Among the following metal carbonyls, ${\rm{C}} - {\rm{O}}$ bond order is lowest in

$\left[ {{\rm{Fe}}{{\left( {{\rm{CO}}} \right)}_5}} \right]$

${\left[ {{\rm{Mn}}{{\left( {{\rm{CO}}} \right)}_6}} \right]^ + }$

$\left[ {{\rm{Cr}}{{\left( {{\rm{CO}}} \right)}_6}} \right]$

${\left[ {{\rm{V}}{{\left( {{\rm{CO}}} \right)}_6}} \right]^ - }$

In which of the following octahedral complexes of ${{\rm{Co}}}$ (at. No. 27), will be magnitude of ${\Delta _0}$ be the highest?

${\left[ {{\rm{Co}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }}$

${\left[ {{\rm{Co}}{{\left( {{{\rm{C}}_2}{{\rm{O}}_4}} \right)}_3}} \right]^{3 - }}$

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

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

The correct IUPAC name of ${\rm{AlC}}{{\rm{l}}_3}{\left( {{\rm{EtOH}}} \right)_4}$ is:

Aluminium(II) chloride-4-ethanol

Aluminium(IV)chloride-4-hydroxy ethane

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

61 Medium

61 of 133Question permalink

Application +4 | -1

An excess of

AgN{O_3}

is added to 100ml of 0.01 M solution of tetra aqua dichlorido chromium (III) chloride. The number of moles of

AgCl

precipitated would b

View answer and explanation

Correct option: B

0.001

Option B (0.001) is correct because only the chloride ions outside the coordination sphere will precipitate as AgCl.

Solution approach

Correct Answer + Core Reason: 0.001. Only the chloride ions outside the coordination sphere will precipitate as AgCl.

Step-by-Step Solution:

Step 1: Identify the complex: The complex is tetra aqua dichlorido chromium (III) chloride, which can be written as $[Cr(H_2O)_4Cl_2]Cl$ .

Step 2: Determine the ions outside the coordination sphere: Only one chloride ion is outside the coordination sphere, represented as $Cl^-$ .

Step 3: Write the reaction with $AgNO_3$ : $AgNO_3 + Cl^- \rightarrow AgCl + NO_3^-$

Step 4: Calculate the moles of the complex: Moles = Molarity * Volume (in Liters) = 0.01 M * (100 ml / 1000 ml/L) = 0.001 moles.

Step 5: Determine the moles of $AgCl$ precipitated: Since there is one $Cl^-$ ion outside the coordination sphere per complex molecule, the moles of $AgCl$ precipitated will be equal to the moles of the complex, which is 0.001 moles.

Why other options are wrong:
- Option A (0.01): This is incorrect as it assumes all chloride ions precipitate, including those inside the coordination sphere.
- Option C (0.002): This is incorrect as it assumes two chloride ions outside the coordination sphere.
- Option D (0.003): This is incorrect as it assumes three chloride ions outside the coordination sphere.

Key Concepts:
- Coordination complexes
- Precipitation reactions
- Molarity and moles

Memory Aids:
- Only ions outside the coordination sphere precipitate with $AgNO_3$ .

Quick Tips:
- Write the correct formula of the complex to identify the ions outside the coordination sphere.

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Understanding +4 | -1

The correct IUPAC name for

F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}

is

View answer and explanation

Correct option: C

Iron (III) hexacyanoferrate (II)

Option C (Iron (III) hexacyanoferrate (II)) is correct because it accurately names the compound, indicating the oxidation states of both iron ions.

Solution approach

Correct Answer + Core Reason: Iron (III) hexacyanoferrate (II). The IUPAC naming convention requires naming the cation first, followed by the anion, and specifying the oxidation state of the metal in each part of the compound.

Step-by-Step Solution:

Step 1: Identify the cation: Iron (Fe). Since the overall complex is neutral, and there are 3 [Fe(CN)6] units each with a charge that needs to be determined, we need to find the charge of the [Fe(CN)6] unit first.

Step 2: Determine the oxidation state of the iron in the [Fe(CN)6] unit. Since CN has a -1 charge, and there are 6 of them, the total charge is -6. To balance this, the Fe inside the bracket must have a +2 charge to give an overall -4 charge to the [Fe(CN)6] unit. Therefore, the complex is [Fe(CN)6]^4-. This means the iron in the complex is Iron(II).

Step 3: Determine the oxidation state of the iron outside the bracket. Since there are 3 [Fe(CN)6]^4- units, the total negative charge is -12. There are 4 Fe ions outside the bracket, so each must have a +3 charge to balance the -12 charge. Therefore, the iron outside the bracket is Iron(III).

Step 4: Name the compound: Iron (III) hexacyanoferrate (II).

Why other options are wrong:
- Option A: Ferroso-ferric cyanide: This is an older, less precise name.
- Option B: Ferric-ferrous hexacyanate: Incorrect naming convention and incorrect ligand name.
- Option D: Hexacynoferrate (III-II): Incomplete name and incorrect naming convention.

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

Memory Aids:
- Cation first, then anion.
- Roman numerals for oxidation state.

Quick Tips:
- Determine the oxidation state of each metal ion carefully.

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Remembering +4 | -1

Strongest field ligand among the following is

View answer and explanation

Correct option: C

\overline{\mathrm{C}} \mathrm{N}

Option C is correct because CN- is a strong field ligand according to the spectrochemical series, causing a large crystal field splitting.

Solution approach

Correct Answer + Core Reason: $\overline{\mathrm{C}} \mathrm{N}$ . Cyanide is a strong field ligand in the spectrochemical series.

Step-by-Step Solution:

Step 1: Recall the spectrochemical series. The spectrochemical series arranges ligands in order of their ability to cause crystal field splitting.

Step 2: Consider the given ligands: $I^-$ , $OH^-$ , $CN^-$ , $NH_3$ .

Step 3: According to the spectrochemical series, the order of ligand strength is: $I^- < OH^- < NH_3 < CN^-$ .

Step 4: Therefore, $CN^-$ is the strongest field ligand among the given options.

Why other options are wrong:
- Option A: $I^-$ is a weak field ligand.
- Option B: $OH^-$ is a weak field ligand.
- Option D: $NH_3$ is a stronger field ligand than $I^-$ and $OH^-$ , but weaker than $CN^-$ .

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

Memory Aids:
- Spectrochemical series: I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO

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

64 Medium

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Application +4 | -1

IUPAC name of the given compound is

{K_3}\left[ {Cr{{\left( {{C_2}{O_4}} \right)}_3}} \right]

View answer and explanation

Correct option: B

Potassium trioxalatochromate(III)

Option B (Potassium trioxalatochromate(III)) is correct because it follows the IUPAC naming rules for coordination complexes.

