CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Hard, Page 3

61 Hard

Application +4 | -1

The conductivity measurement of a coordination compound of Cobalt (III) shows that it dissociates into 3 ions in solution. The compound is

View answer and explanation

Correct option: D

Pentaamminechloridocobalt(III) chloride

Option D (Pentaamminechloridocobalt(III) chloride) is correct because it dissociates into 3 ions: [Co(NH3)5Cl]2+ and 2 Cl- ions.

Solution approach

Correct Answer + Core Reason: Pentaamminechloridocobalt(III) chloride. This compound dissociates into 3 ions in solution: [Co(NH3)5Cl]2+ and 2 Cl- ions.

Step-by-Step Solution:

Step 1: Identify the coordination compounds and their potential dissociation products.

Step 2: Determine the number of ions each compound would produce upon dissociation in solution.

Step 3: Analyze each option:
- A. Hexaamminecobalt(III) chloride: [Co(NH3)6]Cl3 dissociates into [Co(NH3)6]3+ and 3Cl-, giving 4 ions.
- B. Pentaamminesulphatocobalt(III) chloride: [Co(NH3)5SO4]Cl dissociates into [Co(NH3)5SO4]+ and Cl-, giving 2 ions.
- C. Pentaamminechloridocobalt(III) sulphate: [Co(NH3)5Cl]SO4 dissociates into [Co(NH3)5Cl]2+ and SO42-, giving 2 ions.
- D. Pentaamminechloridocobalt(III) chloride: [Co(NH3)5Cl]Cl2 dissociates into [Co(NH3)5Cl]2+ and 2Cl-, giving 3 ions.

Step 4: Select the compound that dissociates into 3 ions.

Final Answer: Pentaamminechloridocobalt(III) chloride

Why other options are wrong:
- Option A: Hexaamminecobalt(III) chloride dissociates into 4 ions, not 3.
- Option B: Pentaamminesulphatocobalt(III) chloride dissociates into 2 ions, not 3.
- Option C: Pentaamminechloridocobalt(III) sulphate dissociates into 2 ions, not 3.
- Option D: Pentaamminechloridocobalt(III) chloride dissociates into 3 ions, making it the correct answer.

Key Concepts:
- Coordination compounds
- Dissociation of complexes in solution
- Ion counting

Memory Aids:
- Remember that counter ions outside the coordination sphere dissociate in solution.

Quick Tips:
- Pay close attention to the placement of ions inside and outside the coordination sphere.

62 Hard

62 of 79Question permalink

Remembering +4 | -1

The increasing order of crystal field splitting strength of the given ligands is

View answer and explanation

Correct option: C

{Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}

The correct answer is C (

{Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}

) because it correctly represents the increasing order of crystal field splitting strength based on the spectrochemical series.

Solution approach

Correct Answer + Core Reason: ${Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}$ . This order follows the spectrochemical series.

Step-by-Step Solution:
Step 1: The question asks for the increasing order of crystal field splitting strength of the given ligands.
Step 2: The crystal field splitting strength is determined by the spectrochemical series.
Step 3: The spectrochemical series is (approximately): $\mathrm{I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < CN^- < CO}$ .
Step 4: Based on the spectrochemical series, the increasing order is ${Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}$ .

Why other options are wrong:
- **Option A ( ${NH}_3 < {Cl}^{-} < {CN}^{-} < {F}^{-} < {CO} < {H}_2 {O}$ ): This is incorrect because it does not follow the spectrochemical series.
- Option B ( ${F}^{-} < {Cl}^{-} < {NH}_3 < {CN}^{-} < {H}_2 {O} < {CO}$ ): This is incorrect because it does not follow the spectrochemical series.
- Option D ( ${CO} < {CN}^{-} < {NH}_3 < {H}_2 {O} < {F}^{-} < {Cl}^{-}$ ): This is incorrect because it shows a decreasing order and does not follow the spectrochemical series.

Key Concepts:
- Crystal field splitting strength
- Spectrochemical series
- Ligand field strength

Memory Aids:
- Memorize the spectrochemical series.

Quick Tips:**
- Use the spectrochemical series to compare ligand field strengths.

63 Hard

63 of 79Question permalink

Analyzing +4 | -1

{CuSO}_4+{KCN} \longrightarrow {X}+({CN})_2

{X}+ \text{excess }{KCN} \longrightarrow {Y}+ \text{byproduct}

{Cd}^{+2}+ \text{excess } {CN}^{-} \longrightarrow {Z}

If

{Y}

and

{Z}

are soluble complexes, then the incorrect statement about the reactions in the following.

View answer and explanation

Correct option: A

In the formation of '

{X}

',

{CuSO}_4

acts as reducing agent

Option A (In the formation of '

{X}

',

{CuSO}_4

acts as reducing agent) is correct because in the reaction,

CuSO_4

is reduced to

CuCN

, hence it acts as an oxidizing agent, not a reducing agent.

Solution approach

Correct Answer + Core Reason: In the formation of ' ${X}$ ', ${CuSO}_4$ acts as reducing agent. $CuSO_4$ is reduced to $CuCN$ , hence it acts as an oxidizing agent.

Step-by-Step Solution:
Step 1: Write the balanced chemical equations for the reactions.
Step 2: Determine the oxidation states of copper in each complex.
Step 3: Analyze the role of $CuSO_4$ in the first reaction.

