The IUPAC name of ${{\rm{K}}_2}\left[ {{\rm{PtC}}{{\rm{l}}_6}} \right]$ is

Potassium hexachloroplatinum(IV)

The name of ${\left[ {{\rm{Pt}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}{\rm{C}}{{\rm{l}}_2}} \right]^{2 + }}{\rm{\;}},{\left[ {{\rm{PtC}}{{\rm{l}}_4}} \right]^{2 - }}$ is

Tetrachloroplatinum (II) tetrammineplatinate

Tetrachloroplatinum (II) dichlorotetraammine platinate

The IUPAC name of the complex $\left[ {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}{\rm{C}}{{\rm{l}}_2}} \right]{\rm{Cl\;}}$ is

Dichloro tetraammine cobalt (III) chloride

Tetraammine dichloro cobalt (II) chloride

Tetraammine dichloro cobalt (IV) chloride

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

31 Easy

31 of 98Question permalink

Application +4 | -1

Oxidation number of

\mathrm{Fe}

in

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

is:

View answer and explanation

Correct option: A

3

Option A (3) is correct because the oxidation state of Fe is calculated based on the charge of the complex and the ligands.

Solution approach

Correct Answer + Core Reason: 3. The oxidation number of Fe is calculated by considering the charges of the other ions in the complex.

Step-by-Step Solution:

Step 1: Write the formula of the complex: ${\mathrm{K}_3}[\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}]$

Step 2: Determine the charge of the complex ion. Since there are three ${\mathrm{K}^ + }$ ions, the complex ion has a charge of -3: ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$

Step 3: Determine the charge of the cyanide ligands. Each ${\mathrm{CN}^ - }$ ligand has a charge of -1, and there are six of them, so the total charge from the ligands is -6.

Step 4: Let x be the oxidation number of Fe. Then, $x + (-6) = -3$

Step 5: Solve for x: $x = -3 + 6 = +3$

Step 6: Final Answer: The oxidation number of Fe is +3.

Why other options are wrong:
- Option B: An oxidation state of +2 would imply an incorrect charge balance within the complex ion.
- Option C: An oxidation state of +10 is impossible for iron in this complex.
- Option D: An oxidation state of +1 would imply an incorrect charge balance within the complex ion.

Key Concepts:
- Oxidation number
- Coordination complexes
- Charge balance

Memory Aids:
- The sum of oxidation numbers in a complex equals the overall charge of the complex.

Quick Tips:
- Always consider the charges of all ligands and counterions when determining oxidation numbers.

32 Easy

32 of 98Question permalink

Application +4 | -1

Which follows EAN rule?

View answer and explanation

Correct option: D

All are correct

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

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

,

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

, and

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

, follow the Effective Atomic Number (EAN) rule.

Solution approach

Correct Answer + Core Reason: All are correct. All the given complexes follow the EAN rule.

Step-by-Step Solution:

Step 1: Define the EAN rule. The Effective Atomic Number (EAN) rule states that the metal atom in a complex tends to surround itself with sufficient ligands such that the resulting effective atomic number is numerically equal to the atomic number of the next noble gas.

Step 2: Check if $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ follows the EAN rule. Fe has an atomic number of 26. Each CO contributes 2 electrons. So, EAN = 26 + 5(2) = 26 + 10 = 36, which is the atomic number of Krypton (Kr). Therefore, it follows the EAN rule.

Step 3: Check if $\mathrm{Ni}{\left( {\mathrm{CO}} \right)_4}$ follows the EAN rule. Ni has an atomic number of 28. Each CO contributes 2 electrons. So, EAN = 28 + 4(2) = 28 + 8 = 36, which is the atomic number of Krypton (Kr). Therefore, it follows the EAN rule.

Step 4: Check if ${\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ follows the EAN rule. Fe has an oxidation state of +2. So, $Fe^{2+}$ has 24 electrons. Each CN contributes 2 electrons. So, EAN = 24 + 6(2) = 24 + 12 = 36, which is the atomic number of Krypton (Kr). Therefore, it follows the EAN rule.

Final Answer: All are correct

Why other options are wrong:
- **Option A ( $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ ): This is not the only correct answer, as the other complexes also follow the EAN rule.
- Option B ( $\mathrm{Ni}{\left( {\mathrm{CO}} \right)_4}$ ): This is not the only correct answer, as the other complexes also follow the EAN rule.
- Option C ( ${\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ ): This is not the only correct answer, as the other complexes also follow the EAN rule.

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

Memory Aids:
- EAN = (Atomic number of metal) - (Oxidation state of metal) + (Electrons donated by ligands)

Quick Tips:**
- Remember that each neutral ligand like CO and $NH_3$ donates 2 electrons.
- Remember that each anionic ligand like CN and Cl donates 2 electrons.

33 Easy

33 of 98Question permalink

Remembering +4 | -1

Ziegler-Natta catalyst is

View answer and explanation

Correct option: C

\mathrm{[A}{\mathrm{l}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_6}} \right)_6} + \mathrm{TiC}{\mathrm{l}_4}]

Option C (

\mathrm{[A}{\mathrm{l}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_6}} \right)_6} + \mathrm{TiC}{\mathrm{l}_4}]

) is correct because Ziegler-Natta catalysts typically consist of a transition metal compound (like

TiCl_4

) and an organometallic compound (like triethylaluminum).

Solution approach

Correct Answer + Core Reason: $\mathrm{[A}{\mathrm{l}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_6}} \right)_6} + \mathrm{TiC}{\mathrm{l}_4}]$ . Ziegler-Natta catalysts are typically a combination of a transition metal compound and an organoaluminum compound.