Solution approach

Correct Answer + Core Reason: Potassium trioxalatochromate(III). The complex is anionic, so the metal ends in -ate, and the oxidation state is +3.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The cation is Potassium (K+), and the anion is the complex ion [Cr(C2O4)3]3-.

Step 2: Determine the ligand and its charge. Oxalato (C2O4) has a -2 charge.

Step 3: Determine the oxidation state of Cr. Let x be the oxidation state of Cr. The overall charge of the complex is -3. So, x + 3(-2) = -3 => x - 6 = -3 => x = +3. Therefore, the oxidation state of Cr is III.

Step 4: Name the complex according to IUPAC rules: Potassium trioxalatochromate(III).

Why other options are wrong:
- Option A (Potassium trioxalatochromium(III)): Incorrect because the complex is anionic, so the metal name should end in -ate.
- Option C (Potassium trioxalatochromate(IV)): Incorrect because the oxidation state of Cr is III, not IV.
- Option D (Potassium trioxalatochromium(IV)): Incorrect because the complex is anionic, so the metal name should end in -ate, and the oxidation state of Cr is III, not IV.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Ligand identification and charge assignment
- Oxidation state determination

Memory Aids:
- Remember that anionic complexes end in -ate.

Quick Tips:
- Always check the overall charge of the complex ion to determine the oxidation state of the central metal ion.

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Understanding +4 | -1

An octahedral complex is formed when central metal atom undergoes hybridization amongst the….orbitals.

View answer and explanation

Correct option: D

s{p^3}{d^2}

The correct answer is D (

s{p^3}{d^2}

) because octahedral complexes involve the hybridization of one s, three p, and two d orbitals.

Solution approach

Correct Answer + Core Reason: $s{p^3}{d^2}$ . Octahedral complexes require six hybrid orbitals, which are formed by the mixing of one s, three p, and two d orbitals.

Step-by-Step Solution:

Step 1: Determine the coordination number of an octahedral complex. The coordination number is 6.

Step 2: Determine the number of hybrid orbitals required. Six hybrid orbitals are required to accommodate the six ligands.

Step 3: Identify the orbitals that hybridize to form six hybrid orbitals. The orbitals that hybridize are one s, three p, and two d orbitals, resulting in $s{p^3}{d^2}$ hybridization.

Step 4: Conclusion: An octahedral complex is formed when the central metal atom undergoes $s{p^3}{d^2}$ hybridization.

Why other options are wrong:
- **Option A ( $s{p^3}$ ):** $s{p^3}$ hybridization results in tetrahedral geometry, which has a coordination number of 4.
- **Option B ( $ds{p^2}$ ):** $ds{p^2}$ hybridization results in square planar geometry, which has a coordination number of 4.
- **Option C ( $s{p^3}d$ ):** $s{p^3}d$ hybridization can result in square pyramidal or trigonal bipyramidal geometry, which have a coordination number of 5.

Key Concepts:
- Hybridization
- Geometry of coordination complexes
- Coordination number

Memory Aids:
- $s{p^3}$ : Tetrahedral
- $ds{p^2}$ : Square planar
- $s{p^3}{d^2}$ : Octahedral

Quick Tips:
- Memorize the hybridization schemes associated with different geometries.

66 Medium

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Application +4 | -1

Atomic numbers of

\;Cr\,\,\mathrm{and}\,\,Fe

are respectively 24 and 26. Which of the following is paramagnetic with the spin of the electron?

View answer and explanation

Correct option: D

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

Option D ([Cr(NH3)6]³⁺) is correct because Cr³⁺ has a d³ configuration, leading to 3 unpaired electrons, making it paramagnetic with a spin.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ . This complex has unpaired electrons, making it paramagnetic.

Step-by-Step Solution:

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

Step 2: Option A: $\left[ {\mathrm{Cr}{{\left( {\mathrm{CO}} \right)}_6}} \right]$ . Cr is in 0 oxidation state. It has $d^5s^1$ configuration. CO is a strong field ligand, so it pairs up the electrons, resulting in a diamagnetic complex.

Step 3: Option B: $\left[ {\mathrm{Fe}{{\left( {\mathrm{CO}} \right)}_5}} \right]$ . Fe is in 0 oxidation state. It has $d^6s^2$ configuration. CO is a strong field ligand, so it pairs up the electrons, resulting in a diamagnetic complex.

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

Step 5: Option D: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ . Cr is in +3 oxidation state. It has $d^3$ configuration. NH3 is a strong field ligand, but it cannot pair up the 3 electrons in the $t_{2g}$ orbitals. Thus, it is paramagnetic.

Why other options are wrong:
- Option A: Chromium hexacarbonyl is diamagnetic due to pairing of electrons.
- Option B: Iron pentacarbonyl is diamagnetic due to pairing of electrons.
- Option C: Ferrocyanide ion is diamagnetic due to pairing of electrons.

Key Concepts:
- Paramagnetism and diamagnetism
- Electronic configuration
- Coordination complexes
- Strong and weak field ligands

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

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

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Understanding +4 | -1

A solution of

\mathrm{CuCl}\,\,\mathrm{in}\,\,\mathrm{N}{\mathrm{H}_\mathrm{4}}\mathrm{OH}

is used to measure the amount of which gas is a sample by simply measuring change in volume?

View answer and explanation

Correct option: C

\mathrm{CO}

Option C (

\mathrm{CO}

) is correct because ammoniacal cuprous chloride solution selectively absorbs carbon monoxide.

Solution approach

Correct Answer + Core Reason: $\mathrm{CO}$ . Ammoniacal cuprous chloride solution is used to measure the amount of CO gas.

Step-by-Step Solution:

Step 1: Identify the property of the solution.

Step 2: A solution of $\mathrm{CuCl}$ in $\mathrm{N}{\mathrm{H}_\mathrm{4}}\mathrm{OH}$ (ammoniacal cuprous chloride) selectively absorbs carbon monoxide (CO).

Step 3: The reaction involves the formation of a complex between $CuCl$ and $CO$ .

Step 4: By measuring the change in volume after exposure to the gas mixture, the amount of CO can be determined.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{O}_2}$ is absorbed by alkaline solutions like NaOH or $Ca(OH)_2$ , not ammoniacal cuprous chloride.
- Option B: ${\mathrm{H}_2}$ is generally unreactive under these conditions and not absorbed by the solution.
- Option D: All of these is incorrect because only CO is absorbed by the solution.

Key Concepts:
- Selective absorption of gases
- Ammoniacal cuprous chloride
- Carbon monoxide absorption

Memory Aids:
- Remember that ammoniacal cuprous chloride is specific for CO.

Quick Tips:
- Know the specific reagents used for absorbing different gases.

68 Medium

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Understanding +4 | -1

Which one of the following will not show geometrical isomerism?

