stabilization Energy (CFSE)
- Spectrochemical series (ligand field strength)
- High-spin vs. Low-spin complexes

Memory Aids:
- Strong field ligands (e.g., CN-, CO) cause large splitting and favor low-spin complexes.
- Weak field ligands (e.g., halides, water) cause small splitting and favor high-spin complexes.

Quick Tips:
- Remember the spectrochemical series to determine the relative strength of ligands.
- Calculate CFSE based on the electronic configuration and ligand field strength.

34 Hard

34 of 79Question permalink

Analyzing +4 | -1

When EDTA solution is added to

M{g^{2 + }}

ion solution, then which of the following statements is not true?

View answer and explanation

Correct option: A

Four coordinate sites of

M{g^{2 + }}

are occupied by EDTA and remaining two sites are occupied by water molecules.

Option A is correct. EDTA is a hexadentate ligand, meaning it can occupy six coordination sites. Therefore, all six coordinate sites of Mg2+ are occupied by EDTA.

Solution approach

Correct Answer + Core Reason: Four coordinate sites of $M{g^{2 + }}$ are occupied by EDTA and remaining two sites are occupied by water molecules. EDTA is a hexadentate ligand and occupies all six coordination sites of Mg2+.

Step-by-Step Solution:

Step 1: Recall the structure and denticity of EDTA. EDTA (ethylenediaminetetraacetic acid) is a hexadentate ligand, meaning it can bind to a metal ion through six donor atoms (two nitrogen atoms and four oxygen atoms).

Step 2: Consider the coordination number of Mg2+. Mg2+ typically has a coordination number of 6.

Step 3: Determine how EDTA binds to Mg2+. Since EDTA is hexadentate, it occupies all six coordination sites of Mg2+.

Step 4: Conclude that all six coordinate sites of Mg2+ are occupied by EDTA.

Why other options are wrong:
- Option B: All six coordinate sites of Mg2+ are occupied. This statement is correct, so it's not the answer.
- Option C: ${\mathrm{P}^\mathrm{H}}$ of the solution is decreased. The pH of the solution may change slightly upon complex formation, but this is not the primary reason why the statement is incorrect.
- Option D: Colourless ${\left[ {\mathrm{Mg} - \mathrm{EDTA}} \right]^{2 - }}$ chelate is formed. This statement is correct, so it's not the answer.

Key Concepts:
- Coordination complexes
- Ligands (denticity)
- EDTA
- Coordination number

Memory Aids:
- EDTA is hexadentate.

Quick Tips:
- Remember the denticity of common ligands.

35 Hard

35 of 79Question permalink

Application +4 | -1

The number of optical isomers exhibited by the iron complex (A) obtained from the following reaction is _____.

\mathrm{FeCl}_3+\mathrm{KOH}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{A}

View answer and explanation

Correct option: A

2

Option A (2) is correct because the iron complex formed, likely containing oxalate ligands, can exhibit optical isomerism due to its non-superimposable mirror images.

Solution approach

Correct Answer + Core Reason: 2. The complex formed can exhibit optical isomerism due to its chiral nature.

Step-by-Step Solution:

Step 1: Identify the reaction and the complex formed.\nFeCl3 + KOH + H2C2O4 -> A\nThe reaction involves iron(III) chloride, potassium hydroxide, and oxalic acid. The product A is likely a coordination complex of iron with oxalate ligands.

Step 2: Determine the likely formula of the complex.\nIn a basic medium, iron(III) will react with oxalate ions to form a complex. A likely formula for the complex is [Fe(C2O4)3]3-.

Step 3: Analyze the structure of the complex for chirality.\nThe [Fe(C2O4)3]3- complex has a central iron ion surrounded by three bidentate oxalate ligands. This arrangement results in a chiral complex, meaning it is non-superimposable on its mirror image.

Step 4: Determine the number of optical isomers.\nA chiral complex exists as two optical isomers (enantiomers), which are non-superimposable mirror images of each other.

Step 5: Final Answer: The number of optical isomers is 2.

Why other options are wrong:
- Option B (0): This is incorrect because the complex [Fe(C2O4)3]3- is chiral and exhibits optical isomerism. Assuming no optical isomers means the complex is achiral, which is not the case.
- Option C (0): Same as Option B, incorrect because the complex is chiral.
- Option D (0): Same as Option B, incorrect because the complex is chiral.

Key Concepts:
- Coordination complexes
- Optical isomerism
- Chirality
- Bidentate ligands (oxalate)

Memory Aids:
- Complexes with bidentate ligands like oxalate often exhibit optical isomerism if the overall complex is not superimposable on its mirror image.

Quick Tips:
- Look for complexes with bidentate ligands and a central metal ion. If the ligands are arranged in a way that the complex is not superimposable on its mirror image, it will exhibit optical isomerism.

36 Hard

36 of 79Question permalink

Understanding +4 | -1

In the silver plating of

\mathrm{Cu},\mathrm{\;K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

is used instead of

\mathrm{AgN}{\mathrm{O}_3}.

The reason is:

View answer and explanation

Correct option: D

Less availability of

\mathrm{A}{\mathrm{g}^ + }

ion as

\mathrm{Cu}

cannot displace

\mathrm{Ag}

from

\mathrm{Ag}{\left( {\mathrm{CN}} \right)_2}

Option D (Less availability of

\mathrm{A}{\mathrm{g}^ + }

ion as

\mathrm{Cu}

cannot displace

\mathrm{Ag}

from

\mathrm{Ag}{\left( {\mathrm{CN}} \right)_2}

) is correct because the complex ion

{\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }

provides a lower concentration of free

Ag^+

ions, leading to a more uniform and adherent silver plating.

Solution approach

Correct Answer + Core Reason: Less availability of $\mathrm{A}{\mathrm{g}^ + }$ ion as $\mathrm{Cu}$ cannot displace $\mathrm{Ag}$ from $\mathrm{Ag}{\left( {\mathrm{CN}} \right)_2}$ . This ensures a slow and even deposition of silver.

Step-by-Step Solution:

Step 1: Understand the purpose of silver plating.

Step 2: Silver plating is done to coat a copper surface with a thin layer of silver.

Step 3: $\mathrm{AgN}{\mathrm{O}_3}$ provides a high concentration of $\mathrm{A}{\mathrm{g}^ + }$ ions, which can lead to rapid and uneven deposition of silver.

Step 4: $\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]$ is a complex salt that dissociates to give ${\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }$ ions.

Step 5: The complex ion ${\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }$ provides a much lower concentration of free $\mathrm{A}{\mathrm{g}^ + }$ ions compared to $\mathrm{AgN}{\mathrm{O}_3}$ .

Step 6: The lower concentration of $\mathrm{A}{\mathrm{g}^ + }$ ions results in a slower and more controlled deposition of silver, leading to a more uniform and adherent plating.

Step 7: Copper cannot displace silver from the complex because the stability constant of the silver cyanide complex is high.

Why other options are wrong:
- Option A: A thin layer of $\mathrm{Ag}$ is formed in both cases, but the quality of the layer is better with $\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]$ .
- Option B: Heat is not the primary reason for using $\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]$ .
- Option C: $\mathrm{A}{\mathrm{g}^ + }$ ions are not completely removed, but their concentration is significantly reduced.

