The correct IUPAC name of ${\rm{KAl}}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_2} \cdot 12{{\rm{H}}_2}{\rm{O}}$ is

Aluminium potassium sulphate-12-water

Potassium aluminate(III) sulphatehydrate

Aluminium(III) potassium sulphate hydrate-12

The colour of ${\left[ {{\rm{Ti}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{3 + }}$ is due to:

Transfer of an electron from one ${\rm{Ti}}$ to another

The IUPAC name of $\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_4}{\rm{C}}{{\rm{l}}_2}} \right]{\rm{Cl}}$ is:

Tetrahydrodichlorochromium(III) chloride

Tetraaquodichlorochromium(I) chloride

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

61 Easy

61 of 98Question permalink

Application +4 | -1

In which of the following complex compounds the oxidation number of the metal is zero?

View answer and explanation

Correct option: B

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

The correct answer is B (

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

). Carbonyl (CO) is a neutral ligand, so if the overall complex has no charge, the metal's oxidation state must be zero.

Solution approach

Correct Answer + Core Reason: $\left[{Cr}({CO})_6\right]$ . CO is a neutral ligand, leading to a zero oxidation state for Cr.

Step-by-Step Solution:

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

Step 2: $\left[{Pt}({NH}_3)_2 {Cl}_2\right]$ : $\mathrm{NH}_3$ is neutral, and Cl has a charge of -1. So, $\mathrm{Pt} + 2(0) + 2(-1) = 0$ , which means $\mathrm{Pt} = +2$ .

Step 3: $\left[{Cr}({CO})_6\right]$ : CO is neutral, so the oxidation state of Cr is 0.

Step 4: $\left[{Cr}({NH}_3)_3 {Cl}_3\right]$ : $\mathrm{NH}_3$ is neutral, and Cl has a charge of -1. So, $\mathrm{Cr} + 3(0) + 3(-1) = 0$ , which means $\mathrm{Cr} = +3$ .

Step 5: $\left[{Cr}(\text { en })_2 {Cl}_2\right]$ : en (ethylenediamine) is neutral, and Cl has a charge of -1. So, $\mathrm{Cr} + 2(0) + 2(-1) = 0$ , which means $\mathrm{Cr} = +2$ .

Why other options are wrong:
- Option A: $\left[{Pt}({NH}_3)_2 {Cl}_2\right]$ : The oxidation state of Pt is +2.
- Option C: $\left[{Cr}({NH}_3)_3 {Cl}_3\right]$ : The oxidation state of Cr is +3.
- Option D: $\left[{Cr}(\text { en })_2 {Cl}_2\right]$ : The oxidation state of Cr is +2.

Key Concepts:
- Oxidation state
- Coordination compounds
- Ligand charge

Memory Aids:
- CO is a neutral ligand.

Quick Tips:
- Remember the common charges of ligands.

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Analyzing +4 | -1

Which possesses tetrahedral shape (

s{p^3} -

hybridization of central atom)?

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because all the given complexes,

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

,

\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]

, and

{\left[ {\mathrm{Cd}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}

, possess tetrahedral shape with

s{p^3}

hybridization.

Solution approach

Correct Answer + Core Reason: All are correct. All the given complexes have tetrahedral shape and $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Analyze ${\left[ {\mathrm{Zn}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : Zn2+ has a d10 configuration. With four ligands, it forms a tetrahedral complex with $sp^3$ hybridization.

Step 2: Analyze $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ : Ni has a 0 oxidation state. CO is a strong field ligand, leading to pairing and $sp^3$ hybridization, resulting in a tetrahedral shape.

Step 3: Analyze ${\left[ {\mathrm{Cd}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : Cd2+ has a d10 configuration. With four ligands, it forms a tetrahedral complex with $sp^3$ hybridization.

Step 4: Since all three complexes have tetrahedral geometry, the correct answer is "All are correct".

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Zn}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : This complex is indeed tetrahedral, but it's not the only correct option.
- Option B: $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ : This complex is indeed tetrahedral, but it's not the only correct option.
- Option C: ${\left[ {\mathrm{Cd}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : This complex is indeed tetrahedral, but it's not the only correct option.

Key Concepts:
- Hybridization
- Geometry of complexes
- Tetrahedral complexes
- d10 configuration
- Strong field ligands

Memory Aids:
- d10 complexes tend to be tetrahedral.

Quick Tips:
- Remember that Zn2+ and Cd2+ usually form tetrahedral complexes.
- CO is a strong field ligand that can lead to pairing and tetrahedral geometry.

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Understanding +4 | -1

The correct IUPAC name of

\mathrm{KAl}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_2} \cdot 12{\mathrm{H}_2}\mathrm{O}

is

View answer and explanation

Correct option: B

Potassium aluminium(III) sulphate-12-water

Option B is correct because it follows IUPAC nomenclature rules, correctly naming the complex salt with proper oxidation state and hydration.

Solution approach

Correct Answer + Core Reason: Potassium aluminium(III) sulphate-12-water is correct because it follows IUPAC naming conventions for coordination compounds.

Step-by-Step Solution:

Step 1: Identify the components of the compound: $\mathrm{KAl}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_2} \cdot 12{\mathrm{H}_2}\mathrm{O}$ consists of potassium (K), aluminium (Al), sulphate ( $\mathrm{SO}_4$ ), and water ( ${\mathrm{H}_2}\mathrm{O}$ ).

Step 2: Determine the oxidation state of aluminium. Since potassium has a +1 charge, sulphate has a -2 charge, and the overall compound is neutral, aluminium must have a +3 charge. This is represented as aluminium(III).

Step 3: Apply IUPAC naming rules. The cation (potassium) is named first, followed by the other cation (aluminium) with its oxidation state in parentheses, then the anion (sulphate), and finally the number of water molecules as a hydrate.

Step 4: Combine the names: Potassium aluminium(III) sulphate-12-water.

Why other options are wrong:
- Option A: Aluminium potassium sulphate-12-water: This is incorrect because it doesn't specify the oxidation state of aluminium, which is crucial in IUPAC nomenclature.
- Option C: Potassium aluminate(III) sulphatehydrate: This is incorrect because "aluminate" is used when aluminium is part of an anionic complex, which is not the case here. Also, it uses "hydrate" instead of specifying the number of water molecules.
- Option D: Aluminium(III) potassium sulphate hydrate-12: This is incorrect because the order of cations is reversed; potassium should come before aluminium in the name.

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

Memory Aids:
- Cation before anion in naming
- Roman numerals for oxidation states

Quick Tips:
- Always specify the oxidation state of metals in coordination compounds.
- Follow the IUPAC naming conventions strictly.