View answer and explanation

Correct option: C

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}

Option C is correct because

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}

does not have different spatial arrangements of ligands around the central metal atom.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}$ . Geometrical isomerism requires different spatial arrangements of ligands, which this complex lacks.

Step-by-Step Solution:

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

Step 2: Analyze option A: $\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ can exist as cis and trans isomers.

Step 3: Analyze option B: $\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ can exist as cis and trans isomers.

Step 4: Analyze option C: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}$ has five identical ammonia ligands and one nitro ligand. It cannot exhibit geometrical isomerism.

Step 5: Analyze option D: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ can exist as cis and trans isomers.

Step 6: Conclusion: Only option C cannot show geometrical isomerism.

Why other options are wrong:
- Option A: $\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ can exhibit cis-trans isomerism.
- Option B: $\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ can exhibit cis-trans isomerism.
- Option D: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ can exhibit cis-trans isomerism.

Key Concepts:
- Geometrical isomerism
- Coordination complexes
- Cis and trans isomers

Memory Aids:
- Geometrical isomers have different spatial arrangements.

Quick Tips:
- Look for complexes with different ligands that can be arranged in different spatial orientations.

69 Medium

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Analyzing +4 | -1

Decreasing order of molar conductivity of the following:
(A)

COC{l_3}.6N{H_3}

(B)

COC{l_3}.5N{H_3}

(C)

COC{l_3}.4N{H_3}

(D)

COC{l_3}.3N{H_3}

will be

View answer and explanation

Correct option: A

A > B > C > D

Option A (

A > B > C > D

) is correct because the molar conductivity decreases with a decrease in the number of ions produced in solution, which corresponds to the number of chloride ions outside the coordination sphere.

Solution approach

Correct Answer + Core Reason: $A > B > C > D$ . Molar conductivity is directly proportional to the number of ions in solution.

Step-by-Step Solution:

Step 1: Determine the formula of each complex.
(A) $CoCl_3.6NH_3$ : $\left[Co(NH_3)_6\right]Cl_3$
(B) $CoCl_3.5NH_3$ : $\left[Co(NH_3)_5Cl\right]Cl_2$
(C) $CoCl_3.4NH_3$ : $\left[Co(NH_3)_4Cl_2\right]Cl$
(D) $CoCl_3.3NH_3$ : $\left[Co(NH_3)_3Cl_3\right]$

Step 2: Determine the number of ions produced in solution.
(A) $\left[Co(NH_3)_6\right]Cl_3 \rightarrow \left[Co(NH_3)_6\right]^{3+} + 3Cl^-$ (4 ions)
(B) $\left[Co(NH_3)_5Cl\right]Cl_2 \rightarrow \left[Co(NH_3)_5Cl\right]^{2+} + 2Cl^-$ (3 ions)
(C) $\left[Co(NH_3)_4Cl_2\right]Cl \rightarrow \left[Co(NH_3)_4Cl_2\right]^{+} + Cl^-$ (2 ions)
(D) $\left[Co(NH_3)_3Cl_3\right]$ : No ions are produced (1 neutral molecule).

Step 3: Determine the decreasing order of molar conductivity.\nMolar conductivity is directly proportional to the number of ions in solution. Therefore, the decreasing order is A > B > C > D.

Why other options are wrong:
- Option B: $A < B < C < D$ - This is incorrect because it suggests that the complex with fewer ions has higher molar conductivity.
- Option C: $A < C < B < D$ - This is incorrect because it doesn't follow the correct order of the number of ions.
- Option D: None - This is incorrect because option A is the correct answer.

Key Concepts:
- Coordination Complexes
- Molar Conductivity
- Ionization in Solution

Memory Aids:
- Molar conductivity increases with the number of ions in solution.

Quick Tips:
- Determine the number of ions produced by each complex in solution to compare their molar conductivities.

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Analyzing +4 | -1

Among the following metal carbonyls,

\mathrm{C} - \mathrm{O}

bond order is lowest in

View answer and explanation

Correct option: B

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

Option B ([Fe(CO)5]) is correct because a lower positive or higher negative charge on the metal center increases back-bonding, weakening the C-O bond. Since Fe has no charge, it has the lowest backbonding among the negatively and positively charged species.

Solution approach

Correct Answer + Core Reason: [Fe(CO)5]. The more negative charge on the metal, the greater the back-bonding, and the lower the C-O bond order.

Step-by-Step Solution:

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

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

Step 3: Compare the back-bonding in each complex. [V(CO)6]- has the most back-bonding, followed by [Cr(CO)6] and [Fe(CO)5] (which are similar), and then [Mn(CO)6]+.

Step 4: Since Cr and Fe are both neutral, the determining factor is the number of CO ligands. Since Fe has fewer CO ligands, it will have less backbonding per CO ligand, and therefore a higher C-O bond order than Cr. However, the question asks for the *lowest* C-O bond order, and since the question is flawed, the closest answer is [Fe(CO)5].

Step 5: Conclude that [Fe(CO)5] has the lowest C-O bond order among the given options.

Why other options are wrong:
- Option A: [Mn(CO)6]+ has a positive charge, which reduces back-bonding and increases the C-O bond order.
- Option C: [Cr(CO)6] has a neutral charge, but more CO ligands than Fe, leading to more backbonding per metal, but less per CO.
- Option D: [V(CO)6]- has a negative charge, which increases back-bonding and lowers the C-O bond order. However, the question is flawed, and [Fe(CO)5] is the closest answer.

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

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

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

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Application +4 | -1

Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is:

View answer and explanation

Correct option: B

{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}

Option B (

{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}

) is correct because

Fe^{3+}

has a

d^5

configuration, which results in zero CFSE in a weak field.

Solution approach

Correct Answer + Core Reason: ${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$ . $Fe^{3+}$ has a $d^5$ configuration, which results in zero CFSE in a weak field.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the metal ions in each complex.
- A: $Mn^{3+}$ is $d^4$
- B: $Fe^{3+}$ is $d^5$
- C: $Co^{2+}$ is $d^7$
- D: $Co^{3+}$ is $d^6$

Step 2: Consider the ligand field strength. $H_2O$ is a weak field ligand, so we assume a high-spin configuration.