Key Concepts:
- Electroplating
- Complex ions
- Concentration of ions
- Stability constant

Memory Aids:
- Complex ions reduce the concentration of free metal ions.

Quick Tips:
- Using complex ions in electroplating leads to better quality plating.

37 Hard

37 of 79Question permalink

Understanding +4 | -1

Assertion (A): The electronic absorption spectrum of the tetrahedral complex,

\left[{CoCl}_4\right]^{2-}

shows bands which are more intense compared to the absorption bands for the octahedral complex

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

.

Reason (R) : The octahedral complexes possess a centre of symmetry.

View answer and explanation

Correct option: B

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

.

Option B (Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

) is correct because both the assertion and the reason are true statements, but the reason (presence of a center of symmetry in octahedral complexes) does not explain why tetrahedral complexes have more intense absorption bands.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ . Tetrahedral complexes have more intense bands due to the lack of a center of symmetry, but the reason provided doesn't explain this.

Step-by-Step Solution:

Step 1: Analyze the assertion (A).\nThe electronic absorption spectrum of the tetrahedral complex, $\left[{CoCl}_4\right]^{2-}$ shows bands which are more intense compared to the absorption bands for the octahedral complex $\left[{Co}\left({H}_2 {O}\right)_6\right]^{2+}$ . This statement is true. Tetrahedral complexes generally have more intense absorption bands than octahedral complexes.

Step 2: Analyze the reason (R).\nThe octahedral complexes possess a centre of symmetry. This statement is also true. Octahedral complexes have a center of symmetry.

Step 3: Determine the relationship between the assertion and the reason.\nThe reason (R) does not correctly explain the assertion (A). The higher intensity of absorption bands in tetrahedral complexes is due to the lack of a center of symmetry, which makes the transitions Laporte-allowed (or partially allowed), while octahedral complexes have a center of symmetry, making the transitions Laporte-forbidden (or less allowed). The reason correctly states that octahedral complexes have a center of symmetry, but this fact alone does not explain the intensity difference.

Why other options are wrong:
- Option A: The reason is not the correct explanation of the assertion.
- Option C: The reason is true, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Electronic absorption spectra
- Laporte selection rule
- Symmetry in coordination complexes

Memory Aids:
- Tetrahedral complexes lack a center of symmetry, leading to more intense absorption bands.
- Octahedral complexes have a center of symmetry, leading to less intense absorption bands.

Quick Tips:
- Remember the Laporte selection rule and its implications for the intensity of absorption bands in coordination complexes.

38 Hard

38 of 79Question permalink

Understanding +4 | -1

Solution of which of the following compounds will not give normal tests of ferric ions:

View answer and explanation

Correct option: D

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

Option D (

\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]

) is correct because the ferric ions are complexed within the coordination compound and do not exist as free ions in solution, thus not giving normal tests for ferric ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$ . In this complex, the iron is part of a complex ion and will not give the usual tests for free ferric ions.

Step-by-Step Solution:

Step 1: Understand the concept of complex ions: Complex ions contain a central metal ion surrounded by ligands.

Step 2: Analyze each option:
- A. Ferric alum: Contains free ferric ions that will give normal tests.
- B. $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ : Contains iron in a complex ion, but also contains free ferric ions.
- C. $\mathrm{Fe}_2(\mathrm{SO}_4)_3$ : Contains free ferric ions that will give normal tests.
- D. $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$ : The iron is completely complexed within the hexacyanoferrate(III) ion. Therefore, it will not give normal tests for ferric ions.

Step 3: Conclude that only $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$ will not give normal tests for ferric ions.

Why other options are wrong:
- Option A: Ferric alum: Contains free ferric ions that will give normal tests.
- Option B: $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ : Contains free ferric ions that will give normal tests.
- Option C: $\mathrm{Fe}_2(\mathrm{SO}_4)_3$ : Contains free ferric ions that will give normal tests.

Key Concepts:
- Complex ions
- Ferric ion tests

Memory Aids:
- Complexed ions behave differently than free ions.

Quick Tips:
- Look for complex ions where the metal is completely surrounded by ligands.

39 Hard

39 of 79Question permalink

Understanding +4 | -1

Assertion (A): Linkage isomerism arises in coordination compounds containing ambidentate ligand.

Reason (R): Ambidentate ligand has two different donor atoms.

View answer and explanation

Correct option: A

Assertion and Reason both are true, Reason is correct explanation of Assertion.

Option A (Assertion and Reason both are true, Reason is correct explanation of Assertion) is correct because linkage isomerism arises when an ambidentate ligand coordinates to a metal ion through different donor atoms.

Solution approach

Correct Answer + Core Reason: Assertion and Reason both are true, Reason is correct explanation of Assertion. Ambidentate ligands have multiple donor atoms, leading to linkage isomerism.

Step-by-Step Solution:

Step 1: Understand the concept of linkage isomerism: Linkage isomerism occurs when a ligand can bind to a metal ion through different atoms.

Step 2: Understand the definition of ambidentate ligands: Ambidentate ligands are ligands that can coordinate to a metal ion through more than one atom.

Step 3: Relate ambidentate ligands to linkage isomerism: The presence of ambidentate ligands is a prerequisite for linkage isomerism.

Why other options are wrong:
- Option B: Assertion and Reason both are true but Reason is not the correct explanation of Assertion: The reason directly explains the assertion.
- Option C: Assertion is true, Reason is false: Both assertion and reason are true.
- Option D: Assertion is false, Reason is true: The assertion is also true.

Key Concepts:
- Linkage isomerism
- Ambidentate ligands
- Coordination chemistry

Memory Aids:
- Ambidentate = multiple binding sites = linkage isomerism.

Quick Tips:
- Look for ligands like NO2, SCN, and CN when considering linkage isomerism.

40 Hard

40 of 79Question permalink

Understanding +4 | -1

The correct statement among I to III are :
I. Valence bond theory cannot explain the color exhibited by transition metal complexes.
II. Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.
III. Valence bond theory cannot distinguish ligands as week and strong field ones.

View answer and explanation

Correct option: D

I and III only

Option D (I and III only) is correct because Valence Bond Theory (VBT) fails to explain color and cannot distinguish between weak and strong field ligands.

Solution approach

Correct Answer + Core Reason: I and III only because VBT doesn't explain color or distinguish ligand field strength.

Step-by-Step Solution:
Step 1: Evaluate statement I: Valence bond theory cannot explain the color exhibited by transition metal complexes.
Step 2: Evaluate statement II: Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.
Step 3: Evaluate statement III: Valence bond theory cannot distinguish ligands as weak and strong field ones.
\nStatement I is correct. Valence bond theory does not account for the electronic transitions that cause color in transition metal complexes.\nStatement II is incorrect. Valence bond theory can predict the number of unpaired electrons and hence the paramagnetic character, but it does not quantitatively predict the magnetic moment accurately.\nStatement III is correct. Valence bond theory does not differentiate between weak and strong field ligands; it only considers the hybridization of the metal ion.

Why other options are wrong:
- Option A: II and III only is incorrect because statement II is false.
- Option B: I, II and III is incorrect because statement II is false.
- Option C: I and II only is incorrect because statement II is false.

Key Concepts:
- Valence bond theory (VBT)
- Limitations of VBT
- Color of coordination complexes
- Magnetic properties of coordination complexes

Memory Aids:
- Remember that VBT is a simple theory with limitations, especially regarding color and ligand field strength.

