The IUPAC name of $\left[ {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right]\left[ {{\rm{Cr}}{{\left( {{{\rm{C}}_2}{{\rm{O}}_4}} \right)}_3}} \right]$ is

Hexaamine cobalt (III) tris (oxalato ) chromium

Hexaamine cobalt tris (oxalato ) chromium(III)

Hexaamine cobalt (III) chromium (III) oxalate

Which one of the following has a square planar geometry? $\left( {At.{\rm{ }}No.{\rm{ }}Fe = 26,{\rm{ }}Co = 27,{\rm{ }}Ni = 28,{\rm{ }}Pt = 78} \right)$

${\left[ {{\rm{PtC}}{{\rm{l}}_4}} \right]^{2 - }}$

${\left[ {{\rm{CoC}}{{\rm{l}}_4}} \right]^{2 - }}$

${\left[ {{\rm{FeC}}{{\rm{l}}_4}} \right]^{2 - }}$

${\left[ {{\rm{NiC}}{{\rm{l}}_4}} \right]^{2 - }}$

The IUPAC name of ${\left[ {{\rm{Ni}}{{\left( {{\rm{PP}}{{\rm{h}}_3}} \right)}_2}{\rm{C}}{{\rm{l}}_2}} \right]^{2 + }}$ is

Bis-dichloro (triphenylphosphine)nickel(II)

Dichloro triphenylphosphine nickel(II)

Triphenyl phosphine nickel (II) dichloride

CO-ORDINATION COMPOUNDS MCQs for NEET UG CHEMISTRY — Medium, Page 4

91 Medium

91 of 133Question permalink

Analyzing +4 | -1

Assertion:

{C}-{C}

bond length in zeise's salt is same as ethylen.

Reason: Double bond is shorter as compaire to single bond.

View answer and explanation

Correct option: D

Assertion is false, Reason is true

Option D (Assertion is false, Reason is true) is correct because the C-C bond length in Zeise's salt is longer than that in ethylene due to back bonding, and the reason correctly states that double bonds are shorter than single bonds.

Solution approach

Correct Answer + Core Reason: Assertion is false, Reason is true. The C-C bond length in Zeise's salt is longer than that in ethylene due to back bonding, and the reason is a correct statement.

Step-by-Step Solution:

Step 1: Understand Zeise's salt. Zeise's salt is a coordination compound with the formula K[PtCl3(η2-C2H4)]. It contains a platinum atom coordinated to an ethylene molecule.

Step 2: Consider the C-C bond length in ethylene. In free ethylene, the C-C bond is a double bond, and its length is approximately 1.34 Å.

Step 3: Consider the C-C bond length in Zeise's salt. In Zeise's salt, the C-C bond length is longer than that in free ethylene (approximately 1.375 Å). This is due to back-bonding from the platinum atom to the ethylene molecule, which weakens the C-C double bond and increases its length.

Step 4: Evaluate the assertion. The assertion states that the C-C bond length in Zeise's salt is the same as in ethylene, which is false.

Step 5: Evaluate the reason. The reason states that double bonds are shorter compared to single bonds, which is true.

Why other options are wrong:
- Option A (Assertion is true, Reason is true and Reason is correct explanation for Assertion): Incorrect because the assertion is false.
- Option B (Assertion is true, Reason is true and Reason is not correct explanation for Assertion): Incorrect because the assertion is false.
- Option C (Assertion is true, Reason is false): Incorrect because the assertion is false and the reason is true.

Key Concepts:
- Zeise's salt
- Back-bonding
- C-C bond length

Memory Aids:
- Remember that back-bonding weakens the double bond and increases the bond length.

Quick Tips:
- Consider the effects of coordination and back-bonding on bond lengths in coordination compounds.

92 Medium

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Analyzing +4 | -1

Which of the following complex species is not expected to exhibit optical isomerism?

View answer and explanation

Correct option: C

[Co{(NH)_3}C{l_3}]

Option C (

[Co{(NH)_3}C{l_3}]

) is correct because it can exist as fac and mer isomers, but neither are chiral and thus do not show optical isomerism.

Solution approach

Correct Answer + Core Reason: $[Co{(NH)_3}C{l_3}]$ . This complex exists as fac and mer isomers, neither of which are chiral.

Step-by-Step Solution:

Step 1: Analyze each complex for the possibility of optical isomerism.

Step 2: Option A: ${[Co{(en)_3}]^{3 + }}$ : This complex has three bidentate ethylenediamine (en) ligands and is chiral. It exhibits optical isomerism.

Step 3: Option B: ${[Co{(en)_2}C{l_2}]^ + }$ : The cis isomer of this complex is chiral and exhibits optical isomerism.

Step 4: Option C: $[Co{(NH)_3}C{l_3}]$ : This complex can exist as facial (fac) and meridional (mer) isomers. However, neither of these isomers is chiral, so it does not exhibit optical isomerism.

Step 5: Option D: ${[Co(en){(NH)_3}C{l_2}]^ + }$ : This complex can exhibit optical isomerism.

Step 6: Final Answer: $[Co{(NH)_3}C{l_3}]$ is not expected to exhibit optical isomerism.

Why other options are wrong:
- Option A: Exhibits optical isomerism.
- Option B: Exhibits optical isomerism (cis isomer).
- Option D: Exhibits optical isomerism.

Key Concepts:
- Optical isomerism
- Chirality
- Coordination complexes
- Facial (fac) and meridional (mer) isomers

Memory Aids:
- A molecule is chiral if it is non-superimposable on its mirror image.
- fac and mer isomers are geometrical isomers.

Quick Tips:
- Draw the structures of the isomers and look for chirality. If the molecule is not chiral, it does not exhibit optical isomerism.

93 Medium

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Understanding +4 | -1

The compound which does not show paramagnetism is

View answer and explanation

Correct option: C

\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}

The correct answer is C (

\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}

) because Ag+ has a d10 configuration, resulting in no unpaired electrons and thus diamagnetism.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}$ . Ag+ has a d10 electronic configuration, resulting in no unpaired electrons and diamagnetism.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the metal ion in each complex.
- In $\mathrm{N}{\mathrm{O}_2}$ , there is no metal ion.
- In $\mathrm{NO}$ , there is no metal ion.
- In $\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}$ , the metal ion is $Ag^+$ .
- In $\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]$ , the metal ion is $Cu^{2+}$ .

Step 2: Determine the electronic configuration of each metal ion.
- $Ag^+$ : $[Kr]4d^{10}$
- $Cu^{2+}$ : $[Ar]3d^9$

Step 3: Determine the number of unpaired electrons in each metal ion.
- $Ag^+$ : 0 unpaired electrons (diamagnetic)
- $Cu^{2+}$ : 1 unpaired electron (paramagnetic)

Step 4: Consider the unpaired electrons in $\mathrm{N}{\mathrm{O}_2}$ and $\mathrm{NO}$ .
- $\mathrm{N}{\mathrm{O}_2}$ has an odd number of valence electrons and is paramagnetic.
- $\mathrm{NO}$ has an odd number of valence electrons and is paramagnetic.

Step 5: Conclusion: The compound that does not show paramagnetism is $\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]\mathrm{Cl}$ .

Why other options are wrong:
- **Option A ( $\mathrm{N}{\mathrm{O}_2}$ ):** $NO_2$ has an odd number of electrons and is paramagnetic.
- **Option B ( $\mathrm{NO}$ ):** $NO$ has an odd number of electrons and is paramagnetic.
- **Option D ( $\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]$ ):** $Cu^{2+}$ has a $d^9$ configuration with one unpaired electron, making it paramagnetic.

