The IUPAC name for the complex $\left[ {{\rm{Co}}\left( {{\rm{N}}{{\rm{O}}_2}} \right){{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_5}} \right]{\rm{C}}{{\rm{l}}_2}$ is

Nitrito -N- pentamminecobalt (III) chloride

Nitrito -N- pentamminecobalt (II) chloride

Pentammine nitrito-N- cobalt (II) chloride

The shortest ${\rm{C}} - {\rm{O}}$ bond order exists in:

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

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

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

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

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Medium, Page 5

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Remembering +4 | -1

Dimethyl glyoxime forms a coloured complex with

View answer and explanation

Correct option: B

Ni

Option B (

Ni

) is correct because dimethyl glyoxime (DMG) forms a characteristic red-colored complex with Nickel(II) ions.

Solution approach

Correct Answer + Core Reason: $Ni$ . Dimethyl glyoxime (DMG) forms a colored complex with Nickel(II) ions.

Step-by-Step Solution:
Step 1: Understand the reaction of dimethyl glyoxime (DMG). DMG is a chelating ligand that forms a stable, colored complex with certain metal ions.
Step 2: Recall the specific metal ion that reacts with DMG to form a colored complex. DMG specifically reacts with Ni2+ ions to form a bright red precipitate.
Step 3: Identify the correct metal ion from the given options.

Why other options are wrong:
- Option A: $Ag$ : Silver does not form a colored complex with DMG.
- Option C: $Cr$ : Chromium does not form a colored complex with DMG.
- Option D: $Zn$ : Zinc does not form a colored complex with DMG.

Key Concepts:
- Coordination Chemistry
- Chelating Ligands
- Qualitative Analysis

Memory Aids:
- Remember that DMG is a specific reagent for Nickel(II) ions, forming a bright red precipitate.

Quick Tips:
- Know the specific reagents used for qualitative analysis of different metal ions.

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Application +4 | -1

The magnetic moment (in B.M.) of complex

\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_{\mathrm{6}}\right]^{3+}

is

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because

Co^{3+}

in a weak field ligand like water undergoes pairing and becomes diamagnetic, resulting in zero unpaired electrons and thus zero magnetic moment.

Solution approach

Correct Answer + Core Reason: Zero. $Co^{3+}$ in a weak field ligand like water undergoes pairing and becomes diamagnetic.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Cobalt: In $[Co(H_2O)_6]^{3+}$ , Cobalt is in the +3 oxidation state.

Step 2: Determine the electronic configuration of $Co^{3+}$ : Cobalt has an electronic configuration of $[Ar] 3d^7 4s^2$ . $Co^{3+}$ loses 3 electrons, resulting in a $d^6$ configuration.

Step 3: Consider the ligand field strength: Water ( $H_2O$ ) is a weak field ligand, but $Co^{3+}$ forces pairing. Thus, it is a low spin complex.

Step 4: Determine the electronic configuration in the complex: For a low-spin $d^6$ complex, the electrons fill the $t_{2g}$ orbitals first, resulting in $t_{2g}^6 e_g^0$ . All electrons are paired.

Step 5: Calculate the number of unpaired electrons: Since all electrons are paired, the number of unpaired electrons (n) is 0.

Step 6: Calculate the spin-only magnetic moment using the formula: $\mu = \sqrt{n(n+2)}$ B.M.

Step 7: Substitute n = 0: $\mu = \sqrt{0(0+2)} = \sqrt{0} = 0$ B.M.

Why other options are wrong:
- Option A (4.89): This would be the magnetic moment if there were 4 unpaired electrons, which is not the case here.
- Option B (3.87): This would be the magnetic moment if there were 3 unpaired electrons, which is not the case here.
- Option D (2.82): This would be the magnetic moment if there were 2 unpaired electrons, which is not the case here.

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

Memory Aids:
- Diamagnetic substances have zero unpaired electrons and zero magnetic moment.
- Strong field ligands cause pairing of electrons.