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Application +4 | -1

The coordination number and oxidation state of

Cr

in

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

are respectively

View answer and explanation

Correct option: A

+6 and +3

Option A (+6 and +3) is correct because in

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

, the coordination number of Cr is 6 (since each oxalate ligand is bidentate), and the oxidation state of Cr is +3.

Solution approach

Correct Answer + Core Reason: +6 and +3. The coordination number is determined by the number of ligands attached to the central metal ion, and the oxidation state is calculated based on the charges of the ligands and the overall charge of the complex.

Step-by-Step Solution:

Step 1: Determine the coordination number of Cr. The complex is ${\mathrm{K}_3}\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]$ . The ligand is oxalate ( $C_2O_4^{2-}$ ), which is a bidentate ligand (it coordinates through two atoms). There are 3 oxalate ligands, so the coordination number is 3 * 2 = 6.

Step 2: Determine the oxidation state of Cr. Let x be the oxidation state of Cr. The overall charge of the complex ion is -3 (from $K_3$ ). Each oxalate ligand has a charge of -2. So, $x + 3(-2) = -3$ .

Step 3: Solve for x: $x - 6 = -3$ , so $x = +3$ .

Final Answer: +6 and +3

Why other options are wrong:
- Option B (3 and 0): The coordination number is incorrect, and the oxidation state is also incorrect.
- Option C (4 and +2): Both the coordination number and the oxidation state are incorrect.
- Option D (3 and +3): The coordination number is incorrect, although the oxidation state is correct.

Key Concepts:
- Coordination Chemistry
- Coordination Number
- Oxidation State
- Bidentate Ligands

Memory Aids:
- Oxalate ( $C_2O_4^{2-}$ ) is a bidentate ligand.

Quick Tips:
- Remember to consider the denticity of the ligands when determining the coordination number.
- Set up an equation to solve for the oxidation state of the metal.

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Application +4 | -1

The effective atomic number of

Cr\mathrm{ }\left( {\mathrm{At}\mathrm{. No}\mathrm{.} = 24} \right)\mathrm{in }\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}\mathrm{\;}

is

View answer and explanation

Correct option: C

33

The correct answer is 33 because the EAN of Cr in

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

is calculated as 24 (atomic number) - 3 (oxidation state) + 2(6) (electrons from ligands) = 33.

Solution approach

Correct Answer + Core Reason: 33. The EAN is calculated as Atomic Number - Oxidation State + Number of electrons donated by ligands.

Step-by-Step Solution:

Step 1: Identify the complex: $\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}$

Step 2: Determine the atomic number of Cr: 24

Step 3: Determine the oxidation state of Cr. Since there are 3 Cl $^-$ ions outside the coordination sphere, the complex ion has a +3 charge. NH $_3$ is a neutral ligand, so the oxidation state of Cr is +3.

Step 4: Determine the number of ligands: 6 NH $_3$ ligands.

Step 5: Each NH $_3$ ligand donates 2 electrons, so the total number of electrons donated by the ligands is $6 \times 2 = 12$ .

Step 6: Calculate the EAN: EAN = Atomic number - Oxidation state + Electrons from ligands = $24 - 3 + 12 = 33$

Why other options are wrong:
- Option A: Incorrect calculation of EAN.
- Option B: Incorrect calculation of EAN.
- Option D: Incorrect calculation of EAN.

Key Concepts:
- Effective Atomic Number (EAN)
- Coordination complexes
- Ligands

Memory Aids:
- EAN = Atomic number - Oxidation state + Electrons from ligands

Quick Tips:
- Remember that each ligand donates a certain number of electrons to the central metal atom.

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Analyzing +4 | -1

How many unpaired electrons are present in

\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4

complex?

View answer and explanation

Correct option: A

3

The correct answer is 3 because in

\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4

, NO is present as

NO^+

which is a strong field ligand, and Fe is in +1 oxidation state, resulting in 3 unpaired electrons.

Solution approach

Correct Answer + Core Reason: 3. The complex has 3 unpaired electrons due to the presence of $NO^+$ and $Fe^+$ .

Step-by-Step Solution:

Step 1: Determine the oxidation state of Fe in the complex $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ .

Step 2: The overall charge of the complex ion is +2 (since it is balanced by $SO_4^{2-}$ ).

Step 3: The charge of $H_2O$ is 0, and NO is present as $NO^+$ . Let the oxidation state of Fe be x. Then, x + 0*5 + (+1) = +2. Therefore, x = +1.

Step 4: The electronic configuration of $Fe^+$ is $3d^7$ .

Step 5: Since $NO^+$ is a strong field ligand, it will cause pairing of electrons. However, even with pairing, there will be 3 unpaired electrons in the $d^7$ configuration.

Why other options are wrong:
- Option B: 0 is incorrect because there are unpaired electrons.
- Option C: 0 is incorrect because there are unpaired electrons.
- Option D: 0 is incorrect because there are unpaired electrons.

Key Concepts:
- Coordination complexes
- Oxidation state
- Electronic configuration
- Strong field ligands
- Unpaired electrons

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

Quick Tips:
- Remember to consider the charge of the ligands when determining the oxidation state of the metal.

67 Easy

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Remembering +4 | -1

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

is called:

View answer and explanation

Correct option: A

Hexaammine cobalt (III) chloride

Option A (Hexaammine cobalt (III) chloride) is correct because it follows the IUPAC nomenclature rules for coordination compounds.

Solution approach

Correct Answer + Core Reason: Hexaammine cobalt (III) chloride. The name follows IUPAC nomenclature rules.

Step-by-Step Solution:

Step 1: Identify the cation and anion: The complex is $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}$ . The cation is $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3+}$ and the anion is $Cl^-$ .

Step 2: Name the cation: The ligands are ammine ( $NH_3$ ). There are six ammine ligands, so it is hexaammine. The central metal ion is cobalt. The oxidation state of cobalt is +3. Therefore, the cation is hexaamminecobalt(III).

Step 3: Name the anion: The anion is chloride.

Step 4: Combine the names: The name of the complex is hexaamminecobalt(III) chloride.

Why other options are wrong:
- Option B: Amino cobalt chloride (III) - Incorrect because "amino" is not the correct name for the $NH_3$ ligand in coordination chemistry. It should be "ammine".
- Option C: Cobalt chloride hexaammine - Incorrect order. The cation is named before the anion.
- Option D: Hexaammine tricobalt chloride - Incorrect. The prefix "tri-" is not used for the counter ion.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Naming ligands
- Determining oxidation state of metal ions

Memory Aids:
- Remember that $NH_3$ as a ligand is called "ammine" (with two m's).

Quick Tips:
- Follow the IUPAC rules for naming coordination compounds systematically.

