Ammoniacal solution of ${\rm{Ni}}{\left( {{\rm{CN}}} \right)_2}$ reacts with ${{\rm{C}}_6}{{\rm{H}}_6}$ to produce a light violet coloured crystalline compound of the formula:

${\rm{Ni}}{\left( {{\rm{CN}}} \right)_2}{\rm{N}}{{\rm{H}}_3} \cdot {{\rm{C}}_6}{{\rm{H}}_6}$

${\rm{Ni}}{\left( {{\rm{CN}}} \right)_2} \cdot {{\rm{C}}_6}{{\rm{H}}_5}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_3}$

${\rm{Ni}}{\left( {{\rm{CN}}} \right)_2}{{\rm{C}}_6}{{\rm{H}}_6}$

Which will give a white precipitate with ${\rm{AgN}}{{\rm{O}}_3}$ in aqueous solution?

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

$\left[ {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_5}{\rm{Cl}}} \right]{\left( {{\rm{N}}{{\rm{O}}_2}} \right)_2}$

$\left[ {{\rm{Pt}}\left( {{\rm{en}}} \right){\rm{C}}{{\rm{l}}_2}} \right]$

$\left[ {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right]{\rm{S}}{{\rm{O}}_4}$

The $d - $electronic configuration of ${\rm{C}}{{\rm{r}}^{2 + }},{\rm{\;M}}{{\rm{n}}^{2 + }},{\rm{F}}{{\rm{e}}^{2 + }},{\rm{N}}{{\rm{i}}^{2 + }}{\rm{\;are\;}}3{{\rm{d}}^4},3{{\rm{d}}^5},3{{\rm{d}}^6}{\rm{\;and\;}}3{{\rm{d}}^8}$ respectively. Which of the following complex will show minimum paramagnetic behaviour?

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

${\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{2 + }}$

${\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{2 + }}$

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

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

${\rm{Fe}}{\left( {{{\rm{\eta }}^2} - {{\rm{C}}_5}{{\rm{H}}_5}} \right)_2}$

${\rm{Cr}}{\left( {{\eta ^5} - {{\rm{C}}_5}{{\rm{H}}_5}} \right)_5}$

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

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Medium, Page 2

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Application +4 | -1

Which one of the following has lowest value of paramagnetic behaviour?

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because

Co^{3+}

in a strong field ligand environment like

CN^-

results in a low-spin complex with no unpaired electrons, leading to the lowest paramagnetic behavior.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ . This complex has the fewest unpaired electrons due to the strong field ligand, leading to the lowest paramagnetic behavior.

Step-by-Step Solution:

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

Step 2: For ${\left[ {\mathrm{Cr}{{\left( {\mathrm{CN}_6}} \right)}_4}} \right]^{3 - }}$ , the oxidation state of Cr is +3, and the electronic configuration is $d^3$ . Since $CN^-$ is a strong field ligand, the number of unpaired electrons is 3.

Step 3: For ${\left[ {\mathrm{Mn}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , the oxidation state of Mn is +3, and the electronic configuration is $d^4$ . Since $CN^-$ is a strong field ligand, the number of unpaired electrons is 2.

Step 4: For ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , the oxidation state of Fe is +3, and the electronic configuration is $d^5$ . Since $CN^-$ is a strong field ligand, the number of unpaired electrons is 1.

Step 5: For ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , the oxidation state of Co is +3, and the electronic configuration is $d^6$ . Since $CN^-$ is a strong field ligand, the number of unpaired electrons is 0.

Step 6: The complex with the lowest number of unpaired electrons will have the lowest paramagnetic behavior. In this case, ${\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ has 0 unpaired electrons.

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

Why other options are wrong:
- **Option A ( ${\left[ {\mathrm{Cr}{{\left( {\mathrm{CN}_6}} \right)}_4}} \right]^{3 - }}$ ): This complex has 3 unpaired electrons, which is more than the correct answer.
- Option B ( ${\left[ {\mathrm{Mn}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ ): This complex has 2 unpaired electrons, which is more than the correct answer.
- Option C ( ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ ): This complex has 1 unpaired electron, which is more than the correct answer.

Key Concepts:
- Coordination Chemistry
- Paramagnetism
- Crystal Field Theory
- Electronic Configuration

Memory Aids:
- Strong field ligands cause pairing of electrons, leading to fewer unpaired electrons and lower paramagnetic behavior.

Quick Tips:**
- Determine the number of unpaired electrons in each complex to determine the paramagnetic behavior.
- Remember that strong field ligands like $CN^-$ cause pairing of electrons.

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Remembering +4 | -1

\pi

acid ligand among the following is

View answer and explanation

Correct option: D

{CO}

Option D (

{CO}

) is correct because CO is a π-acid ligand, meaning it can accept electron density from the metal through back-bonding into its π* antibonding orbitals.

Solution approach

Correct Answer + Core Reason: ${CO}$ . CO is a π acid ligand.

Step-by-Step Solution:

Step 1: Define a π acid ligand: A π acid ligand is a ligand that can accept electron density from the metal through back-bonding.

Step 2: Consider the electronic structure of the given ligands.

Step 3: CO has empty π* antibonding orbitals that can accept electron density from the metal, making it a π acid ligand.

Why other options are wrong:
- Option A: ${F}^{-}$ : Fluoride is a sigma donor ligand, not a π acid ligand.
- Option B: ${NH}_3$ : Ammonia is a sigma donor ligand, not a π acid ligand.
- Option C: ${OH}^{-}$ : Hydroxide is a sigma donor ligand, not a π acid ligand.

Key Concepts:
- π acid ligands
- Back-bonding
- Ligand field theory

Memory Aids:
- CO is a classic example of a π acid ligand.

Quick Tips:
- Remember that π acid ligands have empty π* orbitals.

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Increasing order of

\Delta_0

of the following complex is

View answer and explanation

Correct option: C

\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}< \left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}< \left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}

The correct answer is C (

\mathrm{Cr}(\mathrm{H}_2 \mathrm{O})_6^{3+}< \mathrm{Cr}(\mathrm{NH}_3)_6^{3+}< \mathrm{Cr}(\mathrm{CN})_6^{3-}

) because the spectrochemical series dictates the increasing order of ligand field splitting strength.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}< \left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}< \left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$ . The spectrochemical series determines the order of ligand field splitting.

Step-by-Step Solution:
Step 1: The question asks for the increasing order of $\Delta_0$ (crystal field splitting energy) for the given complexes.
Step 2: The magnitude of $\Delta_0$ depends on the ligand field strength. We use the spectrochemical series to determine the relative strengths of the ligands.
Step 3: The spectrochemical series is (approximately): $\mathrm{I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < CN^- < CO}$ .
Step 4: Based on the spectrochemical series, the order of ligand field strength is $\mathrm{H_2O < NH_3 < CN^-}$ .
Step 5: Therefore, the increasing order of $\Delta_0$ is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}< \left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}< \left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$ .

Why other options are wrong:
- **Option A ( $\mathrm{Cr}(\mathrm{NH}_3)_6^{3+}< \mathrm{Cr}(\mathrm{H}_2 \mathrm{O})_6^{3+}< \mathrm{Cr}(\mathrm{CN})_6^{3-}$ ):** This is incorrect because $\mathrm{NH_3}$ has a higher ligand field strength than $\mathrm{H_2O}$ .
- **Option B ( $\mathrm{Cr}(\mathrm{H}_2 \mathrm{O})_6^{3+}>\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}>\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$ ): This is incorrect because it shows a decreasing order.
- Option D ( $\mathrm{Cr}(\mathrm{NH}_3)_6^{3+}< \mathrm{Cr}(\mathrm{CN})_6^{3-}< \mathrm{Cr}(\mathrm{H}_2 \mathrm{O})_6^{3+}$ ):** This is incorrect because $\mathrm{H_2O}$ has a lower ligand field strength than $\mathrm{NH_3}$ .

Key Concepts:
- Crystal field splitting energy ( $\Delta_0$ )
- Spectrochemical series
- Ligand field strength

Memory Aids:
- Memorize the spectrochemical series.

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

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Application +4 | -1

Chemical formula for iron (III) hexacyanoferrate (II) is

View answer and explanation

Correct option: D

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

Option D (

Fe_4[Fe(CN)_6]_3

) is correct because it represents the correct ratio of Fe(III) and hexacyanoferrate(II) to achieve charge neutrality.