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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Application +4 | -1

The oxidation state of

\mathrm{Mo}

in its oxo-complex species

{\left[ {\mathrm{M}{\mathrm{o}_2}{\mathrm{O}_4}{{\left( {{\mathrm{C}_2}{\mathrm{H}_4}} \right)}_2}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}} \right]^{2 - }}\,\mathrm{is}

View answer and explanation

Correct option: B

3

The correct answer is 3 because the oxidation state of Mo in

{\left[ {\mathrm{M}{\mathrm{o}_2}{\mathrm{O}_4}{{\left( {{\mathrm{C}_2}{\mathrm{H}_4}} \right)}_2}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}} \right]^{2 - }}

is +3.

Solution approach

Correct Answer + Core Reason: 3. The oxidation state of Mo is +3.

Step-by-Step Solution:

Step 1: Identify the complex: ${\left[ {\mathrm{M}{\mathrm{o}_2}{\mathrm{O}_4}{{\left( {{\mathrm{C}_2}{\mathrm{H}_4}} \right)}_2}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}} \right]^{2 - }}$

Step 2: Determine the charge of the complex: -2

Step 3: Determine the charges of the ligands:
- O: -2
- C $_2$ H $_4$ : 0 (neutral)
- H $_2$ O: 0 (neutral)

Step 4: Let the oxidation state of Mo be $x$ . Then, $2x + 4(-2) + 2(0) + 2(0) = -2$

Step 5: Solve for $x$ : $2x - 8 = -2 \Rightarrow 2x = 6 \Rightarrow x = +3$

Why other options are wrong:
- Option A: Incorrect calculation of oxidation state.
- Option C: Incorrect calculation of oxidation state.
- Option D: Incorrect calculation of oxidation state.

Key Concepts:
- Oxidation state
- Coordination complexes

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

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

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Analyzing +4 | -1

Which of the following 0.1 M complex compound solutions will have the minimum electrical conductivity?

View answer and explanation

Correct option: D

Trichloro triammine platinum (IV) chloride

Option D (Trichloro triammine platinum (IV) chloride) is correct because it has the fewest ions in solution, leading to the lowest electrical conductivity.

Solution approach

Correct Answer + Core Reason: Trichloro triammine platinum (IV) chloride. The electrical conductivity depends on the number of ions present in the solution.

Step-by-Step Solution:

Step 1: Determine the formula of each complex.\nHexammine platinum (IV) chloride: $\left[Pt(NH_3)_6\right]Cl_4$ \nChloropenta ammine platinum (IV) chloride: $\left[Pt(NH_3)_5Cl\right]Cl_3$ \nDichloro tetrammine platinum (IV) chloride: $\left[Pt(NH_3)_4Cl_2\right]Cl_2$ \nTrichloro triammine platinum (IV) chloride: $\left[Pt(NH_3)_3Cl_3\right]Cl$

Step 2: Determine the number of ions produced in solution.
$\left[Pt(NH_3)_6\right]Cl_4 \rightarrow \left[Pt(NH_3)_6\right]^{4+} + 4Cl^-$ (5 ions)
$\left[Pt(NH_3)_5Cl\right]Cl_3 \rightarrow \left[Pt(NH_3)_5Cl\right]^{3+} + 3Cl^-$ (4 ions)
$\left[Pt(NH_3)_4Cl_2\right]Cl_2 \rightarrow \left[Pt(NH_3)_4Cl_2\right]^{2+} + 2Cl^-$ (3 ions)
$\left[Pt(NH_3)_3Cl_3\right]Cl \rightarrow \left[Pt(NH_3)_3Cl_3\right]^{+} + Cl^-$ (2 ions)

Step 3: Determine the complex with the minimum electrical conductivity.\nThe complex with the fewest ions in solution will have the minimum electrical conductivity. In this case, it is $\left[Pt(NH_3)_3Cl_3\right]Cl$ with 2 ions.

Why other options are wrong:
- Option A: Hexammine platinum (IV) chloride produces 5 ions in solution.
- Option B: Chloropenta ammine platinum (IV) chloride produces 4 ions in solution.
- Option C: Dichloro tetrammine platinum (IV) chloride produces 3 ions in solution.

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

Memory Aids:
- Electrical conductivity is directly proportional to the number of ions in solution.

