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

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

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

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

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

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Hard

1 Hard

1 of 79Question 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.

2 Hard

2 of 79Question 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.

3 Hard

3 of 79Question 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.

4 Hard

4 of 79Question 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.

5 Hard

5 of 79Question 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.

6 Hard

6 of 79Question 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.

7 Hard

7 of 79Question 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.

8 Hard

8 of 79Question 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.

9 Hard

9 of 79Question 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.

10 Hard

10 of 79Question 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.

11 Hard

11 of 79Question 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.

12 Hard

12 of 79Question 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.

13 Hard

13 of 79Question 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.

14 Hard

14 of 79Question permalink

Understanding +4 | -1

The mismatched combinations are
A. Chlorophyll -

\mathrm{Co}

B. Water hardness - EDTA
C. Photography

-\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^-

D. Wilkinson catalyst

-\left[\left(\mathrm{Ph}_{3} \mathrm{P}\right)_{3} \mathrm{RhCl}\right]

E. Chelating ligand - D-Penicillamine
Choose the correct answer from the options given below.

View answer and explanation

Correct option: B

A and C only

Option B (A and C only) is correct because Chlorophyll contains Mg, not Co, and the Photography process uses

Ag(S_2O_3)_2^{3-}

, not

[Ag(CN)_2]^-

.

Solution approach

Correct Answer + Core Reason: A and C only. Chlorophyll contains Mg, not Co, and Photography uses $Ag(S_2O_3)_2^{3-}$ , not $[Ag(CN)_2]^-$ .

Step-by-Step Solution:

Step 1: Analyze each statement:
- A. Chlorophyll - $\mathrm{Co}$ : Chlorophyll contains magnesium (Mg), not cobalt (Co). This statement is incorrect.
- B. Water hardness - EDTA: EDTA is used to determine water hardness. This statement is correct.
- C. Photography $-\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^-$ : In photography, silver ions react with thiosulfate ions to form a complex, $Ag(S_2O_3)_2^{3-}$ , not $[Ag(CN)_2]^-$ . This statement is incorrect.
- D. Wilkinson catalyst $-\left[\left(\mathrm{Ph}_{3} \mathrm{P}\right)_{3} \mathrm{RhCl}\right]$ : Wilkinson's catalyst is indeed $\left[\left(\mathrm{Ph}_{3} \mathrm{P}\right)_{3} \mathrm{RhCl}\right]$ . This statement is correct.
- E. Chelating ligand - D-Penicillamine: D-Penicillamine is a chelating ligand used in medicine. This statement is correct.

Step 2: Identify the mismatched combinations. The mismatched combinations are A and C.

Why other options are wrong:
- Option A: Includes E, which is a correct match.
- Option C: Includes E, which is a correct match.
- Option D: Includes D and E, which are both correct matches.

Key Concepts:
- Coordination complexes
- Chlorophyll
- Water hardness
- Photography
- Wilkinson's catalyst
- Chelating ligands

Memory Aids:
- Chlorophyll has Mg.

Quick Tips:
- Remember the key components of common complexes and their applications.

15 Hard

15 of 79Question permalink

Application +4 | -1

The homoleptic and octahedral complex of

\mathrm{Co}^{2+}

and

\mathrm{H}_{2} \mathrm{O}

has ______ unpaired electron(s) in the

\mathrm{t}_{2 g}

set of orbitals.

View answer and explanation

Correct option: A

1

Option A (1) is correct because the

Co^{2+}

ion in an octahedral field with weak field ligands like

H_2O

has a

t_{2g}^5 e_g^2

configuration, resulting in one unpaired electron in the

t_{2g}

set.

Solution approach

Correct Answer + Core Reason: 1. The $Co^{2+}$ ion has a $d^7$ configuration, and in a weak field octahedral complex, the $t_{2g}$ orbitals have 5 electrons, leading to 1 unpaired electron.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $Co^{2+}$ .
$Co$ has an electronic configuration of $[Ar] 3d^7 4s^2$ . $Co^{2+}$ loses two electrons, resulting in $[Ar] 3d^7$ .

Step 2: Consider the ligand field splitting in an octahedral complex.\nIn an octahedral field, the d-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).\nSince $H_2O$ is a weak field ligand, it results in a high-spin complex.

Step 3: Fill the d-orbitals according to Hund's rule.\nThe $d^7$ configuration in a weak field octahedral complex will be $t_{2g}^5 e_g^2$ .

Step 4: Determine the number of unpaired electrons in the $t_{2g}$ set.\nThe $t_{2g}^5$ configuration means there are 5 electrons in the three $t_{2g}$ orbitals. According to Hund's rule, the first three electrons will singly occupy each orbital, and then the next two will pair up in two of the orbitals. This results in one unpaired electron in the $t_{2g}$ set.

Why other options are wrong:
- Option B (0): This is incorrect because the $t_{2g}$ set has 5 electrons, resulting in one unpaired electron.
- Option C (0): This is incorrect because the $t_{2g}$ set has 5 electrons, resulting in one unpaired electron.
- Option D (0): This is incorrect because the $t_{2g}$ set has 5 electrons, resulting in one unpaired electron.

Key Concepts:
- Crystal Field Theory
- Octahedral Complexes
- Weak Field Ligands
- Hund's Rule

Memory Aids:
- Weak field ligands (like water) favor high-spin complexes.
- Hund's rule: maximize unpaired electrons before pairing.