68 Easy

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Remembering +4 | -1

Which order is correct in spectrochemical series of ligands?

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{l}^ - } < {\mathrm{F}^ - } < {\left[ {{\mathrm{C}_2}{\mathrm{O}_4}} \right]^{2 - }} < \mathrm{NO}_2^ - < C{\mathrm{N}^ - }

Option A is correct because it represents the ligands arranged in increasing order of their ability to cause d-orbital splitting in a complex.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}^ - } < {\mathrm{F}^ - } < {\left[ {{\mathrm{C}_2}{\mathrm{O}_4}} \right]^{2 - }} < \mathrm{NO}_2^ - < C{\mathrm{N}^ - }$ . This order reflects the increasing field strength of the ligands in the spectrochemical series.

Step-by-Step Solution:

Step 1: Understand the spectrochemical series. The spectrochemical series is an empirical ordering of ligands based on their ability to split d-orbitals in transition metal complexes.

Step 2: Recall the general trends. Strong field ligands (like CN-) cause large splitting, while weak field ligands (like Cl-) cause small splitting.

Step 3: Consider the given ligands. We need to arrange Cl-, F-, [C2O4]2-, NO2-, and CN- in order of increasing field strength.

Step 4: Apply knowledge of common ligands. Generally, halide ions are weak field ligands, followed by oxygen donors, nitrogen donors, and finally carbon donors, which are strong field ligands.

Step 5: Arrange in increasing order: Cl- < F- < [C2O4]2- < NO2- < CN-

Why other options are wrong:
- Option B: Incorrect because CN- is a strong field ligand and should be at the end of the series, not the beginning. Also, the relative positions of Cl-, [C2O4]2-, and F- are incorrect.
- Option C: Incorrect because [C2O4]2- is not the weakest field ligand and the relative positions of Cl-, F-, and [C2O4]2- are incorrect.
- Option D: Incorrect because F- and Cl- are in the wrong order, and [C2O4]2- should not be stronger than CN-.

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

Memory Aids:
- I < Br < SCN- < Cl- < F- < OH- < C2O42- < H2O < NCS- < EDTA < NH3 < en < NO2- < CN- < CO (a simplified version to remember relative positions)

Quick Tips:
- Remember the general trend: halide < oxygen donor < nitrogen donor < carbon donor.
- Strong field ligands generally have pi-acceptor properties.

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Remembering +4 | -1

CO ligand is

View answer and explanation

Correct option: A

Carbonyl

Option A (Carbonyl) is correct because CO is the ligand name for Carbonyl.

Solution approach

Correct Answer + Core Reason: Carbonyl. CO ligand is called Carbonyl.

Step-by-Step Solution:

Step 1: Identify the ligand: CO.

Step 2: Recall the name of the CO ligand in coordination chemistry.

Step 3: The name of the CO ligand is Carbonyl.

Why other options are wrong:
- Option B: Carbonium: Carbonium refers to a carbocation, not a ligand.
- Option C: Carboso: This is not a recognized term in chemistry.
- Option D: Carbanosium: This is not a recognized term in chemistry.

Key Concepts:
- Nomenclature of ligands in coordination chemistry
- Carbonyl ligand

Memory Aids:
- CO is always referred to as Carbonyl in coordination complexes.

Quick Tips:
- Memorize the names of common ligands.

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Understanding +4 | -1

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

is called:

View answer and explanation

Correct option: D

Potassium trioxalatoaluminate(III)

Option D (Potassium trioxalatoaluminate(III)) is correct because it correctly names the complex, indicating the potassium cation, the trioxalato ligands, the aluminate anion, and the oxidation state of aluminum.

Solution approach

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

Step-by-Step Solution:

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

Step 2: Identify the ligands: Oxalato (C2O4) with a charge of -2. There are three oxalato ligands.

Step 3: Identify the central metal: Aluminum (Al).

Step 4: Determine the oxidation state of aluminum: The complex ion has a charge of -3. Since each oxalate ligand has a -2 charge, the total charge from the ligands is -6. Therefore, the oxidation state of aluminum must be +3 to result in a -3 overall charge for the complex ion.

Step 5: Name the complex: Potassium trioxalatoaluminate(III).

Why other options are wrong:
- Option A: Potassium aluminooxalate: Incorrect naming convention; does not specify the number of oxalate ligands or the oxidation state of aluminum.
- Option B: Potassium alumino(III) oxalate: Incorrect naming convention; does not specify the number of oxalate ligands.
- Option C: Potassium trioxalatoaluminate: Missing the oxidation state of aluminum (III).

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

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

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

71 Easy

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Remembering +4 | -1

Which of the following is potassiumferricyanide?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents potassium ferricyanide, where iron is in the +3 oxidation state and is complexed with six cyanide ligands, balanced by three potassium cations.

Solution approach

Correct Answer + Core Reason: ${K}_3\left[{Fe}({CN})_6\right]$ . This is the correct formula for potassium ferricyanide.

Step-by-Step Solution:

Step 1: Identify the components: Potassium (K), Iron (Fe), and Cyanide (CN).

Step 2: Determine the oxidation state of iron in ferricyanide: Iron is in the +3 oxidation state (Fe3+).

Step 3: Recognize that cyanide (CN) has a -1 charge.

Step 4: Determine the number of cyanide ligands: There are six cyanide ligands, giving a total negative charge of -6.

Step 5: Balance the charges with potassium cations: To balance the -6 charge from the cyanide ligands and the +3 charge from iron, three potassium ions (K+) are needed.

Step 6: Write the correct formula: ${K}_3\left[{Fe}({CN})_6\right]$ .

Why other options are wrong:
- Option A: ${K}_4\left[{Fe}({CN})_0\right]$ : This is incorrect because it has no cyanide ligands.
- Option C: ${K}_3\left[{Fe}({CN})_3\right]$ : This is incorrect because it has only three cyanide ligands.
- Option D: ${K}_3\left[{Fe}({CN})_4\right]$ : This is incorrect because it has only four cyanide ligands.

Key Concepts:
- Nomenclature of coordination compounds
- Oxidation states of transition metals
- Charge balancing in ionic compounds

Memory Aids:
- Ferri- indicates iron in the +3 oxidation state.

Quick Tips:
- Always consider the oxidation states and charges when writing the formula of a coordination compound.

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Analyzing +4 | -1

Which is not true of the coordination compound

\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}

?

View answer and explanation

Correct option: C

Exhibits ionisation isomerism

Option C (Exhibits ionisation isomerism) is correct because the counter ion (Cl-) is already outside the coordination sphere, so exchanging it with a ligand inside the sphere does not create a different compound.