Step 4: Reaction 1: $2CuSO_4 + 2KCN \rightarrow 2CuCN + (CN)_2 + K_2SO_4$ . Here, X is $CuCN$ .
Step 5: Reaction 2: $CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$ . Here, Y is $K_3[Cu(CN)_4]$ .
Step 6: Reaction 3: $Cd^{2+} + 4CN^- \rightarrow [Cd(CN)_4]^{2-}$ . Here, Z is $[Cd(CN)_4]^{2-}$ .

Step 7: Oxidation state of Cu in $CuSO_4$ is +2.
Step 8: Oxidation state of Cu in $CuCN$ is +1.
Step 9: Oxidation state of Cu in $K_3[Cu(CN)_4]$ is +1.
Step 10: Oxidation state of Cd in $[Cd(CN)_4]^{2-}$ is +2.

Step 11: In the formation of X ( $CuCN$ ), $CuSO_4$ is reduced from +2 to +1. Therefore, $CuSO_4$ acts as an oxidizing agent, not a reducing agent.
Step 12: X ( $CuCN$ ) and Y ( $K_3[Cu(CN)_4]$ ) have copper in the +1 oxidation state.
Step 13: $Y$ ( $[Cu(CN)_4]^{3-}$ ) and $Z$ ( $[Cd(CN)_4]^{2-}$ ) are colorless complexes because the d-d transition is not possible for $Cu^+$ (d10) and $Cd^{2+}$ (d10) ions.
Step 14: The secondary valency (coordination number) in $Y$ ( $[Cu(CN)_4]^{3-}$ ) is 4, and in $Z$ ( $[Cd(CN)_4]^{2-}$ ) is 4.

Why other options are wrong:
- Option B: Where ' ${X}$ ' and ' ${Y}$ ' have copper in +1 oxidation state - This is correct, so this option is wrong.
- Option C: Where $Y$ and $Z$ are colourless complexes - This is correct, so this option is wrong.
- Option D: Number secondary valencies in $Y$ and $Z$ are 4 - This is correct, so this option is wrong.

Key Concepts:
- Redox Reactions
- Coordination Complexes
- Oxidation States
- Coordination Number

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

Quick Tips:
- Determine the oxidation states of the elements involved in the reaction.
- Identify the oxidizing and reducing agents based on the change in oxidation states.

64 Hard

64 of 79Question permalink

Application +4 | -1

Excess of silver nitrate solution is added to 100 mL of 0.01 M pentaaqua chloro chromium (III) chloride solution. The mass of silver chloride obtained in grams is [Atomic mass of silver is 108].

View answer and explanation

Correct option: A

287 \times {10^{ - 3}}

Option A (287 × 10^-3) is correct because only one chloride ion is outside the coordination sphere, leading to the precipitation of only that one chloride ion as AgCl.

Solution approach

Correct Answer + Core Reason: $287 \times {10^{ - 3}}$ . Only one chloride ion is 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: $2AgNO_3 + 2Cl^- \rightarrow 2AgCl + 2NO_3^-$

Step 3: Moles of complex = 0.01 M * 0.1 L = 0.001 moles.

Step 4: Moles of $AgCl$ formed = 2 * moles of complex = 2 * 0.001 = 0.002 moles.

Step 5: Molar mass of $AgCl$ = 108 + 35.5 = 143.5 g/mol.

Step 6: Mass of $AgCl$ formed = moles * molar mass = 0.002 * 143.5 = 0.287 g = $287 \times {10^{ - 3}}$ g.

Why other options are wrong:
- Option B: This corresponds to only 0.001 moles of AgCl.
- Option C: This corresponds to 0.01 moles of AgCl.
- Option D: This corresponds to 0.02 moles of AgCl, but the units are incorrect.

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.

65 Hard

65 of 79Question permalink

Analyzing +4 | -1

A certain complex ion has the formula

\left[M(\mathrm{en})_2 \mathrm{Br}_2\right]^{+}

where

M

is the metal ion and en

=

ethylene diamine. How many isomers are possible for this ?

View answer and explanation

Correct option: A

3

Option A (3) is correct because the complex can exist as cis, trans, and mer isomers due to the arrangement of the ligands around the metal center.

Solution approach

Correct Answer + Core Reason: 3. The complex $\left[M(en)_2 Br_2\right]^+$ can exhibit cis, trans, and mer isomers.

Step-by-Step Solution:

Step 1: Identify the complex type.\nThe complex is of the type $MA_2B_2$ where A is a bidentate ligand (en) and B is a monodentate ligand (Br).

Step 2: Draw possible isomers.
- *trans* isomer: The two Br ligands are opposite to each other.
- *cis* isomer: The two Br ligands are adjacent to each other. This cis isomer is chiral and exists as a pair of enantiomers.

Step 3: Count the isomers.\nThere is one trans isomer and two cis isomers (enantiomers). Therefore, there are a total of 3 isomers.

Why other options are wrong:
- Option B: 0. Incorrect, isomers exist.
- Option C: 0. Incorrect, isomers exist.
- Option D: 0. Incorrect, isomers exist.

Key Concepts:
- Isomerism in Coordination Compounds
- Geometrical Isomerism (cis-trans)
- Optical Isomerism (enantiomers)
- Bidentate Ligands

Memory Aids:
- *cis*: same side; *trans*: opposite sides

Quick Tips:
- Draw all possible arrangements of the ligands around the metal center.
- Consider the possibility of optical isomers (enantiomers) for cis isomers with bidentate ligands.

66 Hard

66 of 79Question permalink

Understanding +4 | -1

In an octahedral geometry, the pair of

{d}

orbitals involved in

{d}^2 {sp}^3

hybridization is:

View answer and explanation

Correct option: A

{d}_{{x}^2-{y}^2}, {~d}_{{z}^2}

The correct answer is A (

{d}_{{x}^2-{y}^2}, {~d}_{{z}^2}

). In

{d}^2 {sp}^3

hybridization, the

{d}

orbitals involved are

{d}_{{x}^2-{y}^2}

and

{d}_{{z}^2}

because they point directly along the axes.