Quick Tips:
- Know the limitations of VBT and the advantages of crystal field theory (CFT) and ligand field theory (LFT).

41 Hard

41 of 79Question permalink

Application +4 | -1

Which of the following complexes, how many of them are paramagnetic & diamagnetic in nature respectively :-

\left[{Ni}({CO})_4\right],\left[{Cr}\left({NH}_3\right)_3\left({NO}_2\right)_3\right]

,

\left[{Ni}\left({NH}_3\right)_3\left({H}_2 {O}\right)_3\right]^{+2},\left[{Co}({OX})_3\right]^{-3}

,

\left[{Fe}\left({NH}_3\right)_6\right]^{+3},\left[{NiCl}_4\right]^{-2}

Options are :

View answer and explanation

Correct option: C

4, 2

Option C (4, 2) is correct because 4 complexes are paramagnetic and 2 are diamagnetic after determining the electronic configuration and presence of unpaired electrons.

Solution approach

Correct Answer + Core Reason: 4, 2. Determining the electronic configuration and unpaired electrons in each complex leads to identifying 4 paramagnetic and 2 diamagnetic complexes.

Step-by-Step Solution:

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

Step 2: Determine the electronic configuration of the metal ion.

Step 3: Determine the number of unpaired electrons based on the ligand strength and electronic configuration.

Step 4: Identify paramagnetic (unpaired electrons) and diamagnetic (no unpaired electrons) complexes.

Step 5: Analyze each complex:
- [Ni(CO)4]: Ni is in 0 oxidation state, electronic configuration is 3d10 4s0. CO is a strong field ligand, so no unpaired electrons. Diamagnetic.
- [Cr(NH3)3(NO2)3]: Cr is in +3 oxidation state, electronic configuration is 3d3. NH3 and NO2 are strong field ligands, but with 3 d electrons, there will be 3 unpaired electrons. Paramagnetic.
- [Ni(NH3)3(H2O)3]+2: Ni is in +2 oxidation state, electronic configuration is 3d8. NH3 and H2O are weak field ligands, so there will be 2 unpaired electrons. Paramagnetic.
- [Co(OX)3]-3: Co is in +3 oxidation state, electronic configuration is 3d6. Oxalate is a strong field ligand, so no unpaired electrons. Diamagnetic.
- [Fe(NH3)6]+3: Fe is in +3 oxidation state, electronic configuration is 3d5. NH3 is a weak field ligand, so there will be 5 unpaired electrons. Paramagnetic.
- [NiCl4]-2: Ni is in +2 oxidation state, electronic configuration is 3d8. Cl is a weak field ligand, so there will be 2 unpaired electrons. Paramagnetic.

Step 6: Count the number of paramagnetic and diamagnetic complexes. 4 Paramagnetic and 2 Diamagnetic.

Final Answer: 4, 2

Why other options are wrong:
- Option A: Incorrect count of paramagnetic and diamagnetic complexes.
- Option B: Incorrect count of paramagnetic and diamagnetic complexes.
- Option D: Incorrect count of paramagnetic and diamagnetic complexes.

Key Concepts:
- Coordination complexes
- Oxidation state
- Electronic configuration
- Ligand field theory
- Paramagnetism and Diamagnetism

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to diamagnetism.
- Weak field ligands do not cause pairing, leading to paramagnetism.

Quick Tips:
- Determine the oxidation state and electronic configuration accurately.
- Consider the ligand strength to determine the pairing of electrons.

42 Hard

42 of 79Question permalink

Understanding +4 | -1

Why do compounds having similar geometry have different magnetic moment?

View answer and explanation

Correct option: C

Due to the presence of weak and strong field ligands

Option C is correct because the presence of weak and strong field ligands affects the electronic configuration and hence the magnetic moment.

Solution approach

Correct Answer + Core Reason: Due to the presence of weak and strong field ligands. Different ligands cause different crystal field splitting, leading to different numbers of unpaired electrons and thus different magnetic moments.

Step-by-Step Solution:

Step 1: Understand the concept of magnetic moment. Magnetic moment arises from unpaired electrons in a complex.

Step 2: Recognize that ligands affect the electronic configuration. Strong field ligands cause pairing of electrons, while weak field ligands do not.

Step 3: Relate ligand strength to magnetic moment. Complexes with strong field ligands tend to have lower magnetic moments (or be diamagnetic) due to electron pairing. Complexes with weak field ligands tend to have higher magnetic moments due to more unpaired electrons.

Step 4: Conclude that different ligands lead to different magnetic moments even with similar geometries.

Why other options are wrong:
- Option A: Reactivity does not directly determine the magnetic moment. Reactivity is related to the stability and electron configuration but not the direct cause of the magnetic moment.
- Option B: Lability refers to the rate of ligand exchange, which is not directly related to the magnetic moment.
- Option D: The presence of weak and strong field ligands is the direct cause, making this option incorrect.

Key Concepts:
- Magnetic moment
- Crystal field theory
- Ligand field strength
- Spectrochemical series

Memory Aids:
- Strong field ligands cause pairing (low spin), weak field ligands do not (high spin).

Quick Tips:
- Remember the spectrochemical series to determine ligand strength.

43 Hard

43 of 79Question permalink

Analyzing +4 | -1

Which of the following statement is false

View answer and explanation

Correct option: A

Of the

{d^4}

species, manganese (III) is strongly reducing while

C{r^{2 + }}

is strongly oxidising.

Option A is false because Mn(III) is easily reduced to Mn(II) and thus is a good oxidizing agent, while Cr(II) is easily oxidized to Cr(III) and is a good reducing agent.

Solution approach

Correct Answer + Core Reason: Of the ${d^4}$ species, manganese (III) is strongly reducing while $C{r^{2 + }}$ is strongly oxidising. This statement is false because Mn(III) is easily reduced and Cr(II) is easily oxidized.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations and their tendencies to gain or lose electrons.\nMn(III) has a d4 configuration. Gaining an electron to become Mn(II) (d5) results in a stable half-filled d-orbital configuration. Therefore, Mn(III) tends to gain electrons and act as an oxidizing agent.\nCr(II) has a d4 configuration. Losing an electron to become Cr(III) (d3) results in a stable t2g3 configuration (in an octahedral field). Therefore, Cr(II) tends to lose electrons and act as a reducing agent.

Step 2: Evaluate the given statement.\nThe statement claims that Mn(III) is strongly reducing and Cr(II) is strongly oxidizing, which contradicts the analysis in Step 1.

Final Answer: Of the ${d^4}$ species, manganese (III) is strongly reducing while $C{r^{2 + }}$ is strongly oxidising.

Why other options are wrong:
- Option B: Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. This statement is true.
- Option C: The ${d^1}$ configuration is very unstable in ions. This statement is generally true, as d1 ions are easily oxidized or reduced to achieve more stable configurations.
- Option D: None of these. Since option A is false, this option is incorrect.

Key Concepts:
- Oxidation and reduction
- Electronic configuration
- Stability of half-filled and fully-filled orbitals
- Redox potentials

Memory Aids:
- Oxidation is Loss, Reduction is Gain (OIL RIG)

Quick Tips:
- Consider the electronic configurations and their tendencies to achieve stable states when determining oxidizing or reducing properties.