Step-by-Step Solution:

Step 1: Recall the definition and components of a Ziegler-Natta catalyst.

Step 2: Ziegler-Natta catalysts are used for the polymerization of alkenes.

Step 3: They typically consist of a transition metal halide (e.g., $TiCl_4$ ) and an organometallic compound, often an alkylaluminum compound (e.g., $Al(C_2H_5)_3$ ).

Step 4: The formula $\mathrm{[A}{\mathrm{l}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_6}} \right)_6} + \mathrm{TiC}{\mathrm{l}_4}]$ represents such a combination.

Why other options are wrong:
- Option A: ${\left( {\mathrm{P}{\mathrm{h}_3}\mathrm{P}} \right)_3}\mathrm{RhCl}$ : Wilkinson's catalyst, used for hydrogenation.
- Option B: $\mathrm{K}\left[ {\mathrm{PtC}{\mathrm{l}_3}\left( {{\mathrm{C}_2}{\mathrm{H}_4}} \right)} \right]$ : Zeise's salt, an example of an organometallic compound but not a Ziegler-Natta catalyst.
- Option D: $\left[ {\mathrm{Fe}{{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)}_2}} \right]$ : An organometallic compound, but not typically used as a Ziegler-Natta catalyst on its own.

Key Concepts:
- Ziegler-Natta catalyst
- Polymerization of alkenes
- Organometallic compounds

Memory Aids:
- Ziegler-Natta: Think "Zippy Polymerization" to remember its use in polymer synthesis.

Quick Tips:
- Recognize the common components of Ziegler-Natta catalysts: transition metal halides and organoaluminum compounds.

34 Easy

34 of 98Question permalink

Application +4 | -1

Which of the following shows

ds{p^2}

hybridisation?

View answer and explanation

Correct option: D

\mathrm{PtCl}_4^{2 - }

Option D (PtCl4^2-) is correct because Pt in this complex has a coordination number of 4 and forms a square planar complex, which requires dsp² hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{PtCl}_4^{2 - }$ . Platinum in this complex exhibits $ds{p^2}$ hybridization, leading to a square planar geometry.

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: $\mathrm{NiCl}_4^{2 - }$ . Ni is in +2 oxidation state. It has $d^8$ configuration. Since Cl- is a weak field ligand, it does not cause pairing. Thus, it is $sp^3$ hybridized and tetrahedral.

Step 3: Option B: $\mathrm{SC}{\mathrm{l}_4}$ . S has 4 bond pairs and 1 lone pair, so it is $sp^3d$ hybridized and seesaw shaped.

Step 4: Option C: $\mathrm{NH}_4^ +$ . N has 4 bond pairs and no lone pairs, so it is $sp^3$ hybridized and tetrahedral.

Step 5: Option D: $\mathrm{PtCl}_4^{2 - }$ . Pt is in +2 oxidation state. Pt is a 5d series element and forms square planar complexes with strong field ligands like Cl-. It has $ds{p^2}$ hybridization.

Why other options are wrong:
- Option A: Nickel tetrachloride is tetrahedral and $sp^3$ hybridized.
- Option B: Sulfur tetrachloride is seesaw shaped and $sp^3d$ hybridized.
- Option C: Ammonium ion is tetrahedral and $sp^3$ hybridized.

Key Concepts:
- Hybridization of coordination complexes
- Square planar geometry
- Tetrahedral geometry
- Spectrochemical series

Memory Aids:
- Strong field ligands cause pairing of electrons.
- Pt(II) complexes are generally square planar.

Quick Tips:
- Determine the oxidation state and electronic configuration of the metal ion.
- Consider the ligand strength to determine pairing and hybridization.

35 Easy

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Understanding +4 | -1

Which of the following is an organometallic compound?

View answer and explanation

Correct option: D

\mathrm{Ti}{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}

Option D (

\mathrm{Ti}{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}

) is correct because it contains a direct titanium-carbon bond, which is the defining characteristic of an organometallic compound.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ti}{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}$ . This compound contains a direct titanium-carbon bond.

Step-by-Step Solution:

Step 1: Define Organometallic Compound: An organometallic compound contains at least one metal-carbon bond.

Step 2: Analyze the options:
- A. $\mathrm{Ti}{\left( {\mathrm{O}{\mathrm{C}_6}{\mathrm{H}_5}} \right)_4}$ : Titanium tetraphenoxide contains Ti-O bonds, not Ti-C bonds. Hence, not organometallic.
- B. $\mathrm{Ti}{\left( {\mathrm{OCOC}{\mathrm{H}_3}} \right)_4}$ : Titanium tetraacetate contains Ti-O bonds, not Ti-C bonds. Hence, not organometallic.
- C. $\mathrm{Ti}{\left( {\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}$ : Titanium tetraethoxide contains Ti-O bonds, not Ti-C bonds. Hence, not organometallic.
- D. $\mathrm{Ti}{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}$ : Tetraethyltitanium contains Ti-C bonds, hence organometallic.

Step 3: Conclusion: Tetraethyltitanium is an organometallic compound.

Why other options are wrong:
- Option A: $\mathrm{Ti}{\left( {\mathrm{O}{\mathrm{C}_6}{\mathrm{H}_5}} \right)_4}$ : Contains Ti-O bonds, not Ti-C bonds.
- Option B: $\mathrm{Ti}{\left( {\mathrm{OCOC}{\mathrm{H}_3}} \right)_4}$ : Contains Ti-O bonds, not Ti-C bonds.
- Option C: $\mathrm{Ti}{\left( {\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}$ : Contains Ti-O bonds, not Ti-C bonds.