Solution approach

Correct Answer + Core Reason: ${d}_{{x}^2-{y}^2}, {~d}_{{z}^2}$ . These d-orbitals have lobes along the axes.

Step-by-Step Solution:

Step 1: Understand ${d}^2 {sp}^3$ hybridization. It involves one s orbital, three p orbitals, and two d orbitals.

Step 2: Identify the d orbitals involved in octahedral geometry. These are the d orbitals that point directly along the axes.

Step 3: The d orbitals that point along the axes are ${d}_{{x}^2-{y}^2}$ and ${d}_{{z}^2}$ .

Step 4: The other d orbitals ( $d_{xy}, d_{xz}, d_{yz}$ ) point between the axes.

Why other options are wrong:
- Option B: $d_{x z}, d_{x^2-y^2}$ : $d_{xz}$ points between the axes.
- Option C: ${d}_{{z}^2}, {~d}_{{xz}}$ : $d_{xz}$ points between the axes.
- Option D: $d_{x y}, d_{y z}$ : Both $d_{xy}$ and $d_{yz}$ point between the axes.

Key Concepts:
- Hybridization
- d orbitals
- Octahedral geometry

Memory Aids:
- ${d}^2 {sp}^3$ uses ${d}_{{x}^2-{y}^2}$ and ${d}_{{z}^2}$

Quick Tips:
- Remember the orientation of d orbitals.

67 Hard

67 of 79Question permalink

Understanding +4 | -1

\mathrm{F}{\mathrm{e}_2}{\left( {\mathrm{CO}} \right)_9}

is diamagnetic. Which of the following reasons is correct?

View answer and explanation

Correct option: C

Metal-metal (

\mathrm{Fe} - \mathrm{Fe}

) bond in molecule

Option C (Metal-metal (

\mathrm{Fe} - \mathrm{Fe}

) bond in molecule) is correct because the presence of an Fe-Fe bond allows for the pairing of electrons between the two iron atoms, resulting in a diamagnetic complex.

Solution approach

Correct Answer + Core Reason: Metal-metal ( $\mathrm{Fe} - \mathrm{Fe}$ ) bond in molecule. The Fe-Fe bond allows for electron pairing, leading to diamagnetism.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Fe in $\mathrm{F}{\mathrm{e}_2}{\left( {\mathrm{CO}} \right)_9}$ .
2Fe + 9(0) = 0\nFe = 0

Step 2: Determine the electronic configuration of Fe(0).\nFe: [Ar] $3d^6 4s^2$ \nFe(0): [Ar] $3d^8 4s^0$

Step 3: Analyze the structure of $\mathrm{F}{\mathrm{e}_2}{\left( {\mathrm{CO}} \right)_9}$ .
$\mathrm{F}{\mathrm{e}_2}{\left( {\mathrm{CO}} \right)_9}$ has a structure with two Fe atoms bridged by three CO ligands and a direct Fe-Fe bond. Each Fe atom is also bonded to three terminal CO ligands.

Step 4: Explain the diamagnetism.\nEach Fe atom contributes 4 electrons to the Fe-Fe bond. The Fe-Fe bond allows for the pairing of electrons between the two iron atoms. The pairing of electrons results in a diamagnetic complex.

Why other options are wrong:
- **Option A (Presence of one ${\mathrm{CO}}$ as bridge group): While CO ligands can act as bridging ligands, the presence of bridging CO ligands alone does not guarantee diamagnetism.
- Option B (Presence of monodentate ligand): The presence of monodentate ligands does not directly explain the diamagnetism of the complex.
- Option D (Resonance hybridization of ${\mathrm{CO}}$ ): Resonance hybridization of CO ligands does not directly explain the diamagnetism of the complex.

Key Concepts:
- Metal carbonyls
- Metal-metal bonds
- Diamagnetism

Memory Aids:
- Metal-metal bonds can lead to electron pairing and diamagnetism.

Quick Tips:**
- Consider the structure and bonding in metal carbonyls to understand their magnetic properties.

68 Hard

68 of 79Question permalink

Understanding +4 | -1

Arrange the following cobalt complexes in the order of increasing crystal field stabilization energy (CFSE) value. Complexes

\mathop {{{\left[ {Co{F_6}} \right]}^{3 - }}}\limits_{(A)} ,\,\,\,\mathop {{{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]}^{3 + }}}\limits_{(B)} ,\,\,\,\,\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]}^{3 + }}}\limits_{\left( C \right)} \,\,\mathrm{and}\,\,\mathop {{{\left[ {Co{{\left( {en} \right)}_3}} \right]}^{3 + }}}\limits_{\left( D \right)}

Choose the correct option.

View answer and explanation

Correct option: A

A < B < C < D\,

Option A (

A < B < C < D\,

) is correct because the spectrochemical series dictates the ligand field strength order:

F^- < H_2O < NH_3 < en

, hence the increasing CFSE order is

A < B < C < D

.

Solution approach

Correct Answer + Core Reason: $A < B < C < D\,$ because the spectrochemical series dictates the ligand field strength.