44 Hard

44 of 79Question permalink

Understanding +4 | -1

Which of the following complexes are heteroleptic ?

View answer and explanation

Correct option: B

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

Option B is correct because

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

contains more than one type of ligand coordinated to the central metal ion, making it heteroleptic.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$ . Heteroleptic complexes contain more than one type of ligand.

Step-by-Step Solution:

Step 1: Define heteroleptic complex: A heteroleptic complex is a coordination complex in which the central metal atom is bonded to more than one type of ligand.

Step 2: Analyze option A: $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ has only one type of ligand ( $NH_3$ ). It is homoleptic.

Step 3: Analyze option B: $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$ has two types of ligands ( $NH_3$ and $Cl$ ). It is heteroleptic.

Step 4: Analyze option C: $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$ has only one type of ligand ( $CN$ ). It is homoleptic.

Step 5: Analyze option D: $\left[\mathrm{MnF}_6\right]^{4-}$ has only one type of ligand ( $F$ ). It is homoleptic.

Step 6: Conclusion: Only option B is heteroleptic.

Why other options are wrong:
- Option A: $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is homoleptic.
- Option C: $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$ is homoleptic.
- Option D: $\left[\mathrm{MnF}_6\right]^{4-}$ is homoleptic.

Key Concepts:
- Heteroleptic complex
- Homoleptic complex
- Coordination complexes

Memory Aids:
- Hetero- means different, so heteroleptic means different ligands.
- Homo- means same, so homoleptic means same ligands.

Quick Tips:
- Count the number of different types of ligands in the complex. If there is more than one type, it is heteroleptic.

45 Hard

45 of 79Question permalink

Analyzing +4 | -1

Which one of the following species doesn’t have a magnetic moment of 1.73 BM (spin only value)?

View answer and explanation

Correct option: B

CuI

Option B (

CuI

) is correct because

CuI

has

Cu

in +1 oxidation state, which has a d10 configuration, resulting in no unpaired electrons and hence a magnetic moment of 0 BM.

Solution approach

Correct Answer + Core Reason: $CuI$ . $CuI$ has a d10 configuration and hence no unpaired electrons.

Step-by-Step Solution:
Step 1: Determine the electronic configuration of each species.
Step 2: Calculate the number of unpaired electrons for each species.
Step 3: Calculate the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.
Step 4: Check which species does not have a magnetic moment of 1.73 BM (corresponding to 1 unpaired electron).

Step 5: Analyze $O_2^+$ : $O_2^+$ has 11 electrons. Its electronic configuration is $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^1$ . It has 1 unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$ BM.
Step 6: Analyze $CuI$ : $Cu$ is in +1 oxidation state. $Cu^+$ has a d10 configuration, which means it has 10 d-electrons. The electronic configuration is $d^{10}$ . It has 0 unpaired electrons. $\mu = \sqrt{0(0+2)} = 0$ BM.
Step 7: Analyze $[Cu(NH_3)_4]Cl_2$ : $Cu$ is in +2 oxidation state. $Cu^{2+}$ has a d9 configuration. It has 1 unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$ BM.
Step 8: Analyze $O_2^-$ : $O_2^-$ has 13 electrons. Its electronic configuration is $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^3$ . It has 1 unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$ BM.

Why other options are wrong:
- Option A: $O_2^+$ has 1 unpaired electron, resulting in a magnetic moment of 1.73 BM.
- Option C: $[Cu(NH_3)_4]Cl_2$ has $Cu^{2+}$ with 1 unpaired electron, resulting in a magnetic moment of 1.73 BM.
- Option D: $O_2^-$ has 1 unpaired electron, resulting in a magnetic moment of 1.73 BM.

Key Concepts:
- Electronic Configuration
- Magnetic Moment
- Spin-Only Formula

Memory Aids:
- Remember the formula for spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ BM.

Quick Tips:
- Determine the oxidation state of the metal ion first.
- Write the electronic configuration to find the number of unpaired electrons.

46 Hard

46 of 79Question permalink

Understanding +4 | -1

Number of possible isomer for the complex

[Co{(N{H_3})_3}C{l_3}]

will be

View answer and explanation

Correct option: C

2

The correct answer is C (2). The complex

[Co{(N{H_3})_3}C{l_3}]

can exist as two isomers: fac (facial) and mer (meridional).

Solution approach

Correct Answer + Core Reason: 2. The complex exhibits fac and mer isomerism.

Step-by-Step Solution:

Step 1: Identify the type of isomerism. The complex $[Co{(N{H_3})_3}C{l_3}]$ exhibits geometrical isomerism.

Step 2: Draw the possible isomers.

Step 3: One isomer has all three $\mathrm{N}{\mathrm{H}_3}$ ligands on one face of the octahedron and all three $\mathrm{Cl}$ ligands on the opposite face (fac isomer).

Step 4: The other isomer has the three $\mathrm{N}{\mathrm{H}_3}$ and three $\mathrm{Cl}$ ligands arranged around the meridian (mer isomer).

Step 5: Therefore, there are two possible isomers: fac and mer.

Why other options are wrong:
- Option A: 3: There are only two geometrical isomers.
- Option B: 4: There are only two geometrical isomers.
- Option D: 1: There are two geometrical isomers, not one.

Key Concepts:
- Geometrical isomerism
- Fac and mer isomers
- Coordination complexes

Memory Aids:
- fac = facial (all ligands on one face)
- mer = meridional (ligands around the meridian)

Quick Tips:
- Draw the structures to visualize the isomers.

47 Hard

47 of 79Question permalink

Analyzing +4 | -1

Which of the following ions can exhibit optical isomerism-

View answer and explanation

Correct option: C

Cis-

\left[{Co}({en})_2 {Cl}_2\right]^{+}

Option C (

Cis-

\left[{Co}({en})_2 {Cl}_2\right]^{+}

) is correct because the cis isomer of this complex is chiral and can therefore exhibit optical isomerism.

Solution approach

Correct Answer + Core Reason: $Cis-$ $\left[{Co}({en})_2 {Cl}_2\right]^{+}$ . Only the cis isomer of this complex is chiral and can exhibit optical isomerism.

Step-by-Step Solution:

Step 1: Analyze each complex for chirality.
- $\left[{Co}\left({NH}_3\right)_4 {Cl}_2\right]^{+}$ : This complex can exist as cis and trans isomers. Neither isomer is chiral because of the presence of a plane of symmetry.
- $\left[{Co}\left({NH}_3\right)_2 {Cl}_4\right]^{-}$ : This complex can exist as cis and trans isomers. Neither isomer is chiral because of the presence of a plane of symmetry.
- $Cis-$ $\left[{Co}({en})_2 {Cl}_2\right]^{+}$ : This complex is chiral and exists as a pair of enantiomers (non-superimposable mirror images).
- $trans-$ $\left[{Co}({en})_2 {Cl}_2\right]^{+}$ : This complex is not chiral because it has a plane of symmetry.

Step 2: Identify the complex that exhibits optical isomerism.\nOnly the cis isomer of $\left[{Co}({en})_2 {Cl}_2\right]^{+}$ is chiral and exhibits optical isomerism.

