The IUPAC name of ${{\rm{K}}_2}\left[ {{\rm{Ni}}{{\left( {{\rm{CN}}} \right)}_4}} \right]$ is

Potassium tetracyanatonickelate (III)

Potassium tetracyanatonickel (II)

Potassium tetracyanonickel (III)

The IUPAC name of the compound $\left[ {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}{{\left( {{\rm{N}}{{\rm{O}}_3}} \right)}_2}} \right]$ is:

Tetraamminecopper(II) dinitrite

Tetraamminecopper(III) dinitrite

Which of the following organometallic compound is $\sigma \,\,{\rm{and}}\,\,\pi $ bonded?

$\left[ {{\rm{Fe}}{{\left( {{{\rm{\eta }}^5} - {{\rm{C}}_5}{{\rm{H}}_5}} \right)}_2}} \right]$

${\rm{Fe}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}$

${\left[ {{\rm{Co}}{{\left( {{\rm{CO}}} \right)}_5}{\rm{N}}{{\rm{H}}_3}} \right]^{2 + }}$

${\rm{K}}\left[ {{\rm{PtC}}{{\rm{l}}_3}\left( {{{\rm{\eta }}^2} - {{\rm{C}}_2}{{\rm{H}}_4}} \right)} \right]$

The IUPAC name of $\left[ {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right]\left[ {{\rm{NiC}}{{\rm{l}}_4}} \right]{\rm{\;}}$ is

Tetrachloro nickel (II) – tetraammine nickel (II)

Tetraammine nickel (II) –tetrachloro nickel(II)

Tetrachloro nickel (II) –tetraammine nickelate(0)

The IUPAC name of the coordination compound ${{\rm{K}}_3}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]$ is

Tripotassium hexacyanoiron (II)

The IUPAC name of $\left[ {{\rm{CoCl}}\left( {{\rm{N}}{{\rm{O}}_2}} \right){{\left( {{\rm{en}}} \right)}_2}} \right]{\rm{Cl}}$ is:

Chloronitro-$bis$(ethylenediamine) cobaltic(III) chloride

Chloronitro-$bis$(ethylenediamine)cobalt(II) chloride

$Bis - $(ethylenediamine)chloronitrocobalt(III) chloride

The IUPAC name of compound ${{\rm{K}}_3}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_5}{\rm{NO}}} \right]$ is

Pentacyano nitrosyl potassium ferrate(II)

Potassium cyano pentanitrosyl ferrate(II)

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Easy

1 Easy

1 of 98Question permalink

Application +4 | -1

The IUPAC name of

{\mathrm{K}_2}\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]

is

View answer and explanation

Correct option: A

Potassium tetracyanonickelate (II)

Option A (Potassium tetracyanonickelate (II)) is correct because it correctly names the compound, indicating the potassium cation, the tetracyano ligands, the nickelate anion, and the oxidation state of nickel.

Solution approach

Correct Answer + Core Reason: Potassium tetracyanonickelate (II). 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: Cyano (CN) with a charge of -1. There are four cyano ligands.

Step 3: Identify the central metal: Nickel (Ni).

Step 4: Determine the oxidation state of nickel: The complex ion has a charge of -2. Since each cyanide ligand has a -1 charge, the total charge from the ligands is -4. Therefore, the oxidation state of nickel must be +2 to result in a -2 overall charge for the complex ion.

Step 5: Name the complex: Potassium tetracyanonickelate (II).

Why other options are wrong:
- Option B: Potassium tetracyanatonickelate (III): Incorrect oxidation state of nickel (should be II).
- Option C: Potassium tetracyanatonickel (II): Incorrect name for the complex ion (should be nickelate).
- Option D: Potassium tetracyanonickel (III): Incorrect oxidation state of nickel and incorrect name for the complex ion.

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.

2 Easy

2 of 98Question permalink

Application +4 | -1

The pair of compounds having metals in their highest oxidation state is

View answer and explanation

Correct option: B

{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - },\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

Option B (

{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - },\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

) is correct because Mn in

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

and Cr in

\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

are in their highest oxidation states, +7 and +6 respectively.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - },\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ . The oxidation states of Mn and Cr are +7 and +6, respectively, which are their highest possible oxidation states.

Step-by-Step Solution:

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

Step 2: For $\mathrm{Mn}{\mathrm{O}_2}$ , let the oxidation state of Mn be x. Then, $x + 2(-2) = 0$ , so $x = +4$ .

Step 3: For $\mathrm{FeC}{\mathrm{l}_3}$ , the oxidation state of Fe is +3.

Step 4: For ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$ , let the oxidation state of Mn be x. Then, $x + 4(-2) = -1$ , so $x = +7$ .

Step 5: For $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ , let the oxidation state of Cr be x. Then, $x + 2(-2) + 2(-1) = 0$ , so $x = +6$ .

Step 6: For ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , let the oxidation state of Fe be x. Then, $x + 6(-1) = -3$ , so $x = +3$ .

Step 7: For ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_3}} \right]}$ , let the oxidation state of Co be x. Then, $x + 3(-1) = 0$ , so $x = +3$ .

Step 8: For ${\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }}$ , let the oxidation state of Ni be x. Then, $x + 4(-1) = -2$ , so $x = +2$ .

Step 9: For ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^ - }$ , let the oxidation state of Co be x. Then, $x + 4(-1) = -1$ , so $x = +3$ .

Step 10: Identify the pair with metals in their highest oxidation states. Mn can have a maximum oxidation state of +7, and Cr can have a maximum oxidation state of +6. Fe can have a maximum oxidation state of +6, Ni can have a maximum oxidation state of +4, and Co can have a maximum oxidation state of +5.

Step 11: Compare the oxidation states with the maximum possible oxidation states. In option B, Mn is in +7 state and Cr is in +6 state, which are their highest oxidation states.

Why other options are wrong:
- Option A: In $\mathrm{Mn}{\mathrm{O}_2}$ , Mn is in +4 oxidation state, which is not its highest (+7). In $\mathrm{FeC}{\mathrm{l}_3}$ , Fe is in +3 oxidation state, which is not its highest (+6).
- Option C: In ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , Fe is in +3 oxidation state, which is not its highest (+6). In ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_3}} \right]}$ , Co is in +3 oxidation state, which is not its highest (+5).
- Option D: In ${\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }}$ , Ni is in +2 oxidation state, which is not its highest (+4). In ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^ - }$ , Co is in +3 oxidation state, which is not its highest (+5).

Key Concepts:
- Oxidation state calculation
- Maximum oxidation states of transition metals

Memory Aids:
- Remember the common oxidation states of oxygen (-2) and chlorine (-1).
- The sum of oxidation states in a neutral compound is zero, and in an ion, it equals the charge of the ion.

Quick Tips:
- Know the maximum oxidation states of common transition metals.
- Practice oxidation state calculations frequently.

3 Easy

3 of 98Question permalink

Remembering +4 | -1

An imperfect complex of a complex compound is 100% ionized; the compound is called:

View answer and explanation

Correct option: A

Double salt

Option A (Double salt) is correct because double salts dissociate into their constituent ions when dissolved in water, exhibiting 100% ionization.