Solution approach

Correct Answer + Core Reason: Exhibits ionisation isomerism. Ionization isomerism requires exchange of ligands inside and outside the coordination sphere, which doesn't result in a different compound in this case.

Step-by-Step Solution:

Step 1: Analyze the given coordination compound: [Co(en)2Cl2]Cl

Step 2: Determine if it exhibits geometrical isomerism.\nThe complex can exist as cis and trans isomers, where the two Cl ligands are either on the same side (cis) or opposite sides (trans) of the Co ion. Therefore, it exhibits geometrical isomerism.

Step 3: Determine if it exhibits optical isomerism.\nThe cis isomer of [Co(en)2Cl2]Cl is chiral and can exist as two optical isomers (enantiomers). Therefore, it exhibits optical isomerism.

Step 4: Determine if it exhibits ionization isomerism.\nIonization isomerism involves the exchange of ligands inside the coordination sphere with ions outside the coordination sphere. In this case, the Cl- ion outside the coordination sphere could potentially exchange with one of the Cl ligands inside the coordination sphere. However, since both are chloride ions, this exchange would not result in a different compound. Therefore, it does not exhibit ionization isomerism.

Step 5: Determine if it is an octahedral complex.\nCobalt (Co) is a transition metal and typically forms octahedral complexes. The coordination number of Co in this complex is 6 (2 from each en ligand and 2 from the Cl ligands), which confirms that it is an octahedral complex.

Step 6: Final Answer: The statement that is not true is that it exhibits ionization isomerism.

Why other options are wrong:
- Option A (Exhibits geometrical isomerism): The complex can exist as cis and trans isomers, so it does exhibit geometrical isomerism.
- Option B (Exhibits optical isomerism): The cis isomer of the complex is chiral and exhibits optical isomerism.
- Option D (Is an octahedral complex): The coordination number of Co is 6, making it an octahedral complex.

Key Concepts:
- Coordination complexes
- Geometrical isomerism (cis-trans)
- Optical isomerism (chirality)
- Ionization isomerism
- Octahedral complexes

Memory Aids:
- Remember the definitions of different types of isomerism.
- Cis-trans isomerism occurs when ligands are arranged differently around the central metal ion.
- Optical isomerism occurs when a molecule is chiral (non-superimposable on its mirror image).
- Ionization isomerism involves the exchange of ions inside and outside the coordination sphere.

Quick Tips:
- Draw the structures of the possible isomers to determine if geometrical or optical isomerism is possible.
- Consider the coordination number of the central metal ion to determine the geometry of the complex.

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Understanding +4 | -1

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Br}} \right]\mathrm{S}{\mathrm{O}_4}\mathrm{\;}

and

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]

\mathrm{Br}

are examples of which type of isomerism?

View answer and explanation

Correct option: D

Ionisation

The correct answer is D (Ionisation) because the complexes

{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Br}} \right]\mathrm{S}{\mathrm{O}_4}\mathrm{\;}

and

{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]\mathrm{Br}}

exchange ions (Br- and SO42-) in the coordination sphere and outside it.

Solution approach

Correct Answer + Core Reason: Ionisation. The complexes exhibit ionisation isomerism because they exchange ions inside and outside the coordination sphere.

Step-by-Step Solution:

Step 1: Analyze the given complexes. The complexes are ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Br}} \right]\mathrm{S}{\mathrm{O}_4}$ and ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]\mathrm{Br}$ .

Step 2: Identify the difference between the complexes. The difference is that in the first complex, $Br^-$ is inside the coordination sphere and $SO_4^{2-}$ is outside, while in the second complex, $SO_4^{2-}$ is inside the coordination sphere and $Br^-$ is outside.

Step 3: Determine the type of isomerism. This exchange of ions inside and outside the coordination sphere is characteristic of ionisation isomerism.

Step 4: Conclusion: The complexes are examples of ionisation isomerism.

Why other options are wrong:
- Option A (Linkage): Linkage isomerism occurs when a ligand can bind to the metal through different atoms (e.g., $SCN^-$ can bind through S or N). This is not the case here.
- Option B (Optical): Optical isomerism occurs when a complex is chiral (non-superimposable mirror image). These complexes do not necessarily exhibit chirality.
- Option C (Geometrical): Geometrical isomerism (cis/trans) requires different arrangements of ligands around the central metal ion. This is not applicable here as the ligands inside the coordination sphere are different.

Key Concepts:
- Isomerism in coordination complexes
- Ionisation isomerism
- Linkage isomerism
- Geometrical isomerism
- Optical isomerism

Memory Aids:
- Ionisation isomerism: Exchange of ions inside and outside the coordination sphere.

Quick Tips:
- Look for the exchange of ions inside and outside the coordination sphere to identify ionisation isomers.

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74 of 98Question permalink

Analyzing +4 | -1

Which of the following will give maximum number of isomers?

View answer and explanation

Correct option: D

{\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }

Option D (

{\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }

) is correct because SCN- is an ambidentate ligand, leading to linkage isomers in addition to geometric isomers.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }$ . SCN is an ambidentate ligand, which allows for linkage isomerism in addition to geometrical isomerism.

Step-by-Step Solution:

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

Step 2: Option A: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]$ : This complex can exhibit geometrical isomerism (cis and trans). It does not exhibit optical isomerism.

Step 3: Option B: ${\left[ {\mathrm{Ni}\left( {\mathrm{en}} \right){{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : This complex does not exhibit geometrical or optical isomerism because the ethylenediamine (en) ligand occupies adjacent positions, and the four ammonia ligands are equivalent.

Step 4: Option C: $\left[ {\mathrm{Ni}\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right){{\left( {\mathrm{en}} \right)}_2}} \right]$ : This complex does not exhibit geometrical or optical isomerism because the oxalate ligand and two ethylenediamine ligands occupy fixed positions.

Step 5: Option D: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }$ : This complex can exhibit geometrical isomerism (cis and trans). Additionally, the SCN- ligand is an ambidentate ligand, meaning it can bind through either the sulfur (S) or the nitrogen (N) atom. This leads to linkage isomerism. Therefore, it has the potential for a greater number of isomers.

Step 6: Final Answer: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^ + }$ has the highest number of isomers.

Why other options are wrong:
- Option A: Only exhibits geometrical isomerism (cis and trans).
- Option B: Does not exhibit geometrical or optical isomerism.
- Option C: Does not exhibit geometrical or optical isomerism.

Key Concepts:
- Geometrical isomerism (cis-trans)
- Optical isomerism
- Ambidentate ligands (linkage isomerism)
- Coordination complexes

Memory Aids:
- Ambidentate ligands can bind through different atoms, leading to linkage isomers.

Quick Tips:
- Look for ambidentate ligands like SCN-, NO2-, etc., as they can lead to a greater number of isomers.