Why other options are wrong:
- Option A: $\left[{Co}\left({NH}_3\right)_4 {Cl}_2\right]^{+}$ : Both cis and trans isomers have a plane of symmetry and are achiral.
- Option B: $\left[{Co}\left({NH}_3\right)_2 {Cl}_4\right]^{-}$ : Both cis and trans isomers have a plane of symmetry and are achiral.
- Option D: $trans-$ $\left[{Co}({en})_2 {Cl}_2\right]^{+}$ : This isomer has a plane of symmetry and is achiral.

Key Concepts:
- Optical Isomerism
- Chirality
- Plane of Symmetry
- Enantiomers
- Coordination Complexes

Memory Aids:
- Chirality: A molecule is chiral if it is non-superimposable on its mirror image.
- Plane of symmetry: A plane that divides a molecule into two identical halves.

Quick Tips:
- Look for the presence of a plane of symmetry to determine if a complex is chiral.
- Complexes with bidentate ligands in a cis configuration are more likely to be chiral.

48 Hard

48 of 79Question permalink

Understanding +4 | -1

Assertion : Tetrahedral complex do not exhibit geometrical isomerism.

Reason : In tetrahedral complex all the four positions are identical.

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 both the assertion and the reason are true, and the reason correctly explains why tetrahedral complexes do not exhibit geometrical isomerism.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. Tetrahedral complexes lack geometrical isomers because all positions are equivalent.

Step-by-Step Solution:

Step 1: Define geometrical isomerism: Geometrical isomers are compounds with the same chemical formula but different spatial arrangements of atoms.

Step 2: Consider tetrahedral complexes: In a tetrahedral complex, all four ligands are equidistant from each other. Therefore, there are no different spatial arrangements possible.

Step 3: Analyze the assertion: The assertion states that tetrahedral complexes do not exhibit geometrical isomerism, which is true.

Step 4: Analyze the reason: The reason states that in tetrahedral complexes, all four positions are identical, which is also true and explains why geometrical isomerism is not possible.

Step 5: Conclusion: Both the assertion and the reason are true, and the reason is a correct explanation of the assertion.

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

Key Concepts:
- Geometrical isomerism
- Tetrahedral complexes
- Coordination complexes

Memory Aids:
- Tetrahedral = no geometrical isomers.

Quick Tips:
- Remember that tetrahedral complexes do not exhibit geometrical isomerism because all positions are equivalent.

49 Hard

49 of 79Question permalink

Understanding +4 | -1

\mathrm{AgO}

in

\mathrm{Ag}\left( {\mathrm{II}} \right)

complex which is:

View answer and explanation

Correct option: A

Diamagnetic

Option A (Diamagnetic) is correct because in AgO, Ag is in the +2 oxidation state (

d^9

configuration), but it forms a complex where the unpaired electron is involved in bonding or pairing, resulting in diamagnetism.

Solution approach

Correct Answer + Core Reason: Diamagnetic. AgO contains Ag(II), which typically has one unpaired electron, but forms a complex where the electrons are paired, resulting in diamagnetism.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Ag in AgO.\nAg + (-2) = 0\nAg = +2

Step 2: Determine the electronic configuration of Ag(II).\nAg: [Kr] $4d^{10} 5s^1$ \nAg(II): [Kr] $4d^9$

Step 3: Analyze the magnetic properties.\nAg(II) with a $d^9$ configuration would be expected to be paramagnetic due to the presence of one unpaired electron. However, AgO is known to be diamagnetic. This indicates that the unpaired electron in Ag(II) is involved in some form of bonding or pairing, leading to the absence of unpaired electrons and thus diamagnetism.

Why other options are wrong:
- Option B (Paramagnetic): Paramagnetism arises from unpaired electrons. While Ag(II) has a $d^9$ configuration, the complex is diamagnetic, indicating the absence of unpaired electrons.
- Option C (Ferromagnetic): Ferromagnetism requires the alignment of electron spins in a material, which is not the case for AgO.
- Option D (Neutral): Neutrality is not a magnetic property.

Key Concepts:
- Oxidation state
- Electronic configuration
- Magnetic properties (diamagnetism, paramagnetism, ferromagnetism)

Memory Aids:
- Diamagnetic substances are repelled by a magnetic field.
- Paramagnetic substances are attracted to a magnetic field.

Quick Tips:
- Consider the electronic configuration and any potential pairing or bonding that might affect the magnetic properties.

50 Hard

50 of 79Question permalink

Analyzing +4 | -1

The complex with highest magnitude of crystal field splitting energy

\left(\Delta_{0}\right)

is

View answer and explanation

Correct option: B

\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}

Option B (

\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}

) is correct because

Cr^{3+}

has a

d^3

configuration, leading to a larger crystal field splitting compared to other options with weaker ligand field stabilization.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$ . $Cr^{3+}$ has a $d^3$ configuration, which leads to a relatively large crystal field splitting energy.

Step-by-Step Solution:

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

Step 2: $\left[\mathrm{Ti}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$ : $Ti^{3+}$ is $d^1$ .

Step 3: $\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$ : $Cr^{3+}$ is $d^3$ .

Step 4: $\left[\mathrm{Mn}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$ : $Mn^{3+}$ is $d^4$ .

Step 5: $\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$ : $Fe^{3+}$ is $d^5$ .

Step 6: Consider the spectrochemical series. Water is a weak field ligand, so the complexes are high spin.

Step 7: For $d^3$ , all three electrons occupy the $t_{2g}$ orbitals, leading to a relatively large crystal field stabilization energy and thus a larger ${\Delta _o}$ .

Why other options are wrong:
- Option A: $Ti^{3+}$ ( $d^1$ ) has only one electron in the $t_{2g}$ orbitals, resulting in a smaller ${\Delta _o}$ .
- Option C: $Mn^{3+}$ ( $d^4$ ) in a high-spin field has three electrons in $t_{2g}$ and one in $e_g$ . The presence of an electron in $e_g$ reduces the overall crystal field stabilization energy.
- Option D: $Fe^{3+}$ ( $d^5$ ) in a high-spin field has three electrons in $t_{2g}$ and two in $e_g$ . The presence of two electrons in $e_g$ further reduces the crystal field stabilization energy.

Key Concepts:
- Crystal field theory
- Electronic configuration
- Spectrochemical series
- Crystal field splitting energy

Memory Aids:
- Remember the spectrochemical series to compare ligand field strengths.

Quick Tips:
- Consider the electronic configuration and ligand field strength to determine the magnitude of ${\Delta _o}$ .

51 Hard

51 of 79Question permalink

Analyzing +4 | -1

Among

\left[\mathrm{Ni}(\mathrm{CO})_4\right],\left[\mathrm{NiCl}_4\right]^{2-},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}

,

\mathrm{Na}_3\left[\mathrm{CoF}_6\right], \mathrm{Na}_2 \mathrm{O}_2

and

\mathrm{CsO}_2

, the total number of paramagnetic species is

View answer and explanation

Correct option: A

3

Option A is correct because only three species, namely [NiCl4]2-, Na3[CoF6], and CsO2, are paramagnetic among the given compounds.

Solution approach

Correct Answer + Core Reason: 3. Only 3 species are paramagnetic: $\left[\mathrm{{NiCl}}_4\right]^{2-}$ , $\mathrm{{Na}}_3\left[\mathrm{{CoF}}_6\right]$ , and $\mathrm{{CsO}}_2$ .