Solution approach

Correct Answer + Core Reason: Double salt. Double salts completely dissociate into ions in solution.

Step-by-Step Solution:

Step 1: Understand the definition of a double salt.

Step 2: Double salts are ionic compounds formed from the combination of two different salts. When dissolved in water, they dissociate completely into their constituent ions.

Step 3: This complete dissociation leads to 100% ionization of the complex.

Step 4: Complex salts, acid salts, and normal salts do not necessarily undergo 100% ionization in solution.

Why other options are wrong:
- Option B: Complex salts contain complex ions that do not fully dissociate in solution. Therefore, they are not 100% ionized.
- Option C: Acid salts are salts that contain an ionizable hydrogen atom. Their ionization depends on the specific salt and is not necessarily 100%.
- Option D: Normal salts are salts that do not contain ionizable hydrogen or hydroxide ions. Their ionization is also not necessarily 100% and depends on the salt's solubility and dissociation constant.

Key Concepts:
- Double salts
- Complex salts
- Acid salts
- Normal salts
- Ionization

Memory Aids:
- Double salts = Double the ions (complete dissociation)

Quick Tips:
- Remember that double salts are mixtures of two salts and dissociate completely.

4 Easy

4 of 98Question permalink

Understanding +4 | -1

The two compounds pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt(III) chloride represent:

View answer and explanation

Correct option: D

No isomerism

Option D is correct. The two compounds are not isomers because they differ only in the counter-ion (bromide vs. chloride), not in the arrangement of ligands around the central metal ion.

Solution approach

Correct Answer + Core Reason: D. No isomerism. They differ only in the counter-ion.

Step-by-Step Solution:

Step 1: Identify the two compounds: pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt(III) chloride.

Step 2: Write the formulas of the compounds: $\Co(NH_3)_5SO_4\Br$ and $\Co(NH_3)_5SO_4\Cl$ .

Step 3: Observe the difference between the compounds: The only difference is the counter-ion outside the coordination sphere (bromide vs. chloride).

Step 4: Since the compounds only differ in the counter-ion and not in the arrangement of ligands around the central metal ion, they are not isomers.

Why other options are wrong:
- Option A: Linkage isomerism requires an ambidentate ligand, which is not the case here.
- Option B: Ionization isomerism involves the exchange of ligands inside and outside the coordination sphere, which is not the case here.
- Option C: Coordination isomerism requires two or more complex ions, which is not the case here.

Key Concepts:
- Isomerism in coordination compounds
- Coordination sphere
- Counter-ions

Memory Aids:
- Isomers have the same formula but different arrangements.

Quick Tips:
- Check if the compounds have the same formula and different arrangements before classifying them as isomers.

5 Easy

5 of 98Question permalink

Application +4 | -1

The IUPAC name of the compound

\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}{{\left( {\mathrm{N}{\mathrm{O}_3}} \right)}_2}} \right]

is:

View answer and explanation

Correct option: B

Dinitratotetraamminecopper(II)

The correct answer is B (Dinitratotetraamminecopper(II)). The IUPAC name follows the rules for naming coordination compounds: ligands are named alphabetically, followed by the metal with its oxidation state in Roman numerals.

Solution approach

Correct Answer + Core Reason: Dinitratotetraamminecopper(II). Follows IUPAC naming conventions.

Step-by-Step Solution:

Step 1: Identify the ligands and the metal. The ligands are ammine ( $\mathrm{N}{\mathrm{H}_3}$ ) and nitrate ( $\mathrm{N}{\mathrm{O}_3}$ ). The metal is copper (Cu).

Step 2: Determine the oxidation state of the metal. The complex has a charge of 0. Nitrate has a charge of -1, and there are two nitrates, so the total negative charge is -2. Therefore, copper must have an oxidation state of +2.

Step 3: Name the ligands alphabetically. Ammine comes before nitrate.

Step 4: Write the name: Tetraammine dinitrato copper(II).

Why other options are wrong:
- Option A: Cuprammonium nitrate: This is an older, less systematic name.
- Option C: Tetraamminecopper(II) dinitrite: Nitrate is $\mathrm{N}{\mathrm{O}_3}^-$ , not nitrite ( $\mathrm{N}{\mathrm{O}_2}^-$ ).
- Option D: Tetraamminecopper(III) dinitrite: The oxidation state of copper is +2, not +3, and nitrate is $\mathrm{N}{\mathrm{O}_3}^-$ , not nitrite ( $\mathrm{N}{\mathrm{O}_2}^-$ ).

Key Concepts:
- IUPAC nomenclature
- Coordination compounds
- Oxidation state

Memory Aids:
- Ligands named alphabetically, then metal with oxidation state.

Quick Tips:
- Follow the IUPAC naming rules systematically.

6 Easy

6 of 98Question permalink

Understanding +4 | -1

Which of the following organometallic compound is

\sigma \,\,\mathrm{and}\,\,\pi

bonded?

View answer and explanation

Correct option: C

\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]

Option C (

\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]

) is correct because Ferrocene has both sigma and pi bonding between Fe and the cyclopentadienyl rings.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]$ . Ferrocene exhibits both sigma and pi bonding.

Step-by-Step Solution:

Step 1: Define $\sigma$ and $\pi$ bonding in organometallic compounds: $\sigma$ bonding involves direct overlap of orbitals, while $\pi$ bonding involves sideways overlap of p-orbitals.

Step 2: Analyze the options:
- A. $\mathrm{Fe}{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}$ : This involves only sigma bonds between Fe and methyl groups.
- B. ${\left[ {\mathrm{Co}{{\left( {\mathrm{CO}} \right)}_5}\mathrm{N}{\mathrm{H}_3}} \right]^{2 + }}$ : This involves sigma bonds between Co and CO, and Co and $NH_3$ , and pi backbonding between Co and CO.
- C. $\left[ {\mathrm{Fe}{{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)}_2}} \right]$ : Ferrocene has both sigma and pi bonding between Fe and the cyclopentadienyl rings.
- D. $\mathrm{K}\left[ {\mathrm{PtC}{\mathrm{l}_3}\left( {{\mathrm{\eta }^2} - {\mathrm{C}_2}{\mathrm{H}_4}} \right)} \right]$ : Zeise's salt has sigma and pi bonding between Pt and ethylene.

Step 3: Conclusion: Ferrocene is the best answer because the cyclopentadienyl ligands form both sigma and pi bonds with the iron center.

Why other options are wrong:
- Option A: $\mathrm{Fe}{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}$ : Only sigma bonds are present.
- Option B: ${\left[ {\mathrm{Co}{{\left( {\mathrm{CO}} \right)}_5}\mathrm{N}{\mathrm{H}_3}} \right]^{2 + }}$ : While pi backbonding exists between Co and CO, the bonding with $NH_3$ is purely sigma.
- Option D: $\mathrm{K}\left[ {\mathrm{PtC}{\mathrm{l}_3}\left( {{\mathrm{\eta }^2} - {\mathrm{C}_2}{\mathrm{H}_4}} \right)} \right]$ : While Zeise's salt has sigma and pi bonding, Ferrocene is a more classic example of both.