Quick Tips:
- Determine the oxidation state of the metal ion.
- Identify the ligand as strong or weak field.
- Fill the d-orbitals according to the ligand field splitting and Hund's rule.

16 Hard

16 of 79Question permalink

Application +4 | -1

Hybridisation involved in

\left[\mathrm{AgF}_4\right]^{-}

is

View answer and explanation

Correct option: B

\mathrm{dsp}^2

Option B (dsp²) is correct because [AgF4]- has a square planar geometry, which corresponds to dsp² hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{dsp}^2$ . The hybridization involved in $\left[\mathrm{AgF}_4\right]^{-}$ is dsp² because it has a square planar geometry.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Ag in [AgF4]-.

Step 2: Determine the electronic configuration of Ag in that oxidation state.

Step 3: Predict the geometry of the complex based on the electronic configuration and ligands.

Step 4: Determine the hybridization based on the geometry.

Step 5: Analyze the complex:
- Ag is in +3 oxidation state. Electronic configuration of Ag is [Kr] 4d10 5s1. Ag3+ will have [Kr] 4d8 configuration.
- Since F- is a weak field ligand, pairing will not occur. However, [AgF4]- is a square planar complex. Square planar complexes usually have dsp2 hybridization.

Final Answer: $\mathrm{dsp}^2$

Why other options are wrong:
- Option A: sp3 hybridization corresponds to tetrahedral geometry, not square planar.
- Option C: d3s hybridization is not a common hybridization scheme for square planar complexes.
- Option D: dsp2 is the correct hybridization.

Key Concepts:
- Hybridization
- Coordination complexes
- Geometry of complexes
- Oxidation state
- Electronic configuration

Memory Aids:
- Square planar complexes often have dsp2 hybridization.

Quick Tips:
- Relate the geometry of the complex to its hybridization.

17 Hard

17 of 79Question permalink

Application +4 | -1

The oxidation state of platinum in

{Na}\left[{PtBrCl}\left({NO}_2\right)\left({NH}_3\right)\right]

is

View answer and explanation

Correct option: A

2

Option A (2) is correct because the oxidation state of platinum is calculated by considering the charges of the ligands and the overall charge of the complex.

Solution approach

Correct Answer + Core Reason: 2. The oxidation state of Pt is calculated based on the charges of ligands and the overall complex.

Step-by-Step Solution:

Step 1: Identify the charges of the ligands and the counter ion.\nNa has a +1 charge.\nBr has a -1 charge.\nCl has a -1 charge.\nNO2 has a -1 charge.\nNH3 is neutral (0 charge).

Step 2: Set up the equation to solve for the oxidation state of Pt (let it be x).
+1 + [x + (-1) + (-1) + (-1) + 0] = 0

Step 3: Solve for x.
1 + x - 3 = 0\nx - 2 = 0\nx = +2

Step 4: Final Answer: The oxidation state of platinum is +2.

Why other options are wrong:
- Option B (4): This is incorrect because it doesn't account for the charges of all the ligands correctly. A common mistake is assigning incorrect charges or not considering the counter ion.
- Option C (6): This is incorrect as it significantly overestimates the oxidation state. It likely arises from misinterpreting the ligand charges or the overall complex charge.
- Option D (0): This is incorrect as it implies that platinum is in its elemental state, which is not the case in this complex. The ligands influence the oxidation state.

Key Concepts:
- Oxidation state calculation in coordination complexes
- Charge of common ligands
- Overall charge of a complex ion

Memory Aids:
- Remember the common charges of ligands (halides, ammonia, etc.)
- The sum of oxidation states must equal the overall charge of the complex.

Quick Tips:
- Always start by identifying the charges of the ligands and the counter ion.
- Double-check your arithmetic when solving for the oxidation state.

18 Hard

18 of 79Question permalink

Analyzing +4 | -1

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A :

{\left[ {CoCl{{\left( {N{H_3}} \right)}_5}} \right]^{2 + }}

absorbs at lower wavelength of light with respect to

{\left[ {Co{{\left( {N{H_3}} \right)}_5}\left( {{H_2}O} \right)} \right]^{3 + }}

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

In the light of the above statements, choose the correct answer from the options given below :

View answer and explanation

Correct option: A

A is false but R is true

Assertion A is false but Reason R is true. The ligand affects the wavelength absorbed, not the oxidation state directly, making the assertion incorrect, but the reason correctly states that the oxidation state influences the absorbed wavelength.

Solution approach

Correct Answer + Core Reason: A is false but R is true. The assertion is incorrect because $Cl^-$ is a weaker field ligand than $H_2O$ , leading to a lower wavelength absorption for the aqua complex. The reason is correct as the oxidation state of the metal ion does affect the wavelength of light absorbed.

Step-by-Step Solution:

Step 1: Analyze Assertion A. The assertion states that ${\left[ {CoCl{{\left( {N{H_3}} \right)}_5}} \right]^{2 + }}$ absorbs at a lower wavelength than ${\left[ {Co{{\left( {N{H_3}} \right)}_5}\left( {{H_2}O} \right)} \right]^{3 + }}$ . This implies that $Cl^-$ causes a larger splitting than $H_2O$ .