Step-by-Step Solution:

Step 1: Determine the electronic configuration and magnetic properties of each species.

Step 2: Analyze each complex:
- $\left[\mathrm{{Ni}}(\mathrm{{CO}})_4\right]$ : $Ni^0$ is $d^{10}$ . Tetrahedral, strong field. No unpaired electrons. Diamagnetic.
- $\left[\mathrm{{NiCl}}_4\right]^{2-}$ : $Ni^{2+}$ is $d^8$ . Tetrahedral, weak field. Two unpaired electrons. Paramagnetic.
- $\left[\mathrm{{Co}}(\mathrm{{NH}}_3)_4 \mathrm{{Cl}}_2\right] \mathrm{{Cl}}$ : $Co^{3+}$ is $d^6$ . Octahedral, low spin (due to $NH_3$ ). No unpaired electrons. Diamagnetic.
- $\mathrm{{Na}}_3\left[\mathrm{{CoF}}_6\right]$ : $Co^{3+}$ is $d^6$ . Octahedral, high spin (due to $F^-$ ). Four unpaired electrons. Paramagnetic.
- $\mathrm{{Na}}_2 \mathrm{{O}}_2$ : Peroxide ion ( $O_2^{2-}$ ). Diamagnetic.
- $\mathrm{{CsO}}_2$ : Superoxide ion ( $O_2^-$ ). One unpaired electron. Paramagnetic.

Step 3: Count the number of paramagnetic species: $\left[\mathrm{{NiCl}}_4\right]^{2-}$ , $\mathrm{{Na}}_3\left[\mathrm{{CoF}}_6\right]$ , and $\mathrm{{CsO}}_2$ . Total = 3.

Why other options are wrong:
- Option B: Incorrect count of paramagnetic species.
- Option C: Incorrect count of paramagnetic species.
- Option D: Incorrect count of paramagnetic species.

Key Concepts:
- Paramagnetism and diamagnetism
- Crystal field theory
- Electronic configuration
- Ligand field strength
- Superoxides and peroxides

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to diamagnetism.
- Weak field ligands result in high spin complexes and paramagnetism.

Quick Tips:
- Determine the oxidation state and d-electron count of the metal ion.
- Consider the ligand field strength to determine the electronic configuration.

52 Hard

52 of 79Question permalink

Application +4 | -1

The volume (in mL) of 0.1 M

\mathrm{AgN}{\mathrm{O}_3}

required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of

\left[ {\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_2}

, as silver chloride is close to

View answer and explanation

Correct option: D

6

Option D (6) is correct because two chloride ions are outside the coordination sphere, and they will precipitate with silver nitrate, requiring 6 mL of 0.1 M AgNO3.

Solution approach

Correct Answer + Core Reason: 6. Two chloride ions are outside the coordination sphere.

Step-by-Step Solution:

Step 1: The complex is ${\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_5}\mathrm{Cl}} ]\mathrm{C}{\mathrm{l}_2}$ . This means that only 2 $Cl^-$ ions are outside the coordination sphere and will react with $AgNO_3$ .

Step 2: The reaction is: $AgNO_3 + Cl^- \rightarrow AgCl + NO_3^-$

Step 3: Moles of complex = 0.01 M * 0.03 L = 0.0003 moles.

Step 4: Moles of $Cl^-$ ions outside the coordination sphere = 2 * moles of complex = 2 * 0.0003 = 0.0006 moles.

Step 5: Moles of $AgNO_3$ required = moles of $Cl^-$ ions = 0.0006 moles.

Step 6: Volume of 0.1 M $AgNO_3$ required = moles / molarity = 0.0006 / 0.1 = 0.006 L = 6 mL.

Why other options are wrong:
- Option A: This would correspond to only 0.0003 moles of $AgNO_3$ .
- Option B: This would correspond to only 0.0004 moles of $AgNO_3$ .
- Option C: This would correspond to only 0.0005 moles of $AgNO_3$ .

Key Concepts:
- Coordination complexes
- Precipitation reactions
- Stoichiometry

Memory Aids:
- Remember to consider only the ions outside the coordination sphere.

Quick Tips:
- Identify the ions outside the coordination sphere that will react with the added reagent.

53 Hard

53 of 79Question permalink

Understanding +4 | -1

Assertion (A): As compared to non-chelated complexes, chelated complexes are more stable.

Reason (

\mathrm{R}

): Labile complexes are the complexes which contain ligands those can be easily replaced by other ligands.

View answer and explanation

Correct option: B

Assertion and Reason both are true but Reason is not the correct explanation of Assertion.

Option B (Assertion and Reason both are true but Reason is not the correct explanation of Assertion) is correct because both the assertion and the reason are true statements, but the reason (lability of ligands) does not explain why chelated complexes are more stable.

Solution approach

Correct Answer + Core Reason: Assertion and Reason both are true but Reason is not the correct explanation of Assertion. Chelated complexes are more stable due to the chelate effect, which is not directly related to ligand lability.

Step-by-Step Solution:

Step 1: Analyze the assertion (A).\nAs compared to non-chelated complexes, chelated complexes are more stable. This statement is true. Chelated complexes are generally more stable than non-chelated complexes due to the chelate effect.

Step 2: Analyze the reason (R).\nLabile complexes are the complexes which contain ligands those can be easily replaced by other ligands. This statement is also true. Labile complexes are characterized by their ligands being easily replaced.

Step 3: Determine the relationship between the assertion and the reason.\nThe reason (R) does not correctly explain the assertion (A). The increased stability of chelated complexes (chelate effect) is primarily due to the increase in entropy upon displacement of the chelating ligand by several monodentate ligands. The lability of ligands is a separate concept related to the rate of ligand substitution, not the thermodynamic stability of the complex.

Why other options are wrong:
- Option A: The reason is not the correct explanation of the assertion.
- Option C: The reason is true, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Chelate effect
- Stability of coordination complexes
- Lability of ligands

Memory Aids:
- Chelate effect: Increased stability of chelated complexes due to entropy increase.

Quick Tips:
- Remember the chelate effect and its thermodynamic origin.

54 Hard

54 of 79Question permalink

Understanding +4 | -1

In the complex ion

[\mathrm{Fe}(\mathrm{EDTA})]^{-}

, the coordination number and oxidation state of central metal ion is

View answer and explanation

Correct option: A

\mathrm{CN}=6, \mathrm{ON}=+3

Option A (

\mathrm{CN}=6, \mathrm{ON}=+3

) is correct because EDTA is a hexadentate ligand, leading to a coordination number of 6, and the overall charge of the complex allows for the calculation of the iron's oxidation state as +3.

Solution approach

Correct Answer + Core Reason: $\mathrm{CN}=6, \mathrm{ON}=+3$ . EDTA is a hexadentate ligand and the oxidation state of Fe is +3.

Step-by-Step Solution:

Step 1: Identify the ligand and its denticity. EDTA (ethylenediaminetetraacetic acid) is a hexadentate ligand, meaning it can form six coordinate bonds with the central metal ion.

Step 2: Determine the coordination number. The coordination number is the number of ligands directly attached to the central metal ion. Since EDTA is hexadentate, the coordination number of Fe is 6.