Key Concepts:
- Sigma bonding
- Pi bonding
- Organometallic compounds
- Ferrocene
- Zeise's salt

Memory Aids:
- Remember Ferrocene as a classic example of sigma and pi bonding in organometallic chemistry.

Quick Tips:
- Understand the nature of bonding between metals and ligands in organometallic compounds.

7 Easy

7 of 98Question permalink

Application +4 | -1

The IUPAC name of

\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]\mathrm{\;}

is

View answer and explanation

Correct option: C

Tetraammine nickel (II) –tetrachloro nickelate(II)

Option C (Tetraammine nickel (II) –tetrachloro nickelate(II)) is correct because it follows the IUPAC naming conventions for coordination compounds, correctly identifying the cation and anion.

Solution approach

Correct Answer + Core Reason: Tetraammine nickel (II) –tetrachloro nickelate(II). The cation is named first, followed by the anion. The oxidation state of Ni is +2, and the anion is a complex.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The complex is $\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]\mathrm{\;}$ which dissociates into $\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }$ and $\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }$ .

Step 2: Name the cation. The cation is a complex ion with Ni as the central metal and four ammonia ligands. The name is tetraammine nickel(II).

Step 3: Name the anion. The anion is a complex ion with Ni as the central metal and four chloride ligands. The name is tetrachloronickelate(II).

Step 4: Combine the names. The IUPAC name is Tetraammine nickel (II) –tetrachloro nickelate(II).

Why other options are wrong:
- Option A: Tetrachloro nickel (II) – tetraammine nickel (II) - Incorrect order, anion should be named after cation.
- Option B: Tetraammine nickel (II) –tetrachloro nickel(II) - Incorrect, the anionic complex should end with -ate.
- Option D: Tetrachloro nickel (II) –tetraammine nickelate(0) - Incorrect order and incorrect oxidation state.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Naming of complex ions
- Identifying cation and anion
- Oxidation state determination

Memory Aids:
- Cation is named before anion.
- Anionic complexes end with -ate.

Quick Tips:
- Always identify the cation and anion first.
- Determine the oxidation state of the metal.
- Follow IUPAC rules for naming ligands and metal ions.

8 Easy

8 of 98Question permalink

Understanding +4 | -1

Complexes with

C{N^ - }

ligands are usually:

View answer and explanation

Correct option: B

Low spin complexes

The correct answer is B (Low spin complexes) because

C{N^-}

is a strong field ligand, which causes pairing of electrons and results in low spin complexes.

Solution approach

Correct Answer + Core Reason: Low spin complexes. $CN^-$ is a strong field ligand, leading to a large crystal field splitting and favoring low spin complexes.

Step-by-Step Solution:

Step 1: Identify the ligand. The ligand is $CN^-$ .

Step 2: Determine the strength of the ligand based on the spectrochemical series. $CN^-$ is a strong field ligand.

Step 3: Relate the ligand strength to the spin state of the complex. Strong field ligands cause a large crystal field splitting ( $\Delta_o$ ), which is greater than the pairing energy (P). This favors pairing of electrons and results in low spin complexes.

Step 4: Conclusion: Complexes with $CN^-$ ligands are usually low spin complexes.

Why other options are wrong:
- Option A (High spin complexes): High spin complexes are favored by weak field ligands, not strong field ligands like $CN^-$ .
- Option C (Both (1) and (2)): This is incorrect because $CN^-$ only forms low spin complexes.
- Option D (None of these): This is incorrect because $CN^-$ forms low spin complexes.

Key Concepts:
- Spectrochemical series
- Crystal field splitting energy ( $\Delta_o$ )
- Pairing energy (P)
- High spin vs. low spin complexes

Memory Aids:
- Strong field ligands = Low spin complexes
- Weak field ligands = High spin complexes

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

9 Easy

9 of 98Question permalink

Understanding +4 | -1

Which one of the following octahedral complexes will not show geometrical isomerism? (A and B are monodentate ligands)

View answer and explanation

Correct option: B

\left[ {M{A_5}B} \right]

Option B is correct because [MA5B] does not exhibit geometrical isomerism as there is only one possible arrangement of the ligands around the central metal atom.

Solution approach

Correct Answer + Core Reason: $\left[ {M{A_5}B} \right]$ . This complex has only one possible arrangement of ligands.

Step-by-Step Solution:

Step 1: Define geometrical isomerism. Geometrical isomerism occurs when there are different spatial arrangements of ligands around the central metal atom.

Step 2: Analyze each complex:
- $\left[ {M{A_4}{B_2}} \right]$ : Can exhibit cis and trans isomers.
- $\left[ {M{A_5}B} \right]$ : Only one possible arrangement. All A ligands are adjacent to the B ligand.
- $\left[ {M{A_2}{B_4}} \right]$ : Can exhibit cis and trans isomers.
- $\left[ {M{A_3}{B_3}} \right]$ : Can exhibit fac (facial) and mer (meridional) isomers.

Step 3: Identify the complex that does not show geometrical isomerism. $\left[ {M{A_5}B} \right]$ does not show geometrical isomerism because all five A ligands are equivalent, and the position of B is fixed relative to them.

Why other options are wrong:
- Option A: $\left[ {M{A_4}{B_2}} \right]$ exhibits cis and trans isomers.
- Option C: $\left[ {M{A_2}{B_4}} \right]$ exhibits cis and trans isomers.
- Option D: $\left[ {M{A_3}{B_3}} \right]$ exhibits fac and mer isomers.

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

Memory Aids:
- Complexes of the type $MA_5B$ do not show geometrical isomerism.

Quick Tips:
- Draw the possible arrangements of ligands to determine if geometrical isomers exist.

10 Easy

10 of 98Question permalink

Analyzing +4 | -1

Which one of the following complexes is not expected to exhibit isomerism?

View answer and explanation

Correct option: C

\left[{Ni}\left({NH}_3\right)_2 {Cl}_2\right]

Option C ([{Ni}({NH}_3)_2 {Cl}_2]) is correct because it is a square planar complex, and the tetrahedral complex does not exhibit isomerism.

Solution approach

Correct Answer + Core Reason: $\left[{Ni}\left({NH}_3\right)_2 {Cl}_2\right]$ . This complex is square planar and does not exhibit isomerism.

Step-by-Step Solution:

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

Step 2: Option A: $\left[{Ni}\left({NH}_3\right)_4\left({H}_2 {O}\right)_2\right]^{2+}$ : This complex is octahedral and can exhibit geometrical isomerism (cis and trans). It can also exhibit optical isomerism in the cis form.

Step 3: Option B: $\left[{Pt}\left({NH}_3\right)_2 {Cl}_2\right]$ : Platinum(II) complexes are typically square planar and can exhibit geometrical isomerism (cis and trans).

Step 4: Option C: $\left[{Ni}\left({NH}_3\right)_2 {Cl}_2\right]$ : Nickel(II) with these ligands forms a tetrahedral complex. Tetrahedral complexes with two different ligands do not exhibit geometrical isomerism or optical isomerism.