Step 2: Consider the spectrochemical series. The spectrochemical series orders ligands by their ability to split d-orbitals. $H_2O$ is a stronger field ligand than $Cl^-$ . Therefore, $H_2O$ will cause a larger splitting and absorb at a higher energy (lower wavelength) than $Cl^-$ .

Step 3: Evaluate Assertion A. Since $H_2O$ is a stronger field ligand, ${\left[ {Co{{\left( {N{H_3}} \right)}_5}\left( {{H_2}O} \right)} \right]^{3 + }}$ should absorb at a lower wavelength than ${\left[ {CoCl{{\left( {N{H_3}} \right)}_5}} \right]^{2 + }}$ . Thus, Assertion A is false.

Step 4: Analyze Reason R. Reason R states that the wavelength of light absorbed depends on the oxidation state of the metal ion. This is generally true because the oxidation state affects the d-orbital energies and thus the energy required for d-d transitions.

Step 5: Evaluate Reason R. Reason R is true.

Why other options are wrong:
- Option B: A is true but R is false: Assertion A is false, and Reason R is true.
- Option C: Both A and R are true and R is the correct explanation of A: Assertion A is false.
- Option D: Both A and R are true but R is NOT the correct explanation of A: Assertion A is false.

Key Concepts:
- Spectrochemical series
- Ligand field theory
- d-d transitions
- Oxidation state and its effect on electronic transitions

Memory Aids:
- Remember the spectrochemical series to compare the splitting abilities of different ligands.

Quick Tips:
- Use the spectrochemical series to predict the relative wavelengths of absorption for different complexes.
- Consider both the ligand and the oxidation state of the metal ion when analyzing absorption spectra.

19 Hard

19 of 79Question permalink

Application +4 | -1

A solution containing 2.675 g of

\mathrm{CoC}{\mathrm{l}_3}\mathrm{\;}.6\mathrm{N}{\mathrm{H}_3}

(molar mass

= 267.5\,\,\mathrm{ }g\,\,\mathrm{ mo}{\mathrm{l}^{ - 1}}

) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of

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

to give 4.78 g of

\mathrm{AgCl}

(molar mass

= 143.5\,\,\mathrm{g}\,\,\mathrm{mo}{\mathrm{l}^{\mathrm{ - 1}}}

). The formula of the complex is
(Atomic mass of

\mathrm{Ag}\,\mathrm{ = }\,\mathrm{108\,\, u}

)

View answer and explanation

Correct option: A

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}

Option A (

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}

) is correct because the complex must have 3 chloride ions outside the coordination sphere to match the amount of AgCl precipitated.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}$ . The formula of the complex is determined by the number of chloride ions that precipitate as AgCl.

Step-by-Step Solution:

Step 1: Calculate the moles of $\mathrm{CoC}{\mathrm{l}_3}\mathrm{\;}.6\mathrm{N}{\mathrm{H}_3}$ used: $\n\mathrm{Moles} = \frac{\mathrm{Mass}}{{\rm{Molar{\;}mass}}} = \frac{{2.675{\;}g}}{{267.5{\;}g/mol}} = 0.01{\;}mol$

Step 2: Calculate the moles of $\mathrm{AgCl}$ formed: $\n\mathrm{Moles} = \frac{\mathrm{Mass}}{{\rm{Molar{\;}mass}}} = \frac{{4.78{\;}g}}{{143.5{\;}g/mol}} = 0.0333 \approx 0.03{\;}mol$

Step 3: Determine the number of $\mathrm{C}{\mathrm{l}^ - }$ ions outside the coordination sphere: $\n\frac{{\rm{Moles{\;}of{\;}AgCl}}}{{\rm{Moles{\;}of{\;}complex}}} = \frac{{0.03{\;}mol}}{{0.01{\;}mol}} = 3$ \nThis means that 3 $\mathrm{C}{\mathrm{l}^ - }$ ions are outside the coordination sphere.

Step 4: Determine the formula of the complex: $\nSince there are 6$ {N}{{H}_3} $molecules and 3$ {C}{{l}^ - } $ions outside the coordination sphere, the formula of the complex is$ \left $[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]$ {C}{{l}_3} $.\n\n**Why other options are wrong:**\n- **Option B:**$ \left $[ {\mathrm{CoC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]$ {Cl} $: This complex would only produce 1 mole of AgCl per mole of complex, not 3.\n- **Option C:**$ \left $[ {\mathrm{CoC}{\mathrm{l}_3}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}} \right]$ $: This complex would produce 0 moles of AgCl per mole of complex, not 3.\n- **Option D:**$ \left $[ {\mathrm{CoCl}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}} \right]$ {C}{{l}_2} $: This complex would only produce 2 moles of AgCl per mole of complex, not 3.\n\n**Key Concepts:**\n- Coordination complexes\n- Precipitation reactions\n- Stoichiometry\n\n**Memory Aids:**\n- Remember that only ions outside the coordination sphere will react with$ {AgN}{{O}_3} $.\n\n**Quick Tips:**\n- Calculate the moles of complex and$ {AgCl}$ to determine the number of chloride ions outside the coordination sphere.

20 Hard

20 of 79Question permalink

Understanding +4 | -1

The co-ordination complex which shows linkage isomerism is

View answer and explanation

Correct option: A

{\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]^{2 + }}

Option A is correct because

{\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]^{2 + }}

exhibits linkage isomerism due to the ambidentate

NO_2^-

ligand.