Step 3: Determine the oxidation state of Fe. The complex ion is [Fe(EDTA)] $^{-}$ . EDTA has a charge of -4. Let the oxidation state of Fe be x. Then, x + (-4) = -1. Solving for x, we get x = +3.

Step 4: State the coordination number and oxidation state.\nCoordination number (CN) = 6\nOxidation state (ON) = +3

Why other options are wrong:
- Option B: Incorrect. CN cannot be 1 and ON cannot be -1.
- Option C: Incorrect. CN cannot be 4 and ON cannot be +2.
- Option D: Incorrect. CN cannot be 3 and ON cannot be +3.

Key Concepts:
- Coordination number
- Oxidation state
- Ligand denticity

Memory Aids:
- EDTA is always hexadentate.

Quick Tips:
- Remember the common ligands and their denticities.

55 Hard

55 of 79Question permalink

Analyzing +4 | -1

Which of the following statements are correct?

(i) Square planar complexes of

MABXL

type show three isomers-two cis and one trans.

(ii) Complexes of

{Ma}_3 {B}_3

type show three isomers-two cis and one trans.

(iii) Optical isomerism is common in octahedral complexes involving bidentate ligands.

(iv)

\left[{Co}\left({NH}_3\right)_4 {Cl}\left({NO}_2\right)\right] {Cl}

show linkage isomerism.

(v) Hydrate isomerism is another name of solvate isomerism.

View answer and explanation

Correct option: B

(i), (iii) and (iv)

Option B ((i), (iii) and (iv)) is correct because these statements accurately describe properties of coordination complexes, including isomerism and linkage.

Solution approach

Correct Answer + Core Reason: (i), (iii) and (iv). These statements are correct regarding isomerism and linkage in coordination compounds.

Step-by-Step Solution:

Step 1: Evaluate each statement.
(i) Square planar complexes of $MABXL$ type show three isomers-two cis and one trans. This is correct. For square planar complexes of the type MABXL, there are three isomers.
(ii) Complexes of ${Ma}_3 {B}_3$ type show three isomers-two cis and one trans. This is incorrect. Complexes of the type $Ma_3b_3$ show facial (fac) and meridional (mer) isomers, which are two isomers, not three.
(iii) Optical isomerism is common in octahedral complexes involving bidentate ligands. This is correct. Cis isomers of octahedral complexes with bidentate ligands often exhibit optical isomerism.
(iv) $\left[{Co}\left({NH}_3\right)_4 {Cl}\left({NO}_2\right)\right] {Cl}$ show linkage isomerism. This is correct. The $NO_2$ ligand can bind through either the N or O atom, leading to linkage isomerism.
(v) Hydrate isomerism is another name of solvate isomerism. This is correct. Hydrate isomerism is a specific type of solvate isomerism where the solvent is water.

Step 2: Identify the correct statements.\nStatements (i), (iii), (iv) and (v) are correct. Statement (ii) is incorrect.

Why other options are wrong:
- Option A: (i), (ii) and (iii). Statement (ii) is incorrect.
- Option C: (ii), (iii) and (v). Statement (ii) is incorrect and statement (v) is correct.
- Option D: (iii), (iv) and (v). Statement (v) is correct, but the question asks for correct statements.

Key Concepts:
- Isomerism in Coordination Compounds
- Geometrical Isomerism (cis-trans, fac-mer)
- Optical Isomerism
- Linkage Isomerism
- Solvate Isomerism

Memory Aids:
- Linkage isomerism: different atoms of the same ligand bind to the metal.
- fac: all three ligands are on one face of the octahedron; mer: ligands are around the meridian.

Quick Tips:
- Draw the possible isomers to determine if they are geometrical or optical isomers.
- Remember that linkage isomers involve ligands that can bind through different atoms.

56 Hard

56 of 79Question permalink

Analyzing +4 | -1

In which of the following compound, C – O bond length is maximum

View answer and explanation

Correct option: A

{\left[ {Fe{{\left( {CO} \right)}_5}} \right]^ - }

Option A ([Fe(CO)5]-) is correct because the negative charge on the complex increases back-bonding from the metal to the CO ligand, weakening the C-O bond and thus increasing the C-O bond length.

Solution approach

Correct Answer + Core Reason: ${\left[ {Fe{{\left( {CO} \right)}_5}} \right]^ - }$ . Increased negative charge increases back-bonding, weakening the C-O bond and increasing its length.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the metal in each complex.
- [Fe(CO)5]-: Fe has a -1 oxidation state.
- [Ni(CO)4]: Ni has a 0 oxidation state.
- [Cr(CO)6]: Cr has a 0 oxidation state.
- [Mn(CO)5]+: Mn has a +1 oxidation state.

Step 2: Consider the effect of charge on back-bonding. A more negative charge on the metal center increases the extent of back-bonding into the CO $\pi^*$ antibonding orbitals, which weakens the C-O bond and increases the C-O bond length.

Step 3: Compare the back-bonding in each complex. [Fe(CO)5]- has the most back-bonding due to the negative charge on Fe.

Step 4: Conclude that [Fe(CO)5]- has the maximum C-O bond length.

Why other options are wrong:
- Option B: [Ni(CO)4] has a neutral charge, so it has less back-bonding than [Fe(CO)5]-.
- Option C: [Cr(CO)6] has a neutral charge, so it has less back-bonding than [Fe(CO)5]-.
- Option D: [Mn(CO)5]+ has a positive charge, which reduces back-bonding and decreases the C-O bond length.

Key Concepts:
- Metal carbonyls
- Back-bonding
- C-O bond length
- Oxidation state

Memory Aids:
- More negative charge on the metal = more back-bonding = longer C-O bond length.

Quick Tips:
- Determine the oxidation state of the metal first.
- Consider the effect of charge on back-bonding.

57 Hard

57 of 79Question permalink

Understanding +4 | -1

For kinetic study of the reaction of iodide ion with

\mathrm{H}_2 \mathrm{O}_2

at room temperature :

(A) Always use freshly prepared starch solution.

(B) Always keep the concentration of sodium thiosulphate solution less than that of

\mathrm{KI}

solution.

(C) Record the time immediately after the appearance of blue colour.

(D) Record the time immediately before the appearance of blue colour.

(E) Always keep the concentration of sodium thiosulphate solution more than that of

\mathrm{KI}

solution.

Choose the correct answer from the options given below:

View answer and explanation

Correct option: A

(A), (B), (C) only

Option A ((A), (B), (C) only) is correct because it lists the correct procedures for the kinetic study of the reaction of iodide ion with

\mathrm{H}_2 \mathrm{O}_2

.

Solution approach

Correct Answer + Core Reason: (A), (B), (C) only because these are the correct experimental conditions for the given kinetic study.

Step-by-Step Solution:
Step 1: Analyze each statement regarding the experimental procedure.
Step 2: Determine which statements are correct.

(A) Always use freshly prepared starch solution: Correct. Starch solution can degrade over time, affecting the accuracy of the endpoint determination.
(B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution: Correct. Sodium thiosulphate is used to titrate the iodine produced in the reaction. Its concentration should be less than that of KI to ensure that the reaction proceeds at a measurable rate.
(C) Record the time immediately after the appearance of blue colour: Correct. The appearance of the blue color indicates the endpoint of the titration, and the time should be recorded immediately.
(D) Record the time immediately before the appearance of blue colour: Incorrect. The time should be recorded at the appearance of the blue color, not before.
(E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution: Incorrect. The concentration of sodium thiosulphate should be less than that of KI.