Step 5: Option D: $\left[{Ni}(\text { en })_3\right]^{2+}$ : This complex is octahedral and can exhibit optical isomerism.

Step 6: Final Answer: $\left[{Ni}\left({NH}_3\right)_2 {Cl}_2\right]$ is not expected to exhibit isomerism.

Why other options are wrong:
- Option A: Exhibits geometrical isomerism.
- Option B: Exhibits geometrical isomerism.
- Option D: Exhibits optical isomerism.

Key Concepts:
- Geometrical isomerism (cis-trans)
- Optical isomerism
- Coordination complexes
- Square planar complexes
- Tetrahedral complexes
- Octahedral complexes

Memory Aids:
- Square planar complexes can exhibit geometrical isomerism.
- Tetrahedral complexes with two different ligands do not exhibit geometrical or optical isomerism.

Quick Tips:
- Determine the geometry of the complex based on the metal ion and the ligands.
- Draw the structures of the possible isomers to determine if geometrical or optical isomerism is possible.

11 Easy

11 of 98Question permalink

Understanding +4 | -1

The complex

\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4}

furnishes:

View answer and explanation

Correct option: A

5 ions

Option A (5 ions) is correct because the complex dissociates into one complex ion,

[\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}}]^{4+}

, and four chloride ions,

4\mathrm{Cl}^-

.

Solution approach

Correct Answer + Core Reason: 5 ions. The complex dissociates into one complex ion and four chloride ions.

Step-by-Step Solution:

Step 1: Identify the complex ion. In this case, it is $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]$ .

Step 2: Determine the charge of the complex ion. Since there are four chloride ions ( $\mathrm{C}{\mathrm{l}_4}$ ), each with a -1 charge, the complex ion must have a +4 charge: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{4 + }$ .

Step 3: Determine the number of ions produced upon dissociation. The complex dissociates as follows:
$\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4} \to \left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{4 + } + 4\mathrm{C}{\mathrm{l}^ - }$ \nThis yields one complex ion and four chloride ions, for a total of 5 ions.

Why other options are wrong:
- Option B: Incorrect. It suggests 6 ions, which is not the correct number of ions produced upon dissociation.
- Option C: Incorrect. It suggests 4 ions, which only accounts for the chloride ions and not the complex ion.
- Option D: Incorrect. It suggests 2 ions, which is far from the correct number.

Key Concepts:
- Coordination complexes
- Ionization of complexes
- Counter ions

Memory Aids:
- The counter ions outside the brackets dissociate.

Quick Tips:
- Count the number of ions outside the coordination sphere and add 1 (for the complex ion).

12 Easy

12 of 98Question permalink

Application +4 | -1

The IUPAC name of the coordination compound

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

is

View answer and explanation

Correct option: C

Potassium hexacyanoferrate (III)

Option C (Potassium hexacyanoferrate (III)) is correct because it follows the IUPAC naming rules for coordination compounds, correctly identifying the cation and the anionic complex.

Solution approach

Correct Answer + Core Reason: Potassium hexacyanoferrate (III). The compound is an ionic compound where potassium is the cation and the complex ion is the anion. The oxidation state of iron is +3.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The compound is ${\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ \ which dissociates into $3{\mathrm{K}^ + }$ and $\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }$ .

Step 2: Name the cation. The cation is potassium.

Step 3: Name the anion. The anion is a complex ion with Fe as the central metal and six cyanide ligands. The name is hexacyanoferrate(III).

Step 4: Combine the names. The IUPAC name is Potassium hexacyanoferrate (III).

Why other options are wrong:
- Option A: Tripotassium hexacyanoiron (II) - Incorrect, the prefix "tri" is not used for cations, and the oxidation state is incorrect.
- Option B: Potassium hexacyanoiron(II) - Incorrect oxidation state of iron.
- Option D: Potassium hexacyanoferrate (II) - Incorrect oxidation state of iron.

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Naming of complex ions
- Identifying cation and anion
- Oxidation state determination

Memory Aids:
- Cation is named before anion.
- Anionic complexes end with -ate.

Quick Tips:
- Always identify the cation and anion first.
- Determine the oxidation state of the metal.
- Follow IUPAC rules for naming ligands and metal ions.

13 Easy

13 of 98Question permalink

Understanding +4 | -1

Exchange of coordination group by a water molecule in complex molecule results in:

View answer and explanation

Correct option: C

Hydration isomerism

Option C is correct. Hydration isomerism involves the exchange of coordinated groups with water molecules in a complex.

Solution approach

Correct Answer + Core Reason: C. Hydration isomerism. This type of isomerism involves water molecules.

Step-by-Step Solution:

Step 1: Define hydration isomerism: Hydration isomerism occurs when water molecules inside the coordination sphere are exchanged with counter ions outside the coordination sphere.

Step 2: Consider the given scenario: Exchange of a coordination group by a water molecule implies that a ligand is being replaced by $H_2O$ .

Step 3: This is the definition of hydration isomerism.

Why other options are wrong:
- Option A: Ionization isomerism involves the exchange of ions inside and outside the coordination sphere, but not specifically with water.
- Option B: Ligand isomerism involves different arrangements of ligands, but not necessarily involving water.
- Option D: Geometrical isomerism involves different spatial arrangements of ligands around the central metal ion.

Key Concepts:
- Isomerism in coordination compounds
- Hydration isomerism
- Coordination sphere

Memory Aids:
- Hydration: Think of water.

Quick Tips:
- Know the definitions of different types of isomerism in coordination compounds.

14 Easy

14 of 98Question permalink

Remembering +4 | -1

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

is a :

View answer and explanation

Correct option: B

Complex compound

Option B (Complex compound) is correct because

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

is a coordination complex, where the iron ion is surrounded by cyanide ligands, forming a complex ion.

Solution approach

Correct Answer + Core Reason: Complex compound. ${K}_4\left[{Fe}({CN})_6\right]$ is a coordination complex.

Step-by-Step Solution:

Step 1: Analyze the formula: ${K}_4\left[{Fe}({CN})_6\right]$ .

Step 2: Identify the components: Potassium (K), Iron (Fe), and Cyanide (CN).

Step 3: Recognize the presence of a complex ion: The iron ion (Fe) is surrounded by cyanide ligands (CN) within the square brackets, indicating a complex ion.

Step 4: Determine the type of compound: Since it contains a complex ion, it is a complex compound.

Why other options are wrong:
- Option A: Double salt: A double salt is a mixture of two salts that crystallize together, but the individual ions retain their identity in solution. ${K}_4\left[{Fe}({CN})_6\right]$ does not dissociate into simple salts in solution.
- Option C: Acid: This compound is not an acid.
- Option D: Base: This compound is not a base.

Key Concepts:
- Coordination complexes
- Complex ions
- Double salts

Memory Aids:
- Square brackets indicate a complex ion.

Quick Tips:
- Look for the presence of complex ions to identify complex compounds.