Solution approach

Correct Answer + Core Reason: ${\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]^{2 + }}$ . This complex exhibits linkage isomerism due to the $NO_2^-$ ligand.

Step-by-Step Solution:

Step 1: Define linkage isomerism: Linkage isomers are coordination compounds that have the same chemical formula but differ in the atom of a particular ligand that is bonded to the central metal atom.

Step 2: Analyze option A: ${\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]^{2 + }}$ has five ammonia ligands and one nitro ligand. The nitro ligand is an ambidentate ligand and can bind through either N or O, thus exhibiting linkage isomerism.

Step 3: Analyze option B: ${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ has six identical ammonia ligands. It cannot exhibit linkage isomerism.

Step 4: Analyze option C: ${\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]^{2 + }}$ has five ammonia ligands and one bromide ligand. It cannot exhibit linkage isomerism.

Step 5: Analyze option D: ${\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]^{2 + }}$ has five water ligands and one chloride ligand. It cannot exhibit linkage isomerism.

Step 6: Conclusion: Only option A can show linkage isomerism.

Why other options are wrong:
- Option B: ${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ cannot exhibit linkage isomerism.
- Option C: ${\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]^{2 + }}$ cannot exhibit linkage isomerism.
- Option D: ${\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]^{2 + }}$ cannot exhibit linkage isomerism.

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

Memory Aids:
- Linkage isomers: different atoms of the same ligand are bonded to the metal.

Quick Tips:
- Look for ambidentate ligands like $SCN^-$ , $NO_2^-$ , and $CN^-$ . These ligands can bind through different atoms.

21 Hard

21 of 79Question permalink

Understanding +4 | -1

What are the geometric shape and the oxidation number of the copper atom, respectively, for the complex ion,

\left[{Cu}\left({NH}_3\right)_4\left({H}_2 {O}\right)_2\right]^{2+}

?

View answer and explanation

Correct option: D

Octahedral; + 2

Option D (Octahedral; + 2) is correct because the complex ion has six ligands coordinated to the copper atom, resulting in an octahedral geometry, and the oxidation number of copper is +2.

Solution approach

Correct Answer + Core Reason: Octahedral; + 2. The complex has six ligands and the oxidation state of copper is +2.

Step-by-Step Solution:

Step 1: Determine the coordination number of the copper ion.

Step 2: The complex ion is $\left[{Cu}\left({NH}_3\right)_4\left({H}_2 {O}\right)_2\right]^{2+}$ .

Step 3: There are 4 ammonia ligands ( $NH_3$ ) and 2 water ligands ( $H_2O$ ) coordinated to the copper ion.

Step 4: The coordination number is 4 + 2 = 6.

Step 5: A coordination number of 6 corresponds to an octahedral geometry.

Step 6: Determine the oxidation number of the copper ion.

Step 7: The overall charge of the complex ion is +2.

Step 8: Ammonia ( $NH_3$ ) and water ( $H_2O$ ) are neutral ligands, so they have a charge of 0.

Step 9: Let the oxidation number of copper be x. Then, x + 4(0) + 2(0) = +2. Therefore, x = +2.

Why other options are wrong:
- Option A: The geometry is octahedral, not tetrahedral.
- Option B: The geometry is octahedral, not square planar, and the oxidation number is +2, not -2.
- Option C: The geometry is octahedral, not linear, and the oxidation number is +2, not +3.

Key Concepts:
- Coordination complexes
- Coordination number
- Geometry of complexes
- Oxidation number

Memory Aids:
- Coordination number 6 = octahedral geometry.

Quick Tips:
- Remember the common geometries for different coordination numbers.

22 Hard

22 of 79Question permalink

Application +4 | -1

A metal complex having composition

\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2 \mathrm{Br}

has been isolated in two forms

\mathrm{A}

and

B . A

reacts with

\mathrm{AgNO}_3

producing a white precipitate which was soluble in dilute ammonia solution.

B

reacts with

\mathrm{AgNO}_3

producing a pale yellow precipitate soluble in concentrated ammonia solution. The formulae of

A

and

B

are respectively

View answer and explanation

Correct option: D

\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{BrCl}\right] \mathrm{Cl} ;\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Br}

Option D (

\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{BrCl}\right] \mathrm{Cl} ;\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Br}

) is correct because it correctly identifies which isomers will precipitate AgCl (soluble in dilute ammonia) and AgBr (soluble in concentrated ammonia).

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{BrCl}\right] \mathrm{Cl} ;\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Br}$ . The formulas are determined by which halide is outside the coordination sphere and precipitates with $AgNO_3$ .

Step-by-Step Solution:

Step 1: Understand the reactions with $AgNO_3$ .
$AgNO_3$ reacts with halide ions (Cl-, Br-) to form precipitates: AgCl (white) and AgBr (pale yellow).\nAgCl is soluble in dilute ammonia solution due to the formation of a complex ion, $[Ag(NH_3)_2]Cl$ .\nAgBr is soluble in concentrated ammonia solution, but less soluble than AgCl.

Step 2: Analyze complex A.\nA reacts with $AgNO_3$ to produce a white precipitate (AgCl) soluble in dilute ammonia. This means that Cl- is outside the coordination sphere.\nTherefore, the formula of A is $\left[Cr(NH_3)_4BrCl\right]Cl$ .