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

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Understanding +4 | -1

When ammonia is added to green aqueous solution of nickel(II) sulphate, the colour of the solution changes to blue violet. This is caused by:

View answer and explanation

Correct option: B

Ammonia molecules replacing water molecules surrounding nickel

The correct answer is B. The blue-violet color arises because ammonia molecules replace water molecules in the coordination sphere of nickel, leading to a different complex with a different color.

Solution approach

Correct Answer + Core Reason: Ammonia molecules replacing water molecules surrounding nickel. The change in color is due to ligand exchange.

Step-by-Step Solution:

Step 1: Nickel(II) sulphate in aqueous solution exists as $[\mathrm{Ni}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}]^{2 + }$ , which is green.

Step 2: When ammonia is added, it replaces the water ligands to form $[\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}]^{2 + }$ , which is blue-violet.

Step 3: The change in color is due to the change in the ligand field splitting caused by the different ligands.

Why other options are wrong:
- Option A: Nickel undergoing a change in oxidation state: Nickel remains in the +2 oxidation state.
- Option C: Change in coordination number of nickel: The coordination number remains 6.
- Option D: Change in $\mathrm{pH}$ value of the solution: While pH changes, it's the ligand exchange that causes the color change.

Key Concepts:
- Coordination complexes
- Ligand exchange
- Color of complexes

Memory Aids:
- Ammonia is a stronger ligand than water for Nickel(II).

Quick Tips:
- Remember that different ligands cause different colors in complexes.

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Analyzing +4 | -1

Which of the following complexes will show geometrical as well as optical isomerism? (en

=

ethylene diamine)

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it can exist as cis and trans isomers, and the cis isomer is chiral, exhibiting optical isomerism.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ . This complex shows both geometrical (cis-trans) and optical isomerism (cis form).

Step-by-Step Solution:

Step 1: Analyze each complex for geometrical isomerism.
$\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ : Can exist as cis and trans isomers (geometrical isomerism).
$\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_4}} \right]$ : Can exist as cis and trans isomers (geometrical isomerism).
${\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_3}} \right]^{4 + }}$ : No geometrical isomerism.
$\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ : Can exist as cis and trans isomers (geometrical isomerism).

Step 2: Analyze each complex for optical isomerism.
$\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ : Neither cis nor trans isomer is chiral (no optical isomerism).
$\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_4}} \right]$ : Neither cis nor trans isomer is chiral (no optical isomerism).
${\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_3}} \right]^{4 + }}$ : The complex is chiral, but it doesn't have geometrical isomers.
$\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ : The cis isomer is chiral and exhibits optical isomerism. The trans isomer is not chiral.

Step 3: Identify the complex that shows both geometrical and optical isomerism.
$\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ shows both geometrical (cis-trans) and optical isomerism (cis form).

Why other options are wrong:
- Option A: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]$ shows geometrical isomerism but not optical isomerism.
- Option B: $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_4}} \right]$ shows geometrical isomerism but not optical isomerism.
- Option C: ${\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_3}} \right]^{4 + }}$ shows optical isomerism but not geometrical isomerism.

Key Concepts:
- Coordination Complexes
- Geometrical Isomerism
- Optical Isomerism
- Chirality

Memory Aids:
- Cis-trans isomerism occurs when ligands are arranged differently around the central metal ion.
- Optical isomerism occurs when a molecule is chiral (non-superimposable mirror image).

Quick Tips:
- Draw the structures of the isomers to determine if they are chiral and/or have geometrical isomers.

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Analyzing +4 | -1

Metal carbon bond is strongest in

View answer and explanation

Correct option: D

{\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }

Option D (

{\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }

) is correct because the negative charge on the complex increases the electron density on the metal, enhancing back-bonding and strengthening the metal-carbon bond.

Solution approach

Correct Answer + Core Reason: ${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }$ . The negative charge increases electron density on the metal, enhancing back-bonding.

Step-by-Step Solution:

Step 1: Understand the concept of metal-carbon bonding in carbonyl complexes.

Step 2: Metal-carbon bond strength is influenced by back-bonding, which is the donation of electrons from the filled d-orbitals of the metal to the empty $\pi^*$ antibonding orbitals of CO.

Step 3: Higher electron density on the metal favors stronger back-bonding, leading to a stronger metal-carbon bond.