15 Easy

15 of 98Question permalink

Understanding +4 | -1

Complexes with halide ligands are generally:

View answer and explanation

Correct option: A

High spin complexes

Option A (High spin complexes) is correct because halide ligands are weak field ligands, leading to high spin complexes.

Solution approach

Correct Answer + Core Reason: High spin complexes. Halide ligands are weak field ligands.

Step-by-Step Solution:

Step 1: Recall Ligand Field Strength: Halide ligands (like $F^-$ , $Cl^-$ , $Br^-$ , $I^-$ ) are generally weak field ligands.

Step 2: Weak Field Ligands and Splitting: Weak field ligands cause a small crystal field splitting ( $\Delta_o$ ).

Step 3: High Spin vs. Low Spin: When $\Delta_o$ is small, electrons prefer to occupy higher energy d-orbitals rather than pairing up in lower energy orbitals. This leads to high spin complexes.

Step 4: Conclusion: Complexes with halide ligands are generally high spin complexes.

Why other options are wrong:
- Option B (Low spin complexes): Low spin complexes are formed with strong field ligands that cause a large crystal field splitting, forcing electrons to pair up in lower energy orbitals.
- Option C (Both (1) and (2)): Halide ligands generally form high spin complexes, not both high and low spin.
- Option D (None of these): Halide ligands do form a specific type of complex (high spin).

Key Concepts:
- Crystal Field Theory
- High Spin and Low Spin Complexes
- Ligand Field Strength

Memory Aids:
- Weak field ligands = High spin; Strong field ligands = Low spin

Quick Tips:
- Remember that halides are generally weak field ligands.

16 Easy

16 of 98Question permalink

Application +4 | -1

Iron has lowest oxidation state in:

View answer and explanation

Correct option: A

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

The correct answer is A (

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

). In metal carbonyls, the oxidation state of the metal is considered to be zero because CO is a neutral ligand.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ . Carbonyl ligands are neutral, resulting in zero oxidation state for Fe.

Step-by-Step Solution:

Step 1: Determine the oxidation state of iron in each compound.

Step 2: $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ : CO is a neutral ligand, so the oxidation state of Fe is 0.

Step 3: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ : Oxygen has an oxidation state of -2, so $2(\mathrm{Fe}) + 3(-2) = 0$ , which means $\mathrm{Fe} = +3$ .

Step 4: ${\mathrm{K}_2}\mathrm{Fe}{\mathrm{O}_4}$ : Potassium has an oxidation state of +1, and oxygen has an oxidation state of -2, so $2(+1) + \mathrm{Fe} + 4(-2) = 0$ , which means $\mathrm{Fe} = +6$ .

Step 5: $\mathrm{FeS}{\mathrm{O}_4} \cdot {\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4} \cdot 6{\mathrm{H}_2}\mathrm{O}$ : Sulphate ( $\mathrm{S}{\mathrm{O}_4}^{2-}$ ) has a charge of -2, so $\mathrm{Fe} - 2 = 0$ , which means $\mathrm{Fe} = +2$ .

Why other options are wrong:
- Option B: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ : The oxidation state of Fe is +3.
- Option C: ${\mathrm{K}_2}\mathrm{Fe}{\mathrm{O}_4}$ : The oxidation state of Fe is +6.
- Option D: $\mathrm{FeS}{\mathrm{O}_4} \cdot {\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4} \cdot 6{\mathrm{H}_2}\mathrm{O}$ : The oxidation state of Fe is +2.

Key Concepts:
- Oxidation state
- Metal carbonyls
- Ligand charge

Memory Aids:
- CO is a neutral ligand.

Quick Tips:
- Remember the common oxidation states of ligands.

17 Easy

17 of 98Question permalink

Application +4 | -1

Which compound is zero valent metal complex?

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because in this complex, the oxidation state of Ni is zero. CO is a neutral ligand, and the overall charge of the complex is zero.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ . In this complex, Ni has an oxidation state of zero.

Step-by-Step Solution:

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

Step 2: Analyze Option A: $\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{S}{\mathrm{O}_4}$
$SO_4$ has a charge of -2, so the complex ion has a charge of +2. $NH_3$ is a neutral ligand, so the oxidation state of Cu is +2.

Step 3: Analyze Option B: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$
$NH_3$ is neutral and Cl has a charge of -1. Since there are two Cl ligands, the total negative charge is -2. Therefore, the oxidation state of Pt is +2.

Step 4: Analyze Option C: $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ \nCO is a neutral ligand, and the overall charge of the complex is zero. Therefore, the oxidation state of Ni is 0.

Step 5: Analyze Option D: ${\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$
$K_3$ indicates a +3 charge, so the complex ion has a charge of -3. CN has a charge of -1, and there are six CN ligands, so the total negative charge from CN is -6. Therefore, the oxidation state of Fe is +3 (-3 = x - 6, so x = +3).

Step 6: Identify the complex with a zero-valent metal.\nOnly $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ has a metal (Ni) with an oxidation state of 0.

Why other options are wrong:
- Option A: $\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{S}{\mathrm{O}_4}$ The oxidation state of Cu is +2, not zero.
- Option B: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ The oxidation state of Pt is +2, not zero.
- Option D: ${\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ The oxidation state of Fe is +3, not zero.

Key Concepts:
- Coordination Chemistry
- Oxidation State
- Ligands and their charges

Memory Aids:
- Remember the common charges of ligands like $NH_3$ (0), $Cl^-$ (-1), $CN^-$ (-1), $CO$ (0), $SO_4^{2-}$ (-2).

Quick Tips:
- Always start by determining the overall charge of the complex ion.
- Use the known charges of the ligands to calculate the oxidation state of the metal.

18 Easy

18 of 98Question permalink

Application +4 | -1

The IUPAC name of

\left[ {\mathrm{CoCl}\left( {\mathrm{N}{\mathrm{O}_2}} \right){{\left( {\mathrm{en}} \right)}_2}} \right]\mathrm{Cl}

is:

View answer and explanation

Correct option: C

Chloro-

bis

(ethylenediamine)nitrocobalt(III) chloride

The correct answer is C (Chloro-

bis

(ethylenediamine)nitrocobalt(III) chloride) because it follows the IUPAC naming conventions for coordination compounds.

Solution approach

Correct Answer + Core Reason: Chloro- $bis$ (ethylenediamine)nitrocobalt(III) chloride. This follows IUPAC naming conventions.

Step-by-Step Solution:

Step 1: Identify the complex: $\left[ {\mathrm{CoCl}\left( {\mathrm{N}{\mathrm{O}_2}} \right){{\left( {\mathrm{en}} \right)}_2}} \right]\mathrm{Cl}$

Step 2: Identify the ligands: Cl (chloro), NO $_2$ (nitro), and en (ethylenediamine).

Step 3: Determine the oxidation state of Co. Cl outside the bracket has a -1 charge. Inside the bracket, Cl has a -1 charge, NO $_2$ has a -1 charge, and en is neutral. Therefore, the oxidation state of Co is +3.

Step 4: Name the ligands in alphabetical order: chloro, $bis$ (ethylenediamine), nitro.

Step 5: Combine the names: Chloro- $bis$ (ethylenediamine)nitrocobalt(III) chloride.