Step-by-Step Solution:
Step 1: Recall the spectrochemical series.
Step 2: Arrange the ligands in increasing order of field strength.
Step 3: Relate ligand field strength to CFSE.
\nThe spectrochemical series is a list of ligands ordered by their ability to split d-orbitals. A stronger field ligand leads to a larger crystal field splitting and thus a higher CFSE.\nThe order of field strength is: ${F^ - } < {\mathrm{H}_2}\mathrm{O} < \mathrm{N}{\mathrm{H}_3} < en$ \nTherefore, the order of increasing CFSE is: $A < B < C < D$

Why other options are wrong:
- Option B: $B < A < C < D$ is incorrect because fluoride is a weaker field ligand than water.
- Option C: $B < C < D < A$ is incorrect because fluoride is the weakest field ligand and ethylenediamine (en) is the strongest among the given ligands.
- Option D: $C < D < B < A$ is incorrect because ammonia and ethylenediamine are stronger field ligands than water and fluoride.

Key Concepts:
- Crystal field stabilization energy (CFSE)
- Spectrochemical series
- Ligand field strength

Memory Aids:
- Remember the order of common ligands in the spectrochemical series.

Quick Tips:
- Know the spectrochemical series to predict the relative CFSE values of complexes.

69 Hard

69 of 79Question permalink

Analyzing +4 | -1

The equation which is balanced and represents the correct product(s) is

View answer and explanation

Correct option: B

{[CoCl{(N{H_3})_5}]^ + } + 5{H^ + } \to C{O^{2 + }} + 5NH_4^ + + C{l^ - }

Option B (

{[CoCl{(N{H_3})_5}]^ + } + 5{H^ + } \to C{O^{2 + }} + 5NH_4^ + + C{l^ - }

) is correct because it represents a balanced chemical equation with correct products formed when the complex reacts with acid.

Solution approach

Correct Answer + Core Reason: ${[CoCl{(N{H_3})_5}]^ + } + 5{H^ + } \to C{O^{2 + }} + 5NH_4^ + + C{l^ - }$ because it is a balanced equation with correct products.

Step-by-Step Solution:

Step 1: Analyze each equation for balance and correct products.

Step 2: In option A, $L{i_2}O + 2KCl \to 2LiCl + {K_2}O$ , the reaction is not feasible because Li is less electropositive than K. So, K cannot be displaced by Li.

Step 3: In option B, ${[CoCl{(N{H_3})_5}]^ + } + 5{H^ + } \to C{O^{2 + }} + 5NH_4^ + + C{l^ - }$ , the equation is balanced and the products are correct. The ammonia ligands are protonated to form ammonium ions.

Step 4: In option C, ${[Mg{({H_2}O)_6}]^{2 + }} + {(EDTA)^{4 - }}\xrightarrow{{Excess\,\,NaOH}}{[Mg(EDTA)]^{2 + }} + 6{H_2}O$ , the NaOH is not directly involved in the reaction, and the equation is balanced. However, the question asks for the *correct* product, and while the reaction is balanced, the excess NaOH will likely cause precipitation of $Mg(OH)_2$ before EDTA can fully complex with $Mg^{2+}$ .

Step 5: In option D, $CuS{O_4} + 4KCN \to {K_2}[Cu{(CN)_4}] + {K_2}S{O_4}$ , the product should be ${K_3}[Cu{(CN)_4}]$ and $K_2SO_4$ is incorrect. The balanced equation is $2CuSO_4 + 10KCN \to 2K_3[Cu(CN)_4] + 2K_2SO_4 + (CN)_2$

Step 6: Therefore, option B is the only balanced equation with the correct products.

Why other options are wrong:
- Option A: The reaction is not feasible due to the relative electropositivity of Li and K.
- Option C: While balanced, the excess NaOH will interfere with the complex formation.
- Option D: The product is incorrect; it should be ${K_3}[Cu{(CN)_4}]$ and the equation is not balanced.

Key Concepts:
- Balancing chemical equations
- Reactivity series of metals
- Coordination complexes

Memory Aids:
- Remember to balance both atoms and charges in a chemical equation.

Quick Tips:
- Check the feasibility of reactions based on the reactivity series.

70 Hard

70 of 79Question permalink

Analyzing +4 | -1

Assertion : If in

{\left[ {{\text{Co(N}}{{\text{H}}_3}{{\text{)}}_{\text{6}}}} \right]^{ + 3}}\,,\,{\text{N}}{{\text{H}}_{\text{3}}}

is replaced by

{{\text{H}}_{\text{2}}}{\text{O}}

, same wave1ength will be absorbed by the complex :

Reason : It is a high spin species.

View answer and explanation

Correct option: D

If both Assertion & Reason are false

Option D is correct because replacing NH3 with H2O will change the ligand field strength and hence the wavelength absorbed, and [Co(NH3)6]+3 is a low spin complex, not high spin.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are false. Ligand substitution alters absorption wavelength, and the complex is low spin.

Step-by-Step Solution:

Step 1: Analyze the assertion. Replacing $NH_3$ with $H_2O$ as ligands will definitely change the ligand field splitting and hence the wavelength of light absorbed by the complex. $NH_3$ is a stronger field ligand than $H_2O$ .

Step 2: Analyze the reason. $[Co(NH_3)_6]^{3+}$ is a $d^6$ complex. $Co^{3+}$ has a $d^6$ electronic configuration. $NH_3$ is a strong field ligand, so it will cause pairing of electrons, resulting in a low-spin complex.

Step 3: Since both the assertion and the reason are incorrect, the answer is D.

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

Key Concepts:
- Ligand field theory
- Spectrochemical series
- High spin vs. low spin complexes
- Absorption spectra of coordination compounds

Memory Aids:
- Strong field ligands cause pairing and low spin complexes.
- Spectrochemical series: I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO

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

71 Hard

71 of 79Question permalink

Application +4 | -1

Which of the following compound is not colored?