75 Easy

75 of 98Question permalink

Remembering +4 | -1

Which ligand produces a high crystal field splitting (a strong ligand field)?

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because CO, NO2-, and CN- are all strong field ligands that produce high crystal field splitting.

Solution approach

Correct Answer + Core Reason: All are correct. CO, NO2-, and CN- are all strong field ligands that cause significant crystal field splitting.

Step-by-Step Solution:
Step 1: Understand crystal field splitting. Crystal field splitting is the energy difference between d-orbitals in a coordination complex due to the ligand field.
Step 2: Identify strong field ligands. Strong field ligands cause a large crystal field splitting, leading to low-spin complexes.
Step 3: Evaluate each ligand:
- CO (Carbon Monoxide): A very strong field ligand due to strong π-backbonding.
- NO2- (Nitrite): A strong field ligand, stronger than NH3.
- CN- (Cyanide): A strong field ligand due to strong π-backbonding.
Step 4: Conclude that all three ligands are strong field ligands.

Why other options are wrong:
- Option A: $\mathrm{CO}$ : While CO is a strong field ligand, it's not the only one.
- Option B: $\mathrm{NO}_2^ -$ : While NO2- is a strong field ligand, it's not the only one.
- Option C: $\mathrm{C}{\mathrm{N}^ - }$ : While CN- is a strong field ligand, it's not the only one.

Key Concepts:
- Crystal Field Theory
- Spectrochemical Series
- Strong Field Ligands

Memory Aids:
- Remember that ligands with multiple bonds (like CO and CN-) are generally strong field ligands.

Quick Tips:
- Know the common strong field ligands: CO, CN-, NO2-, en, NH3.

76 Easy

76 of 98Question permalink

Remembering +4 | -1

The shape of cobalt hexaammine cation, which has its central cobalt atom surrounded by six ammonia molecules is:

View answer and explanation

Correct option: B

Octahedral

Option B is correct because cobalt hexaammine cation, [Co(NH3)6]+3, has six ligands coordinated to the central cobalt atom, resulting in an octahedral geometry.

Solution approach

Correct Answer + Core Reason: Octahedral. Six ligands around a central metal atom result in an octahedral shape.

Step-by-Step Solution:

Step 1: Determine the coordination number of the central metal atom. In $[Co(NH_3)_6]^{3+}$ , the coordination number of cobalt is 6, as there are six ammonia ligands attached to it.

Step 2: Relate the coordination number to the geometry. A coordination number of 6 typically corresponds to an octahedral geometry.

Step 3: Visualize the structure. Imagine the cobalt atom at the center of an octahedron, with the six ammonia ligands at the vertices.

Why other options are wrong:
- Option A: Tetrahedral geometry corresponds to a coordination number of 4, not 6.
- Option C: Square planar geometry corresponds to a coordination number of 4, not 6, and is usually seen with strong field ligands and d8 metal ions.
- Option D: Trigonal geometry is not a common geometry for coordination complexes.

Key Concepts:
- Coordination number
- Geometry of coordination complexes
- Octahedral geometry

Memory Aids:
- Coordination number 6 = Octahedral
- Coordination number 4 = Tetrahedral or Square Planar

Quick Tips:
- Memorize the common geometries associated with different coordination numbers.

77 Easy

77 of 98Question permalink

Understanding +4 | -1

The colour of

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

is due to:

View answer and explanation

Correct option: C

Excitation of electrons from

d - d

The correct answer is C (Excitation of electrons from

d - d

) because the color of the complex

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

is due to

d-d

transitions.

Solution approach

Correct Answer + Core Reason: Excitation of electrons from $d - d$ . The color of the complex is due to d-d transitions.

Step-by-Step Solution:
Step 1: The question asks for the reason for the color of ${\left[ {\mathrm{Ti}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ .
Step 2: The color of ${\left[ {\mathrm{Ti}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ is due to $d-d$ transitions. The $\mathrm{Ti}^{3+}$ ion has a $d^1$ electronic configuration. The single $d$ electron can be excited to a higher energy $d$ orbital by absorbing light.

Why other options are wrong:
- **Option A (Transfer of an electron from one $\mathrm{Ti}$ to another): This is not the primary reason for the color.
- Option B (Presence of water molecule): Water molecules act as ligands, but the color is due to d-d transitions.
- Option D (Intramolecular vibration): Intramolecular vibrations can contribute to the spectrum, but the main reason is d-d transitions.

Key Concepts:**
- $d-d$ transitions
- Color of transition metal complexes
- Crystal field theory

Memory Aids:
- Remember that $d-d$ transitions are responsible for the color.

Quick Tips:
- Recognize that $d-d$ transitions are the primary cause of color in transition metal complexes.

78 Easy

78 of 98Question permalink

Application +4 | -1

The EAN of

Cr

in

{\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_6}} \right]^{3 - }}

is:

View answer and explanation

Correct option: B

33

Option B (33) is correct because the Effective Atomic Number (EAN) is calculated by adding the metal's atomic number to the electrons donated by the ligands.

Solution approach

Correct Answer + Core Reason: 33. The EAN is calculated by adding the number of electrons of the metal ion and the number of electrons donated by the ligands.

Step-by-Step Solution:

Step 1: Identify the metal and its atomic number. Chromium (Cr) has an atomic number of 24.

Step 2: Determine the oxidation state of the metal. In ${\left[ {\mathrm{Cr}{{\left( {\mathrm{SCN}} \right)}_6}} \right]^{3 - }}$ , the oxidation state of Cr is +3. Therefore, the number of electrons in $Cr^{3+}$ is 24 - 3 = 21.

Step 3: Determine the number of ligands and the number of electrons each ligand donates. There are 6 SCN ligands, and each SCN ligand donates 2 electrons.

Step 4: Calculate the total number of electrons donated by the ligands: 6 ligands * 2 electrons/ligand = 12 electrons.

Step 5: Calculate the EAN: EAN = Number of electrons in metal ion + Number of electrons donated by ligands = 21 + 12 = 33.

Step 6: Final Answer: The EAN of Cr is 33.

Why other options are wrong:
- Option A: 35 is incorrect as it doesn't accurately reflect the electron count.
- Option C: 34 is incorrect as it doesn't accurately reflect the electron count.
- Option D: 37 is incorrect as it doesn't accurately reflect the electron count.

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

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

Quick Tips:
- Remember to account for the charge of the complex when calculating the oxidation state of the metal.

79 Easy

79 of 98Question permalink

Understanding +4 | -1

Which complex cannot ionize in solution?