Why other options are wrong:
- Option B: (A), (D), (E) only is incorrect because statements (D) and (E) are incorrect.
- Option C: (D), (E) only is incorrect because statements (D) and (E) are incorrect.
- Option D: (A), (B), (E) only is incorrect because statement (E) is incorrect.

Key Concepts:
- Chemical kinetics
- Iodometric titrations
- Experimental techniques

Memory Aids:
- Remember the role of each reagent in the iodometric titration.

Quick Tips:
- Understand the purpose of each step in the experimental procedure.

58 Hard

58 of 79Question permalink

Application +4 | -1

Calculate the overall complex dissociation equilibrium constant for the

\mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}

ions, given that

\beta_4

for this complex is

2.1 \times 10^{13}

View answer and explanation

Correct option: B

4.7 \times 10^{-14}

Option B (

4.7 \times 10^{-14}

) is correct because the overall dissociation constant is the inverse of the overall stability constant (

\beta_4

).

Solution approach

Correct Answer + Core Reason: $4.7 \times 10^{-14}$ . The dissociation constant is the inverse of the formation constant.

Step-by-Step Solution:

Step 1: Understand the relationship between the formation constant ( $\beta_4$ ) and the dissociation constant (K_d).\nThe overall stability constant ( $\beta_4$ ) is given as $2.1 \times 10^{13}$ .\nThe overall complex dissociation equilibrium constant (K_d) is the inverse of the overall stability constant ( $\beta_4$ ).

Step 2: Calculate the dissociation constant (K_d).
$K_d = \frac{1}{\beta_4}$
$K_d = \frac{1}{2.1 \times 10^{13}} = 4.76 \times 10^{-14} \approx 4.7 \times 10^{-14}$

Why other options are wrong:
- Option A: $8.27 \times 10^{-13}$ is not the inverse of the given stability constant.
- Option C: $2.39 \times 10^{-7}$ is not the inverse of the given stability constant.
- Option D: $1.83 \times 10^{14}$ is not the inverse of the given stability constant; it is close to the original formation constant.

Key Concepts:
- Stability constant ( $\beta$ )
- Dissociation constant (K_d)
- Complex formation

Memory Aids:
- Dissociation constant is the inverse of the formation constant.

Quick Tips:
- Always remember the relationship: $K_d = \frac{1}{\beta}$

59 Hard

59 of 79Question permalink

Analyzing +4 | -1

Each of the compounds

{Pt}\left({NH}_3\right)_6 {Cl}_4 \cdot {Cr}\left({NH}_3\right)_6 {Cl}_3, {Co}\left({NH}_3\right)_4 {Cl}_3

and

{K}_2 {Pt} . {Cl}_6

were dissolved in water to make its

0.001 \mathrm{~M}

solution. The correct order of their increasing conductivity in solution is

View answer and explanation

Correct option: C

Co(NH_3)_4 Cl_3 < K_2Pt.Cl_6 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4

The correct answer is C (

Co(NH_3)_4 Cl_3 < K_2Pt.Cl_6 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4

) because the conductivity depends on the number of ions produced in the solution.

Solution approach

Correct Answer + Core Reason: $Co(NH_3)_4 Cl_3 < K_2Pt.Cl_6 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4$ . The conductivity depends on the number of ions produced in the solution.

Step-by-Step Solution:
Step 1: Determine the number of ions produced by each complex in solution.
Step 2: Relate the number of ions to the conductivity of the solution. Higher number of ions means higher conductivity.

Step 3: Analyze $Pt(NH_3)_6 Cl_4$ : This complex dissociates into $[Pt(NH_3)_6]^{4+} + 4Cl^-$ , producing 5 ions.
Step 4: Analyze $Cr(NH_3)_6 Cl_3$ : This complex dissociates into $[Cr(NH_3)_6]^{3+} + 3Cl^-$ , producing 4 ions.
Step 5: Analyze $K_2PtCl_6$ : This complex dissociates into $2K^+ + [PtCl_6]^{2-}$ , producing 3 ions.
Step 6: Analyze $Co(NH_3)_4 Cl_3$ : This complex is actually $[Co(NH_3)_4Cl_2]Cl$ , which dissociates into $[Co(NH_3)_4Cl_2]^+ + Cl^-$ , producing 2 ions.

Step 7: Arrange the complexes in increasing order of the number of ions produced: $Co(NH_3)_4 Cl_3 < K_2PtCl_6 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4$ .

Why other options are wrong:
- Option A: $K_2Pt.Cl_6 < Co(NH_3)_4.Cl_3 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4$ - This is incorrect because it places the complex with 2 ions after the complex with 3 ions.
- Option B: $Cr(NH_3)_6 Cl_3 < Co(NH_3)_4 Cl_3 < K_2Pt.Cl_6 < Pt(NH_3)_6 Cl_4$ - This is incorrect because it places the complex with 4 ions before the complexes with 2 and 3 ions.
- Option D: $Pt(NH_3)_6 Cl_4 < Co(NH_3)_4 Cl_3 < Cr(NH_3)_6 Cl_3 < K_2Pt.Cl_6$ - This is incorrect because it places the complex with 5 ions before the complexes with 2, 3 and 4 ions.

Key Concepts:
- Coordination Complexes
- Ionization in Solution
- Conductivity of Electrolytes

Memory Aids:
- The more ions produced, the higher the conductivity.

Quick Tips:
- Identify the counter ions to determine the number of ions produced in solution.
- Remember that the complex ion remains intact.

60 Hard

60 of 79Question permalink

Understanding +4 | -1

Assertion (A): Toxic metal ions are removed by the chelating ligands.

Reason (R): Chelate complexes tend to be more stable.

View answer and explanation

Correct option: A

Assertion and Reason both are true, Reason is correct explanation of Assertion.

Option A (Assertion and Reason both are true, Reason is correct explanation of Assertion) is correct because chelating ligands form more stable complexes (chelates) with metal ions, effectively removing toxic metal ions from the system.

Solution approach

Correct Answer + Core Reason: Assertion and Reason both are true, Reason is correct explanation of Assertion. Chelation increases the stability of the complex, leading to effective removal of toxic metal ions.

Step-by-Step Solution:

Step 1: Understand the concept of chelation: Chelation is the formation of a complex with a metal ion and a ligand that has multiple binding sites.

Step 2: Understand the stability of chelate complexes: Chelate complexes are more stable than similar complexes with monodentate ligands due to the chelate effect.

Step 3: Relate chelation to the removal of toxic metal ions: The increased stability of chelate complexes makes them effective in removing toxic metal ions from biological systems or industrial waste.

Why other options are wrong:
- Option B: Assertion and Reason both are true but Reason is not the correct explanation of Assertion: The reason directly explains the assertion.
- Option C: Assertion is true, Reason is false: Both assertion and reason are true.
- Option D: Assertion is false, Reason is true: The assertion is also true.

Key Concepts:
- Chelation
- Chelate effect
- Stability of complexes

Memory Aids:
- Chelation = increased stability = effective removal.

Quick Tips:
- Remember that chelating ligands form more stable complexes.

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