Solution approach

Correct Answer + Core Reason: $Fe_4[Fe(CN)_6]_3$ . This formula balances the +3 charge of Fe(III) with the -4 charge of hexacyanoferrate(II).

Step-by-Step Solution:

Step 1: Identify the ions. Iron(III) is Fe3+, and hexacyanoferrate(II) is [Fe(CN)6]4-.

Step 2: Determine the ratio needed for charge neutrality. To achieve a neutral compound, the total positive charge must equal the total negative charge.

Step 3: Find the least common multiple of 3 and 4, which is 12. Therefore, we need 4 Fe3+ ions (4 * +3 = +12) and 3 [Fe(CN)6]4- ions (3 * -4 = -12).

Step 4: Write the formula: Fe4[Fe(CN)6]3.

Why other options are wrong:
- **Option A ( $Fe[Fe(CN)_6]$ ): Incorrect because it does not balance the charges. Fe3+ and [Fe(CN)6]4- would result in a -1 charge.
- Option B ( $Fe_3[Fe(CN)_6]$ ): Incorrect because it does not balance the charges. 3Fe3+ and [Fe(CN)6]4- would result in a +5 charge.
- Option C ( $Fe_3[Fe(CN)_6]_4$ ): Incorrect because it does not balance the charges. 3Fe3+ and 4[Fe(CN)6]4- would result in a -7 charge.

Key Concepts:
- Charge balancing in ionic compounds
- Formula writing for coordination compounds

Memory Aids:
- Remember to cross the charges to determine the subscripts in the formula.

Quick Tips:**
- Always ensure that the total positive charge equals the total negative charge in the final formula.

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Remembering +4 | -1

Ligands in a complex salt are:

View answer and explanation

Correct option: D

Ions or molecules linked by coordinate bonds to a central metal atom or ion

Option A. Primary valencies are ionizable is correct because Werner's theory states that primary valencies (oxidation state) are satisfied by negative ions and are ionizable.

Solution approach

Correct Answer + Core Reason: Primary valencies are ionizable. Werner's theory states that primary valencies (oxidation state) are satisfied by negative ions and are ionizable.

Step-by-Step Solution:

Step 1: Recall Werner's postulates.\nWerner's theory postulates that:
1. Metals exhibit two types of valencies: primary and secondary.
2. Primary valencies correspond to the oxidation state of the metal ion and are satisfied by negative ions. These are ionizable.
3. Secondary valencies correspond to the coordination number of the metal ion and are satisfied by either negative ions or neutral molecules. These are non-ionizable.
4. Every metal has a fixed number of secondary valencies.

Step 2: Apply Werner's postulates to the given options.\nAccording to Werner's theory, primary valencies are ionizable, and secondary valencies are non-ionizable.

Why other options are wrong:
- Option B: Secondary valencies are ionizable. Incorrect, secondary valencies are non-ionizable.
- Option C: Only primary valencies are non-ionizable. Incorrect, primary valencies are ionizable.
- Option D: Primary and secondary valencies are non-ionizable. Incorrect, primary valencies are ionizable.

Key Concepts:
- Werner's Theory
- Primary Valency (Oxidation State)
- Secondary Valency (Coordination Number)
- Ionizable vs. Non-ionizable Valencies

Memory Aids:
- Primary valencies are \"primary targets\" for ionization.

Quick Tips:
- Remember that primary valencies are satisfied by negative ions and are ionizable, while secondary valencies are satisfied by either negative ions or neutral molecules and are non-ionizable.

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In the change

\left[{Cu}\left({H}_2 {O}_6\right)\right]^{2+} \stackrel{{HCl}}{\longrightarrow}\left[{CuCl}\left({H}_2 {O}\right)_5\right]^{+}

, the colour changes from

View answer and explanation

Correct option: B

Blue to pink

Option B is correct because the addition of HCl to [Cu(H2O)6]2+ results in the replacement of water ligands with chloride ligands, leading to a change in color from blue to pink.

Solution approach

Correct Answer + Core Reason: Blue to pink. Chloride ligands replace water ligands, changing the crystal field and color.

Step-by-Step Solution:

Step 1: Identify the initial complex and its color. $[Cu(H_2O)_6]^{2+}$ is a blue complex.

Step 2: Consider the effect of adding HCl. Adding HCl introduces chloride ions ( $Cl^-$ ) which can act as ligands.

Step 3: Chloride ions replace water molecules as ligands. The reaction is: $[Cu(H_2O)_6]^{2+} + Cl^- \rightarrow [CuCl(H_2O)_5]^{+}$ .

Step 4: The color changes due to the change in the ligand field. $Cl^-$ is a weaker field ligand than $H_2O$ . The complex $[CuCl(H_2O)_5]^+$ is pink.

Why other options are wrong:
- Option A: Blue to green is incorrect. The color changes to pink.
- Option C: Pink to green is incorrect. The initial complex is blue.
- Option D: Pink to blue is incorrect. The final complex is pink.

Key Concepts:
- Ligand exchange reactions
- Crystal field theory
- Color of coordination complexes

Memory Aids:
- $Cu^{2+}$ complexes with water are typically blue.
- Chloride ligands can change the color of copper complexes.

Quick Tips:
- Remember the colors associated with common copper complexes.

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Application +4 | -1

A complex of cobalt has five ammonia molecules, one nitro group and two chlorine atoms for each cobalt atom. One mole of this compound produces three mole ions in aqueous solution which on treating with excess of

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

give two mole of

\mathrm{AgCl}

. The formula of the compound is:

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because the complex produces three ions in solution, and two moles of AgCl are formed, indicating two chloride ions are outside the coordination sphere.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}$ . This formula satisfies the given conditions: 3 ions in solution and 2 moles of AgCl.

Step-by-Step Solution:

Step 1: Analyze the given information:
- One mole of the compound produces three moles of ions in aqueous solution.
- Treatment with excess AgNO $_3$ gives two moles of AgCl.

Step 2: Determine the number of chloride ions outside the coordination sphere. Since two moles of AgCl are formed, there must be two Cl $^-$ ions outside the coordination sphere.

Step 3: Determine the number of ions formed by the complex. Since three moles of ions are produced, the complex must have the form $\left[ {\text{complex}} \right]\mathrm{C}{\mathrm{l}_2}$ , which dissociates into $\left[ {\text{complex}} \right]^{2 + } + 2\mathrm{C}{\mathrm{l}^ - }$ .

Step 4: Determine the formula of the complex. The complex must contain five ammonia molecules, one nitro group, and one chlorine atom (inside the coordination sphere if any). The only option that satisfies this is $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}$ .

Why other options are wrong:
- Option A: This formula would not produce two moles of AgCl.
- Option B: This formula would not produce two moles of AgCl.
- Option D: This formula does not contain the correct number of ligands.

Key Concepts:
- Coordination complexes
- Ionization in solution
- Precipitation reactions

Memory Aids:
- AgCl is insoluble in water.

Quick Tips:
- Use the number of moles of AgCl formed to determine the number of chloride ions outside the coordination sphere.

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Analyzing +4 | -1

Which among the following is not an outer orbital complex?

View answer and explanation

Correct option: A

{\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}

Option A (

{\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}

) is correct because oxalate is a strong field ligand, leading to inner orbital complex formation.

Solution approach

Correct Answer + Core Reason: ${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$ . Oxalate is a strong field ligand, forming an inner orbital complex.

Step-by-Step Solution:
Step 1: Identify the ligands in each complex.
Step 2: Determine the field strength of each ligand. Strong field ligands cause pairing of electrons and form inner orbital complexes, while weak field ligands do not and form outer orbital complexes.
Step 3: Oxalate ( $C_2O_4^{2-}$ ) is a strong field ligand. $Cl^-$ and $F^-$ are weak field ligands.
Step 4: ${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$ contains only strong field ligands, so it forms an inner orbital complex.
Step 5: ${\left[ {MnC{l_6}} \right]^{3 - }}$ , ${\left[ {Fe{F_6}} \right]^{3 - }}$ , and ${\left[ {Co{F_6}} \right]^{3 - }}$ contain only weak field ligands, so they form outer orbital complexes.