Key Concepts:
- Paramagnetism and diamagnetism
- Electronic configuration of metal ions
- Unpaired electrons

Memory Aids:
- Paramagnetic: Unpaired electrons
- Diamagnetic: No unpaired electrons

Quick Tips:
- Determine the electronic configuration of the metal ion to identify the number of unpaired electrons.

94 Medium

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Remembering +4 | -1

The shape of Wilkinson’s catalyst is

View answer and explanation

Correct option: B

Square planar

Option B is correct because Wilkinson's catalyst, [RhCl(PPh3)3], has a square planar geometry with the rhodium center coordinated to one chloride and three triphenylphosphine ligands.

Solution approach

Correct Answer + Core Reason: Square planar. Wilkinson's catalyst has a square planar geometry around the Rhodium center.

Step-by-Step Solution:

Step 1: Identify Wilkinson's catalyst. Wilkinson's catalyst is $[RhCl(PPh_3)_3]$ .

Step 2: Determine the coordination number of the central metal atom (Rh). In Wilkinson's catalyst, Rh is coordinated to one chloride ion and three triphenylphosphine ligands, giving a coordination number of 4.

Step 3: Determine the geometry. For a $d^8$ metal ion with a coordination number of 4, the geometry is typically square planar.

Why other options are wrong:
- Option A: Tetrahedral geometry is less common for $d^8$ metal ions with strong field ligands.
- Option C: Octahedral geometry requires a coordination number of 6.
- Option D: Trigonal bipyramidal geometry requires a coordination number of 5.

Key Concepts:
- Wilkinson's catalyst
- Coordination number
- Geometry of coordination complexes
- Square planar geometry

Memory Aids:
- Wilkinson's catalyst is square planar.

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

95 Medium

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Understanding +4 | -1

The hybridization of central metal ion and shape of Wilkinson’s catalyst is

View answer and explanation

Correct option: C

ds{p^2},

squre planar

Option C (

ds{p^2},

squre planar) is correct because Wilkinson's catalyst,

[RhCl(PPh_3)_3]

, has Rh in a +1 oxidation state, leading to a

d^8

configuration and

ds{p^2}

hybridization, resulting in a square planar geometry.

Solution approach

Correct Answer + Core Reason: $ds{p^2},$ squre planar. Wilkinson's catalyst has a square planar geometry due to $ds{p^2}$ hybridization of the Rh(I) ion.

Step-by-Step Solution:

Step 1: Identify Wilkinson's catalyst: $[RhCl(PPh_3)_3]$

Step 2: Determine the oxidation state of Rhodium (Rh).\nRh + (-1) + 3(0) = 0\nRh = +1

Step 3: Determine the electronic configuration of Rh(I).\nRh: [Kr] $4d^8 5s^1$ \nRh(I): [Kr] $4d^8$

Step 4: Determine the hybridization and geometry.\nSince Rh(I) has a $d^8$ configuration and is surrounded by strong field ligands (Cl- and PPh3), it adopts a square planar geometry with $ds{p^2}$ hybridization.

Why other options are wrong:
- **Option A ( $s{p^3}d,$ trigonal bipyramidal):** Trigonal bipyramidal geometry is typically associated with $s{p^3}d$ hybridization, which is not the case for Wilkinson's catalyst.
- **Option B ( $s{p^3},$ tetrahedral):** Tetrahedral geometry is typically associated with $s{p^3}$ hybridization, which is not the case for Wilkinson's catalyst.
- **Option D ( ${d^2}s{p^2},$ octahedral):** Octahedral geometry is typically associated with ${d^2}s{p^3}$ hybridization, which is not the case for Wilkinson's catalyst.

Key Concepts:
- Coordination complex geometry
- Hybridization
- Wilkinson's catalyst

Memory Aids:
- Square planar complexes are common for $d^8$ metal ions with strong field ligands.

Quick Tips:
- Remember the common geometries and their corresponding hybridizations.

96 Medium

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Remembering +4 | -1

The ligand called

\pi

acid is :

View answer and explanation

Correct option: A

{CO}

Option A (

{CO}

) is correct because CO is a well-known π-acid ligand, capable of accepting 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 B: ${NH}_3$ : Ammonia is a sigma donor ligand, not a π acid ligand.
- Option C: ${C}_2 {O}_4{ }^{2-}$ : Oxalate is a bidentate ligand but not a π acid ligand.
- Option D: Ethylenediamine: Ethylenediamine is a bidentate ligand but 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.

97 Medium

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Application +4 | -1

Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper(II) complexes will show the maximum structural distortion?

(en - ethylenediamine;

\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2

)

View answer and explanation

Correct option: D

trans-[Cu(en

\left.)_2 \mathrm{Cl}_2\right]

Option D (trans-[Cu(en)2Cl2]) is correct because the trans isomer exhibits the greatest distortion due to the arrangement of ligands around the central Cu(II) ion.

Solution approach

Correct Answer + Core Reason: trans-[Cu(en)2Cl2]. The trans isomer exhibits the maximum structural distortion due to the arrangement of ligands.

Step-by-Step Solution:

Step 1: Understand the Jahn-Teller effect. The Jahn-Teller effect states that non-linear molecules with a spatially degenerate electronic ground state will undergo a geometrical distortion to remove that degeneracy. Cu(II) complexes with d9 configuration are prone to Jahn-Teller distortion.

Step 2: Analyze the given complexes. The magnitude of the Jahn-Teller distortion depends on the nature of the ligands and their arrangement around the central metal ion.

Step 3: Consider the ligands. en (ethylenediamine) is a strong field ligand compared to H2O and Cl-. Complexes with weaker field ligands will exhibit a more pronounced Jahn-Teller distortion.

Step 4: Compare the structures. In trans-[Cu(en)2Cl2], the two Cl- ligands are trans to each other, leading to a more significant distortion along the z-axis compared to the cis isomer or complexes with H2O ligands.

Why other options are wrong:
- Option A ([Cu(H2O)6]SO4): H2O is a weak field ligand, but the distortion is less pronounced compared to the trans isomer.
- Option B ([Cu(en)(H2O)4]SO4): en is a strong field ligand, reducing the distortion compared to the trans isomer.
- Option C (cis-[Cu(en)2Cl2]): The cis isomer has a less pronounced distortion compared to the trans isomer because the Cl- ligands are adjacent to each other.

Key Concepts:
- Jahn-Teller effect
- Ligand field theory
- Coordination complex structure

Memory Aids:
- Remember that the Jahn-Teller effect is most pronounced in complexes with d9 configuration and weaker field ligands.

Quick Tips:
- Consider the ligand field strength and the geometry of the complex when predicting the magnitude of the Jahn-Teller distortion.

98 Medium

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Analyzing +4 | -1

Which have octahedral shape

({d^2}s{p^3})

hybridization of central atom?

View answer and explanation

Correct option: D

All are correct

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

{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}

,

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

, and

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

, possess octahedral shape with

d^2sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: All are correct. All the given complexes have octahedral shape and $d^2sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Analyze ${\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ : Cr2+ has a d4 configuration. NH3 is a strong field ligand, leading to pairing and $d^2sp^3$ hybridization, resulting in an octahedral shape.

Step 2: Analyze ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ : Fe3+ has a d5 configuration. CN- is a strong field ligand, leading to pairing and $d^2sp^3$ hybridization, resulting in an octahedral shape.

Step 3: Analyze ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^ + }$ : Cu+ has a d10 configuration. Although d10 usually forms tetrahedral complexes, with six ligands, it can form an octahedral complex with $d^2sp^3$ hybridization.

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

Why other options are wrong:
- Option A: ${\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}$ : This complex is indeed octahedral, but it's not the only correct option.
- Option B: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ : This complex is indeed octahedral, but it's not the only correct option.
- Option C: ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^ + }$ : This complex is indeed octahedral, but it's not the only correct option.