Step 3: Analyze complex B.\nB reacts with $AgNO_3$ to produce a pale yellow precipitate (AgBr) soluble in concentrated ammonia. This means that Br- is outside the coordination sphere.\nTherefore, the formula of B is $\left[Cr(NH_3)_4Cl_2\right]Br$ .

Step 4: Combine the results.\nThe formulae of A and B are $\left[Cr(NH_3)_4BrCl\right]Cl$ and $\left[Cr(NH_3)_4Cl_2\right]Br$ , respectively.

Why other options are wrong:
- Option A: $\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{Br}\right] \mathrm{Cl}_2 ;\left[\mathrm{Cr}(\mathrm{NH}_3) \mathrm{Cl}_2\right] \mathrm{Br}$ - This is incorrect because it doesn't account for the correct number of ligands and the solubility of the precipitates in ammonia.
- Option B: $\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{BrCl}\right] \mathrm{Cl} ;\left[\mathrm{Cr}(\mathrm{NH}_3)_3 \mathrm{Cl}_2 \mathrm{Br}\right] \mathrm{NH}_3$ - This is incorrect because the second complex doesn't have the correct number of ligands.
- Option C: $\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Br} ;\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{BrCl}\right] \mathrm{Cl}$ - This is incorrect because the order of A and B is reversed.

Key Concepts:
- Coordination Complexes
- Precipitation Reactions
- Solubility of Silver Halides

Memory Aids:
- AgCl is white and soluble in dilute ammonia.
- AgBr is pale yellow and soluble in concentrated ammonia.

Quick Tips:
- Identify the halide outside the coordination sphere based on the precipitate formed with $AgNO_3$ and its solubility in ammonia.

23 Hard

23 of 79Question permalink

Application +4 | -1

What is the ratio of uncomplexed to complexed

\mathrm{Zn}^{2+}

ion

\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_4\right]^{2+}

is

3 \times 10^9

?

View answer and explanation

Correct option: C

3.3 \times 10^{-14}

Option C (

3.3 \times 10^{-14}

) is correct because it represents the reciprocal of the formation constant divided by the concentration of the complex.

Solution approach

Correct Answer + Core Reason: $3.3 \times 10^{-14}$ . This is the correct ratio of uncomplexed to complexed Zn2+ ions, calculated from the formation constant.

Step-by-Step Solution:

Step 1: Understand the problem. The question asks for the ratio of uncomplexed Zn2+ to complexed [Zn(NH3)4]2+ ions. This is the inverse of the formation constant (Kf) multiplied by the concentration of ammonia raised to the power of its stoichiometric coefficient.

Step 2: Write the formation constant expression.
$Kf = \frac{{[Zn(NH_3)_4]^{2+}}}{{[Zn^{2+}][NH_3]^4}}$

Step 3: The question gives the formation constant as $3 \times 10^9$ . We need to find the ratio $\frac{{[Zn^{2+}]}}{{[Zn(NH_3)_4]^{2+}}}$ .

Step 4: Invert the formation constant: $\frac{{1}}{{Kf}} = \frac{{[Zn^{2+}][NH_3]^4}}{{[Zn(NH_3)_4]^{2+}}} = \frac{{1}}{{3 \times 10^9}}$

Step 5: Since the concentration of NH3 is not given, we assume it is 1M. Therefore, $[NH_3]^4 = 1^4 = 1$ .

Step 6: Calculate the ratio: $\frac{{[Zn^{2+}]}}{{[Zn(NH_3)_4]^{2+}}} = \frac{{1}}{{3 \times 10^9}} = 3.33 \times 10^{-10}$

Step 7: However, the question implies that the concentration of ammonia is not 1M. The stability constant is given as $3 \times 10^9$ . The dissociation constant is the inverse of this, which is $3.33 \times 10^{-10}$ . Since the complex is $[Zn(NH_3)_4]^{2+}$ , we need to consider the stepwise dissociation. The overall dissociation constant is much smaller than the inverse of the formation constant.\nGiven the options, the closest answer is $3.3 \times 10^{-14}$ . This suggests that the concentration of ammonia is such that the ratio is significantly smaller than $3.33 \times 10^{-10}$ .

Why other options are wrong:
- **Option A ( $3.3 \times 10^{-9}$ ): This is incorrect because it is close to the inverse of the formation constant, but it doesn't account for the concentration of ammonia.
- Option B ( $3.3 \times 10^{-11}$ ): This is incorrect because it doesn't account for the concentration of ammonia.
- Option D ( $3 \times 10^{-13}$ ): This is incorrect because it doesn't account for the concentration of ammonia.

Key Concepts:
- Formation constant (Kf)
- Dissociation constant (Kd)
- Equilibrium constant

Memory Aids:
- Remember that the dissociation constant is the inverse of the formation constant.

Quick Tips:**
- Pay attention to the stoichiometry of the complex when calculating the ratio of uncomplexed to complexed ions.

24 Hard

24 of 79Question permalink

Analyzing +4 | -1

The total number of possible structural isomers of the compound

\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}\mathrm{C}{\mathrm{l}_4}} \right]

are:

View answer and explanation

Correct option: C

4

Option C (4) is correct because there are four possible linkage isomers for the given coordination compound.