Key Concepts:
- Organometallic compounds
- Metal-carbon bond
- Titanium compounds

Memory Aids:
- Remember that organometallic compounds must have a direct metal-carbon bond.

Quick Tips:
- Identify the presence of a metal-carbon bond to determine if a compound is organometallic.

36 Easy

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Understanding +4 | -1

Which of the following is not considered as an organometallic compound?

View answer and explanation

Correct option: B

cis -

platin

Option B (

cis -

platin) is correct because it does not have a direct metal-carbon bond, which is a requirement for a compound to be classified as organometallic.

Solution approach

Correct Answer + Core Reason: $cis -$ platin. $cis -$ platin does not contain a direct metal-carbon bond.

Step-by-Step Solution:

Step 1: Define Organometallic Compound: An organometallic compound contains at least one metal-carbon bond.

Step 2: Analyze the options:
- A. Grignard reagent ( $RMgX$ ): Contains a Mg-C bond, hence organometallic.
- B. $cis -$ platin ( $[PtCl_2(NH_3)_2]$ ): Contains Pt-Cl and Pt-N bonds, but no Pt-C bond. Hence, not organometallic.
- C. Zeise’s salt ( $K[PtCl_3(\eta^2-C_2H_4)])$ : Contains a Pt-C bond, hence organometallic.
- D. Ferrocene ( $[Fe(\eta^5-C_5H_5)_2]$ ): Contains Fe-C bonds, hence organometallic.

Step 3: Conclusion: $cis -$ platin is not an organometallic compound.

Why other options are wrong:
- Option A: Grignard reagent: Grignard reagents have a direct carbon-magnesium bond, making them organometallic.
- Option C: Zeise’s salt: Zeise's salt has a platinum-carbon bond, making it organometallic.
- Option D: Ferrocene: Ferrocene has iron-carbon bonds, making it organometallic.

Key Concepts:
- Organometallic compounds
- Metal-carbon bond
- Grignard reagent
- Zeise's salt
- Ferrocene
- $cis -$ platin

Memory Aids:
- Remember that organometallic compounds must have a direct metal-carbon bond.

Quick Tips:
- Identify the presence of a metal-carbon bond to determine if a compound is organometallic.

37 Easy

37 of 98Question permalink

Remembering +4 | -1

In

\mathrm{SCN}

ligand if

\mathrm{N}

is attached to central atom, the name of ligand is:

View answer and explanation

Correct option: A

Thiocyanato-N

Option A (Thiocyanato-N) is correct because when N is attached to the central atom in SCN ligand, it's named Thiocyanato-N.

Solution approach

Correct Answer + Core Reason: Thiocyanato-N. The name reflects the nitrogen atom's linkage to the central metal ion.

Step-by-Step Solution:

Step 1: Identify the ligand: The ligand is SCN.

Step 2: Determine the donor atom: The question specifies that N is attached to the central atom.

Step 3: Name the ligand: When N is the donor atom, the ligand is named Thiocyanato-N.

Why other options are wrong:
- Option B: Cyanato-N is incorrect because the ligand is SCN, not CNO.
- Option C: Thiocyanato-S is incorrect because the question specifies that N is attached, not S.
- Option D: Cyanato-S is incorrect because the ligand is SCN, not CNO, and S is not the attached atom.

Key Concepts:
- Nomenclature of coordination compounds
- Ligand naming conventions
- Ambidentate ligands (SCN can bind through S or N)

Memory Aids:
- SCN: Thiocyanato. If N binds, add -N; if S binds, add -S.

Quick Tips:
- Pay close attention to the donor atom specified in the question.

38 Easy

38 of 98Question permalink

Understanding +4 | -1

The IUPAC name of

{\mathrm{K}_2}\left[ {\mathrm{PtC}{\mathrm{l}_6}} \right]

is

View answer and explanation

Correct option: B

Potassium hexachloroplatinate (IV)

Option B (Potassium hexachloroplatinate (IV)) is correct because it correctly names the compound, indicating the potassium cation, the hexachloro ligands, the platinate anion, and the oxidation state of platinum.

Solution approach

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

Step-by-Step Solution:

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

Step 2: Identify the ligands: Chloro (Cl) with a charge of -1. There are six chloro ligands.

Step 3: Identify the central metal: Platinum (Pt).

Step 4: Determine the oxidation state of platinum: The complex ion has a charge of -2. Since each chloride ligand has a -1 charge, the total charge from the ligands is -6. Therefore, the oxidation state of platinum must be +4 to result in a -2 overall charge for the complex ion.

Step 5: Name the complex: Potassium hexachloroplatinate (IV).

Why other options are wrong:
- Option A: Hexachloroplatinate potassium: Incorrect order of naming (cation should come first).
- Option C: Potassium hexachloroplatinate: Missing the oxidation state of platinum (IV).
- Option D: Potassium hexachloroplatinum(IV): Incorrect name for the complex ion (should be platinate).

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.

39 Easy

39 of 98Question permalink

Understanding +4 | -1

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

complex gives

View answer and explanation

Correct option: D

5 ions

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

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

, and four chloride ions,

4\mathrm{Cl}^-

.

Solution approach

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

Step-by-Step Solution:

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

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

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

Why other options are wrong:
- Option A: Incorrect. It suggests 4 ions, which is not the correct number of ions produced upon dissociation.
- Option B: Incorrect. It suggests 3 ions, which is not the correct number of ions produced upon dissociation.
- Option C: Incorrect. It suggests 2 ions, which is not the correct number of ions produced upon dissociation.

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

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

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

40 Easy

40 of 98Question permalink

Understanding +4 | -1

A solution of potassium ferrocyanide would contains-ions

View answer and explanation

Correct option: D

5

Option D is correct because potassium ferrocyanide, K4[Fe(CN)6], dissociates into 5 ions: 4 K+ ions and 1 [Fe(CN)6]4- ion.