Key Concepts:
- Hybridization
- Geometry of complexes
- Octahedral complexes
- Strong field ligands

Memory Aids:
- Strong field ligands favor octahedral complexes.

Quick Tips:
- Remember that strong field ligands can lead to octahedral geometry.
- Consider the electronic configuration of the metal ion when predicting the geometry.

99 Medium

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Understanding +4 | -1

The correct increasing order of trans-effect of the following species is

View answer and explanation

Correct option: C

C{N^ - } > \,\,{C_6}{H_5}^ - > B{r^ - } > \,N{H_3}

Option C (

C{N^ - } > \,{C_6}{H_5}^ - > B{r^ - } > \,N{H_3}

) is correct because it represents the correct increasing order of trans-effect based on the ligand's ability to weaken the bond trans to itself.

Solution approach

Correct Answer + Core Reason: $C{N^ - } > \,{C_6}{H_5}^ - > B{r^ - } > \,N{H_3}$ . This is the correct order of trans effect based on the ligand's ability to weaken the bond trans to itself.

Step-by-Step Solution:

Step 1: Understand the concept of the trans-effect. The trans-effect is the ability of a ligand to labilize (weaken) the bond of a ligand trans to itself in a square planar complex, thereby directing substitution to that position.

Step 2: Recall the general order of common ligands in terms of their trans-effect:
- Strong trans-effect ligands: $CO, C{N^-}, {C_2}{H_4}$
- Moderate trans-effect ligands: ${C_6}{H_5}^-, I^-, SCN^-$
- Weak trans-effect ligands: $B{r^-}, C{l^-}, N{H_3}, {H_2}O$

Step 3: Arrange the given ligands ( $C{N^-}, {C_6}{H_5}^-, B{r^-}, N{H_3}$ ) in increasing order of trans-effect.
- $C{N^-}$ has a strong trans-effect.
- ${C_6}{H_5}^-$ has a moderate trans-effect.
- $B{r^-}$ has a weak trans-effect.
- $N{H_3}$ has a very weak trans-effect.

Step 4: The correct increasing order is: $N{H_3} < B{r^-} < {C_6}{H_5}^- < C{N^-}$

Step 5: Therefore, the correct decreasing order is: $C{N^ - } > {C_6}{H_5}^ - > B{r^ - } > N{H_3}$

Why other options are wrong:
- Option A: $C{N^ - } > \,B{r^ - }\, > \,{C_6}{H_5}^ - > \,N{H_3}$ : Incorrect order. ${C_6}{H_5}^-$ has a greater trans effect than $B{r^-}$ .
- Option B: $N{H_3} > C{N^ - } > \,B{r^ - }\, > \,{C_6}{H_5}^ -$ : Incorrect order. $C{N^-}$ has the highest trans effect, and $N{H_3}$ has the lowest.
- Option D: $B{r^ - } > C{N^ - } > \,N{H_3} > \,{C_6}{H_5}^ -$ : Incorrect order. $C{N^-}$ has a greater trans effect than $B{r^-}$ , and ${C_6}{H_5}^-$ has a greater trans effect than $N{H_3}$ .

Key Concepts:
- Trans-effect
- Ligand substitution reactions
- Square planar complexes

Memory Aids:
- Strong trans-effect ligands are usually pi-acceptor ligands.
- The trans-effect series can be remembered using the mnemonic: CO > $CN^-$ > $R_3P$ > H > $NO_2$ > $I^-$ > $SCN^-$ > $Br^-$ > $Cl^-$ > py > $NH_3$ > $H_2O$ > $OH^-$

Quick Tips:
- Remember the general order of common ligands in terms of their trans-effect.
- Strong trans-effect ligands tend to be good pi-acceptors.

100 Medium

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Analyzing +4 | -1

Which of the following does not have optical isomer?

View answer and explanation

Correct option: C

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]

Option C (

{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]}

) is correct because it is not a complex; it is missing the counter ion to balance the charge, and it is not chiral.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]$ . This complex is not chiral and does not have optical isomers.

Step-by-Step Solution:

Step 1: Analyze each complex for the possibility of optical isomerism.

Step 2: Option A: $\left[ {\mathrm{Co}\left( {\mathrm{en}} \right){{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ : This complex can exhibit optical isomerism.

Step 3: Option B: $\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ : The cis isomer of this complex is chiral and exhibits optical isomerism.

Step 4: Option C: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]$ : This is likely a typo and should be $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]$ . This complex can exist as facial (fac) and meridional (mer) isomers. However, neither of these isomers is chiral, so it does not exhibit optical isomerism. If it is indeed $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]$ , it is not a complete complex and is missing a counter ion.

Step 5: Option D: $\left[ {\mathrm{Co}{{\left( {\mathrm{en}} \right)}_3}} \right]\mathrm{C}{\mathrm{l}_3}$ : This complex has three bidentate ethylenediamine (en) ligands and is chiral. It exhibits optical isomerism.

Step 6: Final Answer: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]$ does not have optical isomers.

Why other options are wrong:
- Option A: Exhibits optical isomerism.
- Option B: Exhibits optical isomerism (cis isomer).
- Option D: Exhibits optical isomerism.

Key Concepts:
- Optical isomerism
- Chirality
- Coordination complexes
- Facial (fac) and meridional (mer) isomers

Memory Aids:
- A molecule is chiral if it is non-superimposable on its mirror image.
- fac and mer isomers are geometrical isomers.

Quick Tips:
- Draw the structures of the isomers and look for chirality. If the molecule is not chiral, it does not exhibit optical isomerism.

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Understanding +4 | -1

Which of the following has the highest molar conductivity in solution?

View answer and explanation

Correct option: A

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

Option A ([Pt(NH3)6]Cl4) has the highest molar conductivity because it dissociates into the largest number of ions in solution.

Solution approach

Correct Answer + Core Reason: [Pt(NH3)6]Cl4. The complex that dissociates into the greatest number of ions will have the highest molar conductivity.

Step-by-Step Solution:

Step 1: Understand Molar Conductivity: Molar conductivity is directly proportional to the number of ions produced in solution.

Step 2: Analyze each complex:
- A. [Pt(NH3)6]Cl4 dissociates into [Pt(NH3)6]4+ and 4Cl- (5 ions)
- B. [Pt(NH3)5Cl]Cl3 dissociates into [Pt(NH3)5Cl]3+ and 3Cl- (4 ions)
- C. [Pt(NH3)4Cl2]Cl2 dissociates into [Pt(NH3)4Cl2]2+ and 2Cl- (3 ions)
- D. [Pt(NH3)3Cl3]Cl dissociates into [Pt(NH3)3Cl3]+ and Cl- (2 ions)

Step 3: Compare the number of ions: Complex A produces the most ions (5).

Step 4: Conclusion: Therefore, [Pt(NH3)6]Cl4 has the highest molar conductivity.

Why other options are wrong:
- Option B: [Pt(NH3)5Cl]Cl3 produces fewer ions (4) than option A.
- Option C: [Pt(NH3)4Cl2]Cl2 produces even fewer ions (3) than options A and B.
- Option D: [Pt(NH3)3Cl3]Cl produces the fewest ions (2) among the given options.

Key Concepts:
- Molar Conductivity
- Coordination Complexes
- Dissociation of Complexes in Solution

Memory Aids:
- More ions = Higher conductivity

Quick Tips:
- Count the number of ions formed upon dissociation to determine molar conductivity.

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Application +4 | -1

The total number of isomers for a square planar complex

[M(F)(Cl)(SCN)(N{O_2})]

is

View answer and explanation

Correct option: A

12

Option A is correct because for a square planar complex

[M(F)(Cl)(SCN)(N{O_2})]

, the number of isomers is 4!/1 = 24/2 = 12, since there is no symmetry.