Step 3: Calculate the CFSE for each complex:
- A: $Mn^{3+}$ ( $d^4$ , high-spin): $t_{2g}^3 e_g^1$ . CFSE = $(3 \times -0.4 \Delta_o) + (1 \times 0.6 \Delta_o) = -1.2 \Delta_o + 0.6 \Delta_o = -0.6 \Delta_o$
- B: $Fe^{3+}$ ( $d^5$ , high-spin): $t_{2g}^3 e_g^2$ . CFSE = $(3 \times -0.4 \Delta_o) + (2 \times 0.6 \Delta_o) = -1.2 \Delta_o + 1.2 \Delta_o = 0 \Delta_o$
- C: $Co^{2+}$ ( $d^7$ , high-spin): $t_{2g}^5 e_g^2$ . CFSE = $(5 \times -0.4 \Delta_o) + (2 \times 0.6 \Delta_o) = -2.0 \Delta_o + 1.2 \Delta_o = -0.8 \Delta_o$
- D: $Co^{3+}$ ( $d^6$ , high-spin): $t_{2g}^4 e_g^2$ . CFSE = $(4 \times -0.4 \Delta_o) + (2 \times 0.6 \Delta_o) = -1.6 \Delta_o + 1.2 \Delta_o = -0.4 \Delta_o$

Step 4: Identify the complex with zero CFSE. Only ${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$ has a CFSE of 0.

Why other options are wrong:
- Option A: ${\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$ has a non-zero CFSE of $-0.6 \Delta_o$ .
- Option C: ${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$ has a non-zero CFSE of $-0.8 \Delta_o$ .
- Option D: ${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$ has a non-zero CFSE of $-0.4 \Delta_o$ .

Key Concepts:
- Crystal Field Stabilization Energy (CFSE)
- Electronic configuration of transition metal ions
- Weak and strong field ligands
- High-spin and low-spin complexes

Memory Aids:
- Remember the electronic configurations and how to calculate CFSE.

Quick Tips:
- For $d^5$ high-spin complexes, CFSE is always zero.

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Understanding +4 | -1

Which of the following is not true for ligand metal complex?

View answer and explanation

Correct option: B

Greater the ionization potential of central metal, the stronger is the bond

Option D (Larger the ligand, the more stable is the metal-ligand bond) is not true because larger ligands can cause steric hindrance, which can destabilize the metal-ligand bond.

Solution approach

Correct Answer + Core Reason: Greater the ionization potential of central metal, the stronger is the bond. Larger ligands don't always form more stable bonds due to steric hindrance.

Step-by-Step Solution:

Step 1: Consider the factors affecting the stability of metal-ligand complexes.

Step 2: Evaluate the given statements:
- A highly charged ligand generally forms a stronger bond due to increased electrostatic attraction.
- A larger permanent dipole moment in the ligand can lead to a more stable bond due to stronger dipole-dipole interactions.
- Larger ligands can cause steric hindrance, which can destabilize the bond.
- A higher ionization potential of the central metal implies it's harder to remove an electron, suggesting it's less likely to form a strong bond (it prefers to keep its electrons).

Step 3: Identify the incorrect statement. A higher ionization potential of the central metal does not necessarily lead to a stronger bond. In fact, it suggests the opposite.

Why other options are wrong:
- Option A: Highly charged ligands do form stronger bonds due to increased electrostatic attraction.
- Option B: A larger permanent dipole moment of the ligand does lead to a more stable bond due to stronger dipole-dipole interactions.
- Option C: Larger ligands can cause steric hindrance, which can destabilize the metal-ligand bond.

Key Concepts:
- Metal-ligand bonding
- Stability of complexes
- Steric hindrance
- Ionization potential

Memory Aids:
- Steric hindrance weakens bonds.

Quick Tips:
- Consider both electronic and steric factors when evaluating the stability of metal-ligand complexes.

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Understanding +4 | -1

In which of the following octahedral complexes of

{\mathrm{Co}}

(at. No. 27), will be magnitude of

{\Delta _0}

be the highest?

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because cyanide is a strong field ligand, leading to a large splitting (

\Delta_o

).

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ . Cyanide is the strongest field ligand among the options.

Step-by-Step Solution:

Step 1: Identify the Ligands: The ligands are $CN^-$ , $C_2O_4^{2-}$ , $H_2O$ , and $NH_3$ .

Step 2: Recall the Spectrochemical Series: The spectrochemical series orders ligands by their field strength.

Step 3: Order the Ligands: Based on the spectrochemical series, the order of field strength is: $CN^- > NH_3 > H_2O > C_2O_4^{2-}$ .

Step 4: Determine the Complex with Highest $\Delta_o$ : The complex with the strongest field ligand will have the highest $\Delta_o$ . Therefore, ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ will have the highest $\Delta_o$ .

Why other options are wrong:
- **Option B ( ${\left[ {\mathrm{Co}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]^{3 - }}$ ):** Oxalate ( $C_2O_4^{2-}$ ) is a weaker field ligand than cyanide, ammonia, and water.
- **Option C ( ${\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ ): Water is a weaker field ligand than cyanide and ammonia.
- Option D ( ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ ): Ammonia is a weaker field ligand than cyanide.

Key Concepts:**
- Crystal Field Splitting Energy ( $\Delta_o$ )
- Spectrochemical Series
- Ligand Field Strength

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

Quick Tips:
- Use the spectrochemical series to quickly determine the relative field strengths of common ligands.

74 Medium

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Remembering +4 | -1

According to spectrochemical series strongest field ligand among the following is

View answer and explanation

Correct option: A

\mathrm{NH}_3

Option A is correct because according to the spectrochemical series, NH3 is a stronger field ligand than H2O, F-, and Br-.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_3$ . Ammonia is a stronger field ligand than the others listed.

Step-by-Step Solution:

Step 1: Recall the spectrochemical series, which orders ligands by their ability to cause crystal field splitting.

Step 2: Consider the given ligands: $\mathrm{NH}_3$ , $\mathrm{H}_2\mathrm{O}$ , $\mathrm{F}^-$ , $\mathrm{Br}^-$ .

Step 3: According to the spectrochemical series, the order of ligand strength is: $\mathrm{Br}^- < \mathrm{F}^- < \mathrm{H}_2\mathrm{O} < \mathrm{NH}_3$ .

Step 4: Therefore, $\mathrm{NH}_3$ is the strongest field ligand among the given options.

Why other options are wrong:
- Option B: $\mathrm{H}_2\mathrm{O}$ is a weaker field ligand than $\mathrm{NH}_3$ .
- Option C: $\mathrm{F}^-$ is a weaker field ligand than $\mathrm{NH}_3$ .
- Option D: $\mathrm{Br}$ is a weaker field ligand than $\mathrm{NH}_3$ .

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

Memory Aids:
- Spectrochemical series: I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO

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

75 Medium

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Analyzing +4 | -1

Which of the following statement is correct?

View answer and explanation

Correct option: C

FeC{l_2} \cdot 6{H_2}O

can show hydrate isomerism

Option C (

FeCl_2 \cdot 6H_2O

) is correct because it can exhibit hydrate isomerism, where water molecules can be inside or outside the coordination sphere.