Why other options are wrong:
- Option A: Incorrect order of ligands and incorrect naming of cobalt.
- Option B: Incorrect oxidation state of cobalt.
- Option D: Incorrect order of ligands.

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

Memory Aids:
- Remember the IUPAC naming rules for coordination compounds.

Quick Tips:
- Always name the ligands in alphabetical order, followed by the metal and its oxidation state.

19 Easy

19 of 98Question permalink

Application +4 | -1

The oxidation state of

\mathrm{Cr}

in

{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{\;C}{\mathrm{l}_2}} \right]^ + }\,\mathrm{is}

View answer and explanation

Correct option: D

3

Option D (3) is correct because the oxidation state of Cr in

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

is +3, calculated by considering the neutral NH3 ligands and the -1 charge of each Cl ligand, and the overall +1 charge of the complex.

Solution approach

Correct Answer + Core Reason: 3. The oxidation state of Cr is calculated based on the charges of the ligands and the overall charge of the complex.

Step-by-Step Solution:

Step 1: Identify the charges of the ligands. $NH_3$ is a neutral ligand (charge = 0), and $Cl$ has a -1 charge.

Step 2: Set up an equation to solve for the oxidation state of Cr. Let x be the oxidation state of Cr. The overall charge of the complex is +1. So, $x + 4(0) + 2(-1) = +1$ .

Step 3: Solve for x: $x - 2 = +1$ , so $x = +3$ .

Final Answer: 3

Why other options are wrong:
- Option A (0): This is incorrect because it doesn't account for the charges of the chloride ligands and the overall charge of the complex.
- Option B (1): This is incorrect because it doesn't accurately account for the charges of the chloride ligands.
- Option C (2): This is incorrect because the calculation does not result in +2 as the oxidation state.

Key Concepts:
- Coordination Chemistry
- Oxidation State
- Neutral Ligands

Memory Aids:
- Ammonia ( $NH_3$ ) is a neutral ligand.
- Chloride (Cl) has a -1 charge.

Quick Tips:
- Remember the common charges of ligands.
- Set up an equation to solve for the oxidation state of the metal.

20 Easy

20 of 98Question permalink

Analyzing +4 | -1

Assertion(A)

{Ni}({CO})_4

is diamagnetic in nature.

Reason(R) Ni-atom undergoes

s p^3

hybridization.

View answer and explanation

Correct option: B

If both (A) and (R) are correct but (R) is not correct explanation for (A).

Option B (If both (A) and (R) are correct but (R) is not correct explanation for (A)) is correct because both the assertion and reason are true, but the reason does not explain why Ni(CO)4 is diamagnetic.

Solution approach

Correct Answer + Core Reason: If both (A) and (R) are correct but (R) is not correct explanation for (A). Ni(CO)4 is diamagnetic due to the pairing of electrons, and Ni undergoes sp3 hybridization.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nAssertion (A): ${Ni}({CO})_4$ is diamagnetic in nature.\nNi has an electronic configuration of [Ar] 3d⁸ 4s². In ${Ni}({CO})_4$ , Ni is in the 0 oxidation state. CO is a strong field ligand, which causes pairing of electrons. The electronic configuration becomes 3d¹⁰, resulting in no unpaired electrons. Therefore, ${Ni}({CO})_4$ is diamagnetic. Assertion (A) is correct.

Step 2: Analyze the reason.\nReason (R): Ni-atom undergoes $s p^3$ hybridization.\nIn ${Ni}({CO})_4$ , Ni undergoes $s p^3$ hybridization to accommodate the four CO ligands. This hybridization leads to a tetrahedral geometry. Reason (R) is also correct.

Step 3: Determine if the reason is the correct explanation for the assertion.\nThe $s p^3$ hybridization explains the tetrahedral geometry of the complex but does not directly explain why it is diamagnetic. The diamagnetic nature is due to the pairing of electrons caused by the strong field ligand CO.

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

Key Concepts:
- Diamagnetism
- Hybridization
- Coordination complexes
- Strong field ligands

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to diamagnetism.

Quick Tips:
- Consider the electronic configuration and ligand field strength to determine the magnetic properties of a complex.

21 Easy

21 of 98Question permalink

Understanding +4 | -1

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

are a pair of ….. Isomers.

View answer and explanation

Correct option: A

Ionisation

Option A (Ionisation) is correct because the given compounds exchange ions in solution, leading to different counter-ions.

Solution approach

Correct Answer + Core Reason: Ionisation. The isomers differ by the counter-ion that is coordinated to the metal center vs. acting as a free ion in solution.

Step-by-Step Solution:

Step 1: Identify the compounds: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]\mathrm{Br}$ and $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{\;Br}} \right]\mathrm{S}{\mathrm{O}_4}$ .

Step 2: Observe the difference: The sulfate ( $\mathrm{SO}_4$ ) and bromide (Br) ions have exchanged positions between the coordination sphere and the counter-ion.

Step 3: Define Ionisation Isomerism: Ionisation isomers are coordination compounds that have the same molecular formula but give different ions in solution.

Step 4: Apply the definition: In this case, one complex will produce $Br^-$ ions in solution, while the other will produce $SO_4^{2-}$ ions.

Step 5: Conclusion: Therefore, these are ionisation isomers.

Why other options are wrong:
- Option B (Ligand): Ligand isomers involve different ligands coordinating to the metal center, which is not the case here.
- Option C (Coordination): Coordination isomers involve the exchange of ligands between the cationic and anionic coordination entities, which is not applicable here as there is only one complex ion in each compound.
- Option D (Hydrate): Hydrate isomers involve different numbers of water molecules inside and outside the coordination sphere, which is not the case here.

Key Concepts:
- Isomerism in Coordination Compounds
- Ionisation Isomerism

Memory Aids:
- Ionisation isomers exchange ions.

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

22 Easy

22 of 98Question permalink

Understanding +4 | -1

Each metal possesses:

View answer and explanation

Correct option: D

All of the above

The correct answer is D (All of the above) because metals in coordination complexes possess primary valencies satisfied by anions, secondary valencies satisfied by donor molecules, and a coordination number.

Solution approach

Correct Answer + Core Reason: All of the above. Metals possess primary valencies, secondary valencies, and a coordination number.

Step-by-Step Solution:
Step 1: The question asks what each metal possesses in coordination complexes.
Step 2: According to Werner's theory:
- Primary valencies are satisfied by anions.
- Secondary valencies are satisfied by donor molecules (ligands).
- Coordination number is the number of ligands attached to the metal ion.
Step 3: Therefore, each metal possesses all of the above.

Why other options are wrong:
- Option A (Primary valencies satisfied by anions only): This is only partially correct.
- Option B (Secondary valencies satisfied by donor molecules): This is only partially correct.
- Option C (Coordination number): This is only partially correct.
- Option D (All of the above): This is the correct answer.

Key Concepts:
- Werner's theory
- Primary valency
- Secondary valency
- Coordination number

Memory Aids:
- Remember Werner's postulates.

Quick Tips:
- Understand the definitions of primary and secondary valencies.