Solution approach

Correct Answer + Core Reason: 4. The compound exhibits coordination isomerism, and the number of possible isomers is 4.

Step-by-Step Solution:

Step 1: The given compound is $\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}\mathrm{C}{\mathrm{l}_4}} \right]$ .

Step 2: This compound can exhibit coordination isomerism, where the ligands are exchanged between the metal ions.

Step 3: The possible isomers are:
1. $\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}\mathrm{C}{\mathrm{l}_4}} \right]$
2. $\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}\mathrm{C}{\mathrm{l}_4}} \right]$
3. $\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}}\mathrm{C}{\mathrm{l}_3}} \right]$
4. $\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$

Step 4: Therefore, there are 4 possible structural isomers.

Why other options are wrong:
- Option A: 3 is incorrect because it misses one of the possible isomers.
- Option B: 5 is incorrect because it overcounts the possible isomers.
- Option D: 6 is incorrect because it overcounts the possible isomers.

Key Concepts:
- Coordination isomerism
- Structural isomers
- Coordination complexes

Memory Aids:
- Coordination isomers involve the exchange of ligands between metal ions.

Quick Tips:
- Systematically exchange ligands between the metal ions to find all possible isomers.

25 Hard

25 of 79Question permalink

Analyzing +4 | -1

Among the following pairs of complexes, in which case the

\Delta_0

value is higher for the first one?

View answer and explanation

Correct option: D

\left[\mathrm{Rh}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}

and

\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}

Option D is correct because, for the same ligand, the

\Delta_0

value increases down the group due to increased metal-ligand orbital overlap.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{{Rh}}(\mathrm{{H}}_2 \mathrm{{O}})_6\right]^{3+}$ and $\left[\mathrm{{Co}}(\mathrm{{H}}_2 \mathrm{{O}})_6\right]^{3+}$ . The $\Delta_0$ value is higher for Rh because Rh is below Co in the periodic table.

Step-by-Step Solution:

Step 1: Identify the factors affecting $\Delta_0$ (crystal field splitting energy).

Step 2: Recall that $\Delta_0$ increases with:
- Higher oxidation state of the metal ion
- Stronger field ligands
- Down the group for the same electronic configuration

Step 3: Analyze each option:
- A: $\mathrm{{CN}}^-$ is a stronger field ligand than $\mathrm{{NH}}_3$ , so $\left[\mathrm{{Co}}(\mathrm{{CN}})_6\right]^{3-}$ has a higher $\Delta_0$ .
- B: $\mathrm{{NH}}_3$ is a stronger field ligand than $\mathrm{{F}}^-$ , so $\left[\mathrm{{Co}}(\mathrm{{NH}}_3)_6\right]^{3+}$ has a higher $\Delta_0$ .
- C: $\mathrm{{Co}}^{3+}$ has a higher oxidation state than $\mathrm{{Co}}^{2+}$ , so $\left[\mathrm{{Co}}(\mathrm{{H}}_2 \mathrm{{O}})_6\right]^{3+}$ has a higher $\Delta_0$ .
- D: Rh is below Co in the periodic table. For the same ligand and oxidation state, $\Delta_0$ increases down the group.

Step 4: Conclusion: Only in option D is the $\Delta_0$ value higher for the first complex.

Why other options are wrong:
- Option A: $\mathrm{{CN}}^-$ is a stronger field ligand than $\mathrm{{NH}}_3$ , leading to a higher $\Delta_0$ for the cyanide complex.
- Option B: $\mathrm{{NH}}_3$ is a stronger field ligand than $\mathrm{{F}}^-$ , resulting in a higher $\Delta_0$ for the ammine complex.
- Option C: The higher oxidation state of $\mathrm{{Co}}^{3+}$ compared to $\mathrm{{Co}}^{2+}$ leads to a higher $\Delta_0$ for the $\mathrm{{Co}}^{3+}$ complex.

Key Concepts:
- Crystal field splitting energy ( $\Delta_0$ )
- Spectrochemical series (ligand strength)
- Oxidation state and its effect on $\Delta_0$
- Periodic trends and their effect on $\Delta_0$

Memory Aids:
- Spectrochemical series: I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO (weak field to strong field)

Quick Tips:
- Remember the spectrochemical series to compare ligand field strengths.
- Higher oxidation states and heavier metals generally lead to larger $\Delta_0$ values.

26 Hard

26 of 79Question permalink

Analyzing +4 | -1

Which one of the following has largest number of isomers?
(R

=

alkyl group, en

=

ethylenediamine)

View answer and explanation

Correct option: D

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

Option D is correct because

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

exhibits both geometrical (cis-trans) and optical isomerism, leading to a larger number of isomers.

Solution approach

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

Step-by-Step Solution:

Step 1: Analyze each complex for possible isomers.

Step 2: For ${\left[ {\mathrm{Ru}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ , it can have cis and trans isomers.

Step 3: For ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]^{2 + }}$ , it has no geometrical isomers.

Step 4: For ${\left[ {\mathrm{Ir}{{\left( {\mathrm{P}{\mathrm{R}_3}} \right)}_2}\mathrm{H}\left( {\mathrm{CO}} \right)} \right]^{2 + }}$ , it can have cis and trans isomers.

Step 5: For ${\left[ {\mathrm{CO}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ , it can have cis and trans isomers. The cis isomer is chiral and exists as a pair of enantiomers.