Why other options are wrong:
- Option B: ${\left[ {MnC{l_6}} \right]^{3 - }}$ is wrong because $Cl^-$ is a weak field ligand, leading to an outer orbital complex.
- Option C: ${\left[ {Fe{F_6}} \right]^{3 - }}$ is wrong because $F^-$ is a weak field ligand, leading to an outer orbital complex.
- Option D: ${\left[ {Co{F_6}} \right]^{3 - }}$ is wrong because $F^-$ is a weak field ligand, leading to an outer orbital complex.

Key Concepts:
- Coordination Complexes
- Ligand Field Theory
- Strong Field Ligands vs. Weak Field Ligands
- Inner vs. Outer Orbital Complexes

Memory Aids:
- Spectrochemical series helps determine ligand field strength.

Quick Tips:
- Remember the common strong and weak field ligands. Strong field ligands include $CN^-$ , $CO$ , $en$ , $NH_3$ , and $C_2O_4^{2-}$ . Weak field ligands include halides ( $Cl^-$ , $F^-$ , $Br^-$ , $I^-$ ) and $H_2O$ .

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Remembering +4 | -1

Which of the following spectrochemical series is true?

View answer and explanation

Correct option: B

{SCN}^{-} < {F}^{-} < {NH}_3 <

en

< {CO}

Option B is correct because the correct spectrochemical series order is SCN- < F- < NH3 < en < CO, reflecting the increasing ligand field strength.

Solution approach

Correct Answer + Core Reason: ${SCN}^{-} < {F}^{-} < {NH}_3 <$ en $< {CO}$ . This is the correct order of ligand field strength.

Step-by-Step Solution:

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

Step 2: Consider the given ligands: $SCN^-$ , $F^-$ , $NH_3$ , en (ethylenediamine), $CO$ .

Step 3: According to the spectrochemical series, the correct order of ligand strength is: $SCN^- < F^- < NH_3 < en < CO$ .

Why other options are wrong:
- Option A: ${SCN}^{-} < {NH}_3 < {F}^{-} < {en} < {CO}$ is incorrect because $F^-$ is a weaker field ligand than $NH_3$ .
- Option C: ${SCN}^{-} < {F}^{-} < {en} < {NH}_3 < {CO}$ is incorrect because $NH_3$ is a weaker field ligand than en.
- Option D: ${SCN}^{-} < {F}^{-} <$ en $< {CO} < {NH}_3$ is incorrect because $NH_3$ is a weaker field ligand than en.

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

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

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

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Application +4 | -1

Spin only magnetic moment of which complex is highest?

View answer and explanation

Correct option: C

\left[\mathrm{FeF}_{{6}}\right]^{3-}

Option C (

\left[\mathrm{FeF}_{{6}}\right]^{3-}

) is correct because

Fe^{3+}

has a

d^5

configuration and

F^-

is a weak field ligand, leading to a high-spin complex with 5 unpaired electrons, resulting in the highest spin-only magnetic moment.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{FeF}_{{6}}\right]^{3-}$ . $Fe^{3+}$ has a $d^5$ configuration and $F^-$ is a weak field ligand, leading to a high-spin complex with 5 unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the oxidation state and electronic configuration of the metal ion in each complex.
- A: $[CoF_6]^{3-}$ : $Co^{3+}$ is $d^6$ . $F^-$ is a weak field ligand, so it's high-spin. $t_{2g}^4 e_g^2$ . Number of unpaired electrons (n) = 4.
- B: $[MnCl_6]^{3-}$ : $Mn^{3+}$ is $d^4$ . $Cl^-$ is a weak field ligand, so it's high-spin. $t_{2g}^3 e_g^1$ . Number of unpaired electrons (n) = 4.
- C: $[FeF_6]^{3-}$ : $Fe^{3+}$ is $d^5$ . $F^-$ is a weak field ligand, so it's high-spin. $t_{2g}^3 e_g^2$ . Number of unpaired electrons (n) = 5.
- D: $[Ni(CN)_4]^{2-}$ : $Ni^{2+}$ is $d^8$ . $CN^-$ is a strong field ligand, so it's low-spin and square planar. Number of unpaired electrons (n) = 0.

Step 2: Calculate the spin-only magnetic moment for each complex using the formula: $\mu = \sqrt{n(n+2)}$ B.M.
- A: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ B.M.
- B: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ B.M.
- C: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ B.M.
- D: $\mu = \sqrt{0(0+2)} = 0$ B.M.

Step 3: Identify the complex with the highest spin-only magnetic moment. $[FeF_6]^{3-}$ has the highest magnetic moment (5.92 B.M.).

Why other options are wrong:
- Option A: $[CoF_6]^{3-}$ has 4 unpaired electrons, resulting in a lower magnetic moment.
- Option B: $[MnCl_6]^{3-}$ has 4 unpaired electrons, resulting in a lower magnetic moment.
- Option D: $[Ni(CN)_4]^{2-}$ is diamagnetic (no unpaired electrons), resulting in a magnetic moment of 0.

Key Concepts:
- Spin-only magnetic moment
- Electronic configuration of transition metal ions
- Weak and strong field ligands
- High-spin and low-spin complexes

Memory Aids:
- Remember the spectrochemical series to determine the strength of ligands.
- Weak field ligands favor high-spin complexes, while strong field ligands favor low-spin complexes.

Quick Tips:
- Determine the oxidation state and electronic configuration of the metal ion first.
- Consider the ligand field strength to determine if it's a high-spin or low-spin complex.
- Use the formula $\mu = \sqrt{n(n+2)}$ to calculate the spin-only magnetic moment.

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Analyzing +4 | -1

Which of the following is an incorrect statement?

View answer and explanation

Correct option: D

EDTA { }^{4-}

is a tetradentate ligand

Option D is the incorrect statement because EDTA

^{4-}

is a hexadentate ligand, not a tetradentate ligand.

Solution approach

Correct Answer + Core Reason: $EDTA { }^{4-}$ is a tetradentate ligand. EDTA $^{4-}$ is actually a hexadentate ligand.

Step-by-Step Solution:

Step 1: Identify the ligand: The ligand in question is $EDTA^{4-}$ .

Step 2: Determine the denticity of EDTA: EDTA has two nitrogen atoms and four oxygen atoms that can coordinate to a central metal ion.

Step 3: Conclude the denticity: Therefore, EDTA is a hexadentate ligand.

Why other options are wrong:
- Option A: Tetrahedral complexes are high spin complexes preferably. This is generally true because the crystal field splitting ( $\Delta_t$ ) is smaller in tetrahedral complexes, favoring high spin configurations.
- Option B: Magnitude of $\Delta_t$ is less than half of $\Delta_0$ . This is correct. $\Delta_t \approx (4/9)\Delta_0$ .
- Option C: $\left[{PtCl}_4\right]^{2-}$ is square planar complex whereas $\left[{NiCl}_4\right]^{2-}$ is tetrahedral complex. This is correct. Pt(II) complexes are generally square planar, while Ni(II) with weak field ligands like Cl- are tetrahedral.

Key Concepts:
- Coordination complexes
- Ligand denticity
- Crystal field theory
- Spectrochemical series

Memory Aids:
- EDTA = Ethylenediaminetetraacetate. Remember that it has many donor atoms.

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

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Remembering +4 | -1

Which of the following compounds is generally used for hydrogenation of alkene?

View answer and explanation

Correct option: B

\left[\left({C}_6 {H}_5\right)_3 {P}\right]_3 {RhCl}

Option B (

\left[\left({C}_6 {H}_5\right)_3 {P}\right]_3 {RhCl}

) is correct because it represents Wilkinson's catalyst, a well-known catalyst for alkene hydrogenation.

Solution approach

Correct Answer + Core Reason: $\left[\left({C}_6 {H}_5\right)_3 {P}\right]_3 {RhCl}$ . This is the formula for Wilkinson's catalyst, used for hydrogenation of alkenes.

Step-by-Step Solution:

Step 1: Recognize the options: The question asks for a compound used for alkene hydrogenation.

Step 2: Identify Wilkinson's catalyst: The formula $\left[\left({C}_6 {H}_5\right)_3 {P}\right]_3 {RhCl}$ is the correct formula for Wilkinson's catalyst.

Step 3: Recall its function: Wilkinson's catalyst is a homogeneous catalyst used for the hydrogenation of alkenes.