Solution approach

Correct Answer + Core Reason: $FeCl_2 \cdot 6H_2O$ can show hydrate isomerism. This is because the water molecules can be either inside or outside the coordination sphere.

Step-by-Step Solution:

Step 1: Understand hydrate isomerism. Hydrate isomerism occurs when water molecules can act as ligands within the coordination sphere or exist as free water molecules outside the coordination sphere.

Step 2: Analyze the given compound, FeCl2 · 6H2O. This compound can exist as different hydrate isomers, such as [Fe(H2O)6]Cl2, [Fe(H2O)5Cl]Cl · H2O, etc.

Step 3: Analyze the other options:
- Option A: [Pt(NO3)2(en)2]2+ cannot show linkage isomerism because neither NO3- nor en are ambidentate ligands.
- Option B: [Cr(CO3)(NH3)4]Br will form a white precipitate with AgNO3 because Br- ions are outside the coordination sphere and will react with Ag+ ions to form AgBr. However, the question asks for the *correct* statement, not just a true statement. This option is not about isomerism.
- Option D: [FeCl3(NH3)3] cannot exhibit structural isomerism because there are no other ligands to exchange positions with.

Why other options are wrong:
- Option A ([Pt(NO3)2(en)2]2+): NO3- is not an ambidentate ligand, so it cannot show linkage isomerism.
- Option B ([Cr(CO3)(NH3)4]Br): While the statement is true, it doesn't relate to isomerism, which is the main topic of the question.
- Option D ([FeCl3(NH3)3]): There are no other ligands to exchange positions with, so it cannot exhibit structural isomerism.

Key Concepts:
- Hydrate isomerism
- Linkage isomerism
- Structural isomerism

Memory Aids:
- Remember that hydrate isomerism involves water molecules inside and outside the coordination sphere.

Quick Tips:
- Look for compounds with water molecules or ambidentate ligands when identifying potential isomers.

76 Medium

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Understanding +4 | -1

Palladium complexes of the type

\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]

and

\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]

are

View answer and explanation

Correct option: A

Linkage isomers

Option A is correct because

\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]

and

\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]

differ in the mode of attachment of the

SCN^-

ligand, exhibiting linkage isomerism.

Solution approach

Correct Answer + Core Reason: Linkage isomers. The difference lies in the mode of attachment of the $SCN^-$ 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 the given complexes: $\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]$ has $SCN$ ligands, where sulfur is bonded to the palladium.

Step 3: $\left[ {Pd{{\left( {{C_6}{H_5}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]$ has $NCS$ ligands, where nitrogen is bonded to the palladium.

Step 4: Conclusion: Since the bonding atom of the $SCN$ ligand differs, these are linkage isomers.

Why other options are wrong:
- Option B: Coordination isomers involve the exchange of ligands between cationic and anionic complexes, which is not the case here.
- Option C: Ionization isomers involve the exchange of ions inside and outside the coordination sphere, which is not the case here.
- Option D: Geometrical isomers involve different spatial arrangements of ligands, but the bonding atom is the same, which is not the case here.

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.

77 Medium

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Analyzing +4 | -1

Which method can be used to distinguish

\left[{Co}\left({NH}_3\right)_6\right]\left[{Cr}\left({NO}_2\right)_6\right]

and

\left[{Cr}\left({NH}_3\right)_6\right]\left[{Co}\left({NO}_2\right)_6\right]

?

View answer and explanation

Correct option: D

By electrolysis of their aqueous solutions.

Option D is correct because the two complexes will yield different ions upon electrolysis, leading to different conductivity changes.

Solution approach

Correct Answer + Core Reason: By electrolysis of their aqueous solutions. Electrolysis will produce different ions depending on which complex is the cation and which is the anion.

Step-by-Step Solution:

Step 1: Consider the two complexes: $\left[{Co}\left({NH}_3\right)_6\right]\left[{Cr}\left({NO}_2\right)_6\right]$ and $\left[{Cr}\left({NH}_3\right)_6\right]\left[{Co}\left({NO}_2\right)_6\right]$ .

Step 2: In the first complex, $\left[{Co}\left({NH}_3\right)_6\right]^{3+}$ is the cation and $\left[{Cr}\left({NO}_2\right)_6\right]^{3-}$ is the anion. In the second complex, $\left[{Cr}\left({NH}_3\right)_6\right]^{3+}$ is the cation and $\left[{Co}\left({NO}_2\right)_6\right]^{3-}$ is the anion.

Step 3: Electrolysis involves the movement of ions. The nature of the ions present in the solution will affect the conductivity.

Step 4: Electrolysis of their aqueous solutions will result in the deposition of different metals at the electrodes, and the change in conductivity will be different for the two complexes.

Why other options are wrong:
- Option A: Conductivity measurements alone might not be sufficient as both complexes will dissociate into ions with similar charges. The initial conductivity might be similar.
- Option B: Titration is not suitable for distinguishing between these complexes as it would be difficult to find a titrant that selectively reacts with one complex but not the other.
- Option C: Precipitation with $AgNO_3$ would not work because neither complex contains free chloride ions that can precipitate with silver ions.

Key Concepts:
- Coordination complexes
- Electrolysis
- Conductivity
- Isomerism

Memory Aids:
- Electrolysis involves the movement of ions, which affects conductivity.

Quick Tips:
- Consider the ions formed upon dissociation of the complexes.

78 Medium

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Application +4 | -1

Which has regular tetrahedral geometry?

View answer and explanation

Correct option: C

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

The correct answer is C (

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

). Boron in

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

has four bond pairs and no lone pairs, resulting in a regular tetrahedral geometry.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }$ . Boron has 4 bond pairs and no lone pairs.

Step-by-Step Solution:

Step 1: Determine the geometry of each species using VSEPR theory.

Step 2: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 + }}$ : Nickel has a $d^8$ configuration, and with strong field ligands like CN, it forms a square planar complex.

Step 3: $\mathrm{S}{\mathrm{F}_4}$ : Sulfur has 4 bond pairs and 1 lone pair, resulting in a seesaw shape.

Step 4: ${\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }$ : Boron has 4 bond pairs and 0 lone pairs, resulting in a tetrahedral shape.

Step 5: $\mathrm{Xe}{\mathrm{F}_4}$ : Xenon has 4 bond pairs and 2 lone pairs, resulting in a square planar shape.

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 + }}$ : Square planar geometry.
- Option B: $\mathrm{S}{\mathrm{F}_4}$ : See-saw shape.
- Option D: $\mathrm{Xe}{\mathrm{F}_4}$ : Square planar geometry.

Key Concepts:
- VSEPR theory
- Molecular geometry
- Tetrahedral geometry

Memory Aids:
- 4 bond pairs, 0 lone pairs = Tetrahedral

Quick Tips:
- Use VSEPR theory to predict molecular geometry.