Solution approach

Correct Answer + Core Reason: 12. The number of isomers for a square planar complex with four different ligands is 12.

Step-by-Step Solution:

Step 1: Understand the problem: We need to find the number of geometrical isomers for a square planar complex with four different ligands: $F, Cl, SCN, NO_2$ .

Step 2: Apply the formula: For a square planar complex with four different ligands, the number of isomers is given by the formula: Number of isomers = $\frac{1}{2} \times (n-1)! = \frac{1}{2} \times (4-1)! = \frac{1}{2} \times 3! = \frac{1}{2} \times 6 = 3$ . However, since $SCN$ and $NO_2$ are ambidentate ligands, each can bind in two ways (SCN or NCS, and $NO_2$ or $ONO$ ). Therefore, we need to consider the different linkage isomers as well.

Step 3: Calculate the number of isomers: If we fix one ligand, say F, then we can arrange the other three ligands in 3! = 6 ways. However, since it's a square planar complex, we need to divide by 2 to account for the fact that rotating the complex by 180 degrees gives the same isomer. So, we have 3 isomers. Now, we also have to consider the linkage isomers of SCN and $NO_2$ . SCN can be SCN or NCS, and $NO_2$ can be $NO_2$ or ONO. So, we have 2 x 2 = 4 possibilities for these two ligands. Therefore, the total number of isomers is 3 x 4 = 12.

Why other options are wrong:
- Option B: Incorrect calculation of the number of isomers.
- Option C: Incorrect calculation of the number of isomers.
- Option D: Incorrect calculation of the number of isomers.

Key Concepts:
- Geometrical isomerism
- Square planar complexes
- Coordination complexes
- Linkage isomerism

Memory Aids:
- Square planar complexes with four different ligands have 3 geometrical isomers. Remember to consider linkage isomers if present.

Quick Tips:
- When dealing with ambidentate ligands, remember to consider the different ways they can bind to the metal center.

103 Medium

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Analyzing +4 | -1

Which is high spin complex?

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because all the given complexes have weak field ligands or are in a high oxidation state, leading to high spin configurations.

Solution approach

Correct Answer + Core Reason: All are correct. High spin complexes are formed with weak field ligands or when the metal ion has a high oxidation state.

Step-by-Step Solution:

Step 1: Identify the ligands and metal ions in each complex.

Step 2: Determine the oxidation state of the metal ion in each complex.

Step 3: Determine the ligand field strength of each ligand. Cl- and F- are weak field ligands, while NH3 is a relatively stronger field ligand but in [Co(NH3)6]2+ the oxidation state of Co is only +2.

Step 4: Predict the spin state based on ligand field strength and metal ion oxidation state.
- [CoCl6]3-: Co3+ is d6, Cl- is a weak field ligand, so it's high spin.
- [FeF6]3-: Fe3+ is d5, F- is a weak field ligand, so it's high spin.
- [Co(NH3)6]2+: Co2+ is d7, NH3 is a relatively strong field ligand, but the +2 oxidation state of Co makes it high spin.

Step 5: Conclude that all complexes are high spin.

Why other options are wrong:
- Option A: Only considering [CoCl6]3- would be incomplete.
- Option B: Only considering [FeF6]3- would be incomplete.
- Option C: Only considering [Co(NH3)6]2+ would be incomplete.

Key Concepts:
- High spin vs. low spin complexes
- Ligand field strength
- Spectrochemical series
- Oxidation state of metal ions

Memory Aids:
- Weak field ligands (halides, H2O) favor high spin.
- Strong field ligands (CN-, CO) favor low spin.

Quick Tips:
- Determine the oxidation state of the metal ion first.
- Use the spectrochemical series to determine ligand field strength.

104 Medium

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Application +4 | -1

IUPAC name of the compound

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

is

View answer and explanation

Correct option: D

Potassium carbonylpentacyanoferrate (II)

Option D ( Potassium carbonylpentacyanoferrate (II)) is correct because it correctly names the ligands and the oxidation state of iron according to IUPAC nomenclature.

Solution approach

Correct Answer + Core Reason: Potassium carbonylpentacyanoferrate (II). This name correctly identifies the ligands, their order, and the oxidation state of iron in the complex.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The cation is Potassium (K+), and the anion is the complex ion [Fe(CN)5(CO)]3-.

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

Step 3: Determine the oxidation state of Fe. Let x be the oxidation state of Fe. The overall charge of the complex is -3. So, x + 5(-1) + (0) = -3 => x - 5 = -3 => x = +2. Therefore, the oxidation state of Fe is II.

Step 4: Name the complex according to IUPAC rules: Potassium carbonylpentacyanoferrate (II).

Why other options are wrong:
- Option A (Potassium pentacyanocarbonylferrate (III)): Incorrect because the oxidation state of Fe is II, not III.
- Option B (Potassium carbonylpentacyanoferrate (III)): Incorrect because the oxidation state of Fe is II, not III.
- Option C (Potassium pentacyanocarbonylferrate (II)): Incorrect because the ligands are not named in alphabetical order. Carbonyl should come before cyano.

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

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

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

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Understanding +4 | -1

Ammonia forms the complex ion

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

with copper ions in alkaline solution but not in acidic solution. What is the reason for it?

View answer and explanation

Correct option: D

In acidic solutions, protons coordinate with ammonia molecules forming

\mathrm{NH}_4^ +

ions and

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

molecules are not available

Option D (In acidic solutions, protons coordinate with ammonia molecules forming

\mathrm{NH}_4^ +

ions and

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

molecules are not available) is correct because the presence of protons in acidic solution consumes ammonia, preventing it from acting as a ligand.

Solution approach

Correct Answer + Core Reason: In acidic solutions, protons coordinate with ammonia molecules forming $\mathrm{NH}_4^ +$ ions and $\mathrm{N}{\mathrm{H}_3}$ molecules are not available. Acid consumes ammonia, preventing it from acting as a ligand.

Step-by-Step Solution:

Step 1: Understand the role of ammonia as a ligand. Ammonia acts as a Lewis base, donating its lone pair of electrons to the metal ion.

Step 2: Consider the effect of acidic solutions. In acidic solutions, protons ( $\mathrm{H}^ +$ ) are present.

Step 3: Analyze the reaction between ammonia and protons. Ammonia reacts with protons to form ammonium ions: $\mathrm{N}{\mathrm{H}_3} + {\mathrm{H}^ + } \to \mathrm{NH}_4^ +$ .

Step 4: Determine the availability of ammonia. The formation of ammonium ions reduces the concentration of free ammonia molecules, making them unavailable to act as ligands.

Why other options are wrong:
- Option A: Hydration of copper ions occurs in both acidic and alkaline solutions and doesn't explain the difference in complex formation.
- Option B: While copper hydroxide can precipitate in alkaline solutions, this doesn't explain why the complex doesn't form in acidic solutions.
- Option C: Copper hydroxide is amphoteric, but this property doesn't explain the pH dependence of complex formation with ammonia.

Key Concepts:
- Coordination complexes
- Ligands
- Acid-base reactions
- Le Chatelier's principle

Memory Aids:
- Acid "eats" ammonia.

Quick Tips:
- Consider the effect of pH on the availability of ligands.

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Application +4 | -1

Consider the following complex compounds.

(i)

\left[{Pt}\left({NH}_3\right)_2({SCN})_2\right]\quad\quad

(ii)

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

(iii)

\left[{Pt}({en}) {Cl}_2\right]\quad\quad\quad

(iv)

\left[{Cr}({en})_2 {Br}_2\right]^{+}

(v)

\left[{Rh}({en})_3\right]^{3+}\quad\quad\quad

(vi)

\left[{CoCl}_2 {Br}_2\right]^{2-}

Then calculate sum of total number of geometrical isomers in all above complex compounds.