23 Easy

23 of 98Question permalink

Remembering +4 | -1

Carnallite in solution in water shows the properties of

View answer and explanation

Correct option: A

{\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{l}^ - }

Option A (

{\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{l}^ - }

) is correct because Carnallite is

KCl.MgCl_2.6H_2O

, which dissociates into

K^+

,

Mg^{2+}

, and

Cl^-

ions in water.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{l}^ - }$ . Carnallite is $KCl.MgCl_2.6H_2O$ , which dissociates into these ions in water.

Step-by-Step Solution:

Step 1: Identify the chemical formula of Carnallite: $KCl.MgCl_2.6H_2O$ .

Step 2: Determine the ions formed when Carnallite dissolves in water.

Step 3: $KCl$ dissociates into $K^+$ and $Cl^-$ .

Step 4: $MgCl_2$ dissociates into $Mg^{2+}$ and $2Cl^-$ .

Step 5: Therefore, the solution contains $K^+$ , $Mg^{2+}$ , and $Cl^-$ ions.

Why other options are wrong:
- Option B: ${\mathrm{K}^ + },\mathrm{C}{\mathrm{l}^ - },\mathrm{SO}_4^{2 - },\mathrm{B}{\mathrm{r}^ - }$ : Carnallite does not contain sulfate or bromide ions.
- Option C: ${\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{CO}_3^{2 - }$ : Carnallite does not contain carbonate ions.
- Option D: ${\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{l}^ - },\mathrm{B}{\mathrm{r}^ - }$ : Carnallite does not contain bromide ions.

Key Concepts:
- Chemical formula of Carnallite
- Dissociation of ionic compounds in water

Memory Aids:
- Remember the formula of Carnallite: $KCl.MgCl_2.6H_2O$ .

Quick Tips:
- Know the common mineral formulas and their dissociation products.

24 Easy

24 of 98Question permalink

Application +4 | -1

The oxidation number of

\mathrm{Fe}

in

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

View answer and explanation

Correct option: C

2

The correct answer is 2 because the oxidation state of Fe in

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

is +2.

Solution approach

Correct Answer + Core Reason: 2. The oxidation state of Fe is +2.

Step-by-Step Solution:

Step 1: Identify the complex: ${\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$

Step 2: Determine the oxidation state of K: +1

Step 3: Determine the oxidation state of CN: -1

Step 4: Let the oxidation state of Fe be $x$ . The overall charge of the complex is 0.

Step 5: Set up the equation: $4(+1) + x + 6(-1) = 0$

Step 6: Solve for $x$ : $4 + x - 6 = 0 \Rightarrow x = +2$

Why other options are wrong:
- Option A: Incorrect. This would imply an incorrect charge balance within the complex.
- Option B: Incorrect. This would imply an incorrect charge balance within the complex.
- Option D: Incorrect. Oxidation states are usually positive for the metal in coordination compounds.

Key Concepts:
- Oxidation state
- Coordination complexes

Memory Aids:
- Remember the common oxidation states of elements.

Quick Tips:
- Always balance the charges in the complex to find the oxidation state of the metal.

25 Easy

25 of 98Question permalink

Analyzing +4 | -1

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

en

\rightarrow

complex

({X})

. The correct statement about the complex

(X)

is-

View answer and explanation

Correct option: D

(1) and (2) both

Option D ((1) and (2) both) is correct because after the reaction, the complex becomes

\left[\mathrm{Fe}({en})_3\right]^{2+}

, which is a low spin complex and diamagnetic due to the strong field ligand en.

Solution approach

Correct Answer + Core Reason: (1) and (2) both. The complex formed is low spin and diamagnetic due to the strong field ligand.

Step-by-Step Solution:

Step 1: Write the reaction.
$\left[\mathrm{Fe}({en})_2\left({H}_2 {O}\right)_2\right]^{2+}+$ en $\rightarrow \left[\mathrm{Fe}({en})_3\right]^{2+} + 2H_2O$

Step 2: Determine the electronic configuration of Fe(II).\nFe(II) has a $d^6$ configuration.

Step 3: Analyze the ligand field strength.\nEthylenediamine (en) is a strong field ligand. Therefore, it will cause pairing of electrons.

Step 4: Determine the spin state of the complex.\nSince en is a strong field ligand, the complex $\left[\mathrm{Fe}({en})_3\right]^{2+}$ will be a low spin complex.

Step 5: Determine the magnetic properties of the complex.\nIn a low spin $d^6$ complex, all electrons are paired. Therefore, the complex is diamagnetic.

Step 6: Check for geometrical isomerism.\nThe complex $\left[\mathrm{Fe}({en})_3\right]^{2+}$ does not show geometrical isomerism because all the ligands are the same.

Why other options are wrong:
- Option A: It is a low spin complex. This is correct, but not the complete answer.
- Option B: It is diamagnetic. This is correct, but not the complete answer.
- Option C: It shows geometrical isomerism. This is incorrect. $\left[\mathrm{Fe}({en})_3\right]^{2+}$ does not show geometrical isomerism.

Key Concepts:
- Coordination Complexes
- Ligand Field Theory
- Spin State
- Magnetic Properties
- Isomerism

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to low spin complexes.
- Diamagnetic substances have all electrons paired.

Quick Tips:
- Identify the ligand field strength to determine the spin state and magnetic properties of the complex.

26 Easy

26 of 98Question permalink

Analyzing +4 | -1

The pair of the compounds in which both the metals are in the higher possible oxidation state is

View answer and explanation

Correct option: A

\mathrm{CrO}_2 \mathrm{Cl}_2, \mathrm{MnO}_4^{-}

Option A (

\mathrm{CrO}_2 \mathrm{Cl}_2, \mathrm{MnO}_4^{-}

) is correct because Cr and Mn exhibit their highest possible oxidation states (+6 and +7, respectively) in these compounds.

Solution approach

Correct Answer + Core Reason: $\mathrm{CrO}_2 \mathrm{Cl}_2, \mathrm{MnO}_4^{-}$ . In both these compounds, the metals are in their highest possible oxidation states.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Cr in CrO2Cl2. Let x be the oxidation state of Cr. The charge of O is -2, and the charge of Cl is -1. So, x + 2(-2) + 2(-1) = 0 => x - 4 - 2 = 0 => x = +6. The maximum oxidation state of Cr is +6.

Step 2: Determine the oxidation state of Mn in MnO4-. Let y be the oxidation state of Mn. The charge of O is -2. So, y + 4(-2) = -1 => y - 8 = -1 => y = +7. The maximum oxidation state of Mn is +7.

Step 3: Analyze the other options:
- Option B: Co in [Co(CN)6]3- has an oxidation state of +3 (max is +4), and Mn in MnO3 is +6 (max is +7).
- Option C: Ti in TiO3 has an oxidation state of +6 (max is +4), and Mn in MnO2 is +4 (max is +7).
- Option D: Fe in [Fe(CN)6]3- has an oxidation state of +3 (max is +6), and Co in [Co(CN)6]3- has an oxidation state of +3 (max is +4).