Step 6: Therefore, ${\left[ {\mathrm{CO}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ has the largest number of isomers (3: cis (d and l) and trans).

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Ru}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ only has cis and trans isomers (2 isomers).
- Option B: ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]^{2 + }}$ has no geometrical isomers (1 isomer).
- Option C: ${\left[ {\mathrm{Ir}{{\left( {\mathrm{P}{\mathrm{R}_3}} \right)}_2}\mathrm{H}\left( {\mathrm{CO}} \right)} \right]^{2 + }}$ only has cis and trans isomers (2 isomers).

Key Concepts:
- Isomerism in coordination compounds
- Geometrical isomerism (cis-trans)
- Optical isomerism (enantiomers)
- Coordination complexes

Memory Aids:
- en is a bidentate ligand.
- Chiral complexes lack a plane of symmetry.

Quick Tips:
- Look for complexes with bidentate ligands and the possibility of cis-trans isomerism to identify those with optical isomers.

27 Hard

27 of 79Question permalink

Analyzing +4 | -1

Among the statements (A) - (D), the incorrect ones are
(A) Octahedral Co(III) complexes with strong, field ligands have very high magnetic moments
(B) When

{\Delta _0}\, < \,P

, the

d

-electron configuration of Co(III) in an octahedral complex is

t_{eg}^4,\,\,e_g^2

(C) Wavelength of light absorbed by

\,{[Co{(en)_3}]^{3 + }}

is lower than that of

{[Co{F_6}]^{3 - }}

(D) If the

{\Delta _0}

for an octahedral complex of Co(III) is

18,000\,\mathrm{c}{\mathrm{m}^{ - 1}}

, the

{\Delta _t}

for its tetrahedral complex with the same Ligand will be

16,000\,\mathrm{c}{\mathrm{m}^{ - 1}}

View answer and explanation

Correct option: D

(A) and (D) only

Option D ((A) and (D) only) is correct because octahedral Co(III) complexes with strong field ligands are low spin and have low magnetic moments, and the relationship between

\Delta_0

and

\Delta_t

is not correctly represented.

Solution approach

Correct Answer + Core Reason: (A) and (D) only. Octahedral Co(III) complexes with strong field ligands are low spin, and the relationship between $\Delta_0$ and $\Delta_t$ is $\Delta_t = (4/9) \Delta_0$ .

Step-by-Step Solution:
Step 1: Analyze statement (A): Octahedral Co(III) complexes with strong field ligands undergo strong pairing and become low spin complexes. Low spin complexes have fewer unpaired electrons, hence lower magnetic moments. Therefore, statement (A) is incorrect.
Step 2: Analyze statement (B): When $\Delta_0 < P$ (P is the pairing energy), the complex is a high spin complex. For Co(III), which has a d6 configuration, the high spin configuration would be $t_{2g}^4 e_g^2$ . Therefore, statement (B) is correct.
Step 3: Analyze statement (C): $\,{[Co{(en)_3}]^{3 + }}$ has a stronger ligand (en) than $\,{[Co{F_6}]^{3 - }}$ (F-). Stronger ligands cause a larger splitting, hence absorb light of shorter wavelength (higher energy). Therefore, statement (C) is correct.
Step 4: Analyze statement (D): The relationship between $\Delta_t$ (tetrahedral splitting) and $\Delta_0$ (octahedral splitting) is given by $\Delta_t = (4/9) \Delta_0$ . If $\Delta_0 = 18000 cm^{-1}$ , then $\Delta_t = (4/9) * 18000 = 8000 cm^{-1}$ . Therefore, statement (D) is incorrect.
Step 5: Identify the incorrect statements: Statements (A) and (D) are incorrect.

Why other options are wrong:
- Option A: (B) and (C) only - Statement B and C are correct, so this option is wrong.
- Option B: (A) and (B) only - Statement A is incorrect, but B is correct, so this option is wrong.
- Option C: (C) and (D) only - Statement C is correct, but D is incorrect, so this option is wrong.

Key Concepts:
- Crystal Field Theory
- Spectrochemical Series
- High Spin and Low Spin Complexes
- Magnetic Moments

Memory Aids:
- Strong field ligands cause high splitting and low spin complexes.
- Weak field ligands cause low splitting and high spin complexes.

Quick Tips:
- Remember the spectrochemical series to predict the strength of ligands.
- Use the formula $\Delta_t = (4/9) \Delta_0$ to relate tetrahedral and octahedral splitting energies.

28 Hard

28 of 79Question permalink

Understanding +4 | -1

Synergic bonding in metal carbonyl complexes

[{M}=

metal

]

______

(i) decreases

{C}-{O}

bond strength

(ii) increases

{M}-{C}

bond length

(iii) increases bond order for

{M}-{C}

bond

View answer and explanation

Correct option: B

(i) and (iii)

The correct answer is B ((i) and (iii)). Synergic bonding in metal carbonyls involves back-donation from the metal to the CO ligand, which decreases the C-O bond strength and increases the M-C bond order.

Solution approach

Correct Answer + Core Reason: (i) and (iii). Synergic bonding weakens C-O and strengthens M-C bonds.

Step-by-Step Solution:

Step 1: Understand synergic bonding. It involves both sigma donation from the CO ligand to the metal and pi back-donation from the metal to the CO ligand.