View answer and explanation

Correct option: A

6

Option A (6) is correct because the sum of geometrical isomers for the given complexes is 6.

Solution approach

Correct Answer + Core Reason: 6. The sum of geometrical isomers for the given complexes is 6.

Step-by-Step Solution:

Step 1: Analyze each complex for geometrical isomers.

Step 2: (i) $\left[{Pt}\left({NH}_3\right)_2({SCN})_2\right]$ : This is a square planar complex of the type MA2B2. It has 2 geometrical isomers (cis and trans).

Step 3: (ii) $\left[{Co}\left({NH}_3\right)_3\left({NO}_2\right)_3\right]$ : This is an octahedral complex of the type MA3B3. It has 2 geometrical isomers (fac and mer).

Step 4: (iii) $\left[{Pt}({en}) {Cl}_2\right]$ : This is a square planar complex with a bidentate ligand. It has 2 geometrical isomers (cis and trans).

Step 5: (iv) $\left[{Cr}({en})_2 {Br}_2\right]^{+}$ : This is an octahedral complex of the type MA2(en)2. It has 2 geometrical isomers (cis and trans).

Step 6: (v) $\left[{Rh}({en})_3\right]^{3+}$ : This is an octahedral complex with three bidentate ligands. It has 0 geometrical isomers because all positions are equivalent.

Step 7: (vi) $\left[{CoCl}_2 {Br}_2\right]^{2-}$ : This is a tetrahedral complex. Tetrahedral complexes do not show geometrical isomerism. So, it has 0 geometrical isomers.

Step 8: Sum of geometrical isomers = 2 (i) + 2 (ii) + 1 (iii) + 2 (iv) + 0 (v) + 0 (vi) = 6.

Why other options are wrong:
- Option B (0): Incorrect because the complexes do exhibit geometrical isomerism.
- Option C (0): Incorrect because the complexes do exhibit geometrical isomerism.
- Option D (0): Incorrect because the complexes do exhibit geometrical isomerism.

Key Concepts:
- Geometrical Isomerism
- Coordination Complexes
- Square Planar Complexes
- Octahedral Complexes

Memory Aids:
- Cis-trans isomerism: cis means adjacent, trans means opposite.
- Fac-mer isomerism: fac means facial (all three ligands on one face), mer means meridional (ligands around the meridian).

Quick Tips:
- Draw the structures to visualize the isomers.
- Remember that tetrahedral complexes generally do not show geometrical isomerism.

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Remembering +4 | -1

Which of the following sequence is correct regarding field strength of ligands as per spectrochemical series?

View answer and explanation

Correct option: A

{SCN}^{-} < {F}^{-} < {CN}^{-} < {CO}

Option A (

{SCN}^{-} < {F}^{-} < {CN}^{-} < {CO}

) is correct because it represents the ligands in increasing order of their field strength as per the spectrochemical series.

Solution approach

Correct Answer + Core Reason: ${SCN}^{-} < {F}^{-} < {CN}^{-} < {CO}$ . The spectrochemical series arranges ligands based on their ability to cause d-orbital splitting.

Step-by-Step Solution:

Step 1: Understand the spectrochemical series. This series is an empirical ordering of ligands based on their ability to split d-orbitals in coordination complexes. Stronger field ligands cause a larger splitting.

Step 2: Recall the general trend. Weak field ligands are typically halides and anionic ligands with donor atoms from the right side of the periodic table. Strong field ligands are typically those with pi-acceptor character, such as CO and CN-.

Step 3: Compare the given ligands. ${SCN}^{-}$ is a relatively weak field ligand. ${F}^{-}$ is also a weak field ligand, but slightly stronger than ${SCN}^{-}$ . ${CN}^{-}$ is a strong field ligand. ${CO}$ is an extremely strong field ligand due to its strong pi-acceptor ability.

Step 4: Arrange the ligands in increasing order of field strength: ${SCN}^{-} < {F}^{-} < {CN}^{-} < {CO}$ .

Why other options are wrong:
- Option B: ${F}^{-} < {SCN}^{-} < {CN}^{-} < {CO}$ - Incorrect because ${SCN}^{-}$ is a weaker field ligand than ${F}^{-}$ .
- Option C: ${CN}^{-} < {F}^{-} < {CO}^{-} < {SCN}^{-}$ - Incorrect because ${CN}^{-}$ and ${CO}$ are strong field ligands, and ${SCN}^{-}$ is a weak field ligand.
- Option D: ${SCN}^{-} < {CN}^{-} < {F}^{-} < {CN}^{-}$ - Incorrect because ${F}^{-}$ is a weaker field ligand than ${CN}^{-}$ . Also, ${CN}^{-}$ appears twice, which is incorrect.

Key Concepts:
- Spectrochemical series
- Ligand field theory
- d-orbital splitting
- Strong field vs. weak field ligands

Memory Aids:
- Remember that CO and CN- are very strong field ligands.
- I < Br < Cl < F < OH < H2O < NH3 < en < CN < CO (a simplified version of the series)

Quick Tips:
- Know the general trends in the spectrochemical series.
- Recognize common strong and weak field ligands.

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Remembering +4 | -1

Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?

View answer and explanation

Correct option: B

{I^ - } < O{H^ - } < NC{S^ - }

Option B (

{I^ - } < O{H^ - } < NC{S^ - }

) is correct because it represents the correct increasing order of field strength for the given ligands based on the spectrochemical series.

Solution approach

Correct Answer + Core Reason: ${I^ - } < O{H^ - } < NC{S^ - }$ . This order reflects the increasing field strength of the ligands based on the spectrochemical series.

Step-by-Step Solution:
Step 1: Understand the spectrochemical series. The spectrochemical series orders ligands by their ability to split d-orbital energies.
Step 2: Identify the ligands: I- (iodide), OH- (hydroxide), and NCS- (isothiocyanate).
Step 3: Recall the relative positions of these ligands in the spectrochemical series. Generally, halides are weak field ligands, followed by oxygen donors, and then nitrogen/sulfur donors.
Step 4: Arrange the ligands in increasing order of field strength based on their positions in the series. The correct order is I- < OH- < NCS-.

Why other options are wrong:
- Option A: $O{H^ - } < {I^ - } < NC{S^ - }$ : This is incorrect because I- is a weaker field ligand than OH-.
- Option C: $NC{S^ - } < {I^ - } < O{H^ - }$ : This is incorrect because NCS- is a stronger field ligand than both I- and OH-.
- Option D: $O{H^ - } < NC{S^ - } < {I^ - }$ : This is incorrect because I- is a weaker field ligand than both OH- and NCS-.

Key Concepts:
- Spectrochemical Series
- Ligand Field Strength

Memory Aids:
- Remember the general trend: Halides < Oxygen donors < Nitrogen/Sulfur donors.

Quick Tips:
- Memorize the relative positions of common ligands in the spectrochemical series.

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Understanding +4 | -1

The existence of two different coloured complexes with the composition of

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

is due to:

View answer and explanation

Correct option: D

Ionisation isomersim

Option D is correct. The existence of two different colored complexes is due to Ionisation isomerism.

Solution approach

Correct Answer + Core Reason: D. Ionisation isomersim. Different ions are exchanged inside and outside the coordination sphere.

Step-by-Step Solution:

Step 1: The question states the existence of two different colored complexes with the composition of $\ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} ^ +$ .

Step 2: Consider the possible isomers. Since the question mentions different colors, it suggests different electronic environments around the Cobalt ion.

Step 3: Ionization isomerism involves the exchange of ligands inside the coordination sphere with ions outside the coordination sphere. This can lead to different electronic environments and thus different colors.