View answer and explanation

Correct option: A

\left[ {\mathrm{CoC}{\mathrm{l}_3}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}} \right]

Option A is correct. The complex

[CoCl_3(NH_3)_3]

does not ionize in solution because all the ions are within the coordination sphere.

Solution approach

Correct Answer + Core Reason: A. $\ {\mathrm{CoC}{\mathrm{l}_3}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}}$ . All ions are within the coordination sphere.

Step-by-Step Solution:

Step 1: Analyze the complex $\CoCl_3(NH_3)_3\$ : All the chloride ions are inside the coordination sphere.

Step 2: Since all the ions are inside the coordination sphere, the complex will not ionize in solution.

Step 3: Analyze the other options:
- B. ${\mathrm{K}_4}\ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}$ will ionize to give $4K^+$ and $\Fe(CN)_6\^{4-}$
- C. ${\mathrm{K}_2}\ {\mathrm{Pt}\left( {{\mathrm{F}_6}} \right)}$ will ionize to give $2K^+$ and $\PtF_6\^{2-}$
- D. $\ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \mathrm{C}{\mathrm{l}_4}$ will ionize to give $\Pt(NH_3)_6\^{4+}$ and $4Cl^-$

Why other options are wrong:
- Option B: The complex ionizes in solution.
- Option C: The complex ionizes in solution.
- Option D: The complex ionizes in solution.

Key Concepts:
- Coordination complexes
- Ionization of complexes
- Coordination sphere

Memory Aids:
- If all ions are inside the brackets, it won't ionize.

Quick Tips:
- Look for ions outside the coordination sphere to determine if a complex will ionize.

80 Easy

80 of 98Question permalink

Understanding +4 | -1

In triethylenediamine cobalt(III) chloride the coordination number of cobalt is:

View answer and explanation

Correct option: C

6

Option C (6) is correct because triethylenediamine (en) is a bidentate ligand, meaning each en molecule coordinates through two nitrogen atoms. Thus, three en molecules coordinate to the cobalt(III) ion, resulting in a coordination number of 6.

Solution approach

Correct Answer + Core Reason: 6. Triethylenediamine (en) is a bidentate ligand, and three of them coordinate to the cobalt(III) ion.

Step-by-Step Solution:

Step 1: Identify the ligand. In this case, it is triethylenediamine (en), also known as ethylenediamine.

Step 2: Determine the denticity of the ligand. Ethylenediamine is a bidentate ligand, meaning it can bind to the metal ion through two donor atoms (nitrogen atoms in this case).

Step 3: Determine the number of ligands. The complex is triethylenediamine cobalt(III) chloride, which means there are three ethylenediamine ligands.

Step 4: Calculate the coordination number. Since each ethylenediamine ligand is bidentate, the coordination number is $3 \times 2 = 6$ .

Why other options are wrong:
- Option A: Incorrect. It doesn't account for the bidentate nature of ethylenediamine.
- Option B: Incorrect. It's an arbitrary number and doesn't represent the coordination number.
- Option D: Incorrect. It's an arbitrary number and doesn't represent the coordination number.

Key Concepts:
- Coordination number
- Coordination complexes
- Bidentate ligands
- Denticity

Memory Aids:
- Bidentate ligands bind through two atoms.

Quick Tips:
- Remember to consider the denticity of the ligand when calculating the coordination number.

81 Easy

81 of 98Question permalink

Remembering +4 | -1

Chlorophyll is a coordination compound having central atom of:

View answer and explanation

Correct option: B

Mg

Option B (

Mg

) is correct because the central atom in chlorophyll is magnesium (Mg).

Solution approach

Correct Answer + Core Reason: $Mg$ . Chlorophyll is a coordination compound with magnesium as the central metal atom.

Step-by-Step Solution:

Step 1: Recall the structure of chlorophyll.

Step 2: Chlorophyll is a porphyrin ring complex.

Step 3: The central metal atom in chlorophyll is magnesium (Mg).

Why other options are wrong:
- Option A: $Ca$ : Calcium is important in biological systems, but not the central atom in chlorophyll.
- Option C: $Na$ : Sodium is an important electrolyte, but not the central atom in chlorophyll.
- Option D: $K$ : Potassium is an important electrolyte, but not the central atom in chlorophyll.

Key Concepts:
- Coordination compounds
- Chlorophyll structure
- Porphyrin ring

Memory Aids:
- Chlorophyll: Remember "Mg" for magnesium in chlorophyll.

Quick Tips:
- Know the central metal atoms in important biological coordination compounds.

82 Easy

82 of 98Question permalink

Understanding +4 | -1

The IUPAC name of

\left[ {\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}

is:

View answer and explanation

Correct option: B

Tetraaquodichlorochromium(III) chloride

Option B (Tetraaquodichlorochromium(III) chloride) is correct because it follows the IUPAC naming conventions for coordination complexes, correctly naming the ligands, central metal, oxidation state, and counter ion.

Solution approach

Correct Answer + Core Reason: Tetraaquodichlorochromium(III) chloride. This is the correct IUPAC name.

Step-by-Step Solution:

Step 1: Identify the complex ion: $\left[ {\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]^{+}$ .

Step 2: Name the ligands: H2O is aqua, and there are four of them (tetraaqua). Cl is chloro, and there are two of them (dichloro).

Step 3: Name the central metal ion: Chromium (Cr).

Step 4: Determine the oxidation state of chromium: The complex has a +1 charge. Water is neutral, and each chloride ligand has a -1 charge, so the total negative charge from the ligands is -2. Therefore, the oxidation state of chromium is +3.

Step 5: Name the counter ion: Chloride (Cl-).

Step 6: Combine the names: Tetraaquodichlorochromium(III) chloride.

Why other options are wrong:
- Option A: Tetrahydrodichlorochromium(III) chloride: This is incorrect because H2O is named aqua, not hydro, in coordination complexes.
- Option C: Tetraaquodichlorochromium(I) chloride: This is incorrect because the oxidation state of chromium is +3, not +1.
- Option D: None of the above: Option B is the correct name.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Naming complex ions
- Determining oxidation states

Memory Aids:
- Remember that H2O is named aqua in coordination complexes.

Quick Tips:
- Follow the IUPAC rules for naming coordination compounds systematically.

83 Easy

83 of 98Question permalink

Understanding +4 | -1

Type of isomerism shown by

\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{\;N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}\,\mathrm{is}

View answer and explanation

Correct option: D

Linkage

Option D is correct. The complex exhibits linkage isomerism because the

NO_2

ligand can coordinate through either the nitrogen or oxygen atom.

Solution approach

Correct Answer + Core Reason: D. Linkage. $NO_2$ can bind through N or O.

Step-by-Step Solution:

Step 1: Identify the complex: $\Cr(NH_3)_5NO_2\Cl_2$ .