Solution approach

Correct Answer + Core Reason: 5. Potassium ferrocyanide dissociates into 5 ions in solution.

Step-by-Step Solution:

Step 1: Write the formula for potassium ferrocyanide: $K_4[Fe(CN)_6]$ .

Step 2: Determine the ions formed upon dissociation in water. $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$

Step 3: Count the number of ions. There are 4 potassium ions ( $K^+$ ) and 1 ferrocyanide ion ( $[Fe(CN)_6]^{4-}$ ), for a total of 5 ions.

Why other options are wrong:
- Option A: 2 is incorrect. It only considers the potassium and ferrocyanide ions as a single entity each.
- Option B: 3 is incorrect. It doesn't accurately count all the ions.
- Option C: 4 is incorrect. It doesn't accurately count all the ions.

Key Concepts:
- Dissociation of complex salts
- Ion formation in solution

Memory Aids:
- Complex ions remain intact during dissociation.

Quick Tips:
- Carefully count the number of ions formed upon dissociation of a complex salt.

41 Easy

41 of 98Question permalink

Understanding +4 | -1

The shape of the complex

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

is:

View answer and explanation

Correct option: D

Linear

The correct answer is D (Linear) because the complex

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

has a coordination number of 2, resulting in a linear geometry.

Solution approach

Correct Answer + Core Reason: Linear. Silver (Ag+) with a coordination number of 2 typically forms linear complexes.

Step-by-Step Solution:

Step 1: Identify the central metal ion and its oxidation state. Here, it's Ag+.

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

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

Step 4: Predict the geometry based on the coordination number. A coordination number of 2 typically results in a linear geometry.

Step 5: Conclusion: The shape of the complex is linear.

Why other options are wrong:
- Option A (Octahedral): Octahedral geometry requires a coordination number of 6.
- Option B (Square planar): Square planar geometry typically occurs with a coordination number of 4, especially with strong field ligands.
- Option C (Tetrahedral): Tetrahedral geometry requires a coordination number of 4.

Key Concepts:
- Coordination number
- Geometry of coordination complexes

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

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

42 Easy

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Understanding +4 | -1

The compound

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

and

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

are examples of:

View answer and explanation

Correct option: B

Linkage isomers

Option B is correct. The complexes are linkage isomers because the

NO_2

ligand can coordinate through either the nitrogen (nitro) or the oxygen (nitrito) atom.

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the two complexes: $\Co(NO_2)(NH_3)_5\Cl_2$ and $\Co(ONO)(NH_3)_5\Cl_2$ .

Step 2: Observe the difference between the complexes: The only difference is the way the $NO_2$ ligand is attached to the cobalt ion. In one complex, it is attached through the nitrogen atom (nitro), and in the other, it is attached through the oxygen atom (nitrito).

Step 3: Recall the definition of linkage isomerism: Linkage isomerism occurs when an ambidentate ligand coordinates to the metal ion through different atoms.

Step 4: Since $NO_2$ is an ambidentate ligand and it is coordinating through different atoms in the two complexes, they are linkage isomers.

Why other options are wrong:
- Option A: Geometrical isomers have different spatial arrangements of ligands, which is not the case here.
- Option C: Ligand isomers are not a standard type of isomerism.
- Option D: Ionization isomers involve the exchange of ligands inside and outside the coordination sphere, which is not the case here.

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

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

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

43 Easy

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Understanding +4 | -1

The name of

{\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]^{2 + }}\mathrm{\;},{\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }}

is

View answer and explanation

Correct option: A

Tetramminedichloroplatinum(IV) tetrachloro platinate(II)

Option A (Tetramminedichloroplatinum(IV) tetrachloro platinate(II)) is correct because it accurately names both the cationic and anionic complexes, including the correct oxidation states of platinum in each.

Solution approach

Correct Answer + Core Reason: Tetramminedichloroplatinum(IV) tetrachloro platinate(II). 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: [Pt(NH3)4Cl2]2+.

Step 2: Name the cationic complex: Tetraammine dichloro platinum(IV). The oxidation state of Pt is +4 because NH3 is neutral and Cl is -1, so +4 - 2 = +2 (the overall charge).

Step 3: Identify the anionic complex: [PtCl4]2−.

Step 4: Name the anionic complex: Tetrachloro platinate(II). The oxidation state of Pt is +2 because each Cl is -1, so +2 - 4 = -2 (the overall charge).

Step 5: Combine the names: Tetramminedichloroplatinum(IV) tetrachloro platinate(II).

Why other options are wrong:
- Option B: Dichloroplatinum (IV) tetrachloroplatinate: Missing the ammine ligands and incorrect naming of the first complex.
- Option C: Tetrachloroplatinum (II) tetrammineplatinate: Incorrect order of naming the complexes (anion before cation) and missing the dichloro ligands in the first complex.
- Option D: Tetrachloroplatinum (II) dichlorotetraammine platinate: Incorrect order of naming the complexes (anion before cation).

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.

44 Easy

44 of 98Question permalink

Application +4 | -1

Among the following, identify the species with an atom of +6 oxidation state:

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the oxidation state of Cr in this compound is +6.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ . The oxidation state of Cr is +6, calculated based on the oxidation states of O and Cl.

Step-by-Step Solution:

Step 1: Write the formula of the compound: $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$

Step 2: Assign oxidation numbers to known elements. Oxygen (O) has an oxidation number of -2, and chlorine (Cl) has an oxidation number of -1.