Step 4: Geometrical isomerism (cis/trans) could also lead to different colors, but Ionisation isomerism is a more direct explanation given the context.

Why other options are wrong:
- Option A: Linkage isomerism requires an ambidentate ligand, which is not explicitly mentioned.
- Option B: Geometrical isomerism is possible, but ionization isomerism is a better fit for the description.
- Option C: Coordination isomerism requires two or more complex ions, which is not the case here.

Key Concepts:
- Isomerism in coordination compounds
- Ionization isomerism
- Geometrical isomerism

Memory Aids:
- Ionization: Think of ions exchanging places.

Quick Tips:
- Consider the possible types of isomerism based on the given information.

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Remembering +4 | -1

Correct order of increasing field strength of ligands is

View answer and explanation

Correct option: A

{I^ - } < C{l^ - } < {H_2}O < C{N^ - }

Option A (

{I^ - } < C{l^ - } < {H_2}O < C{N^ - }

) is the correct order because it follows the spectrochemical series, which ranks ligands by their field strength.

Solution approach

Correct Answer + Core Reason: ${I^ - } < C{l^ - } < {H_2}O < C{N^ - }$ . The spectrochemical series determines the order of ligand field strength.

Step-by-Step Solution:

Step 1: Understand the spectrochemical series. This series arranges ligands in order of their ability to split the d-orbitals of a metal ion.

Step 2: Recall the general order of common ligands. Weak field ligands (like halides) are on one end, and strong field ligands (like cyanide) are on the other.

Step 3: Apply the series to the given ligands. ${I^-}$ and ${Cl^-}$ are weak field halides. ${H_2O}$ is a moderate field ligand, and ${CN^-}$ is a strong field ligand.

Step 4: Arrange the ligands in increasing order of field strength: ${I^ - } < C{l^ - } < {H_2}O < C{N^ - }$

Why other options are wrong:
- Option B: Incorrect because it places ${H_2O}$ before ${Cl^-}$ , which is not consistent with the spectrochemical series.
- Option C: Incorrect because it places both ${Cl^-}$ and ${I^-}$ out of order and ${H_2O}$ before them.
- Option D: Incorrect because it places ${Cl^-}$ before ${H_2O}$ and ${I^-}$ before ${CN^-}$ , which is not consistent with the spectrochemical series.

Key Concepts:
- Spectrochemical series
- Ligand field strength
- Coordination complexes

Memory Aids:
- Strong field ligands cause large splitting; weak field ligands cause small splitting.

Quick Tips:
- Memorize the general trend of the spectrochemical series to quickly identify the relative strengths of common ligands.

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Analyzing +4 | -1

Assertion :

\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{2+}

complex is less stable than

\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{2+}

complex.

Reason :

\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{2+}

complex shows a chelate effect.

View answer and explanation

Correct option: D

Assertion is false, but Reason is true

Option D is correct. The assertion is false because the complex with ethylenediamine (en) is more stable due to the chelate effect, and the reason is true because the complex with en exhibits the chelate effect.

Solution approach

Correct Answer + Core Reason: D. Assertion is false, but Reason is true

Step-by-Step Solution:

Step 1: Analyze the Assertion: $\Pt(en)_2Cl_2\^{2+}$ complex is less stable than $\Pt(NH_3)_4Cl_2\^{2+}$ complex.

Step 2: Consider the chelate effect: Ethylenediamine (en) is a bidentate ligand, meaning it can form two coordinate bonds with the central metal ion. This leads to the chelate effect, which increases the stability of the complex.

Step 3: The complex $\Pt(en)_2Cl_2\^{2+}$ has two en ligands, resulting in a significant chelate effect. The complex $\Pt(NH_3)_4Cl_2\^{2+}$ has only monodentate ligands (NH3), so it does not exhibit the chelate effect.

Step 4: Therefore, the complex $\Pt(en)_2Cl_2\^{2+}$ is more stable than $\Pt(NH_3)_4Cl_2\^{2+}$ . The assertion is false.

Step 5: Analyze the Reason: $\Pt(en)_2Cl_2\^{2+}$ complex shows a chelate effect.

Step 6: As explained above, the complex $\Pt(en)_2Cl_2\^{2+}$ does exhibit the chelate effect. The reason is true.

Step 7: Conclusion: The assertion is false, but the reason is true.

Why other options are wrong:
- Option A: The assertion is false.
- Option B: The assertion is false.
- Option C: The reason is true.

Key Concepts:
- Chelate effect
- Denticity of ligands
- Stability of coordination complexes

Memory Aids:
- Chelates are more stable.

Quick Tips:
- Remember that chelating ligands increase the stability of complexes due to the chelate effect.

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Understanding +4 | -1

Indicate the complex ion which shows geometrical isomerism

View answer and explanation

Correct option: A

{\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + }

Option A is correct because

{\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + }

can exist as cis and trans isomers, showing geometrical isomerism.

Solution approach

Correct Answer + Core Reason: ${\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + }$ . This complex can exist as cis and trans isomers.

Step-by-Step Solution:

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

Step 2: Analyze option A: ${\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + }$ has four water ligands and two chloride ligands. It can exist as cis and trans isomers.

Step 3: Analyze option B: $\left[ {pt{{\left( {N{H_3}} \right)}_3}Cl} \right]$ has three ammonia ligands and one chloride ligand. It cannot exhibit geometrical isomerism.

Step 4: Analyze option C: ${\left[ {CO{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ has six identical ammonia ligands. It cannot exhibit geometrical isomerism.

Step 5: Analyze option D: ${\left[ {CO{{\left( {CN} \right)}_5}\left( {NC} \right)} \right]^{3 - }}$ can show linkage isomerism but not geometrical isomerism since it has 5 identical CN ligands and one NC ligand.

Step 6: Conclusion: Only option A can show geometrical isomerism.

Why other options are wrong:
- Option B: $\left[ {pt{{\left( {N{H_3}} \right)}_3}Cl} \right]$ cannot exhibit geometrical isomerism.
- Option C: ${\left[ {CO{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ cannot exhibit geometrical isomerism.
- Option D: ${\left[ {CO{{\left( {CN} \right)}_5}\left( {NC} \right)} \right]^{3 - }}$ cannot exhibit geometrical isomerism.

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

Memory Aids:
- Geometrical isomers have different spatial arrangements.

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

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Understanding +4 | -1

From the stability constant (hypothetical values) given below, predict which is the strongest ligand?

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{u}^2} + 4\mathrm{CN}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}\mathrm{\;};\left( {K = 2.0 \times {{10}^{27}}} \right)

Option B (

\mathrm{C}{\mathrm{u}^2} + 4\mathrm{CN}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}\mathrm{\;};\left( {K = 2.0 \times {{10}^{27}}} \right)

) is correct because the stability constant (K) is highest for this reaction, indicating the strongest ligand.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{u}^2} + 4\mathrm{CN}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}\mathrm{\;};\left( {K = 2.0 \times {{10}^{27}}} \right)$ . The strongest ligand will have the highest stability constant.

Step-by-Step Solution:

Step 1: Understand the concept of stability constant (K). The stability constant is a measure of the strength of the bond between the metal ion and the ligand in a complex. A higher stability constant indicates a stronger bond and a more stable complex.

Step 2: Compare the stability constants (K) given for each reaction.