Step 2: Look for ambidentate ligands: $NO_2$ is an ambidentate ligand, meaning it can coordinate to the metal ion through either the nitrogen atom (nitro) or the oxygen atom (nitrito).

Step 3: Since the ligand can bind in two different ways, the complex exhibits linkage isomerism.

Why other options are wrong:
- Option A: Optical isomerism requires a chiral complex, which this complex may or may not be depending on the specific arrangement of ligands, but linkage isomerism is the primary type of isomerism exhibited due to the $NO_2$ ligand.
- Option B: Ionisation isomerism involves the exchange of ligands inside and outside the coordination sphere. This is not the primary type of isomerism exhibited here.
- Option C: Geometrical isomerism involves different spatial arrangements of ligands around the central metal ion. While possible, linkage isomerism is the primary type of isomerism exhibited due to the $NO_2$ ligand.

Key Concepts:
- Isomerism in coordination compounds
- Linkage isomerism
- Ambidentate ligands

Memory Aids:
- Linkage: Think of the different ways a ligand can link to the metal.

Quick Tips:
- Identify ambidentate ligands to determine if linkage isomerism is possible.

84 Easy

84 of 98Question permalink

Remembering +4 | -1

Chromium carbonyl is:

View answer and explanation

Correct option: C

\mathrm{Cr}{\left( {\mathrm{CO}} \right)_6}

Option C is correct because chromium carbonyl has the formula Cr(CO)6, where chromium is bonded to six carbonyl ligands.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\left( {\mathrm{CO}} \right)_6}$ . This formula satisfies the 18-electron rule for stability.

Step-by-Step Solution:

Step 1: Determine the oxidation state of chromium. In chromium carbonyl, chromium is in the 0 oxidation state.

Step 2: Apply the 18-electron rule. Chromium has 6 valence electrons. Each CO ligand contributes 2 electrons. Therefore, to achieve 18 electrons, chromium needs 6 CO ligands: $6 + 6(2) = 18$ .

Step 3: Write the formula based on the number of CO ligands. The formula is $Cr(CO)_6$ .

Why other options are wrong:
- Option A: $\mathrm{Cr}{\left( {\mathrm{CO}} \right)_4}$ would have only 6 + 4(2) = 14 electrons, violating the 18-electron rule.
- Option B: $\mathrm{Cr}{\left( {\mathrm{CO}} \right)_5}$ would have only 6 + 5(2) = 16 electrons, violating the 18-electron rule.
- Option D: One of the other options is correct.

Key Concepts:
- 18-electron rule
- Metal carbonyls
- Coordination chemistry

Memory Aids:
- 18-electron rule: Metals tend to form complexes that have 18 valence electrons.

Quick Tips:
- Use the 18-electron rule to predict the formulas of metal carbonyls and other organometallic compounds.

85 Easy

85 of 98Question permalink

Understanding +4 | -1

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]

possesses:

View answer and explanation

Correct option: D

Octahedral geometry

The correct answer is D (Octahedral geometry) because the complex

{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]}

has a coordination number of 6, leading to an octahedral shape.

Solution approach

Correct Answer + Core Reason: Octahedral geometry. Cobalt (Co) with four ammonia (NH3) and two chloride (Cl) ligands results in a coordination number of 6, hence octahedral geometry.

Step-by-Step Solution:

Step 1: Identify the central metal ion and its oxidation state. Here, it's Cobalt (Co).

Step 2: Determine the number and type of ligands attached to the central metal ion. There are four NH3 ligands and two Cl ligands.

Step 3: Determine the coordination number. The coordination number is the number of ligands directly attached to the central metal ion, which is 4 + 2 = 6 in this case.

Step 4: Predict the geometry based on the coordination number. A coordination number of 6 typically results in an octahedral geometry.

Step 5: Conclusion: The geometry of the complex is octahedral.

Why other options are wrong:
- Option A (Square planar geometry): Square planar geometry typically occurs with a coordination number of 4, especially with strong field ligands, which is not the case here.
- Option B (Tetrahedral geometry): Tetrahedral geometry requires a coordination number of 4, not 6.
- Option C (Tetrahedral nature): This is similar to option B and incorrect because the coordination number is 6.

Key Concepts:
- Coordination number
- Geometry of coordination complexes

Memory Aids:
- Coordination number 2 = Linear
- Coordination number 4 = Tetrahedral or Square Planar
- Coordination number 6 = Octahedral

Quick Tips:
- Memorize the common geometries associated with different coordination numbers.

86 Easy

86 of 98Question permalink

Remembering +4 | -1

Which of the following is colorless?

View answer and explanation

Correct option: A

\mathrm{Ni}(\mathrm{CO})_4

Option A is correct because Ni(CO)4 is a tetrahedral complex with a d10 configuration, resulting in no d-d transitions and thus is colorless.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ni}(\mathrm{CO})_4$ . This complex has a tetrahedral geometry and a d10 configuration, making it colorless.

Step-by-Step Solution:

Step 1: Determine the oxidation state and electronic configuration of the metal in each complex.
- $\mathrm{Ni}(\mathrm{CO})_4$ : Ni is in the 0 oxidation state, with a $d^{10}$ configuration. It is tetrahedral.
- $\mathrm{Fe}(\mathrm{CO})_5$ : Fe is in the 0 oxidation state, with a $d^8$ configuration. It is trigonal bipyramidal.
- $\mathrm{V}(\mathrm{CO})_5$ : V is in the 0 oxidation state, with a $d^5$ configuration. This complex is unstable.

Step 2: Relate the electronic configuration and geometry to color. Color arises from d-d transitions. Complexes with $d^0$ , $d^{10}$ configurations or those that do not have suitable energy gaps for d-d transitions are often colorless.

Step 3: Since $\mathrm{Ni}(\mathrm{CO})_4$ has a $d^{10}$ configuration and is tetrahedral, there are no d-d transitions, making it colorless.

Why other options are wrong:
- Option B: $\mathrm{Fe}(\mathrm{CO})_5$ has a $d^8$ configuration and can exhibit d-d transitions, making it colored.
- Option C: $\mathrm{V}(\mathrm{CO})_5$ is unstable, but if it were stable, it would have a $d^5$ configuration and could exhibit d-d transitions, making it colored.
- Option D: Only $\mathrm{Ni}(\mathrm{CO})_4$ is colorless.

Key Concepts:
- d-d transitions
- Electronic configuration
- Color of coordination complexes
- Crystal field theory

Memory Aids:
- $d^0$ and $d^{10}$ complexes are often colorless.

Quick Tips:
- Consider the electronic configuration and geometry of the complex when predicting its color.