View answer and explanation

Correct option: B

{Na}_2 {Cd} \cdot {Cl}_4

Option B (

{Na}_2 {Cd} \cdot {Cl}_4

) is correct because compounds with filled d-orbitals are generally colorless.

Solution approach

Correct Answer + Core Reason: ${Na}_2 {Cd} \cdot {Cl}_4$ . Compounds with filled d-orbitals are generally colorless.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations of the metal ions in each compound.

Step 2: ${Cu}_2 {O}$ contains $Cu^+$ which can form colored compounds due to charge transfer transitions.

Step 3: ${Na}_2 {Cd} \cdot {Cl}_4$ contains $Cd^{2+}$ which has a $d^{10}$ configuration. Filled d-orbitals generally result in colorless compounds.

Step 4: ${FeSO}_4$ contains $Fe^{2+}$ which has a $d^6$ configuration. It is colored due to d-d transitions.

Step 5: ${VCl}_3$ contains $V^{3+}$ which has a $d^2$ configuration. It is colored due to d-d transitions.

Step 6: Therefore, ${Na}_2 {Cd} \cdot {Cl}_4$ is the only colorless compound among the given options.

Why other options are wrong:
- Option A: ${Cu}_2 {O}$ is colored due to charge transfer transitions involving $Cu^+$ .
- Option C: ${FeSO}_4$ is colored due to d-d transitions in $Fe^{2+}$ ion.
- Option D: ${VCl}_3$ is colored due to d-d transitions in $V^{3+}$ ion.

Key Concepts:
- Color of transition metal compounds
- d-d transitions
- Charge transfer transitions
- Electronic configuration

Memory Aids:
- Filled or empty d-orbitals often lead to colorless compounds.

Quick Tips:
- Consider the electronic configuration of the metal ion to predict the color of the compound.

72 Hard

72 of 79Question permalink

Application +4 | -1

The magnetic moment is measured in Bohr Magneton (BM).

Spin only magnetic moment of

\mathrm{Fe}

in

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

and

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

complexes respectively is:

View answer and explanation

Correct option: C

5.92 B.M. and 1.732 B.M.

Option C (5.92 B.M. and 1.732 B.M.) is correct because it accurately calculates the spin-only magnetic moment based on the number of unpaired electrons in each complex.

Solution approach

Correct Answer + Core Reason: 5.92 B.M. and 1.732 B.M. The magnetic moment is determined by the number of unpaired electrons in each complex.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Fe in each complex.\nIn $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ , Fe is in the +3 oxidation state.\nIn $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ , Fe is in the +3 oxidation state.

Step 2: Determine the electronic configuration of Fe $^{3+}$ in each complex.\nFe $^{3+}$ has a $d^5$ configuration.

Step 3: Determine the number of unpaired electrons in each complex.\nIn $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ , $\mathrm{H}_2\mathrm{O}$ is a weak field ligand, so the complex is high spin. The electronic configuration is $t_{2g}^3 e_g^2$ , with 5 unpaired electrons.\nIn $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ , $\mathrm{CN}^-$ is a strong field ligand, so the complex is low spin. The electronic configuration is $t_{2g}^5 e_g^0$ , with 1 unpaired electron.

Step 4: Calculate the spin-only magnetic moment using the formula: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.\nFor $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ : $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.\nFor $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ : $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732$ BM.

Why other options are wrong:
- Option A: Incorrect because it assumes the same number of unpaired electrons in both complexes, neglecting the effect of the ligand field strength.
- Option B: Incorrect magnetic moment values for both complexes. The high-spin complex has 5 unpaired electrons, not a value corresponding to 3.87 B.M.
- Option D: Incorrect magnetic moment values for both complexes. The low-spin complex has 1 unpaired electron, not a value corresponding to 6.92 B.M.

Key Concepts:
- Crystal Field Theory
- High-spin and Low-spin complexes
- Spin-only magnetic moment
- Ligand field strength

Memory Aids:
- Strong field ligands cause pairing of electrons (low spin).
- Weak field ligands do not cause pairing of electrons (high spin).

Quick Tips:
- Remember the spectrochemical series to determine ligand field strength.
- Use the formula $\mu = \sqrt{n(n+2)}$ to calculate the spin-only magnetic moment.

73 Hard

73 of 79Question permalink

Analyzing +4 | -1

Complex

X

of composition

Cr{\left( {{H_2}O} \right)_6}C{l_n}

has a spin only magnetic moment of 3.83 BM. It reacts with

AgN{O_3}

and shows geometrical isomerism. The IUPAC nomenclature of

X

is

View answer and explanation

Correct option: D

tetraaquadichlorido - chromium (III) chloride dihydrate

Option D (tetraaquadichlorido - chromium (III) chloride dihydrate) is correct because it matches the given information about magnetic moment, reaction with

AgNO_3

, and geometrical isomerism.

Solution approach

Correct Answer + Core Reason: tetraaquadichlorido - chromium (III) chloride dihydrate. The complex matches the properties given: spin-only magnetic moment of 3.83 BM indicates $Cr^{3+}$ , reaction with $AgNO_3$ indicates two $Cl^-$ ions outside the coordination sphere, and geometrical isomerism is possible with $cis-$ and $trans-$ isomers of $[Cr(H_2O)_4Cl_2]^+$ .

Step-by-Step Solution:

Step 1: Determine the oxidation state of Chromium.\nSpin-only magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.
$3.83 = \sqrt{n(n+2)}$
$n(n+2) \approx 15$
$n \approx 3$ \nSince Chromium has 3 unpaired electrons, its oxidation state is +3 ( $Cr^{3+}$ ).