Step 3: Identify the reaction with the highest stability constant.
- A: $\mathrm{C}{\mathrm{u}^{2 + }} + 4\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}\mathrm{\;\;};\left( {K = 4.5 \times {{10}^{11}}} \right)$
- B: $\mathrm{C}{\mathrm{u}^2} + 4\mathrm{CN}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}\mathrm{\;};\left( {K = 2.0 \times {{10}^{27}}} \right)$
- C: $\mathrm{C}{\mathrm{u}^{2 + }} + 2\mathrm{en}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{en}} \right)}_2}} \right]^{2 + }}\mathrm{\;};\mathrm{\;\;}\left( {K = 3.0 \times {{10}^{15}}} \right)$
- D: $\mathrm{C}{\mathrm{u}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }}\mathrm{\;};\left( {K = 9.5 \times {{10}^8}} \right)$

Step 4: The highest stability constant is $2.0 \times {{10}^{27}}$ for the reaction with $CN^-$ .

Step 5: Conclude that $CN^-$ is the strongest ligand.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}^{2 + }} + 4\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}\mathrm{\;\;};\left( {K = 4.5 \times {{10}^{11}}} \right)$ : The stability constant is lower than that of $CN^-$ .
- Option C: $\mathrm{C}{\mathrm{u}^{2 + }} + 2\mathrm{en}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{en}} \right)}_2}} \right]^{2 + }}\mathrm{\;};\mathrm{\;\;}\left( {K = 3.0 \times {{10}^{15}}} \right)$ : The stability constant is lower than that of $CN^-$ .
- Option D: $\mathrm{C}{\mathrm{u}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }}\mathrm{\;};\left( {K = 9.5 \times {{10}^8}} \right)$ : The stability constant is lower than that of $CN^-$ .

Key Concepts:
- Stability constant (K)
- Ligand strength
- Complex stability

Memory Aids:
- Higher K = Stronger ligand = More stable complex

Quick Tips:
- The ligand with the highest stability constant is the strongest ligand.

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Understanding +4 | -1

The IUPAC name of

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]

is

View answer and explanation

Correct option: B

Hexaamine cobalt (III) tris (oxalato ) chromate(III)

Option B (Hexaamine cobalt (III) tris (oxalato ) chromate(III)) is correct because it correctly names both the cationic and anionic complexes, including the correct oxidation states of cobalt and chromium.

Solution approach

Correct Answer + Core Reason: Hexaamine cobalt (III) tris (oxalato ) chromate(III). The IUPAC naming convention requires naming the cation first, followed by the anion, and specifying the oxidation state of the metal in each complex.

Step-by-Step Solution:

Step 1: Identify the cationic complex: [Co(NH3)6].

Step 2: Name the cationic complex: Hexaamine cobalt(III). The oxidation state of Co is +3 because NH3 is neutral and the overall charge of the complex is +3.

Step 3: Identify the anionic complex: [Cr(C2O4)3].

Step 4: Name the anionic complex: Tris(oxalato)chromate(III). The oxidation state of Cr is +3 because each oxalate ligand has a -2 charge, and there are three of them, so the total negative charge is -6. The overall charge of the complex is -3, so Cr must be +3.

Step 5: Combine the names: Hexaamine cobalt (III) tris (oxalato ) chromate(III).

Why other options are wrong:
- Option A: Hexaamine cobalt (III) tris (oxalato ) chromium: Missing the oxidation state of chromium and incorrect name for the anionic complex (should be chromate).
- Option C: Hexaamine cobalt tris (oxalato ) chromium(III): Missing the oxidation state of cobalt and incorrect name for the anionic complex (should be chromate).
- Option D: Hexaamine cobalt (III) chromium (III) oxalate: Incorrect naming of the anionic complex (should include "tris(oxalato)" and "chromate").

Key Concepts:
- IUPAC nomenclature of coordination compounds
- Naming cationic and anionic complexes
- Oxidation state determination

Memory Aids:
- Cation is named before the anion.
- Use "ate" suffix for anionic complexes.

Quick Tips:
- Always determine the oxidation state of the metal in both cationic and anionic complexes.

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Analyzing +4 | -1

Consider the following statements
I. Chain and position isomerism are not possible together between two isomers
II. Tautomerism is a chemical phenomenon which is catalysed by acid as well as base
III. Tautomers are always metamers
IV. Tautomers are always functional isomers
Select the correct answer by using the codes given below

View answer and explanation

Correct option: D

I, II and IV are correct

Option D (I, II and IV are correct) is correct because chain and position isomerism are mutually exclusive, tautomerism is acid/base catalyzed, and tautomers are functional isomers.

Solution approach

Correct Answer + Core Reason: I, II and IV are correct. Chain and position isomerism are not possible together, tautomerism is acid/base catalyzed, and tautomers are functional isomers.

Step-by-Step Solution:

Step 1: Analyze statement I: Chain and position isomerism involve different arrangements of carbon atoms and substituents, respectively. They cannot occur simultaneously between two isomers.

Step 2: Analyze statement II: Tautomerism involves the migration of a proton, which can be catalyzed by both acids and bases.

Step 3: Analyze statement III: Tautomers are not always metamers. Metamers have different alkyl groups attached to the same functional group, which is not always the case for tautomers.

Step 4: Analyze statement IV: Tautomers are functional isomers because they have different functional groups (e.g., keto and enol forms).

Step 5: Based on the analysis, statements I, II, and IV are correct.

Why other options are wrong:
- Option A: Only III is correct: This is incorrect because statements I, II, and IV are also correct.
- Option B: III and IV are correct: This is incorrect because statement I and II are also correct, and III is incorrect.
- Option C: I, II and III are correct: This is incorrect because statement III is incorrect.

Key Concepts:
- Isomerism
- Chain isomerism
- Position isomerism
- Tautomerism
- Metamers
- Functional isomers
- Acid-base catalysis

Memory Aids:
- Tautomers are functional isomers.
- Chain and position isomers are mutually exclusive.

Quick Tips:
- Understand the definitions of different types of isomers.
- Remember that tautomerism involves proton migration and can be catalyzed by acids and bases.

116 Medium

116 of 133Question permalink

Remembering +4 | -1

The common name of

{K}\left[{PtCl}_3\left(n^2 {C}_2 {H}_4\right)\right.]

is

View answer and explanation

Correct option: B

Zeise's salt

Option B (Zeise's salt) is correct because

{K}{PtCl}_3\left(n^2 {C}_2 {H}_4\right)

is the chemical formula for Zeise's salt, a classic example of an organometallic compound.

Solution approach

Correct Answer + Core Reason: Zeise's salt. ${K}{PtCl}_3\left(n^2 {C}_2 {H}_4\right)$ is the formula for Zeise's salt.

Step-by-Step Solution:

Step 1: Recognize the formula: The formula ${K}{PtCl}_3\left(n^2 {C}_2 {H}_4\right)$ represents a specific chemical compound.

Step 2: Recall the common name: This compound is known as Zeise's salt.

Why other options are wrong:
- Option A: Potassium salt: While Zeise's salt is a potassium salt, this is not its common name.
- Option C: Complex salt: While Zeise's salt is a complex salt, this is not its common name.
- Option D: None of these: Zeise's salt is a well-known compound with a specific common name.

Key Concepts:
- Zeise's salt
- Organometallic compounds
- Coordination complexes

Memory Aids:
- Associate the formula ${K}{PtCl}_3\left(n^2 {C}_2 {H}_4\right)$ with Zeise's salt.

Quick Tips:
- Learn the common names and formulas of important organometallic compounds.

117 Medium

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Application +4 | -1

The EAN of platinum in potassium hexachloroplatinate (IV) is:

View answer and explanation

Correct option: B

86

Option B (86) is correct because the EAN is calculated by adding the number of electrons in the metal ion and the number of electrons donated by the ligands.

Solution approach

Correct Answer + Core Reason: 86. The EAN is calculated by adding the electrons of the metal and ligands to reach the nearest noble gas configuration.

Step-by-Step Solution:

Step 1: Identify the metal and its oxidation state: Platinum (Pt) in Potassium hexachloroplatinate (IV) has an oxidation state of +4, so it is $Pt^{4+}$ .