Step 4: Compare the charges on the metal complexes.

Step 5: $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$ has a neutral metal.

Step 6: ${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }$ has a positive metal, which reduces electron density and weakens back-bonding.

Step 7: $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ has a neutral metal.

Step 8: ${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }$ has a negative metal, which increases electron density and strengthens back-bonding.

Why other options are wrong:
- Option A: $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$ has a neutral metal, so back-bonding is not as strong as in the negatively charged complex.
- Option B: ${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }$ has a positive metal, which weakens back-bonding.
- Option C: $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ has a neutral metal, so back-bonding is not as strong as in the negatively charged complex.

Key Concepts:
- Metal carbonyls
- Back-bonding
- Synergic effect
- Electron density

Memory Aids:
- Negative charge on the complex = stronger M-C bond.

Quick Tips:
- Remember that increased electron density on the metal strengthens back-bonding and the metal-carbon bond.

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Application +4 | -1

The IUPAC name for the complex

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

is

View answer and explanation

Correct option: D

Pentaammine nitrito-N- cobalt (III) chloride

The correct answer is D (Pentaammine nitrito-N- cobalt (III) chloride) because it follows the IUPAC naming conventions for coordination compounds, correctly identifying the nitro ligand as nitrito-N.

Solution approach

Correct Answer + Core Reason: Pentaammine nitrito-N- cobalt (III) chloride. This follows IUPAC naming conventions.

Step-by-Step Solution:

Step 1: Identify the complex: $\left[ {\mathrm{Co}\left( {\mathrm{N}{\mathrm{O}_2}} \right){{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}} \right]\mathrm{C}{\mathrm{l}_2}$

Step 2: Identify the ligands: NH $_3$ (ammine) and NO $_2$ (nitrito-N).

Step 3: Determine the oxidation state of Co. There are two Cl $^-$ ions outside the coordination sphere, so the complex ion has a +2 charge. Inside the coordination sphere, NH $_3$ is neutral, and NO $_2$ has a -1 charge. Therefore, the oxidation state of Co is +3.

Step 4: Name the ligands in alphabetical order: ammine, nitrito-N.

Step 5: Combine the names: Pentaammine nitrito-N- cobalt (III) chloride.

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

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.

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Understanding +4 | -1

The IUPAC name of

\mathrm{N}{\mathrm{a}_3}\left[ {\mathrm{Co}{{\left( {\mathrm{ONO}} \right)}_6}} \right]

is:

View answer and explanation

Correct option: B

Sodium hexanitritocobaltate(III)

Option B (Sodium hexanitritocobaltate(III)) is correct because it correctly names the compound, indicating the sodium cation, the hexanitrito ligands, the cobaltate anion, and the oxidation state of cobalt.

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the cation: Sodium (Na).

Step 2: Identify the ligands: Nitrito (ONO) with a charge of -1. There are six nitrito ligands.

Step 3: Identify the central metal: Cobalt (Co).

Step 4: Determine the oxidation state of cobalt: The complex ion has a charge of -3. Since each nitrito ligand has a -1 charge, the total charge from the ligands is -6. Therefore, the oxidation state of cobalt must be +3 to result in a -3 overall charge for the complex ion.

Step 5: Name the complex: Sodium hexanitritocobaltate(III).

Why other options are wrong:
- Option A: Sodium cobaltinitrite: This is a common name, not the IUPAC name.
- Option C: Sodium hexanitrocobalt(III): Incorrect ligand name (should be nitrito, not nitro) and incorrect name for the complex ion (should be cobaltate).
- Option D: Sodium hexanitritocobaltate(II): Incorrect oxidation state of cobalt (should be III).

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

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

Quick Tips:
- Always specify the oxidation state of the metal using Roman numerals.
- Pay attention to the linkage isomerism (nitro vs. nitrito).

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Understanding +4 | -1

Which one is organometalllic compound?

View answer and explanation

Correct option: D

Methyl lithium

The correct answer is D (Methyl lithium). An organometallic compound contains a direct metal-carbon bond. Methyl lithium (CH3Li) has a direct Li-C bond.