87 Easy

87 of 98Question permalink

Remembering +4 | -1

Wilkinson’s catalyst,

{\left( {\mathrm{P}{\mathrm{h}_3}\mathrm{P}} \right)_3}\mathrm{RhCl}

is used for

View answer and explanation

Correct option: C

Hydrogenation of alkenes

Option C (Hydrogenation of alkenes) is correct because Wilkinson's catalyst is primarily used for the hydrogenation of alkenes.

Solution approach

Correct Answer + Core Reason: Hydrogenation of alkenes. Wilkinson's catalyst is a well-known catalyst for alkene hydrogenation.

Step-by-Step Solution:

Step 1: Identify the catalyst: Wilkinson's catalyst is ${\left( {\mathrm{P}{\mathrm{h}_3}\mathrm{P}} \right)_3}\mathrm{RhCl}$ .

Step 2: Recall the function of Wilkinson's catalyst: It is a homogeneous catalyst used for hydrogenation reactions, specifically for alkenes.

Step 3: Conclusion: Therefore, the correct answer is hydrogenation of alkenes.

Why other options are wrong:
- Option A: Hydrogenation of oils: While Wilkinson's catalyst *can* hydrogenate oils (which contain alkenes), it's not its primary or most common application.
- Option B: Hydrogenation of alkynes: Alkynes can be hydrogenated, but Wilkinson's catalyst is not the most efficient or commonly used catalyst for this purpose. Other catalysts are preferred.
- Option D: Polymerization of alkenes: Wilkinson's catalyst is used for hydrogenation, not polymerization.

Key Concepts:
- Wilkinson's catalyst
- Hydrogenation
- Alkene reactions
- Homogeneous catalysis

Memory Aids:
- Think of Wilkinson's catalyst as a "hydrogenation helper" specifically for alkenes.

Quick Tips:
- Remember the specific applications of common catalysts like Wilkinson's catalyst.

88 Easy

88 of 98Question permalink

Application +4 | -1

A complex of platinum, ammonia and chlorine produces four ions per molecule in the solution. The structure consistent with the observation is:

View answer and explanation

Correct option: C

\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3}

Option C ([Pt(NH3)5Cl]Cl3) is correct because it dissociates into one complex ion [Pt(NH3)5Cl]3+ and three chloride ions, resulting in a total of four ions.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3}$ . This complex dissociates into four ions in solution.

Step-by-Step Solution:

Step 1: Analyze each complex to determine the number of ions it produces upon dissolution.

Step 2: Option A: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_4} \rightleftharpoons {\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{4 + }} + 4\mathrm{C}{\mathrm{l}^ - }$ produces 5 ions.

Step 3: Option B: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_4}} \right]$ does not ionize, producing only 1 complex.

Step 4: Option C: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3} \rightleftharpoons {\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]^{3 + }} + 3\mathrm{C}{\mathrm{l}^ - }$ produces 4 ions.

Step 5: Option D: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{C}{\mathrm{l}_2} \rightleftharpoons {\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]^{2 + }} + 2\mathrm{C}{\mathrm{l}^ - }$ produces 3 ions.

Step 6: Based on the analysis, only Option C produces four ions per molecule in solution.

Why other options are wrong:
- Option A: This complex produces 5 ions, not 4.
- Option B: This complex does not ionize and produces only 1 complex.
- Option D: This complex produces 3 ions, not 4.

Key Concepts:
- Coordination complexes
- Ionization of complexes in solution
- Determining the number of ions produced by a complex

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

Quick Tips:
- Identify the coordination sphere and the counter ions to determine the number of ions produced.

89 Easy

89 of 98Question permalink

Understanding +4 | -1

The coordination number of

Cr

in

\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_3}} \right]\mathrm{C}{\mathrm{l}_3}\,\mathrm{is}:

View answer and explanation

Correct option: C

6

Option C (6) is correct because the coordination number is the number of ligands directly attached to the central metal ion, which in this case is 3 NH3 and 3 H2O, totaling 6.

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the central metal ion. In this case, it is $Cr$ .

Step 2: Identify the ligands directly bonded to the metal ion. These are $3\mathrm{N}{\mathrm{H}_3}$ and $3{\mathrm{H}_2}\mathrm{O}$ .

Step 3: Determine the coordination number. The coordination number is the sum of the number of ligands directly bonded to the metal ion: $3 + 3 = 6$ .

Why other options are wrong:
- Option A: Incorrect. It only counts the number of one type of ligand (e.g., only NH3).
- Option B: Incorrect. It's an arbitrary number and doesn't represent the number of ligands.
- Option D: Incorrect. It's an arbitrary number and doesn't represent the number of ligands.

Key Concepts:
- Coordination number
- Coordination complexes
- Ligands

Memory Aids:
- Coordination number = number of direct bonds to the metal.

Quick Tips:
- Only count ligands *directly* bonded to the central metal ion.

90 Easy

90 of 98Question permalink

Analyzing +4 | -1

The complex ion which has no ‘d’-electron in the central metal atom is:

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because Mn in MnO4- has a +7 oxidation state, resulting in a d0 configuration, meaning no d-electrons.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$ . The complex ion which has no ‘d’-electron in the central metal atom is ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$ because Mn is in +7 oxidation state, resulting in d0 configuration.

Step-by-Step Solution:

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

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

Step 3: Identify the complex with no d-electrons.

Step 4: Analyze each complex:
- A. [MnO4]-: Mn is in +7 oxidation state. Electronic configuration of Mn is [Ar] 3d5 4s2. Mn7+ will have [Ar] configuration, so no d-electrons.
- B. [Co(NH3)6]3+: Co is in +3 oxidation state. Electronic configuration of Co is [Ar] 3d7 4s2. Co3+ will have [Ar] 3d6 configuration, so it has d-electrons.
- C. [Fe(CN)6]3-: Fe is in +3 oxidation state. Electronic configuration of Fe is [Ar] 3d6 4s2. Fe3+ will have [Ar] 3d5 configuration, so it has d-electrons.
- D. [Cr(H2O)6]3+: Cr is in +3 oxidation state. Electronic configuration of Cr is [Ar] 3d5 4s1. Cr3+ will have [Ar] 3d3 configuration, so it has d-electrons.

Final Answer: ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$

Why other options are wrong:
- Option B: [Co(NH3)6]3+ has d-electrons.
- Option C: [Fe(CN)6]3- has d-electrons.
- Option D: [Cr(H2O)6]3+ has d-electrons.

Key Concepts:
- Coordination complexes
- Oxidation state
- Electronic configuration

Memory Aids:
- Remember the electronic configurations of common transition metals.

Quick Tips:
- Determine the oxidation state accurately to find the correct electronic configuration.

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