Step 2: Determine the number of electrons in the neutral metal atom: Platinum (Pt) has an atomic number of 78, so it has 78 electrons.

Step 3: Determine the number of electrons in the metal ion: $Pt^{4+}$ has lost 4 electrons, so it has 78 - 4 = 74 electrons.

Step 4: Determine the number of ligands and their electron donation: There are 6 chloride ligands ( $Cl^-$ ), each donating 2 electrons, for a total of 6 * 2 = 12 electrons.

Step 5: Calculate the EAN: EAN = Number of electrons in metal ion + Number of electrons donated by ligands = 74 + 12 = 86.

Why other options are wrong:
- Option A (46): This is incorrect as it doesn't account for the electrons from the metal and ligands properly.
- Option C (36): This is incorrect as it doesn't account for the electrons from the metal and ligands properly.
- Option D (84): This is incorrect as it doesn't account for the electrons from the metal and ligands properly.

Key Concepts:
- Effective Atomic Number (EAN) rule
- Coordination complexes
- Oxidation states
- Ligand electron donation

Memory Aids:
- EAN = (Atomic number of metal - oxidation state) + (2 * coordination number)

Quick Tips:
- Remember to consider the charge of the metal ion when calculating the number of electrons.
- Each ligand donates a certain number of electrons, usually 2 for simple ligands like Cl-, NH3, etc.

118 Medium

118 of 133Question permalink

Application +4 | -1

Spin only magnetic moment value of

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

ion is

View answer and explanation

Correct option: A

2.83 BM

Option A (2.83 BM) is correct because

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

has a

Mn^{3+}

ion with a

d^4

configuration. Since

CN^-

is a strong field ligand, it will pair up the electrons, resulting in 2 unpaired electrons and a magnetic moment of approximately 2.83 BM.

Solution approach

Correct Answer + Core Reason: 2.83 BM. The magnetic moment is determined by the number of unpaired electrons in the complex.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Mn in the complex ${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }}$ . The overall charge of the complex is -3. Each CN ligand has a charge of -1. So, $Mn + 6(-1) = -3$ , which means $Mn = +3$ .

Step 2: Determine the electronic configuration of $Mn^{3+}$ . Mn has an atomic number of 25, so its electronic configuration is $[Ar] 3d^5 4s^2$ . When it loses 3 electrons to become $Mn^{3+}$ , the electronic configuration becomes $[Ar] 3d^4$ .

Step 3: Consider the ligand field splitting. $CN^-$ is a strong field ligand. In an octahedral complex, the d-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy). Because $CN^-$ is a strong field ligand, it will cause pairing of electrons.

Step 4: Fill the d-orbitals. For a $d^4$ ion in a strong field (low spin), the electrons will fill the orbitals as $t_{2g}^4 e_g^0$ . This gives 2 unpaired electrons.

Step 5: Calculate the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)} BM$ , where n is the number of unpaired electrons. Since n = 2, $\mu = \sqrt{2(2+2)} BM = \sqrt{8} BM \approx 2.83 BM$ .

Final Answer: 2.83 BM

Why other options are wrong:
- Option B (3.87 BM): This would be the magnetic moment if there were approximately 3 unpaired electrons. This is incorrect because there are 2 unpaired electrons.
- Option C (4.9 BM): This would be the magnetic moment if there were approximately 4 unpaired electrons. This would be the case if $CN^-$ was a weak field ligand, but it is a strong field ligand.
- Option D (1.73 BM): This would be the magnetic moment if there were approximately 1 unpaired electron. This is incorrect because there are 2 unpaired electrons.

Key Concepts:
- Coordination Chemistry
- Crystal Field Theory
- Spin-only Magnetic Moment
- Electronic Configuration

Memory Aids:
- Strong field ligands lead to low-spin complexes, while weak field ligands lead to high-spin complexes.

Quick Tips:
- Always determine the oxidation state and electronic configuration of the metal ion first.
- Consider the ligand strength to determine whether the complex is high-spin or low-spin.
- Use the formula $\mu = \sqrt{n(n+2)} BM$ to calculate the spin-only magnetic moment.

119 Medium

119 of 133Question permalink

Application +4 | -1

Which one of the following has a square planar geometry?

\left( {At.\mathrm{ }No.\mathrm{ }Fe = 26,\mathrm{ }Co = 27,\mathrm{ }Ni = 28,\mathrm{ }Pt = 78} \right)

View answer and explanation

Correct option: D

{\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }}

Option D ([PtCl4]^2-) is correct because Pt(II) complexes with strong field ligands like Cl- tend to form square planar geometries due to dsp² hybridization.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }}$ . Platinum (II) complexes are generally square planar.

Step-by-Step Solution:

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

Step 2: Option A: ${\left[ {\mathrm{CoC}{\mathrm{l}_4}} \right]^{2 - }}$ . Co is in +2 oxidation state. It has $d^7$ configuration. Since Cl- is a weak field ligand, it forms a tetrahedral complex.

Step 3: Option B: ${\left[ {\mathrm{FeC}{\mathrm{l}_4}} \right]^{2 - }}$ . Fe is in +2 oxidation state. It has $d^6$ configuration. Since Cl- is a weak field ligand, it forms a tetrahedral complex.

Step 4: Option C: ${\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }}$ . Ni is in +2 oxidation state. It has $d^8$ configuration. Since Cl- is a weak field ligand, it forms a tetrahedral complex.

Step 5: Option D: ${\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }}$ . Pt is in +2 oxidation state. Pt is a 5d series element and forms square planar complexes with strong field ligands like Cl-.

Why other options are wrong:
- Option A: Cobalt tetrachloride is tetrahedral.
- Option B: Iron tetrachloride is tetrahedral.
- Option C: Nickel tetrachloride is tetrahedral.

Key Concepts:
- Square planar geometry
- Tetrahedral geometry
- Coordination complexes
- Electronic configuration

Memory Aids:
- Pt(II) complexes are generally square planar.

Quick Tips:
- Platinum complexes are often square planar, especially with strong field ligands.

120 Medium

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Understanding +4 | -1

The IUPAC name of

{\left[ {\mathrm{Ni}{{\left( {\mathrm{PP}{\mathrm{h}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^{2 + }}

is

View answer and explanation

Correct option: B

Dichloro bis (triphenylphosphine)nickel(II)

Option B (Dichloro bis (triphenylphosphine)nickel(II)) is correct because it correctly names the compound, indicating the dichloro and bis(triphenylphosphine) ligands, and the oxidation state of nickel.

Solution approach

Correct Answer + Core Reason: Dichloro bis (triphenylphosphine)nickel(II). The IUPAC naming convention requires naming the ligands alphabetically, using "bis" for two complex ligands, and specifying the oxidation state of the metal.

Step-by-Step Solution:

Step 1: Identify the ligands: Dichloro (Cl) and bis(triphenylphosphine) (PPh3).

Step 2: Identify the central metal: Nickel (Ni).

Step 3: Determine the oxidation state of nickel: Since the overall charge of the complex is +2, and each chloride ligand has a -1 charge, the nickel must have a +2 oxidation state.

Step 4: Name the complex: Dichloro bis (triphenylphosphine)nickel(II).

Why other options are wrong:
- Option A: Bis-dichloro (triphenylphosphine)nickel(II): Incorrect naming convention; dichloro should come first alphabetically.
- Option C: Dichloro triphenylphosphine nickel(II): Missing the "bis" prefix for two triphenylphosphine ligands.
- Option D: Triphenyl phosphine nickel (II) dichloride: Incorrect order of naming and missing the "bis" prefix.

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

Memory Aids:
- Alphabetical order of ligands.
- Use "bis", "tris", etc. for multiple complex ligands.

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

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