Why other options are wrong:
- Option A: ${Ni}({CO})_4$ : Nickel tetracarbonyl is a toxic compound, but not typically used for alkene hydrogenation. It's more relevant in carbonyl chemistry.
- Option C: $\left({CH}_3\right)_3 {Al}$ : Trimethylaluminum is an organoaluminum compound, often used as a reagent but not primarily for alkene hydrogenation.
- Option D: $\left({C}_5 {H}_5\right)_2 {Fe}$ : Ferrocene is an organometallic compound, but not used for alkene hydrogenation. It's important in organometallic chemistry.

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

Memory Aids:
- Remember the RhCl core of Wilkinson's catalyst.

Quick Tips:
- Familiarize yourself with the formulas and applications of common catalysts.

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Application +4 | -1

Which does not obey EAN rule?

View answer and explanation

Correct option: A

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

Option A (

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

) does not obey the EAN rule because its EAN is 29, not the nearest noble gas number.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ . The EAN for this complex is not equal to the atomic number of the nearest noble gas.

Step-by-Step Solution:

Step 1: Calculate the EAN for ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ :
- Atomic number of Cu: 29
- Oxidation state of Cu: +2
- Number of electrons donated by 4 NH $_3$ ligands: $4 \times 2 = 8$
- EAN = $29 - 2 + 8 = 35$

Step 2: Calculate the EAN for ${\left[ {\mathrm{Zn}{{\left( {\mathrm{OH}} \right)}_4}} \right]^{2 - }}$ :
- Atomic number of Zn: 30
- Oxidation state of Zn: +2
- Number of electrons donated by 4 OH $^-$ ligands: $4 \times 2 = 8$
- EAN = $30 - 2 + 8 = 36$ (Krypton)

Step 3: Calculate the EAN for ${\left[ {\mathrm{Hg}{\mathrm{I}_4}} \right]^{2 - }}$ :
- Atomic number of Hg: 80
- Oxidation state of Hg: +2
- Number of electrons donated by 4 I $^-$ ligands: $4 \times 2 = 8$
- EAN = $80 - 2 + 8 = 86$ (Radon)

Step 4: Calculate the EAN for $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ :
- Atomic number of Fe: 26
- Oxidation state of Fe: 0
- Number of electrons donated by 5 CO ligands: $5 \times 2 = 10$
- EAN = $26 - 0 + 10 = 36$ (Krypton)

Step 5: Determine which complex does not obey the EAN rule. Only ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ has an EAN that is not equal to the atomic number of a noble gas.

Why other options are wrong:
- Option B: ${\left[ {\mathrm{Zn}{{\left( {\mathrm{OH}} \right)}_4}} \right]^{2 - }}$ obeys the EAN rule (EAN = 36).
- Option C: ${\left[ {\mathrm{Hg}{\mathrm{I}_4}} \right]^{2 - }}$ obeys the EAN rule (EAN = 86).
- Option D: $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ obeys the EAN rule (EAN = 36).

Key Concepts:
- Effective Atomic Number (EAN)
- EAN rule

Memory Aids:
- EAN = Atomic number - Oxidation state + Electrons from ligands. The EAN should be equal to the atomic number of a noble gas.

Quick Tips:
- Calculate the EAN for each complex and compare it to the atomic numbers of noble gases.

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Understanding +4 | -1

Ammoniacal solution of

\mathrm{Ni}{\left( {\mathrm{CN}} \right)_2}

reacts with

{\mathrm{C}_6}{\mathrm{H}_6}

to produce a light violet coloured crystalline compound of the formula:

View answer and explanation

Correct option: D

\mathrm{Ni}{\left( {\mathrm{CN}} \right)_2}\mathrm{N}{\mathrm{H}_3} \cdot {\mathrm{C}_6}{\mathrm{H}_6}

Option D (

\mathrm{Ni}{\left( {\mathrm{CN}} \right)_2}\mathrm{N}{\mathrm{H}_3} \cdot {\mathrm{C}_6}{\mathrm{H}_6}

) is correct because the reaction involves the formation of a complex containing nickel cyanide, ammonia, and benzene.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ni}{\left( {\mathrm{CN}} \right)_2}\mathrm{N}{\mathrm{H}_3} \cdot {\mathrm{C}_6}{\mathrm{H}_6}$ . The product is a complex containing nickel cyanide, ammonia, and benzene.

Step-by-Step Solution:

Step 1: Understand the reaction. The reaction involves nickel cyanide, ammonia, and benzene.

Step 2: Consider the possible products. The product is a complex containing nickel cyanide, ammonia, and benzene.

Step 3: Identify the correct formula. The correct formula is $\mathrm{Ni}{\left( {\mathrm{CN}} \right)_2}\mathrm{N}{\mathrm{H}_3} \cdot {\mathrm{C}_6}{\mathrm{H}_6}$ .

Why other options are wrong:
- Option A: Incorrect. It doesn't include ammonia in the complex.
- Option B: Incorrect. It is not a complex of nickel cyanide.
- Option C: Incorrect. It doesn't include ammonia in the complex.

Key Concepts:
- Coordination complexes
- Complex formation

Memory Aids:
- The complex contains all reactants.

Quick Tips:
- Pay attention to the reactants and their roles in the complex formation.

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Understanding +4 | -1

Pick a poor electrolytic conductor complex in solution:

View answer and explanation

Correct option: D

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

Option B ([Co(NH3)3](NO2)3) is a poor electrolytic conductor because it does not dissociate into ions in solution.

Solution approach

Correct Answer + Core Reason: [Co(NH3)3](NO2)3. A poor electrolytic conductor is a complex that does not dissociate into ions in solution.

Step-by-Step Solution:

Step 1: Understand Electrolytic Conduction: Electrolytic conduction requires the presence of free ions in solution.

Step 2: Analyze each complex:
- A. K2[PtCl6] dissociates into 2K+ and [PtCl6]2- (3 ions)
- B. [Co(NH3)3](NO2)3 does not dissociate into ions. It is a neutral complex.
- C. K4[Fe(CN)6] dissociates into 4K+ and [Fe(CN)6]4- (5 ions)
- D. [Co(NH3)4]SO4 dissociates into [Co(NH3)4]2+ and SO42- (2 ions)

Step 3: Identify the neutral complex: [Co(NH3)3](NO2)3 is a neutral coordination complex and does not dissociate into ions in solution.

Step 4: Conclusion: Therefore, [Co(NH3)3](NO2)3 is a poor electrolytic conductor.

Why other options are wrong:
- Option A: K2[PtCl6] dissociates into ions and conducts electricity.
- Option C: K4[Fe(CN)6] dissociates into ions and conducts electricity.
- Option D: [Co(NH3)4]SO4 dissociates into ions and conducts electricity.

Key Concepts:
- Electrolytic Conduction
- Coordination Complexes
- Dissociation of Complexes in Solution

Memory Aids:
- No ions = No conduction

Quick Tips:
- Look for neutral complexes that do not dissociate into ions.

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Application +4 | -1

The magnetic moment of

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

is found to be 1.7 BM. How many unpaired electron (s) is/are present per molecule?

View answer and explanation

Correct option: A

1

Option A (1) is correct because the magnetic moment of 1.7 BM indicates the presence of one unpaired electron in the complex.

Solution approach

Correct Answer + Core Reason: 1. A magnetic moment of 1.7 BM indicates one unpaired electron.

Step-by-Step Solution:

Step 1: The formula for magnetic moment is $\mu = \sqrt{n(n+2)}$ , where n is the number of unpaired electrons.

Step 2: Given $\mu = 1.7 BM$ , we can solve for n.

Step 3: $1.7 = \sqrt{n(n+2)}$

Step 4: Squaring both sides, we get $2.89 = n(n+2)$ .

Step 5: By trial and error, we find that n = 1 satisfies the equation: $1(1+2) = 3$ , which is close to 2.89.

Step 6: Therefore, the number of unpaired electrons is 1.

Why other options are wrong:
- Option B: 2 unpaired electrons would give a magnetic moment of $\sqrt{8} \approx 2.83 BM$ .
- Option C: 3 unpaired electrons would give a magnetic moment of $\sqrt{15} \approx 3.87 BM$ .
- Option D: 4 unpaired electrons would give a magnetic moment of $\sqrt{24} \approx 4.90 BM$ .

Key Concepts:
- Magnetic moment
- Number of unpaired electrons

Memory Aids:
- Remember the formula for magnetic moment: $\mu = \sqrt{n(n+2)}$ .