Step 2: Determine the number of chloride ions outside the coordination sphere.\nThe complex reacts with $AgNO_3$ , meaning that $Cl^-$ ions are present outside the coordination sphere. The number of moles of $AgCl$ precipitated will indicate the number of $Cl^-$ ions outside.\nSince the complex shows geometrical isomerism, it must have the form $[Cr(H_2O)_4Cl_2]Cl.2H_2O$ or $[Cr(H_2O)_3Cl_3] . 3H_2O$ etc. But since it reacts with $AgNO_3$ , there must be $Cl^-$ outside the coordination sphere. The magnetic moment suggests $Cr^{3+}$ , so we have $[Cr(H_2O)_4Cl_2]Cl.2H_2O$ .

Step 3: Determine the IUPAC name.\nThe IUPAC name is tetraaquadichloridochromium(III) chloride dihydrate.

Why other options are wrong:
- Option A: tetraaquadichlorido chromium (IV) chloride dihydrate. Incorrect oxidation state of Chromium (IV).
- Option B: hexaaqua chromium (III) chloride. Does not account for the chloride ions inside the coordination sphere and the dihydrate.
- Option C: dichloridotetraaqua - chromium (IV) chloride dihydrate. Incorrect oxidation state of Chromium (IV).

Key Concepts:
- IUPAC Nomenclature of Coordination Compounds
- Oxidation State Calculation
- Magnetic Moment Calculation
- Geometrical Isomerism

Memory Aids:
- IUPAC naming rules: ligands alphabetically, then metal, then oxidation state in Roman numerals, then counter ions.
- Spin-only magnetic moment formula: $\mu = \sqrt{n(n+2)}$ BM

Quick Tips:
- Determine the oxidation state first using the magnetic moment.
- Use the reaction with $AgNO_3$ to determine the number of chloride ions outside the coordination sphere.
- Consider the possibility of geometrical isomers.

74 Hard

74 of 79Question permalink

Analyzing +4 | -1

For the following molecules; if

x

is the total number of

p

-orbitals used in hybridisation of each central atom, then what is the value of

\frac{\sum {x}}{6}

?

{XeO}_2 {~F}_2, {~B}_2 {H}_6,\left[{NiCl}_4\right]^{-2},\left[{PtCl}_4\right]^{-2}, {~N}\left({SiH}_3\right)_3, {H}_3 {BO}_3

View answer and explanation

Correct option: A

3

Option A (3) is correct because the sum of p-orbitals used in hybridization for each central atom is 18, and 18/6 = 3.

Solution approach

Correct Answer + Core Reason: 3. The sum of p-orbitals used in hybridization for each central atom is 18, and 18/6 = 3.

Step-by-Step Solution:
Step 1: Determine the hybridization of each central atom.
Step 2: Count the number of p-orbitals involved in the hybridization of each central atom.
Step 3: Sum the number of p-orbitals for all central atoms.
Step 4: Divide the sum by 6.

Step 5: ${XeO}_2 {~F}_2$ : Xe has 8 valence electrons. Steric number = 4 (2 bonds to O and 2 bonds to F). Hybridization is $sp^3$ . Number of p-orbitals = 3.
Step 6: ${~B}_2 {H}_6$ : Each B has 3 valence electrons. Each B is $sp^3$ hybridized. Number of p-orbitals per B = 3. Total p-orbitals = 3 + 3 = 6.
Step 7: $\left[{NiCl}_4\right]^{-2}$ : Ni is in +2 oxidation state. Ni2+ has d8 configuration. It is tetrahedral, so hybridization is $sp^3$ . Number of p-orbitals = 3.
Step 8: $\left[{PtCl}_4\right]^{-2}$ : Pt is in +2 oxidation state. Pt2+ has d8 configuration. It is square planar, so hybridization is $dsp^2$ . Number of p-orbitals = 2.
Step 9: ${~N}\left({SiH}_3\right)_3$ : N has 5 valence electrons. It is $sp^2$ hybridized. Number of p-orbitals = 2.
Step 10: ${H}_3 {BO}_3$ : B has 3 valence electrons. It is $sp^2$ hybridized. Number of p-orbitals = 2.

Step 11: Sum of p-orbitals = 3 + 6 + 3 + 2 + 2 + 2 = 18.
Step 12: $\frac{\sum {x}}{6} = \frac{18}{6} = 3$ .

Why other options are wrong:
- Option B: 0 - This is incorrect because the sum of p-orbitals divided by 6 is not 0.
- Option C: 0 - This is incorrect because the sum of p-orbitals divided by 6 is not 0.
- Option D: 0 - This is incorrect because the sum of p-orbitals divided by 6 is not 0.

Key Concepts:
- Hybridization
- Steric Number
- Coordination Complexes

Memory Aids:
- Steric number = number of lone pairs + number of sigma bonds.

Quick Tips:
- Determine the steric number to find the hybridization.
- Remember the shapes associated with different hybridizations.

75 Hard

75 of 79Question permalink

Analyzing +4 | -1

Among the following complexes (

A-F

),

\begin{aligned} & {K}_3\left[{Fe}({CN})_6\right] ({A}),\left[{Co}\left({NH}_3\right)_6\right] {Cl}_3 (B)\\ & {Na}_3\left[{Co}(\text { oxalate })_3\right]({C}),\left[{Ni}_2\left({H}_2 {O}\right)_6\right] {Cl}_2 (D),\\ & {~K}_2\left[{Pt}({CN})_4\right]({E}) \text { and }\left[{Zn}\left({H}_2 {O}\right)_6\right]\left({NO}_3\right)_2({~F}) \end{aligned}

The diamagnetic complexes are:

View answer and explanation

Correct option: C

B, C, E, F

Option C is correct because B, C, E, and F are all diamagnetic complexes, meaning they have no unpaired electrons.