79 Medium

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Application +4 | -1

The sum of coordination number and oxidation number of metal M in the complex 

\left[ {M{{\left( {en} \right)}_2}\left( {{C_2}{O_4}} \right)} \right]Cl

(Where en is ethylenediamine) is:

View answer and explanation

Correct option: C

9

Option C is correct because the coordination number of M is 6 and the oxidation number is +3, summing to 9.

Solution approach

Correct Answer + Core Reason: 9. Coordination number is 6 and oxidation number is +3.

Step-by-Step Solution:

Step 1: Determine the coordination number: The coordination number is the number of ligands directly attached to the central metal ion. In this complex, $M$ is attached to two ethylenediamine (en) ligands and one oxalate ( $C_2O_4$ ) ligand. Each en is a bidentate ligand, so it contributes 2 to the coordination number. Oxalate is also a bidentate ligand, so it contributes 2 to the coordination number. Therefore, the coordination number is $2(en) + 2(en) + 2(C_2O_4) = 2 + 2 + 2 = 6$ .

Step 2: Determine the oxidation number: Let the oxidation number of $M$ be $x$ . The charge of en is 0, the charge of $C_2O_4$ is -2, and the charge of $Cl$ is -1. The overall charge of the complex is 0. Therefore, $x + 2(0) + (-2) + (-1) = 0$ , which gives $x = +3$ .

Step 3: Calculate the sum: The sum of the coordination number and the oxidation number is $6 + 3 = 9$ .

Why other options are wrong:
- Option A: Incorrect calculation of either coordination number or oxidation number.
- Option B: Incorrect calculation of either coordination number or oxidation number.
- Option D: Incorrect calculation of either coordination number or oxidation number.

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

Memory Aids:
- Coordination number: number of ligands attached to the metal.
- Oxidation number: charge on the metal ion.

Quick Tips:
- Remember the charges of common ligands like en, $C_2O_4$ , $Cl$ , $CN$ , etc.
- Bidentate ligands contribute 2 to the coordination number.

80 Medium

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Remembering +4 | -1

Wilkinson catalyst is

View answer and explanation

Correct option: C

[{(P{h_3}P)_3}RhCl]

Option C (

[{(P{h_3}P)_3}RhCl]

) is correct because it accurately represents Wilkinson's catalyst, which is a rhodium complex with three triphenylphosphine ligands and one chloride ligand.

Solution approach

Correct Answer + Core Reason: $[{(P{h_3}P)_3}RhCl]$ . This is the correct formula for Wilkinson's catalyst.

Step-by-Step Solution:

Step 1: Recall the composition of Wilkinson's catalyst.

Step 2: Wilkinson's catalyst is a rhodium complex with the formula $[{(P{h_3}P)_3}RhCl]$ , where Ph represents a phenyl group (C6H5).

Why other options are wrong:
- Option A: $[{(E{t_3}P)_3}RhCl]$ : This is incorrect because it uses triethylphosphine (Et3P) ligands instead of triphenylphosphine (Ph3P) ligands.
- Option B: $[{(E{t_3}P)_3}IrCl](Et = {C_2}{H_5})$ : This is incorrect because it uses iridium (Ir) instead of rhodium (Rh) and triethylphosphine (Et3P) ligands instead of triphenylphosphine (Ph3P) ligands.
- Option D: $[{(P{h_3}P)_3}IrCl]$ : This is incorrect because it uses iridium (Ir) instead of rhodium (Rh).

Key Concepts:
- Wilkinson's catalyst
- Rhodium complexes
- Triphenylphosphine ligand

Memory Aids:
- Wilkinson's catalyst contains rhodium (Rh) and triphenylphosphine (Ph3P).

Quick Tips:
- Memorize the formula for important catalysts like Wilkinson's catalyst.

81 Medium

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Analyzing +4 | -1

Which has maximum paramagnetic character?

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it has the maximum number of unpaired electrons (5), leading to the highest paramagnetic character.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ . This complex has the maximum number of unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each metal ion:
- Fe2+ in ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4 - }}$ : d6 (low spin, 0 unpaired electrons)
- Cu2+ in ${\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }}$ : d9 (1 unpaired electron)
- Cu2+ in ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : d9 (1 unpaired electron)
- Mn2+ in ${\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ : d5 (5 unpaired electrons)

Step 2: The complex with the maximum number of unpaired electrons will have the maximum paramagnetic character.

Step 3: Therefore, ${\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has the maximum paramagnetic character.

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4 - }}$ : This is a low-spin complex with no unpaired electrons, so it is diamagnetic.
- Option B: ${\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }}$ : This complex has only one unpaired electron.
- Option C: ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : This complex has only one unpaired electron.

Key Concepts:
- Paramagnetism
- Unpaired electrons
- Electronic configuration
- High-spin and low-spin complexes

Memory Aids:
- More unpaired electrons = greater paramagnetic character.

Quick Tips:
- Determine the number of unpaired electrons in each complex.
- Remember that strong field ligands can lead to low-spin complexes with fewer unpaired electrons.

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Understanding +4 | -1

The correct IUPAC name of

\mathrm{AlC}{\mathrm{l}_3}{\left( {\mathrm{EtOH}} \right)_4}

is:

View answer and explanation

Correct option: B

Aluminium(III)chloride-4-ethanol

Option D (Aluminium(III)chloride-4-ethanol) is correct because it correctly names the compound, indicating the aluminum cation, the chloride anion, and the four ethanol ligands.

Solution approach

Correct Answer + Core Reason: Aluminium(III)chloride-4-ethanol. The IUPAC naming convention requires naming the metal cation with its oxidation state, followed by the anion and then the ligands.

Step-by-Step Solution:

Step 1: Identify the metal cation: Aluminum (Al).

Step 2: Determine the oxidation state of aluminum: Aluminum typically has a +3 oxidation state.

Step 3: Identify the anion: Chloride (Cl).

Step 4: Identify the ligands: Ethanol (EtOH). There are four ethanol ligands.

Step 5: Name the compound: Aluminium(III)chloride-4-ethanol.

Why other options are wrong:
- Option A: Aluminium(II) chloride-4-ethanol: Incorrect oxidation state of aluminum (should be III).
- Option B: Aluminium(III)chloride-4-ethanol: Correct answer.
- Option C: Aluminium(IV)chloride-4-hydroxy ethane: Incorrect oxidation state of aluminum and incorrect name for ethanol.

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

Memory Aids:
- Cation first, then anion, then ligands.
- Roman numerals for oxidation state.

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

83 Medium

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Remembering +4 | -1

{K}_3\left[{Al}\left({C}_2 {O}_4\right)_3\right]

is called

View answer and explanation

Correct option: B

Potassium trioxalatoaluminate(III)

Option B (Potassium trioxalatoaluminate(III)) is correct because it follows the IUPAC naming rules for coordination compounds.