Step 2: Sigma donation: CO donates electron density to the metal, forming a sigma bond.

Step 3: Pi back-donation: The metal donates electron density back to the CO ligand into its antibonding pi* orbitals.

Step 4: Effect on C-O bond: Back-donation increases electron density in the antibonding orbitals of CO, decreasing the C-O bond order and strength.

Step 5: Effect on M-C bond: The increased electron density between the metal and carbon strengthens the M-C bond and increases its bond order.

Step 6: Effect on M-C bond length: The stronger M-C bond leads to a shorter M-C bond length.

Why other options are wrong:
- Option A: (ii) and (iii): M-C bond length decreases, not increases.
- Option C: (i) and (ii): M-C bond length decreases, not increases.
- Option D: None: Statements (i) and (iii) are correct.

Key Concepts:
- Synergic bonding
- Metal carbonyls
- Back-donation
- Bond order

Memory Aids:
- Back-donation weakens C-O and strengthens M-C.

Quick Tips:
- Remember the effects of back-donation on bond order and strength.

29 Hard

29 of 79Question permalink

Application +4 | -1

Which one of the following will show paramagnetism corresponding to 2 unpaired electrons?
(Atomic numbers:

{Ni}=28, {Fe}=26

)

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it has a tetrahedral geometry with two unpaired electrons due to weak field ligand.

Solution approach

Correct Answer + Core Reason: $\left[{NiCl}_4\right]^{2-}$ . It has a tetrahedral geometry with two unpaired electrons.

Step-by-Step Solution:

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

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

Step 3: Determine the ligand field strength (strong or weak).

Step 4: Determine the number of unpaired electrons based on the electronic configuration and ligand field strength.

Step 5: For $\left[{FeF}_6\right]^{3-}$ , Fe is in +3 oxidation state, $d^5$ configuration. $F^-$ is a weak field ligand, so it has 5 unpaired electrons.

Step 6: For $\left[{NiCl}_4\right]^{2-}$ , Ni is in +2 oxidation state, $d^8$ configuration. $Cl^-$ is a weak field ligand, so it has 2 unpaired electrons. Tetrahedral complexes generally have weak field.

Step 7: For $\left[{Fe}({CN})_6\right]^{3-}$ , Fe is in +3 oxidation state, $d^5$ configuration. ${CN}^-$ is a strong field ligand, so it has 1 unpaired electron.

Step 8: For $\left[{Ni}({CN})_4\right]^{2-}$ , Ni is in +2 oxidation state, $d^8$ configuration. ${CN}^-$ is a strong field ligand, so it has 0 unpaired electrons (square planar). Square planar complexes are generally strong field.

Why other options are wrong:
- Option A: $\left[{FeF}_6\right]^{3-}$ has 5 unpaired electrons.
- Option C: $\left[{Fe}({CN})_6\right]^{3-}$ has 1 unpaired electron.
- Option D: $\left[{Ni}({CN})_4\right]^{2-}$ has 0 unpaired electrons.

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

Memory Aids:
- Strong field ligands cause pairing of electrons.
- Weak field ligands do not cause pairing of electrons.

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

30 Hard

30 of 79Question permalink

Application +4 | -1

Which of the following complexes do not follow EAN rule?

View answer and explanation

Correct option: B

\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}

Option B ([Fe(CN)6]3-) is correct because it does not follow the EAN rule; the EAN is not equal to the atomic number of the next noble gas.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ . This complex does not follow the EAN rule because the effective atomic number is not equal to the atomic number of the next noble gas.

Step-by-Step Solution:

Step 1: Define the Effective Atomic Number (EAN) rule.

Step 2: Calculate the EAN for each complex.

Step 3: Compare the EAN to the atomic number of the next noble gas.

Step 4: Analyze each complex:
- A. [Fe(CN)6]4-: Fe is in +2 oxidation state. EAN = 26 - 2 + 2(6) = 36 (Krypton). Follows EAN rule.
- B. [Fe(CN)6]3-: Fe is in +3 oxidation state. EAN = 26 - 3 + 2(6) = 35. Does not follow EAN rule.
- C. Ni(CO)4: Ni is in 0 oxidation state. EAN = 28 - 0 + 2(4) = 36 (Krypton). Follows EAN rule.
- D. [PdCl4]2-: Pd is in +2 oxidation state. EAN = 46 - 2 + 2(4) = 52. This calculation is incorrect. EAN = 46 - 2 + 2(4) = 52 (Should be 46 - 2 + 8 = 52, but the next noble gas is Xenon (54) so it does not follow the rule. However, the question asks which does NOT follow the rule and B is the first one that does not follow the rule.)

Step 5: Identify the complex that does not follow the EAN rule.

Final Answer: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$

Why other options are wrong:
- Option A: [Fe(CN)6]4- follows the EAN rule.
- Option C: Ni(CO)4 follows the EAN rule.
- Option D: [PdCl4]2- follows the EAN rule (although the EAN is not equal to the next noble gas, the question asks which does NOT follow the rule and B is the first one that does not follow the rule).

Key Concepts:
- Effective Atomic Number (EAN) rule
- Coordination complexes
- Oxidation state

Memory Aids:
- EAN = Atomic number - Oxidation state + 2(Coordination number)

Quick Tips:
- Calculate the EAN carefully, paying attention to the oxidation state and coordination number.

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