Solution approach

Correct Answer + Core Reason: Methyl lithium. It contains a direct metal-carbon bond.

Step-by-Step Solution:

Step 1: Define organometallic compounds. They contain at least one metal-carbon bond.

Step 2: Examine the given options.

Step 3: Lithium acetate: $\mathrm{CH}_3\mathrm{COOLi}$ - Lithium is bonded to oxygen, not carbon.

Step 4: Lithium methoxide: $\mathrm{CH}_3\mathrm{OLi}$ - Lithium is bonded to oxygen, not carbon.

Step 5: Lithium dimethyl amide: $\mathrm{(CH}_3{\mathrm{)}_2}\mathrm{NLi}$ - Lithium is bonded to nitrogen, not carbon.

Step 6: Methyl lithium: $\mathrm{CH}_3\mathrm{Li}$ - Lithium is directly bonded to carbon.

Why other options are wrong:
- Option A: Lithium acetate: Lithium is bonded to oxygen.
- Option B: Lithium methoxide: Lithium is bonded to oxygen.
- Option C: Lithium dimethyl amide: Lithium is bonded to nitrogen.

Key Concepts:
- Organometallic compounds
- Metal-carbon bond

Memory Aids:
- Organometallic = Metal + Carbon bond

Quick Tips:
- Look for direct metal-carbon bonds to identify organometallic compounds.

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Understanding +4 | -1

Which of the following shell, form an outer octahedral complex?

View answer and explanation

Correct option: B

{d^8}

Option B (

{d^8}

) is correct because

{d^8}

configuration favors outer orbital octahedral complexes due to electronic repulsions and the need to accommodate all electrons.

Solution approach

Correct Answer + Core Reason: ${d^8}$ . ${d^8}$ configuration favors outer orbital complexes.

Step-by-Step Solution:

Step 1: Understand Outer Octahedral Complexes: Outer orbital octahedral complexes involve $sp^3d^2$ hybridization, using outer d-orbitals.

Step 2: Consider the Electronic Configurations:
- $d^4$ : Can form both inner and outer orbital complexes depending on the ligand strength.
- $d^6$ : Can form both inner and outer orbital complexes depending on the ligand strength.
- $d^8$ : Typically forms outer orbital complexes to minimize electron-electron repulsion.

Step 3: Analyze the $d^8$ Configuration: In an octahedral field, the $d^8$ configuration has 6 electrons in the $t_{2g}$ orbitals and 2 electrons in the $e_g$ orbitals. Adding more electrons would lead to high energy and instability in inner orbital complexes. Therefore, it favors outer orbital complexes.

Step 4: Conclusion: The $d^8$ configuration is most likely to form an outer octahedral complex.

Why other options are wrong:
- **Option A ( ${d^4}$ ):** $d^4$ can form both inner and outer orbital complexes depending on the ligand field strength. With strong field ligands, it forms inner orbital complexes; with weak field ligands, it forms outer orbital complexes.
- **Option C ( ${d^6}$ ):** Similar to $d^4$ , $d^6$ can form both inner and outer orbital complexes depending on the ligand field strength.
- Option D (None of these): $d^8$ is a valid configuration that forms outer orbital complexes.

Key Concepts:
- Coordination Chemistry
- Inner and Outer Orbital Complexes
- Hybridization
- Crystal Field Theory

Memory Aids:
- $d^8$ often leads to outer orbital complexes due to electron repulsion.

Quick Tips:
- Remember the electronic configurations that favor inner vs. outer orbital complexes.

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Application +4 | -1

The total number of possible isomers of the complex compound

\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_4\right]\left[{Pt}^{\mathrm{II}} {Cl}_4\right]

is

View answer and explanation

Correct option: D

4

Option D (4) is correct because the complex can exhibit linkage and coordination isomerism, leading to a total of 4 isomers.

Solution approach

Correct Answer + Core Reason: 4. The complex exhibits coordination and linkage isomerism.

Step-by-Step Solution:

Step 1: Identify the complex: $\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_4\right]\left[{Pt}^{\mathrm{II}} {Cl}_4\right]$

Step 2: Consider coordination isomerism: The ligands can be exchanged between the two metal centers, leading to $\left[{Cu}^{\mathrm{II}} {Cl}_4\right]\left[{Pt}^{\mathrm{II}} \left({NH}_3\right)_4\right]$ . This gives one additional isomer.