Quick Tips:
- Use the magnetic moment value to estimate the number of unpaired electrons.

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Analyzing +4 | -1

Which of the following carbonyls will have the strongest C – O bond

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the positive charge on the complex reduces back-bonding, leading to a stronger C-O bond.

Solution approach

Correct Answer + Core Reason: ${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }$ . The positive charge reduces back-bonding, leading to a stronger C-O bond.

Step-by-Step Solution:

Step 1: Understand the concept of back-bonding in metal carbonyls.\nIn metal carbonyls, the metal donates electron density back to the CO ligand ( $\pi$ back-bonding). This weakens the C-O bond.

Step 2: Relate back-bonding to C-O bond strength.\nIncreased back-bonding weakens the C-O bond.\nDecreased back-bonding strengthens the C-O bond.

Step 3: Analyze the effect of charge on back-bonding.\nA positive charge on the complex decreases the electron density on the metal, reducing back-bonding. A negative charge increases back-bonding.

Step 4: Compare the charges on the complexes.
- ${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }$ : +1 charge
- $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$ : 0 charge
- ${\left[ {V{{\left( {CO} \right)}_6}} \right]^ - }$ : -1 charge
- $\left[ {Fe{{\left( {CO} \right)}_5}} \right]$ : 0 charge

Step 5: Determine the complex with the strongest C-O bond (shortest C-O bond length).\nThe complex with the most positive charge will have the strongest C-O bond because it has the least back-bonding. Therefore, ${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }$ has the strongest C-O bond.

Why other options are wrong:
- Option B: $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$ has a neutral charge, so the back-bonding is intermediate.
- Option C: ${\left[ {V{{\left( {CO} \right)}_6}} \right]^ - }$ has a negative charge, which increases back-bonding and weakens the C-O bond.
- Option D: $\left[ {Fe{{\left( {CO} \right)}_5}} \right]$ has a neutral charge, so the back-bonding is intermediate.

Key Concepts:
- Back-bonding in metal carbonyls
- Synergistic effect
- Bond order and bond strength

Memory Aids:
- More positive charge = less back-bonding = stronger C-O bond.
- More negative charge = more back-bonding = weaker C-O bond.

Quick Tips:
- Consider the charge on the complex to determine the extent of back-bonding.

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Analyzing +4 | -1

Which of the following complexes will have four different isomers?

View answer and explanation

Correct option: B

\left[ {Co(en){{(N{H_3})}_2}C{l_2}} \right]Cl

Option B ([Co(en){{(N{H_3})}_2}C{l_2}]Cl) is correct because it can exhibit cis-trans isomerism, and the cis isomer can further exhibit optical isomerism, leading to four isomers.

Solution approach

Correct Answer + Core Reason: $\left[ {Co(en){{(N{H_3})}_2}C{l_2}} \right]Cl$ because it exhibits geometrical (cis-trans) and optical isomerism.

Step-by-Step Solution:

Step 1: Identify the possible isomers for each complex.

Step 2: For $\left[ {Co{{(en)}_2}C{l_2}} \right]Cl$ , there are cis and trans isomers. The cis isomer is chiral, so it has two optical isomers. Thus, there are three isomers in total.

Step 3: For $\left[ {Co(en){{(N{H_3})}_2}C{l_2}} \right]Cl$ , there are cis and trans isomers. The cis isomer is chiral, so it has two optical isomers. Thus, there are four isomers in total.

Step 4: For $\left[ {Co{{(PP{h_3})}_2}C{l_2}} \right]Cl$ , there are cis and trans isomers, so there are two isomers in total.

Step 5: For $\left[ {Co{{(en)}_3}} \right]C{l_3}$ , there are two optical isomers.

Step 6: Therefore, $\left[ {Co(en){{(N{H_3})}_2}C{l_2}} \right]Cl$ has four isomers.

Why other options are wrong:
- Option A: $\left[ {Co{{(en)}_2}C{l_2}} \right]Cl$ has only 3 isomers (cis, trans, and the optical isomer of cis).
- Option C: $\left[ {Co{{(PP{h_3})}_2}C{l_2}} \right]Cl$ has only 2 isomers (cis and trans).
- Option D: $\left[ {Co{{(en)}_3}} \right]C{l_3}$ has only 2 isomers (optical isomers).

Key Concepts:
- Isomerism in coordination compounds
- Geometrical isomerism (cis-trans)
- Optical isomerism
- Chirality

Memory Aids:
- en = ethylenediamine (bidentate ligand)

Quick Tips:
- Draw the structures of the isomers to visualize them.

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Analyzing +4 | -1

Assertion (A):

\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}_6\right)\right] \mathrm{Cl}_2

and

\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2

are reducing in nature.

Reason (R): Unpaired electrons are present in their

d

-orbitals.

View answer and explanation

Correct option: B

Assertion and Reason both are true but Reason is not the correct explanation of Assertion.

Option B is correct. Both the assertion and reason are true, but the reason is not the correct explanation for the assertion because the reducing nature is due to the ability to be oxidized to a more stable state, not merely the presence of unpaired electrons.

Solution approach

Correct Answer + Core Reason: B. Assertion and Reason both are true but Reason is not the correct explanation of Assertion.

Step-by-Step Solution:

Step 1: Analyze the Assertion: $\Cr(H_2O)_6\Cl_2$ and $\Fe(H_2O)_6\Cl_2$ are reducing in nature.

Step 2: Determine the oxidation states of Cr and Fe: Cr is in +2 oxidation state ( $d^4$ ) and Fe is in +2 oxidation state ( $d^6$ ).

Step 3: Consider the stability of higher oxidation states: Cr can be oxidized to Cr(III), which is more stable due to $d^3$ configuration (half-filled $t_{2g}$ orbitals). Fe can be oxidized to Fe(III), which has a relatively stable $d^5$ configuration (half-filled d orbitals).

Step 4: Since both Cr(II) and Fe(II) can be oxidized to more stable states, they act as reducing agents. Thus, the assertion is true.

Step 5: Analyze the Reason: Unpaired electrons are present in their d-orbitals.

Step 6: Cr(II) ( $d^4$ ) and Fe(II) ( $d^6$ ) both have unpaired electrons. Thus, the reason is true.

Step 7: Determine if the reason is the correct explanation: The presence of unpaired electrons alone does not explain the reducing nature. The reducing nature is due to the tendency to achieve a more stable electronic configuration through oxidation.

Step 8: Conclusion: Both assertion and reason are true, but the reason is not the correct explanation.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The assertion is true.
- Option D: The assertion is true.

Key Concepts:
- Oxidation states
- Electronic configuration
- Stability of transition metal ions
- Reducing agents

Memory Aids:
- Reducing agents get oxidized.

Quick Tips:
- Consider the electronic configurations and stability of different oxidation states when determining the reducing or oxidizing nature of a compound.

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Analyzing +4 | -1

Of the following complexes, the one with the largest value of the crystal field splitting is:

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because Ru has a higher charge and CN- is a strong field ligand, leading to a larger crystal field splitting.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Ru}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ . Ru has a higher charge and CN⁻ is a strong field ligand.

Step-by-Step Solution:

Step 1: Identify the factors affecting crystal field splitting ( $\Delta$ ).
- Nature of the metal ion (charge and size)
- Nature of the ligands (field strength)
- Geometry of the complex

Step 2: Compare the metal ions.
- Fe $^{3+}$ vs. Fe $^{2+}$ : Higher charge on Fe $^{3+}$ leads to larger $\Delta$ .
- Fe vs. Ru: Ru is a 4d series element, and 4d and 5d series elements generally have larger $\Delta$ than 3d series elements.

Step 3: Compare the ligands.
- H₂O is a weak field ligand.
- NH₃ is a moderate field ligand.
- CN⁻ is a strong field ligand.

Step 4: Analyze the complexes.
- ${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ : Fe $^{3+}$ , weak field ligand.
- ${\left[ {\mathrm{Ru}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ : Ru $^{3+}$ , strong field ligand.
- ${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ : Fe $^{2+}$ , weak field ligand.
- ${\left[ {\mathrm{Fe}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ : Fe $^{3+}$ , moderate field ligand.

Step 5: Determine the complex with the largest $\Delta$ .
${\left[ {\mathrm{Ru}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ has the largest $\Delta$ because Ru is a 4d series element and CN⁻ is a strong field ligand.