Solution approach

Correct Answer + Core Reason: $B, C, E, F$ . These complexes have no unpaired electrons and are therefore diamagnetic.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the metal ions in each complex.

Step 2: Determine the ligand field strength and pairing of electrons.

Step 3: Identify the complexes with no unpaired electrons (diamagnetic).

Step 4: Analyze each complex:
- A: $K_3[Fe(CN)_6]$ : $Fe^{3+}$ 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. Paramagnetic.
- B: $[Co(NH_3)_6]Cl_3$ : $Co^{3+}$ is $d^6$ . $NH_3$ is a strong field ligand, so it pairs up electrons. Electronic configuration is $t_{2g}^6 e_g^0$ . No unpaired electrons. Diamagnetic.
- C: $Na_3[Co(oxalate)_3]$ : $Co^{3+}$ is $d^6$ . Oxalate is a moderate field ligand, but $Co^{3+}$ is generally low spin. Electronic configuration is $t_{2g}^6 e_g^0$ . No unpaired electrons. Diamagnetic.
- D: $[Ni(H_2O)_6]Cl_2$ : $Ni^{2+}$ is $d^8$ . $H_2O$ is a weak field ligand. Electronic configuration is $t_{2g}^6 e_g^2$ . Two unpaired electrons. Paramagnetic.
- E: $K_2[Pt(CN)_4]$ : $Pt^{2+}$ is $d^8$ . $CN^-$ is a strong field ligand, and Pt is a 4d/5d series metal, so it's square planar and diamagnetic. No unpaired electrons. Diamagnetic.
- F: $[Zn(H_2O)_6](NO_3)_2$ : $Zn^{2+}$ is $d^{10}$ . All electrons are paired. No unpaired electrons. Diamagnetic.

Step 5: Identify the diamagnetic complexes: B, C, E, F.

Why other options are wrong:
- Option A: Includes complex A, which is paramagnetic.
- Option B: Includes complex A, which is paramagnetic.
- Option D: Includes complex D, which is paramagnetic.

Key Concepts:
- Diamagnetism and paramagnetism
- Crystal field theory
- Electronic configuration
- Ligand field strength

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to diamagnetism.
- $d^{10}$ and $d^0$ configurations are always diamagnetic.

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

76 Hard

76 of 79Question permalink

Remembering +4 | -1

[{EDTA}]^{4-}

is a

View answer and explanation

Correct option: D

Hexadentate ligand

Option D (Hexadentate ligand) is correct because EDTA can bind to a central metal ion through six donor atoms.

Solution approach

Correct Answer + Core Reason: Hexadentate ligand. EDTA has six donor atoms (2 nitrogen and 4 oxygen) that can coordinate to a central metal ion.

Step-by-Step Solution:

Step 1: Recall the structure of EDTA: EDTA (Ethylenediaminetetraacetic acid) has two nitrogen atoms and four oxygen atoms that can donate lone pairs.

Step 2: Determine the denticity: Denticity refers to the number of donor atoms a ligand uses to bind to a central metal ion.

Step 3: Conclude the denticity of EDTA: Since EDTA has six donor atoms, it is a hexadentate ligand.

Why other options are wrong:
- Option A: Monodentate ligand is incorrect because EDTA has more than one donor atom.
- Option B: Bidentate ligand is incorrect because EDTA has more than two donor atoms.
- Option C: Quadridentate ligand is incorrect because EDTA has more than four donor atoms.

Key Concepts:
- Denticity of ligands
- EDTA structure and properties
- Coordination chemistry

Memory Aids:
- EDTA: Ethylenediaminetetraacetic acid (6 binding sites = Hexadentate).

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

77 Hard

77 of 79Question permalink

Understanding +4 | -1

For a

{d^4}

metal ion in an octahedral field, the correct electronic configuration is

View answer and explanation

Correct option: A

t_{2g}^3e_g^1\,\,\,\mathrm{when}\,\,\mathrm{ }{\Delta _0} < P

Option A is correct because when Δ₀ < P, pairing energy is higher, leading to high spin configuration with electrons occupying both t2g and eg orbitals.

Solution approach

Correct Answer + Core Reason: $t_{2g}^3e_g^1\,\,\,\mathrm{when}\,\,\mathrm{ }{\Delta _0} < P$ . When the crystal field splitting energy (Δ₀) is less than the pairing energy (P), the electrons will singly occupy all the d orbitals before pairing up in the lower energy t2g orbitals.

Step-by-Step Solution:

Step 1: Identify the electronic configuration of the metal ion. The metal ion is d⁴.

Step 2: Determine the relationship between Δ₀ and P. Given Δ₀ < P, it's a high-spin complex.

Step 3: Fill the orbitals according to Hund's rule. First, fill the t2g orbitals with one electron each, then fill the eg orbitals.

Step 4: The electronic configuration will be $t_{2g}^3e_g^1$ .