Step 3: Let x be the oxidation number of Cr. Then, $x + 2(-2) + 2(-1) = 0$

Step 4: Solve for x: $x - 4 - 2 = 0$ , so $x = +6$

Step 5: Final Answer: The oxidation state of Cr is +6.

Why other options are wrong:
- Option A: In ${\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }$ , the oxidation state of Mn is +7.
- Option B: In ${\left[ {\mathrm{Cr}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}$ , the oxidation state of Cr is +3.
- Option C: In ${\left[ {\mathrm{Ni}{\mathrm{F}_6}} \right]^{2 - }}$ , the oxidation state of Ni is +4.

Key Concepts:
- Oxidation number
- Rules for assigning oxidation numbers

Memory Aids:
- Oxygen is usually -2, halogens are usually -1 (except when bonded to oxygen or fluorine).

Quick Tips:
- Remember that the sum of oxidation numbers in a neutral compound is zero, and in an ion, it equals the charge of the ion.

45 Easy

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Analyzing +4 | -1

Which complex is likely to show optical activity?

View answer and explanation

Correct option: C

Cis - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}

Option C (

Cis - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}

) is correct because the cis isomer of this complex lacks a plane of symmetry and is therefore chiral.

Solution approach

Correct Answer + Core Reason: $Cis - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}$ . The cis isomer of this complex is chiral and lacks a plane of symmetry.

Step-by-Step Solution:

Step 1: Analyze each complex for the presence of a plane of symmetry.

Step 2: Option A: $Trans - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]^ + }$ : The trans isomer has a plane of symmetry and is therefore achiral.

Step 3: Option B: ${\left[ {\mathrm{Cr}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}$ : This complex is octahedral and has multiple planes of symmetry, making it achiral.

Step 4: Option C: $Cis - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}$ : The cis isomer does not have a plane of symmetry and is therefore chiral. It exhibits optical activity.

Step 5: Option D: $Trans - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}$ : The trans isomer has a plane of symmetry and is therefore achiral.

Step 6: Final Answer: $Cis - {\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}{{\left( {\mathrm{en}} \right)}_2}} \right]^{3 + }}$ is likely to show optical activity.

Why other options are wrong:
- Option A: The trans isomer has a plane of symmetry and is achiral.
- Option B: The complex is octahedral and has multiple planes of symmetry, making it achiral.
- Option D: The trans isomer has a plane of symmetry and is achiral.

Key Concepts:
- Optical activity
- Chirality
- Plane of symmetry
- Coordination complexes
- Cis and trans isomers

Memory Aids:
- A molecule is chiral if it is non-superimposable on its mirror image.
- A molecule with a plane of symmetry is achiral.

Quick Tips:
- Draw the structures of the isomers and look for a plane of symmetry. If there is no plane of symmetry, the molecule is likely chiral and optically active.

46 Easy

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Remembering +4 | -1

The ion which exhibits green colour

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{i}^{2 + }}

Option D (

\mathrm{N}{\mathrm{i}^{2 + }}

) is correct because

Ni^{2+}

ions in aqueous solution typically exhibit a green color due to d-d transitions.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{i}^{2 + }}$ . $Ni^{2+}$ ions typically exhibit a green color in aqueous solution.

Step-by-Step Solution:
Step 1: Understand the origin of color in transition metal ions. The color arises from d-d transitions, where electrons in d-orbitals absorb light and transition to higher energy d-orbitals.
Step 2: Consider the electronic configurations of the given ions:
- Cu2+ (Copper(II)): Typically blue or blue-green in aqueous solution.
- Mn2+ (Manganese(II)): Pale pink or almost colorless in aqueous solution.
- Co2+ (Cobalt(II)): Pink in aqueous solution.
- Ni2+ (Nickel(II)): Green in aqueous solution.
Step 3: Identify the ion that exhibits a green color. Based on the typical colors, Ni2+ exhibits a green color.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}^{2 + }}$ : Cu2+ ions are typically blue or blue-green, not green.
- Option B: $\mathrm{M}{\mathrm{n}^{2 + }}$ : Mn2+ ions are typically pale pink or almost colorless.
- Option C: $\mathrm{C}{\mathrm{o}^{2 + }}$ : Co2+ ions are typically pink.

Key Concepts:
- d-d Transitions
- Color of Transition Metal Ions
- Electronic Configuration

Memory Aids:
- Remember the common colors of transition metal ions in solution.

Quick Tips:
- Know the common colors associated with common transition metal ions.

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Remembering +4 | -1

All ligands are:

View answer and explanation

Correct option: B

Lewis base

Option B is correct because ligands are Lewis bases as they donate electron pairs to the central metal atom to form coordinate bonds.

Solution approach

Correct Answer + Core Reason: Lewis base. Ligands donate electron pairs to the central metal atom.

Step-by-Step Solution:

Step 1: Define a Lewis acid. A Lewis acid is a substance that accepts an electron pair.

Step 2: Define a Lewis base. A Lewis base is a substance that donates an electron pair.

Step 3: Consider the role of ligands in coordination complexes. Ligands donate electron pairs to the central metal atom to form coordinate bonds. Therefore, ligands act as Lewis bases.

Why other options are wrong:
- Option A: Ligands donate electron pairs, so they are not Lewis acids.
- Option C: While some ligands can be neutral molecules, they still act as Lewis bases by donating electron pairs.
- Option D: Ligands are Lewis bases.

Key Concepts:
- Lewis acids and bases
- Coordination complexes
- Ligand bonding

Memory Aids:
- Ligands = Lewis bases = Electron pair donors

Quick Tips:
- Remember that ligands donate electron pairs to the central metal atom.