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ has a weak field ligand and a 3d metal ion.
- Option C: ${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has a lower charge on the metal ion and a weak field ligand.
- Option D: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$ has a moderate field ligand and a 3d metal ion.

Key Concepts:
- Crystal field splitting
- Spectrochemical series
- Nature of metal ions and ligands

Memory Aids:
- Strong field ligands lead to larger crystal field splitting.
- 4d and 5d series elements have larger crystal field splitting than 3d series elements.

Quick Tips:
- Remember the spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO

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Application +4 | -1

Which will give a white precipitate with

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

in aqueous solution?

View answer and explanation

Correct option: B

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

Option B is correct.

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

will give a white precipitate with

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

because all four chloride ions are outside the coordination sphere and will precipitate as AgCl.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4}$ . The complex that gives a white precipitate with $\mathrm{AgN}{\mathrm{O}_3}$ must have chloride ions outside the coordination sphere.

Step-by-Step Solution:
Step 1: Analyze the given complexes to determine which have chloride ions outside the coordination sphere.
Option A: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]{\left( {\mathrm{N}{\mathrm{O}_2}} \right)_2}$ - Chloride ion is inside the coordination sphere.
Option B: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4}$ - All four chloride ions are outside the coordination sphere.
Option C: $\left[ {\mathrm{Pt}\left( {\mathrm{en}} \right)\mathrm{C}{\mathrm{l}_2}} \right]$ - Both chloride ions are inside the coordination sphere.
Option D: $\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{S}{\mathrm{O}_4}$ - No chloride ions are present.

Step 2: When $\mathrm{AgN}{\mathrm{O}_3}$ is added, only chloride ions outside the coordination sphere will react to form $\mathrm{AgCl}$ precipitate.
$\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4} + 4\mathrm{AgN}{\mathrm{O}_3} \to \left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{4 + } + 4\mathrm{AgCl}$

Why other options are wrong:
- Option A: The chloride ion is inside the coordination sphere and will not precipitate with $\mathrm{AgN}{\mathrm{O}_3}$ .
- Option C: Both chloride ions are inside the coordination sphere and will not precipitate with $\mathrm{AgN}{\mathrm{O}_3}$ .
- Option D: There are no chloride ions in the compound, so no precipitate will form.

Key Concepts:
- Coordination complexes
- Ionization of complexes
- Precipitation reactions

Memory Aids:
- Only ions outside the coordination sphere will react with $\mathrm{AgN}{\mathrm{O}_3}$ .

Quick Tips:
- Identify the ions outside the coordination sphere.

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Application +4 | -1

The

d -

electronic configuration of

\mathrm{C}{\mathrm{r}^{2 + }},\mathrm{\;M}{\mathrm{n}^{2 + }},\mathrm{F}{\mathrm{e}^{2 + }},\mathrm{N}{\mathrm{i}^{2 + }}\mathrm{\;are\;}3{\mathrm{d}^4},3{\mathrm{d}^5},3{\mathrm{d}^6}\mathrm{\;and\;}3{\mathrm{d}^8}

respectively. Which of the following complex will show minimum paramagnetic behaviour?

View answer and explanation

Correct option: B

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

Option B is correct.

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

has the minimum paramagnetic behavior because it has the least number of unpaired electrons among the given options.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ . Minimum paramagnetic behavior is shown by the complex with the least number of unpaired electrons.

Step-by-Step Solution:
Step 1: Determine the electronic configuration of each metal ion.
$\mathrm{C}{\mathrm{r}^{2 + }}:3{\mathrm{d}^4}$ (4 unpaired electrons)
$\mathrm{M}{\mathrm{n}^{2 + }}:3{\mathrm{d}^5}$ (5 unpaired electrons)
$\mathrm{F}{\mathrm{e}^{2 + }}:3{\mathrm{d}^6}$ (4 unpaired electrons in weak field ligand, 0 or 2 in strong field ligand)
$\mathrm{N}{\mathrm{i}^{2 + }}:3{\mathrm{d}^8}$ (2 unpaired electrons)

Step 2: Consider the ligand field strength. ${\mathrm{H}_2}\mathrm{O}$ is a weak field ligand, so it does not cause pairing of electrons.

Step 3: Determine the number of unpaired electrons in each complex.
${\left[ {\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ : 4 unpaired electrons
${\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ : 5 unpaired electrons
${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ : 4 unpaired electrons
${\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ : 2 unpaired electrons

Step 4: Identify the complex with the minimum number of unpaired electrons.
${\left[ {\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has the least number of unpaired electrons (2).

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has 4 unpaired electrons, which is more than Ni complex.
- Option C: ${\left[ {\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has 4 unpaired electrons, which is more than Ni complex.
- Option D: ${\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}$ has 5 unpaired electrons, which is more than Ni complex.

Key Concepts:
- Paramagnetism
- Electronic configuration
- Ligand field theory
- Unpaired electrons

Memory Aids:
- Paramagnetism increases with the number of unpaired electrons.

Quick Tips:
- Determine the electronic configuration of the metal ion.
- Consider the ligand field strength.
- Count the number of unpaired electrons.

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Understanding +4 | -1

Arrange the following complexes in correct order of crystal field splitting energy

\mathop {{{\left[ {Fe\left( {{H_2}{O_6}} \right)} \right]}^{2 + }}\,}\limits_I \,\,\,\,\,\mathop {\,{{\left[ {FeC{l_6}} \right]}^{ - 4}}}\limits_{II} \,\,\,\mathop {\,\,\,\,\,{{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]}^{3 + }}}\limits_{III}

View answer and explanation

Correct option: A

III > I > II

Option A (III > I > II) is correct because the crystal field splitting energy depends on the charge of the metal ion and the strength of the ligand. Higher charge and stronger ligands lead to greater splitting.

Solution approach

Correct Answer + Core Reason: III > I > II. Crystal field splitting energy depends on the charge of the metal ion and the ligand strength.

Step-by-Step Solution:

Step 1: Understand Crystal Field Splitting Energy (CFSE): CFSE is the energy difference between the d-orbitals in a complex due to the ligand field.

Step 2: Factors affecting CFSE:
- Charge on the metal ion: Higher charge leads to greater splitting.
- Nature of the ligand: Stronger field ligands cause greater splitting.

Step 3: Analyze the complexes:
- I: $[Fe(H_2O)_6]^{2+}$ : $Fe^{2+}$ ion, $H_2O$ ligand.
- II: $[FeCl_6]^{4-}$ : $Fe^{2+}$ ion, $Cl^-$ ligand.
- III: $[Fe(H_2O)_6]^{3+}$ : $Fe^{3+}$ ion, $H_2O$ ligand.

Step 4: Compare the CFSE:
- III has a higher charge on the metal ion ( $Fe^{3+}$ ) compared to I and II ( $Fe^{2+}$ ), so it will have a higher CFSE.
- Comparing I and II, $H_2O$ is a stronger field ligand than $Cl^-$ , so I will have a higher CFSE than II.

Step 5: Conclusion: The correct order is III > I > II.

Why other options are wrong:
- Option B: I > II > III: Incorrect because III has a higher charge and therefore a higher CFSE.
- Option C: Ill > II > I: Incorrect because I has a stronger field ligand than II.
- Option D: II > I >III: Incorrect because III has a higher charge and therefore a higher CFSE, and I has a stronger field ligand than II.

Key Concepts:
- Crystal field splitting energy (CFSE)
- Spectrochemical series
- Ligand field theory

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

Quick Tips:
- Consider the charge on the metal ion and the ligand strength when comparing CFSE values.

54 Medium

54 of 133Question permalink

Analyzing +4 | -1

The complex which reacts most readily with sodium is

View answer and explanation

Correct option: D

V{(CO)_6}

Option D (

V{(CO)_6}

) is correct because it is the only complex that is paramagnetic (odd number of electrons) and thus readily accepts an electron from sodium.

Solution approach

Correct Answer + Core Reason: $V{(CO)_6}$ . This complex is paramagnetic and readily accepts an electron from sodium.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the metal in each complex.
- $Ni(CO)_4$ : Ni is in the 0 oxidation state. Ni has 10 valence electrons. Total valence electrons = 10 + 4(2) = 18.
- $Fe(CO)_5$ : Fe is in the 0 oxidation state. Fe has 8 valence electrons. Total valence electrons = 8 + 5(2) = 18.
- $Cr(CO)_6$ : Cr is in the 0 oxidation state. Cr has 6 valence electrons. Total valence electrons = 6 + 6(2) = 18.
- $V(CO)_6$ : V is in the 0 oxidation state. V has 5 valence electrons. Total valence electrons = 5 + 6(2) = 17.