Why other options are wrong:
- Option B ([Co(CN)6]3-, MnO3): Co is not in its highest possible oxidation state.
- Option C (TiO3, MnO2): Neither Ti nor Mn are in their highest possible oxidation states. TiO3 is not a stable compound. Assuming it is $TiO_3^{2-}$ , then Ti has +4 oxidation state which is its highest possible oxidation state. Mn in $MnO_2$ has +4 oxidation state, while its highest possible oxidation state is +7.
- Option D ([Fe(CN)6]3-, [Co(CN)6]3-): Neither Fe nor Co are in their highest possible oxidation states.

Key Concepts:
- Oxidation state determination
- Maximum oxidation states of transition metals

Memory Aids:
- Remember the maximum oxidation states of common transition metals.

Quick Tips:
- Always check the maximum possible oxidation state of the metal before concluding that it is in its highest oxidation state.

27 Easy

27 of 98Question permalink

Remembering +4 | -1

Estimation of calcium and magnesium is done by

View answer and explanation

Correct option: A

EDTA

Option A (EDTA) is correct because EDTA (ethylenediaminetetraacetic acid) is used in complexometric titrations to estimate the concentration of calcium and magnesium ions.

Solution approach

Correct Answer + Core Reason: EDTA. EDTA forms stable complexes with calcium and magnesium ions, allowing for their estimation via titration.

Step-by-Step Solution:

Step 1: Recall the methods used for estimating calcium and magnesium.

Step 2: EDTA (ethylenediaminetetraacetic acid) is a chelating agent that forms stable complexes with metal ions, including calcium and magnesium.

Step 3: The reaction between EDTA and these ions is used in complexometric titrations to determine their concentrations.

Why other options are wrong:
- Option B: Oxalate: Oxalate can precipitate calcium, but is not typically used for direct estimation of both calcium and magnesium simultaneously via titration.
- Option C: Phosphate: Phosphate can precipitate both calcium and magnesium, but is not typically used for direct estimation via titration.
- Option D: None of these: EDTA is a suitable method.

Key Concepts:
- Complexometric titration
- EDTA as a chelating agent

Memory Aids:
- EDTA: Remember "EDTA Estimates Divalent Atoms" to associate it with calcium and magnesium estimation.

Quick Tips:
- EDTA is a versatile reagent in analytical chemistry for metal ion determination.

28 Easy

28 of 98Question permalink

Application +4 | -1

The IUPAC name of compound

{\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NO}} \right]

is

View answer and explanation

Correct option: D

Potassium pentacyanonitrosyl ferrate (II)

Option D (Potassium pentacyanonitrosyl ferrate (II)) is correct because it correctly names the compound, indicating the potassium cation, the pentacyano and nitrosyl ligands, the ferrate anion, and the oxidation state of iron.

Solution approach

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

Step-by-Step Solution:

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

Step 2: Identify the ligands: Cyano (CN) and Nitrosyl (NO). There are five cyano ligands and one nitrosyl ligand.

Step 3: Identify the central metal: Iron (Fe).

Step 4: Determine the oxidation state of iron: The complex ion has a charge of -3. Cyano has a -1 charge, and nitrosyl (NO) is considered neutral in this case. Therefore, the oxidation state of iron must be +2 to result in a -3 overall charge for the complex ion. (-5 from CN + 0 from NO + Fe = -3, therefore Fe = +2).

Step 5: Name the complex: Potassium pentacyanonitrosyl ferrate (II).

Why other options are wrong:
- Option A: Pentacyano nitrosyl potassium ferrate(II): Incorrect order of naming (cation should come first).
- Option B: Potassium cyano pentanitrosyl ferrate(II): Incorrect ligand name and number.
- Option C: Potassium pentacyanonitrosyl ferrate (III): Incorrect oxidation state of iron (should be II).

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.
- Remember that NO is considered neutral in this complex.

29 Easy

29 of 98Question permalink

Remembering +4 | -1

{\left[ {\mathrm{EDTA}} \right]^{4 - }}{\mkern 1mu} \mathrm{is}{\mkern 1mu} {\mkern 1mu} \mathrm{a}

View answer and explanation

Correct option: D

Hexadentate ligand

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

Solution approach

Correct Answer + Core Reason: Hexadentate ligand. EDTA has six donor atoms that can bind to a central metal ion.

Step-by-Step Solution:
Step 1: Understand the definition of denticity.
Step 2: Analyze the structure of EDTA.
Step 3: Count the number of donor atoms in EDTA.

Step 4: Denticity refers to the number of donor atoms in a single ligand that bind to a central metal ion.
Step 5: EDTA (ethylenediaminetetraacetic acid) has the formula $[\mathrm{EDTA}]^{4 - }$ .
Step 6: The structure of EDTA has two nitrogen atoms and four oxygen atoms that can donate electron pairs to a central metal ion.
Step 7: Therefore, EDTA has 6 donor atoms and is a hexadentate ligand.

Why other options are wrong:
- Option A: Monodentate ligand - A monodentate ligand has only one donor atom.
- Option B: Bidentate ligand - A bidentate ligand has two donor atoms.
- Option C: Quadridentate ligand - A quadridentate ligand has four donor atoms.

Key Concepts:
- Ligands
- Denticity
- Coordination Complexes

Memory Aids:
- EDTA = Ethylenediamine Tetraacetic Acid (6 donor atoms)

Quick Tips:
- Count the number of atoms with lone pairs that can bind to the metal ion.
- Remember common examples of ligands with different denticities.

30 Easy

30 of 98Question permalink

Understanding +4 | -1

Which statement about coordination number of a cation is true?

View answer and explanation

Correct option: D

For most cations, the coordination number depends on the size, structure and charge of the ligands

Option D (For most cations, the coordination number depends on the size, structure and charge of the ligands) is correct because these factors influence how many ligands can effectively coordinate to the metal center.

Solution approach

Correct Answer + Core Reason: For most cations, the coordination number depends on the size, structure and charge of the ligands. These factors affect the number of ligands that can coordinate.

Step-by-Step Solution:

Step 1: Define coordination number. The coordination number is the number of ligands directly bonded to the central metal ion.

Step 2: Consider factors affecting coordination number. These include the size and charge of the metal ion, as well as the size, structure, and charge of the ligands.

Step 3: Evaluate the given statements. The coordination number is not solely determined by achieving a rare gas electron configuration.

Step 4: Recognize that ligand size and charge influence steric hindrance and electrostatic interactions, thus affecting the number of ligands that can coordinate.

Why other options are wrong:
- Option A: Many metal ions exhibit multiple coordination numbers depending on the ligands present.
- Option B: While the coordination number *is* the number of ligands bonded, this option doesn't explain *why* the coordination number is what it is. Option D provides the reason.
- Option C: The tendency to achieve a rare gas configuration (effective atomic number rule) can influence, but doesn't solely determine, the coordination number.

Key Concepts:
- Coordination number
- Coordination complexes
- Ligand properties

Memory Aids:
- Think of ligands crowding around the metal; size and charge matter.

Quick Tips:
- Consider steric hindrance and electrostatic interactions when determining coordination number.

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