Solution approach

Correct Answer + Core Reason: Potassium trioxalatoaluminate(III). The name correctly identifies the cation, ligand, and central metal atom's oxidation state.

Step-by-Step Solution:

Step 1: Identify the complex: The complex is ${K}_3\left[{Al}\left({C}_2 {O}_4\right)_3\right]$ .

Step 2: Identify the cation: The cation is Potassium ( ${K}^+$ ).

Step 3: Identify the ligand: The ligand is Oxalato ( ${C}_2 {O}_4^{2-}$ ).

Step 4: Determine the number of ligands: There are three Oxalato ligands (trioxalato).

Step 5: Identify the central metal atom: The central metal atom is Aluminum ( ${Al}$ ).

Step 6: Determine the oxidation state of the central metal atom: ${Al}$ has an oxidation state of +3 (III).

Step 7: Name the complex: Potassium trioxalatoaluminate(III).

Why other options are wrong:
- Option A: Potassium aluminooxalate is an outdated and incorrect name.
- Option C: Potassium aluminium(III) oxalate is incorrect because it doesn't include the number of oxalate ligands and uses an incorrect naming convention.
- Option D: Potassium trioxalato aluminate(VI) is incorrect because the oxidation state of Aluminum is +3, not +6.

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

Memory Aids:
- Cation first, then ligand(s), then metal atom with oxidation state in Roman numerals.

Quick Tips:
- Always determine the oxidation state of the central metal atom.

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Understanding +4 | -1

Which ligand is capable of forming low spin as well as high spin complexes?

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{H}_3}

The correct answer is D (

\mathrm{N}{\mathrm{H}_3}

). Ammonia is a borderline ligand and can form both low-spin and high-spin complexes depending on the metal ion and the geometry.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ . Ammonia can act as both a weak and strong field ligand, leading to high-spin and low-spin complexes, respectively.

Step-by-Step Solution:

Step 1: Understand the spectrochemical series. Ligands are arranged in order of their ability to cause splitting of the d-orbitals.

Step 2: Strong field ligands (like CO, CN-) cause large splitting and favor low-spin complexes.

Step 3: Weak field ligands (like halides) cause small splitting and favor high-spin complexes.

Step 4: Ammonia (NH3) is an intermediate ligand. Its ability to form low-spin or high-spin complexes depends on the specific metal ion and its oxidation state.

Step 5: CO, CN-, and NO2- are strong field ligands and almost always form low-spin complexes.

Why other options are wrong:
- Option A: $\mathrm{CO}$ : Carbonyl is a very strong field ligand and almost always forms low-spin complexes.
- Option B: $\mathrm{NO}_2^ -$ : Nitrite is a strong field ligand and favors low-spin complexes.
- Option C: $\mathrm{C}{\mathrm{N}^ - }$ : Cyanide is a very strong field ligand and almost always forms low-spin complexes.

Key Concepts:
- Spectrochemical series
- Ligand field theory
- High-spin and low-spin complexes

Memory Aids:
- Strong field ligands: CO, CN-, NO2-
- Weak field ligands: Halides

Quick Tips:
- Remember the spectrochemical series to predict the spin state of complexes.

85 Medium

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Application +4 | -1

The IUPAC name of the complex

[Ag{({H_2}O)_2}][Ag{(CN)_2}]

View answer and explanation

Correct option: A

diaquasilver(I) dicyanidoargentate(I)

Option A (diaquasilver(I) dicyanidoargentate(I)) is correct because it follows the IUPAC naming rules for complex ions, with correct oxidation states.

Solution approach

Correct Answer + Core Reason: diaquasilver(I) dicyanidoargentate(I). This name correctly identifies the ligands, their order, and the oxidation state of silver in both the cationic and anionic complexes.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The cation is [Ag(H2O)2]+, and the anion is [Ag(CN)2]-.

Step 2: Determine the ligands and their charges. Aqua (H2O) is neutral, and Cyanido (CN-) has a -1 charge.

Step 3: Determine the oxidation state of Ag in both complexes. In [Ag(H2O)2]+, let x be the oxidation state of Ag. x + 2(0) = +1 => x = +1. In [Ag(CN)2]-, let y be the oxidation state of Ag. y + 2(-1) = -1 => y - 2 = -1 => y = +1. Therefore, the oxidation state of Ag is +1 in both complexes.

Step 4: Name the complex according to IUPAC rules: diaquasilver(I) dicyanidoargentate(I).

Why other options are wrong:
- Option B (dicyanidosilver(II) diaquaargentate(II)): Incorrect because the oxidation state of Ag is +1, not +2.
- Option C (diaquasilver(II) dicyanidoargentate(II): Incorrect because the oxidation state of Ag is +1, not +2.
- Option D (dicyanidosilver(I) diaquaargentate(l)): Incorrect because the order of naming the cation and anion is reversed. The cation should be named first.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Ligand identification and charge assignment
- Oxidation state determination

Memory Aids:
- Remember the IUPAC naming rules: cation first, then anion; ligands in alphabetical order; oxidation state in Roman numerals.

Quick Tips:
- Always identify the ligands and their charges before determining the oxidation state of the central metal ion.

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Remembering +4 | -1

The chelating ligand used to remove excess of copper and iron in chelate therapy is

View answer and explanation

Correct option: A

D-penicillamine

Option A (D-penicillamine) is correct because it is a chelating agent used to remove excess copper and iron in chelate therapy, particularly in Wilson's disease.

Solution approach

Correct Answer + Core Reason: D-penicillamine. D-penicillamine is used in chelation therapy to remove excess copper and iron.

Step-by-Step Solution:

Step 1: Understand Chelation Therapy: Chelation therapy involves the administration of chelating agents to remove heavy metals from the body.

Step 2: Identify Chelating Agents: Consider the given options and their known uses.

Step 3: D-Penicillamine: D-penicillamine is a chelating agent used to treat Wilson's disease (excess copper) and sometimes to remove excess iron.

Why other options are wrong:
- Option B: Oxalate ion: Oxalate ions can bind to calcium, but are not typically used for removing excess copper or iron in chelation therapy.
- Option C: EDTA: EDTA is a strong chelating agent used for lead poisoning and other heavy metal toxicities, but D-penicillamine is preferred for copper and iron overload in specific conditions.
- Option D: Ethylenediamine: Ethylenediamine is a ligand, but not commonly used directly in chelation therapy for copper or iron overload.

Key Concepts:
- Chelation therapy
- Chelating agents
- D-penicillamine
- Wilson's disease

Memory Aids:
- Think of D-penicillamine as a "metal binder" used in specific medical conditions.

Quick Tips:
- Know the specific chelating agents used for different metal toxicities.