48 Easy

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Application +4 | -1

A complex compound in which the oxidation number of a metal is zero, is

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because in this complex, the oxidation state of Ni is zero since CO is a neutral ligand.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$ . The oxidation state of Ni is zero because CO is a neutral ligand.

Step-by-Step Solution:

Step 1: Identify the complex: $\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]$

Step 2: Determine the charge of the complex: 0

Step 3: Determine the charge of the ligand CO: 0 (neutral ligand)

Step 4: Let the oxidation state of Ni be $x$ . Then, $x + 4(0) = 0$ , so $x = 0$ .

Step 5: Check the oxidation states of the metals in the other options:
- Option A: Fe in ${\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ is +2.
- Option B: Fe in ${\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ is +3.
- Option D: Pt in $\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}$ is +2.

Why other options are wrong:
- Option A: Fe has an oxidation state of +2.
- Option B: Fe has an oxidation state of +3.
- Option D: Pt has an oxidation state of +2.

Key Concepts:
- Oxidation state
- Neutral ligand

Memory Aids:
- CO is a neutral ligand.

Quick Tips:
- Remember the common neutral ligands and their charges.

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Understanding +4 | -1

The type of isomerism shown by the complex

\left[ {\mathrm{Co}\,\mathrm{C}{\mathrm{l}_\mathrm{2}}\,{{\left( {\mathrm{en}} \right)}_\mathrm{2}}} \right]

is

View answer and explanation

Correct option: B

Geometrical isomerism

Option B is correct. The complex exhibits geometrical isomerism because the two chloride ligands and two ethylenediamine ligands can be arranged in cis or trans configurations.

Solution approach

Correct Answer + Core Reason: B. Geometrical isomerism. The complex can exist as cis or trans isomers.

Step-by-Step Solution:

Step 1: Identify the complex: $\CoCl_2(en)_2\$ .

Step 2: Determine the possible isomers: The complex has two chloride ligands and two ethylenediamine (en) ligands. These ligands can be arranged in different spatial arrangements around the central cobalt ion.

Step 3: Draw the possible isomers: The complex can exist as a cis isomer (where the two chloride ligands are on the same side) or a trans isomer (where the two chloride ligands are on opposite sides).

Step 4: Since the complex can exist as cis and trans isomers, it exhibits geometrical isomerism.

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

Key Concepts:
- Isomerism in coordination compounds
- Geometrical isomerism
- Cis and trans isomers

Memory Aids:
- Geometrical: Think of the different shapes the complex can form.

Quick Tips:
- Draw the possible isomers to determine if geometrical isomerism is possible.

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Application +4 | -1

In sodium tetrafluorooxochromate(….),

\mathrm{N}{\mathrm{a}_3}\left[ {\mathrm{Cr}\left( \mathrm{O} \right){\mathrm{F}_4}} \right]

the left out place should be filled with which of the following roman numerals?

View answer and explanation

Correct option: B

III

Option B (III) is correct because the oxidation state of Cr in the complex must be calculated to balance the charges.

Solution approach

Correct Answer + Core Reason: III. The oxidation state of Cr must be +3 to balance the charges of the complex ion.

Step-by-Step Solution:

Step 1: Determine the charge of the complex ion. Since the overall compound is neutral (Na3[Cr(O)F4]), and Na has a +1 charge, the complex ion [Cr(O)F4] has a -3 charge.

Step 2: Assign oxidation states to the ligands. Oxygen (O) has an oxidation state of -2, and Fluorine (F) has an oxidation state of -1.

Step 3: Let x be the oxidation state of Cr. Set up the equation: x + (-2) + 4(-1) = -3

Step 4: Solve for x: x - 2 - 4 = -3 => x - 6 = -3 => x = +3

Step 5: Therefore, the oxidation state of Cr is +3, which is represented by the Roman numeral III.

Why other options are wrong:
- Option A (VI): This is incorrect because it would result in an unbalanced charge for the complex ion.
- Option C (IV): This is incorrect because it would result in an unbalanced charge for the complex ion.
- Option D (None of these): This is incorrect because the oxidation state of Cr is indeed III.

Key Concepts:
- Oxidation state determination
- Charge balancing in coordination complexes

Memory Aids:
- Remember the common oxidation states of elements like Na, O, and F.

Quick Tips:
- Always start by determining the charge of the complex ion before calculating the oxidation state of the central metal ion.

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Analyzing +4 | -1

Assertion : On cooling a freezing mixture, colour of the mixture turns to pink from deep blue for a reaction.

{Co}\left({H}_2 {O}\right)_6^{2+}{ }_{(a q)}+4 {Cl}_{(a q)}^{-} \rightleftharpoons

{CoCl}_4^{2-}(a q)+6 {H}_2 {O}_{(l)}

.

Reason : Reaction is endothermic so on cooling, the reaction moves to backward direction.

View answer and explanation

Correct option: A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

The correct answer is A because the reaction is endothermic, and according to Le Chatelier's principle, cooling favors the reverse reaction, shifting the equilibrium towards the reactants and changing the color from blue to pink.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and the reason is the correct explanation of the assertion. Cooling an endothermic reaction shifts the equilibrium to the left.

Step-by-Step Solution:

Step 1: Analyze the given reaction: ${Co}\left({H}_2 {O}\right)_6^{2+}{ }_{(a q)}+4 {Cl}_{(a q)}^{-} \rightleftharpoons$ ${CoCl}_4^{2-}(a q)+6 {H}_2 {O}_{(l)}$ .

Step 2: The assertion states that cooling the mixture turns the color from deep blue to pink.

Step 3: The reason states that the reaction is endothermic, so cooling favors the backward direction.