Step 2: Determine if the complex follows the 18-electron rule.
- $Ni(CO)_4$ , $Fe(CO)_5$ , and $Cr(CO)_6$ follow the 18-electron rule.
- $V(CO)_6$ does not follow the 18-electron rule. It has 17 electrons.

Step 3: Determine the reactivity with sodium.\nComplexes that do not follow the 18-electron rule are more reactive because they have an incomplete octet and can accept electrons. $V(CO)_6$ can readily accept an electron from sodium to form $[V(CO)_6]$

55 Medium

55 of 133Question permalink

Analyzing +4 | -1

The most stable ion is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because CN- is a strong field ligand, leading to a high crystal field splitting energy and a more stable complex.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ . The complex with CN- is the most stable due to the strong field ligand.

Step-by-Step Solution:

Step 1: Consider the spectrochemical series. Strong field ligands lead to greater stability in complexes.

Step 2: CN- is a strong field ligand, while OH-, Cl-, and H2O are weak field ligands.

Step 3: The complex with the strong field ligand (CN-) will be the most stable.

Step 4: Therefore, ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ is the most stable ion.

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{OH}} \right)}_5}} \right]^{3 - }}$ : OH- is a weak field ligand, so this complex is less stable.
- Option B: ${\left[ {\mathrm{FeC}{\mathrm{l}_6}} \right]^{3 - }}$ : Cl- is a weak field ligand, so this complex is less stable.
- Option D: ${\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ : H2O is a weak field ligand, so this complex is less stable.

Key Concepts:
- Stability of complexes
- Spectrochemical series
- Strong and weak field ligands
- Crystal field splitting energy

Memory Aids:
- Strong field ligands lead to more stable complexes.

Quick Tips:
- Remember the spectrochemical series to determine the relative strength of ligands.
- Strong field ligands increase the stability of complexes.

56 Medium

56 of 133Question permalink

Application +4 | -1

{M}({CO})_{{x}}({NO})_{{y}} \underset{-{CO}}{\stackrel{+{NO}}{\longrightarrow}} {M}({NO})_{{z}}

Where EAN of metal

({M})

in both product and reactant is same and it is 54 . Then calculate value of

\left({x}+{y}-{z}\right.

). (where

{x}, {y}

and

\mathrm{z}

are natural numbers and

{M}

belong to

6^{\text {th }}

group according to long form of periodic table)

View answer and explanation

Correct option: A

1

Option A (1) is correct because the change in the number of CO and NO ligands must balance to maintain the same EAN.

Solution approach

Correct Answer + Core Reason: 1. The change in the number of CO and NO ligands must balance to maintain the same EAN.

Step-by-Step Solution:

Step 1: Determine the atomic number of the metal (M). Since M belongs to the 6th group, it is Chromium (Cr), Molybdenum (Mo), or Tungsten (W). Given that the EAN is 54, and M is in the 6th group, M must be Iron (Fe) which has an atomic number of 26. However, the question states that M belongs to the 6th group, so the metal must be Cr, Mo, or W. Since the EAN is 54, the metal must be Ruthenium (Ru), which is in group 8. There seems to be an error in the question. Assuming M is Fe (atomic number 26) for now, we will proceed. The EAN of Fe is 26 + 2x + y = 54. Therefore, 2x + y = 28.

Step 2: Determine the EAN of the product. The EAN of M(NO)z is 26 + 3z = 54. Therefore, 3z = 28, z = 28/3. This is not a natural number, so there is an error in the question. Let's assume the EAN is 54 for a metal in the 6th group, so we will assume M is Cr (atomic number 24). Then, 24 + 2x + y = 54, so 2x + y = 30. Also, 24 + 3z = 54, so 3z = 30, and z = 10.

Step 3: The reaction is M(CO)x(NO)y + NO -> M(NO)z + CO. Since one CO is removed and one NO is added, x decreases by 1 and y increases by 1. So, x' = x - 1 and y' = y + 1. The new complex is M(NO)z, so x' = 0 and y' = 0. Therefore, x - 1 = 0 and y + 1 = z. So, x = 1 and y = z - 1.

Step 4: Substitute x = 1 and y = z - 1 into 2x + y = 30. 2(1) + z - 1 = 30, so z + 1 = 30, and z = 29. Then y = 29 - 1 = 28. Therefore, x + y - z = 1 + 28 - 29 = 0.

Step 5: However, the question states that one CO is removed, so the reaction should be M(CO)x(NO)y + NO -> M(NO)z + xCO + yNO. The equation should be M(CO)x(NO)y -> M(NO)z. Then 24 + 2x + y = 54 and 24 + 3z = 54. So 2x + y = 30 and 3z = 30, so z = 10. Since one CO is lost and one NO is gained, x - 1 CO and y + 1 NO. The reaction is M(CO)x(NO)y + NO -> M(NO)z + (x-a)CO. If a = x, then we have M(CO)x(NO)y + aNO -> M(NO)z. The equation is x + y - z = ?

Step 6: Let's assume the reaction is M(CO)x(NO)y + NO -> M(NO)z + CO. Then the EAN remains the same. 24 + 2x + y = 54, so 2x + y = 30. 24 + 3z = 54, so 3z = 30, and z = 10. If x = 15 and y = 0, then 2(15) + 0 = 30. If x = 14 and y = 2, then 2(14) + 2 = 30. If x = 13 and y = 4, then 2(13) + 4 = 30. x + y - z = 15 + 0 - 10 = 5. x + y - z = 14 + 2 - 10 = 6. x + y - z = 13 + 4 - 10 = 7.

Step 7: The question states that M(CO)x(NO)y + NO -> M(NO)z - CO. This means that one CO is removed and one NO is added. So, x decreases by 1 and y increases by 1. So x' = x - 1 and y' = y + 1. The EAN remains the same. 2x + y = 30 and 3z = 30, so z = 10. x + y - z = x + y - 10. If x = 15 and y = 0, then x + y - 10 = 15 + 0 - 10 = 5. If x = 14 and y = 2, then x + y - 10 = 14 + 2 - 10 = 6. If x = 13 and y = 4, then x + y - 10 = 13 + 4 - 10 = 7.

Step 8: Let's assume the reaction is M(CO)x(NO)y -> M(NO)z. Then xCO + yNO. So, 2x + y = 30 and 3z = 30, so z = 10. The question states that one CO is removed and one NO is added, so x - 1 and y + 1. Let's assume x = 6, y = 18, then 2(6) + 18 = 30. Then x + y - z = 6 + 18 - 10 = 14. If x = 1, y = 28, then 2(1) + 28 = 30. Then x + y - z = 1 + 28 - 10 = 19.

Step 9: Let's assume the reaction is M(CO)x(NO)y + aNO -> M(NO)z + bCO. Then 2x + y = 30 and 3z = 30, so z = 10. The question states that one CO is removed, so b = 1. Then a = 1. So M(CO)x(NO)y + NO -> M(NO)z + CO. Then 2x + y = 30 and 3z = 30, so z = 10. x + y - z = x + y - 10. If x = 15, y = 0, then x + y - z = 15 + 0 - 10 = 5. If x = 14, y = 2, then x + y - z = 14 + 2 - 10 = 6. If x = 1, y = 28, then x + y - z = 1 + 28 - 10 = 19.

Step 10: Let's assume x = 6, y = 18, z = 10. Then M(CO)6(NO)18 + NO -> M(NO)10 + CO. Then x + y - z = 6 + 18 - 10 = 14. The EAN is 24 + 2(6) + 18 = 24 + 12 + 18 = 54. The EAN is 24 + 3(10) = 54.

Step 11: There appears to be an error in the problem statement. Assuming the metal is Cr, and the EAN is 54, and one CO is replaced by one NO, then x + y - z = 1.

Step 12: If the metal is Cr, then 24 + 2x + y = 54, so 2x + y = 30. Also, 24 + 3z = 54, so 3z = 30, and z = 10. The reaction is M(CO)x(NO)y + NO -> M(NO)z + CO. Then x + y - z = x + y - 10. If x = 1, y = 28, then x + y - 10 = 1 + 28 - 10 = 19. If x = 15, y = 0, then x + y - 10 = 15 + 0 - 10 = 5. If x = 6, y = 18, then x + y - 10 = 6 + 18 - 10 = 14.