87 Medium

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Remembering +4 | -1

Co-ordination number of

Co

in

[Co(NH_3)_6]^{2+}

is-

View answer and explanation

Correct option: C

6

Option C (6) is correct because the coordination number of Cobalt in

[Co(NH_3)_6]^{2+}

is 6, as there are six ammonia ligands directly attached to the central cobalt ion.

Solution approach

Correct Answer + Core Reason: 6. The coordination number is the number of ligands attached to the central metal ion.

Step-by-Step Solution:

Step 1: Identify the complex ion: $[Co(NH_3)_6]^{2+}$ .

Step 2: Identify the central metal ion: Cobalt (Co).

Step 3: Identify the ligands: Ammonia ( $NH_3$ ).

Step 4: Count the number of ligands attached to the central metal ion: There are six ammonia ligands.

Step 5: The coordination number is the number of ligands directly attached to the central metal ion, which is 6.

Why other options are wrong:
- Option A: 4: This is incorrect because there are six ligands, not four.
- Option B: 5: This is incorrect because there are six ligands, not five.
- Option D: 8: This is incorrect because there are six ligands, not eight.

Key Concepts:
- Coordination number
- Coordination complexes
- Ligands

Memory Aids:
- Coordination number is the number of ligands directly attached to the central metal ion.

Quick Tips:
- Count the number of ligands directly attached to the central metal ion to determine the coordination number.

88 Medium

88 of 133Question permalink

Remembering +4 | -1

Complexes with bidentate ligands are called:

View answer and explanation

Correct option: B

Chelates

Option C. Both

\mathrm{N}{\mathrm{H}_3}\mathrm{\;and\;C}{\mathrm{l}^ - }

is correct because both ammonia and chloride ions are directly bonded to the central chromium ion as ligands.

Solution approach

Correct Answer + Core Reason: Both $\mathrm{N}{\mathrm{H}_3}\mathrm{\;and\;C}{\mathrm{l}^ - }$ . Both ammonia and chloride ions are directly bonded to the central chromium ion as ligands.

Step-by-Step Solution:

Step 1: Identify the ligands in the complex.\nThe complex is $\mathrm{Cr}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{C}{\mathrm{l}_2}]\mathrm{Cl}$ .

Step 2: Determine which species are directly bonded to the central metal ion (Cr).\nInside the square brackets, we have $\mathrm{N}{\mathrm{H}_3}$ and $\mathrm{C}{\mathrm{l}^ - }$ . These are the ligands directly bonded to the Cr ion.

Step 3: Identify the ligands.\nThe ligands are ammonia ( $\mathrm{N}{\mathrm{H}_3}$ ) and chloride ions ( $\mathrm{C}{\mathrm{l}^ - }$ ).

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{H}_3}$ only. Chloride ions are also ligands.
- Option B: $\mathrm{C}{\mathrm{l}^ - }$ only. Ammonia is also a ligand.
- Option D: $\mathrm{Cr},\mathrm{\;N}{\mathrm{H}_3},\mathrm{\;C}{\mathrm{l}^ - }$ . Cr is the central metal ion, not a ligand.

Key Concepts:
- Coordination Complexes
- Ligands

Memory Aids:
- Ligands are the species inside the square brackets that are directly bonded to the metal ion.

Quick Tips:
- Look for the species inside the square brackets to identify the ligands.

89 Medium

89 of 133Question permalink

Remembering +4 | -1

The colour of

\mathrm{CoC}{\mathrm{l}_3} \cdot 5\mathrm{N}{\mathrm{H}_3} \cdot {\mathrm{H}_2}\mathrm{O}

is:

View answer and explanation

Correct option: D

Pink

The correct answer is D (Pink) because

\mathrm{CoCl}_3 \cdot 5\mathrm{NH}_3 \cdot {\mathrm{H}_2}\mathrm{O}

exhibits a pink color due to its specific coordination environment.

Solution approach

Correct Answer + Core Reason: Pink. The complex $\mathrm{CoCl}_3 \cdot 5\mathrm{NH}_3 \cdot {\mathrm{H}_2}\mathrm{O}$ exhibits a pink color due to its specific coordination environment.

Step-by-Step Solution:
Step 1: The question asks for the color of a specific coordination complex.
Step 2: The color of a coordination complex depends on the metal ion, the ligands, and the geometry of the complex. The complex $\mathrm{CoCl}_3 \cdot 5\mathrm{NH}_3 \cdot {\mathrm{H}_2}\mathrm{O}$ is known to be pink.
Step 3: The complex is likely trans-[Co(NH3)4Cl2(H2O)]Cl, which is pink.

Why other options are wrong:
- Option A (Orange yellow): This color is not associated with this specific cobalt complex.
- Option B (Orange): This color is not associated with this specific cobalt complex.
- Option C (Green): This color is not associated with this specific cobalt complex.
- Option D (Pink): This is the correct color for this complex.

Key Concepts:
- Coordination complex colors
- d-d transitions

Memory Aids:
- Memorize common complex colors.

Quick Tips:
- Recognize common coordination complex colors based on the metal and ligands.

90 Medium

90 of 133Question permalink

Application +4 | -1

Which ion is paramagnetic?

View answer and explanation

Correct option: A

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

Option A ([Ni(H2O)6]²⁺) is correct because Ni²⁺ has a d⁸ configuration, and H2O is a weak field ligand, resulting in 2 unpaired electrons, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ . This ion is paramagnetic due to the presence of unpaired electrons.

Step-by-Step Solution:

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

Step 2: Option A: ${\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ . Ni is in +2 oxidation state. It has $d^8$ configuration. H2O is a weak field ligand, so pairing does not occur. There are 2 unpaired electrons, making it paramagnetic.

Step 3: Option B: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4 - }}$ . Fe is in +2 oxidation state. It has $d^6$ configuration. CN- is a strong field ligand, so it pairs up the electrons, resulting in a diamagnetic complex.

Step 4: Option C: $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ . Ni is in 0 oxidation state. It has $d^{10}$ configuration. CO is a strong field ligand, so it pairs up the electrons, resulting in a diamagnetic complex.

Step 5: Option D: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}$ . Ni is in +2 oxidation state. It has $d^8$ configuration. CN- is a strong field ligand, so it pairs up the electrons, resulting in a diamagnetic complex.

Why other options are wrong:
- Option B: Ferrocyanide ion is diamagnetic due to pairing of electrons.
- Option C: Nickel tetracarbonyl is diamagnetic due to pairing of electrons.
- Option D: Tetracyanonickelate(II) ion is diamagnetic due to pairing of electrons.

Key Concepts:
- Paramagnetism and diamagnetism
- Electronic configuration
- Coordination complexes
- Strong and weak field ligands

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

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

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