Step 3: Consider the possibility of linkage isomerism. However, neither complex has ambidentate ligands, so linkage isomerism is not possible within the individual complexes.

Step 4: Consider the original complex and the coordination isomer. The original complex is $\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_4\right]\left[{Pt}^{\mathrm{II}} {Cl}_4\right]$ . The coordination isomer is $\left[{Cu}^{\mathrm{II}} {Cl}_4\right]\left[{Pt}^{\mathrm{II}} \left({NH}_3\right)_4\right]$ .

Step 5: Consider the possibility of geometrical isomers within each complex. Since both are square planar, $\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_4\right]$ and $\left[{Pt}^{\mathrm{II}} {Cl}_4\right]$ have no geometrical isomers. Similarly, $\left[{Cu}^{\mathrm{II}} {Cl}_4\right]$ and $\left[{Pt}^{\mathrm{II}} \left({NH}_3\right)_4\right]$ have no geometrical isomers.

Step 6: Consider the possibility of ionization isomerism. This is not applicable here.

Step 7: Consider the possibility of hydrate isomerism. This is not applicable here.

Step 8: Consider the possibility of ligand isomerism. This is not applicable here.

Step 9: The complex $\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_4\right]\left[{Pt}^{\mathrm{II}} {Cl}_4\right]$ can also form the complex $\left[{Cu}^{\mathrm{II}}\left({NH}_3\right)_3 {Cl}\right]\left[{Pt}^{\mathrm{II}} {Cl}_3({NH}_3)\right]$ . This gives two more isomers.

Step 10: Total isomers = 1 (original) + 1 (coordination) + 2 (mixed ligand) = 4

Why other options are wrong:
- Option A (3): Incorrect because it doesn't consider all possible coordination isomers.
- Option B (6): Incorrect because it overestimates the number of possible isomers, possibly including incorrect combinations or geometrical isomers that don't exist.
- Option C (5): Incorrect because it misses one of the possible coordination isomers.

Key Concepts:
- Coordination Isomerism
- Isomerism in Coordination Compounds
- Square Planar Complexes

Memory Aids:
- Remember to consider all types of isomerism: coordination, linkage, ionization, hydrate, geometrical, and optical.

Quick Tips:
- Systematically analyze each type of isomerism to avoid missing any possibilities.

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Analyzing +4 | -1

The shortest

\mathrm{C} - \mathrm{O}

bond order exists in:

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the negative charge on the complex increases back-bonding, which weakens the C-O bond and thus shortens the C-O bond length.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{V}{{\left( {\mathrm{CO}} \right)}_6}} \right]^ - }$ . The negative charge increases back-bonding, weakening the C-O bond and shortening the C-O bond length.

Step-by-Step Solution:

Step 1: Understand the concept of back-bonding in metal carbonyls.\nIn metal carbonyls, there is a synergistic effect: the CO ligand donates electron density to the metal ( $\sigma$ bonding), and the metal donates electron density back to the CO ligand ( $\pi$ back-bonding).

Step 2: Relate back-bonding to C-O bond order and bond length.\nIncreased back-bonding weakens the C-O bond (decreases bond order) and increases the M-C bond strength. A weaker C-O bond has a longer bond length.\nDecreased back-bonding strengthens the C-O bond (increases bond order) and decreases the M-C bond strength. A stronger C-O bond has a shorter bond length.

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

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

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

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Mn}{{\left( {\mathrm{CO}} \right)}_6}} \right]^ + }$ has a positive charge, which decreases back-bonding and strengthens the C-O bond (longer bond length).
- Option B: $\left[ {\mathrm{Fe}{{\left( {\mathrm{CO}} \right)}_5}} \right]$ has a neutral charge, so the back-bonding is intermediate.
- Option C: $\left[ {\mathrm{Cr}{{\left( {\mathrm{CO}} \right)}_6}} \right]$ has a neutral charge, so the back-bonding is intermediate.

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

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

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

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Co-ordination compounds practice · Medium

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