Why other options are wrong:
- Option B: $t_{2g}^3e_g^1\,\,\,\mathrm{when }\,\,{\Delta _0} > P$ : This is incorrect because when Δ₀ > P, it's a low-spin complex, and the electrons will pair up in the t2g orbitals before occupying the eg orbitals.
- Option C: $t_{2g}^4e_g^0\,\,\,\mathrm{when}\,\,\mathrm{ }{\Delta _0} < P$ : This is incorrect because when Δ₀ < P, electrons will occupy both t2g and eg orbitals to minimize pairing energy.
- Option D: $t_g^2e_{2g}^2\,\,\,\mathrm{when }\,\,{\Delta _0} < P$ : This option uses incorrect notation for the orbitals and doesn't represent the correct filling of orbitals in a high-spin complex.

Key Concepts:
- Crystal field theory
- High-spin and low-spin complexes
- Hund's rule
- d-orbital splitting in octahedral fields

Memory Aids:
- Remember that when Δ₀ < P, it's a high-spin complex, and electrons will singly occupy all orbitals before pairing.

Quick Tips:
- Always consider the relationship between Δ₀ and P when determining the electronic configuration of a complex.

78 Hard

78 of 79Question permalink

Understanding +4 | -1

{K}_4\left[{Fe}({CN})_6\right]

reacts with

{FeCl}_3

to form:

View answer and explanation

Correct option: B

{KFe}\left[{Fe}({CN})_6\right]

Option B (

{KFe}\left[{Fe}({CN})_6\right]

) is correct because

{K}_4\left[{Fe}({CN})_6\right]

reacts with

{FeCl}_3

to form Prussian blue, which has the formula

{KFe}\left[{Fe}({CN})_6\right]

.

Solution approach

Correct Answer + Core Reason: ${KFe}\left[{Fe}({CN})_6\right]$ . The reaction forms Prussian blue, which has the formula ${KFe}\left[{Fe}({CN})_6\right]$ .

Step-by-Step Solution:
Step 1: Identify the reactants: ${K}_4\left[{Fe}({CN})_6\right]$ contains $\mathrm{Fe}^{2+}$ and ${FeCl}_3$ contains $\mathrm{Fe}^{3+}$ .
Step 2: Understand the reaction: The reaction between ${K}_4\left[{Fe}({CN})_6\right]$ and ${FeCl}_3$ forms Prussian blue, a mixed-valence compound.
Step 3: Write the reaction: ${K}_4\left[{Fe}({CN})_6\right] + {FeCl}_3 \rightarrow {KFe}\left[{Fe}({CN})_6\right] + KCl$
Step 4: The product is ${KFe}\left[{Fe}({CN})_6\right]$ , also written as $\mathrm{Fe}_4[{\mathrm{Fe}(\mathrm{CN})_6}]_3 \cdot x\mathrm{H}_2\mathrm{O}$ (Prussian blue).

Why other options are wrong:
- Option A: ${K}_3\left[{Fe}({CN})_5 {Cl}\right]$ is not the product of this reaction.
- Option C: ${K}_4\left[{Fe}({CN})_3 {Cl}_3\right]$ is not the product of this reaction.
- Option D: ${K}_3\left[{Fe}({CN})_6\right]$ is not the product of this reaction. It would require oxidation of the iron in the complex, which isn't happening here.

Key Concepts:
- Complex reactions
- Prussian blue formation
- Mixed-valence compounds

Memory Aids:
- Remember that the reaction between ferrocyanide and ferric ions forms Prussian blue.

Quick Tips:
- Recognize the formation of Prussian blue in reactions involving iron complexes.

79 Hard

79 of 79Question permalink

Analyzing +4 | -1

Assertion : In a mixture of

{\text{Cd(II)}}

and

{\text{Cu(II)}}

,

{\text{(C}}{{\text{d}}^{ + 2}}{\text{)}}

gets precipitated in presence of

\mathrm{KCN}

by H2S.

Reason : The stability constant of

{\left[ {{\text{(Cu(CN}}{{\text{)}}_{\text{4}}}} \right]^{ - 3}}

is greater than

{\left[ {{\text{(Cu(CN}}{{\text{)}}_{\text{4}}}} \right]^{ - 2}}

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 in the presence of KCN, Cu(II) forms a stable complex, preventing its precipitation as CuS, while Cd(II) remains free to precipitate as CdS.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. The stability constant difference allows selective precipitation.

Step-by-Step Solution:

Step 1: Understand the chemistry involved. KCN forms complexes with both Cd(II) and Cu(II).

Step 2: Write the complex formation reactions:
- $Cu^{2+} + 4CN^- \rightleftharpoons [Cu(CN)_4]^{2-}$
- $Cd^{2+} + 4CN^- \rightleftharpoons [Cd(CN)_4]^{2-}$

Step 3: Consider the stability constants. The stability constant of $[Cu(CN)_4]^{2-}$ is significantly higher than that of $[Cd(CN)_4]^{2-}$ .

Step 4: When $H_2S$ is added, it reacts with free metal ions to form sulfides. Since Cu(II) is strongly complexed by $CN^-$ , its concentration of free $Cu^{2+}$ is very low, preventing the formation of CuS. Cd(II) is less strongly complexed, so sufficient $Cd^{2+}$ is available to form CdS.

Step 5: The reason states the stability constant of the copper cyanide complex is greater, which is correct and explains why copper does not precipitate.

Why other options are wrong:
- Option B: The reason correctly explains the assertion, so it's not just a true statement but an explanation.
- Option C: The assertion is true, and the reason is also true and explains the assertion.
- Option D: Both the assertion and the reason are true.

Key Concepts:
- Complex formation
- Stability constants
- Selective precipitation
- Solubility product

Memory Aids:
- Higher stability constant means stronger complex formation.

Quick Tips:
- Consider the relative stability constants when dealing with selective precipitation problems.

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