Step 4: In an endothermic reaction, heat is absorbed. Therefore, cooling the reaction will shift the equilibrium towards the reactants (backward direction) according to Le Chatelier's principle.

Step 5: The backward reaction corresponds to the formation of ${Co}\left({H}_2 {O}\right)_6^{2+}$ , which is pink, while the forward reaction corresponds to the formation of ${CoCl}_4^{2-}$ , which is blue.

Step 6: Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: The reason is true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Le Chatelier's principle
- Endothermic reactions
- Equilibrium
- Color changes in coordination complexes

Memory Aids:
- Le Chatelier's principle: If a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Quick Tips:
- Remember that cooling favors exothermic reactions, while heating favors endothermic reactions.

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Understanding +4 | -1

Which of the following ligand has lowest

{\mathrm{\Delta }_o}

value?

View answer and explanation

Correct option: C

{\mathrm{F}^ - }

Option C (

{\mathrm{F}^ - }

) is correct because fluoride is a weak field ligand and thus has the lowest

{\mathrm{\Delta }_o}

value.

Solution approach

Correct Answer + Core Reason: ${\mathrm{F}^ - }$ . Fluoride is a weak field ligand.

Step-by-Step Solution:

Step 1: Recall the Spectrochemical Series: The spectrochemical series arranges ligands in order of their ability to cause splitting of the d-orbitals (i.e., their field strength).

Step 2: Identify the Ligands: The ligands are $\mathrm{C}{\mathrm{N}^ - }$ , $\mathrm{CO}$ , ${\mathrm{F}^ - }$ , and $\mathrm{N}{\mathrm{H}_3}$ .

Step 3: Order according to Spectrochemical Series: The general order is $\mathrm{CO} > \mathrm{C}{\mathrm{N}^ - } > \mathrm{N}{\mathrm{H}_3} > {\mathrm{F}^ - }$ .

Step 4: Determine the lowest ${\mathrm{\Delta }_o}$ value: Since ${\mathrm{F}^ - }$ is the weakest field ligand, it will have the lowest ${\mathrm{\Delta }_o}$ value.

Why other options are wrong:
- **Option A ( $\mathrm{C}{\mathrm{N}^ - }$ ):** Cyanide is a strong field ligand and causes a large splitting, hence a high ${\mathrm{\Delta }_o}$ value.
- **Option B ( $\mathrm{CO}$ ):** Carbon monoxide is a very strong field ligand, even stronger than cyanide, and thus has a very high ${\mathrm{\Delta }_o}$ value.
- **Option D ( $\mathrm{N}{\mathrm{H}_3}$ ): Ammonia is a moderate field ligand, stronger than fluoride but weaker than cyanide and carbon monoxide.

Key Concepts:**
- Spectrochemical Series
- Crystal Field Splitting Energy ( $\Delta_o$ )
- Ligand Field Strength

Memory Aids:
- Spectrochemical series: I- < Br- < SCN- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO (memorize a simplified version)

Quick Tips:
- Remember the general trend of the spectrochemical series to quickly determine the relative field strengths of common ligands.

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Remembering +4 | -1

cis

-\left[\mathrm{PtC}{l}_2\left(\mathrm{NH}_3\right)_2\right]

is used

View answer and explanation

Correct option: C

As anticancer agent

Option C (As anticancer agent) is correct because cis-platin, which is cis-[PtCl

_2

(NH

_3

)

_2

], is a well-known anticancer drug.

Solution approach

Correct Answer + Core Reason: As anticancer agent. cis-platin is a known anticancer drug.

Step-by-Step Solution:

Step 1: Recognize the compound: The compound is cis $-\left[\mathrm{PtC}{l}_2\left(\mathrm{NH}_3\right)_2\right]$ , also known as cisplatin.

Step 2: Recall the uses of cisplatin: Cisplatin is a widely used anticancer drug.

Step 3: Relate the compound to its use: Therefore, cis $-\left[\mathrm{PtC}{l}_2\left(\mathrm{NH}_3\right)_2\right]$ is used as an anticancer agent.

Why other options are wrong:
- Option A: As catalyst for hydrogenation of alkene - While platinum compounds can be catalysts, cisplatin is not primarily used for alkene hydrogenation.
- Option B: In photography to dissolve undecomposed $\mathrm{AgBr}$ - This is typically done with sodium thiosulfate, not cisplatin.
- Option D: To measure hardness of water - Hardness of water is typically measured using EDTA titration, not cisplatin.

Key Concepts:
- Coordination compounds
- Applications of coordination compounds
- Anticancer drugs

Memory Aids:
- Remember that cisplatin is a platinum-based anticancer drug.

Quick Tips:
- Know the common applications of important coordination compounds.

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Understanding +4 | -1

Colour of transition metal complexes can be explained by:

View answer and explanation

Correct option: C

d - d

transition

The correct answer is C (

d - d

transition) because the color of transition metal complexes arises from the excitation of electrons between

d

orbitals.

Solution approach

Correct Answer + Core Reason: $d - d$ transition. The color arises from electronic transitions between d-orbitals.

Step-by-Step Solution:
Step 1: The question asks for the reason for the color of transition metal complexes.
Step 2: Transition metal complexes exhibit color due to $d-d$ transitions. Electrons in the $d$ orbitals absorb light energy and get excited to higher energy $d$ orbitals.
Step 3: The energy of the absorbed light corresponds to the energy difference between the $d$ orbitals, and the color observed is the complementary color of the absorbed light.

Why other options are wrong:
- **Option A (Completely filled $d -$ orbitals):** Complexes with completely filled $d$ orbitals are generally colorless.
- **Option B (Vacant $d -$ orbitals):** Complexes with vacant $d$ orbitals are generally colorless.
- Option D (None of the above): $d-d$ transition is the reason.