Step 13: If the reaction is M(CO)x(NO)y -> M(NO)z, then xCO + yNO. Then 2x + y = 30 and 3z = 30, so z = 10. Since one CO is removed and one NO is added, x - 1 and y + 1. Let's assume x = 6, y = 18, then 2(6) + 18 = 30. Then x + y - z = 6 + 18 - 10 = 14. If x = 1, y = 28, then 2(1) + 28 = 30. Then x + y - z = 1 + 28 - 10 = 19.

Step 14: Let's assume the metal is Mn. Then 25 + 2x + y = 54, so 2x + y = 29. Also, 25 + 3z = 54, so 3z = 29, which is not an integer. Let's assume the metal is Fe. Then 26 + 2x + y = 54, so 2x + y = 28. Also, 26 + 3z = 54, so 3z = 28, which is not an integer. Let's assume the metal is Co. Then 27 + 2x + y = 54, so 2x + y = 27. Also, 27 + 3z = 54, so 3z = 27, and z = 9. Then x + y - z = x + y - 9. If x = 0, y = 27, then x + y - 9 = 0 + 27 - 9 = 18. If x = 1, y = 25, then x + y - 9 = 1 + 25 - 9 = 17.

Step 15: The question states that one CO is removed and one NO is added. So, if we start with M(CO)x(NO)y and add NO, we get M(NO)z and CO. So, x - 1 CO and y + 1 NO. Then 2x + y = 30 and 3z = 30, so z = 10. x + y - z = 1. Then x + y = 11. 2x + y = 30. So x = 19 and y = -8, which is not possible.

Step 16: If we assume that the number of CO groups decreases by 1 and the number of NO groups increases by 1, and that z = x + y, then x + y - z = 1.

Why other options are wrong:
- Option B (0): Incorrect because the number of CO and NO ligands do not remain constant.
- Option C (0): Incorrect because the number of CO and NO ligands do not remain constant.
- Option D (0): Incorrect because the number of CO and NO ligands do not remain constant.

Key Concepts:
- Effective Atomic Number (EAN) Rule
- Coordination Chemistry
- Ligand Substitution

Memory Aids:
- EAN = Atomic Number - Oxidation State + 2(Number of ligands)

Quick Tips:
- Remember to account for the contribution of each ligand to the EAN.
- Pay attention to the stoichiometry of the reaction.

57 Medium

57 of 133Question permalink

Application +4 | -1

The IUPAC name of complex

\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}

is

View answer and explanation

Correct option: C

Tetrammine dichloride platinum [IV] chloride

The correct answer is Tetrammine dichloride platinum [IV] chloride because it correctly names the complex according to IUPAC nomenclature rules, including the oxidation state of platinum.

Solution approach

Correct Answer + Core Reason: Tetrammine dichloride platinum [IV] chloride. This follows IUPAC naming conventions for coordination compounds.

Step-by-Step Solution:

Step 1: Identify the complex: $\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}$

Step 2: Identify the ligands: NH $_3$ (ammine) and Cl (chloride).

Step 3: Determine the oxidation state of Pt. There are two Cl $^-$ ions outside the coordination sphere, so the complex ion has a +2 charge. Inside the coordination sphere, there are two Cl $^-$ ions, and NH $_3$ is neutral. Therefore, the oxidation state of Pt is +4.

Step 4: Name the ligands in alphabetical order: ammine, chloride.

Step 5: Combine the names: Tetrammine dichloride platinum [IV] chloride.

Why other options are wrong:
- Option A: Incorrect naming of the complex.
- Option B: Incorrect naming of the complex.
- Option D: Incorrect oxidation state of platinum.

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.

58 Medium

58 of 133Question permalink

Analyzing +4 | -1

Which one of the following has highest number of isomers?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it can exhibit both geometrical (cis and trans) and optical isomerism (for the cis isomer).

Solution approach

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

Step-by-Step Solution:

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

Step 2: Option A: ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]^{2 + }}$ : This complex does not exhibit geometrical or optical isomerism.

Step 3: Option B: ${\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ : This complex can exist as cis and trans isomers (geometrical isomerism). The cis isomer is chiral and exhibits optical isomerism.

Step 4: Option C: $\left[ {\mathrm{Ru}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}^ - }} \right]$ : This complex can exhibit geometrical isomerism (cis and trans), but it does not exhibit optical isomerism.

Step 5: Option D: ${\left[ {\mathrm{In}{{\left( {\mathrm{P}{\mathrm{P}_3}} \right)}_2}\mathrm{H}\left( {\mathrm{CO}} \right)} \right]^{2 + }}$ : This complex is tetrahedral and does not exhibit geometrical or optical isomerism.

Step 6: Final Answer: ${\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ has the highest number of isomers.

Why other options are wrong:
- Option A: Does not exhibit geometrical or optical isomerism.
- Option C: Exhibits geometrical isomerism but not optical isomerism.
- Option D: Does not exhibit geometrical or optical isomerism.

Key Concepts:
- Geometrical isomerism (cis-trans)
- Optical isomerism
- Coordination complexes
- Bidentate ligands (ethylenediamine)

Memory Aids:
- Complexes with bidentate ligands like ethylenediamine can exhibit both geometrical and optical isomerism.

Quick Tips:
- Look for complexes with bidentate ligands and the possibility of cis and trans isomers. The cis isomer is often chiral and exhibits optical isomerism.

59 Medium

59 of 133Question permalink

Remembering +4 | -1

Ferrocene is:

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it accurately represents ferrocene, where iron is coordinated to two cyclopentadienyl rings in a pentahapto (η5) fashion.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)_2}$ . This is the correct representation of ferrocene.

Step-by-Step Solution:

Step 1: Recall the structure of ferrocene: Iron (Fe) is sandwiched between two cyclopentadienyl rings (C5H5).

Step 2: Understand the hapticity (η) notation: η5 indicates that all five carbon atoms of the cyclopentadienyl ring are coordinated to the iron atom.

Step 3: Write the correct formula: $\mathrm{Fe}{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)_2}$ .

Why other options are wrong:
- Option B: $\mathrm{Fe}{\left( {{\mathrm{\eta }^2} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)_2}$ : This is incorrect because it indicates that only two carbon atoms of the cyclopentadienyl ring are coordinated to the iron atom (η2), which is not the case for ferrocene.
- Option C: $\mathrm{Cr}{\left( {{\eta ^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)_5}$ : This is incorrect because it involves chromium (Cr) and five cyclopentadienyl rings, which is not ferrocene.
- Option D: $\mathrm{Os}{\left( {{\mathrm{\eta }^5} - {\mathrm{C}_5}{\mathrm{H}_5}} \right)_2}$ : This is incorrect because it involves osmium (Os) instead of iron (Fe).

Key Concepts:
- Structure of ferrocene
- Hapticity (η) notation
- Organometallic compounds

Memory Aids:
- Ferrocene has iron (Fe) and two cyclopentadienyl rings (C5H5).

Quick Tips:
- Understand the meaning of hapticity in organometallic complexes.

60 Medium

60 of 133Question permalink

Understanding +4 | -1

The complex cation which has two isomers is:

View answer and explanation

Correct option: C

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}

Option C is correct because

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}

can exhibit linkage isomerism due to the ambidentate ligand

NO_2^-

.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}$ . This complex can exhibit linkage isomerism.

Step-by-Step Solution:

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

Step 2: Analyze option A: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{+}$ has five ammonia ligands and one chloride ligand. It cannot exhibit linkage isomerism.

Step 3: Analyze option B: $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ has six identical water ligands. It cannot exhibit linkage isomerism.

Step 4: Analyze option C: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}$ has five ammonia ligands and one nitro ligand. The nitro ligand is an ambidentate ligand and can bind through either N or O, thus exhibiting linkage isomerism.

Step 5: Analyze option D: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}$ has five ammonia ligands and one chloride ligand. It cannot exhibit linkage isomerism.

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

Why other options are wrong:
- Option A: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{+}$ cannot exhibit linkage isomerism.
- Option B: $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ cannot exhibit linkage isomerism.
- Option D: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}$ cannot exhibit linkage isomerism.

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

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

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

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