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

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

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

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Remembering +4 | -1

Which is the strongest field ligand?

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{N}^ - }

Option A (

\mathrm{C}{\mathrm{N}^ - }

) is correct because cyanide (

CN^−

) is the strongest field ligand among the given options based on the spectrochemical series.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{N}^ - }$ . Cyanide is the strongest field ligand among the given options.

Step-by-Step Solution:
Step 1: Recall the spectrochemical series.
Step 2: Compare the ligand field strengths of the given ligands.

Step 3: The spectrochemical series is a list of ligands ordered by their ability to split d-orbital energies. A simplified version is:
$I^- < Br^- < S^{2-} < SCN^- < Cl^- < NO_3^- < N_3^- < F^- < OH^- < C_2O_4^{2-} \approx H_2O < NCS^- < CH_3CN < py < NH_3 < en < bipy < phen < NO_2^- < CN^- < CO$

Step 4: Comparing the given ligands: $\mathrm{C}{\mathrm{N}^ - }$ , $\mathrm{NO}_2^ -$ , $\mathrm{N}{\mathrm{H}_3}$ , and $en$ (ethylenediamine).
Step 5: Based on the spectrochemical series, the order of increasing field strength is: $\mathrm{N}{\mathrm{H}_3} < en < \mathrm{NO}_2^ - < \mathrm{C}{\mathrm{N}^ - }$ .
Step 6: Therefore, $\mathrm{C}{\mathrm{N}^ - }$ is the strongest field ligand.

Why other options are wrong:
- Option B: $\mathrm{NO}_2^ -$ - Nitrite is a strong field ligand, but not as strong as cyanide.
- Option C: $\mathrm{N}{\mathrm{H}_3}$ - Ammonia is a moderate field ligand.
- Option D: $en$ - Ethylenediamine is a strong field ligand, but not as strong as cyanide.

Key Concepts:
- Spectrochemical Series
- Ligand Field Strength
- Crystal Field Splitting

Memory Aids:
- Remember the general trend of the spectrochemical series.

Quick Tips:
- Know the relative positions of common ligands in the spectrochemical series.
- Strong field ligands cause large crystal field splitting.

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Understanding +4 | -1

Which of the following is not an organometallic compound?

View answer and explanation

Correct option: A

\mathrm{NaO}{\mathrm{C}_2}{\mathrm{H}_5}

Option A (

\mathrm{NaO}{\mathrm{C}_2}{\mathrm{H}_5}

) is correct because it is an ionic compound (sodium ethoxide) with no direct metal-carbon bond, thus not an organometallic compound.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaO}{\mathrm{C}_2}{\mathrm{H}_5}$ . This compound does not have a direct metal-carbon bond.

Step-by-Step Solution:

Step 1: Define Organometallic Compound: An organometallic compound contains at least one metal-carbon bond.

Step 2: Analyze the options:
- A. $\mathrm{NaO}{\mathrm{C}_2}{\mathrm{H}_5}$ : Sodium ethoxide is an ionic compound with a Na-O bond, not a Na-C bond. Hence, not organometallic.
- B. ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{Al}$ : Trimethylaluminum contains an Al-C bond, hence organometallic.
- C. ${\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}\mathrm{Pb}$ : Tetraethyllead contains a Pb-C bond, hence organometallic.
- D. $R\mathrm{Mg}X$ : Grignard reagent contains a Mg-C bond, hence organometallic.

Step 3: Conclusion: Sodium ethoxide is not an organometallic compound.

Why other options are wrong:
- Option B: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{Al}$ : Contains an Al-C bond, making it organometallic.
- Option C: ${\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}\mathrm{Pb}$ : Contains a Pb-C bond, making it organometallic.
- Option D: $R\mathrm{Mg}X$ : Contains a Mg-C bond, making it organometallic.

Key Concepts:
- Organometallic compounds
- Metal-carbon bond
- Sodium ethoxide
- Trimethylaluminum
- Tetraethyllead
- Grignard reagent

Memory Aids:
- Remember that organometallic compounds must have a direct metal-carbon bond.

Quick Tips:
- Identify the presence of a metal-carbon bond to determine if a compound is organometallic.

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Application +4 | -1

Which one of the following does not give a white precipitate with silver nitrate solution?

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because in this complex, all the chloride ions are coordinated inside the coordination sphere and hence do not ionize to give free chloride ions that can react with silver nitrate to form a white precipitate.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]$ . This complex does not have any chloride ions outside the coordination sphere to react with $AgNO_3$ .

Step-by-Step Solution:

Step 1: Understand the reaction with silver nitrate.\nSilver nitrate ( $AgNO_3$ ) reacts with free chloride ions ( $Cl^−$ ) in solution to form a white precipitate of silver chloride ( $AgCl$ ).

Step 2: Analyze Option A: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}$ \nThis complex has 3 chloride ions outside the coordination sphere. It will produce 3 moles of $AgCl$ precipitate.
$\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3} + 3AgNO_3 \rightarrow \left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]{{\left( {NO_3} \right)}_3} + 3AgCl(s)$

Step 3: Analyze Option B: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_2}$ \nThis complex has 2 chloride ions outside the coordination sphere. It will produce 2 moles of $AgCl$ precipitate.
$\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_2} + 2AgNO_3 \rightarrow \left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]{{\left( {NO_3} \right)}_2} + 2AgCl(s)$

Step 4: Analyze Option C: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ \nThis complex has 1 chloride ion outside the coordination sphere. It will produce 1 mole of $AgCl$ precipitate.
$\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl} + AgNO_3 \rightarrow \left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]NO_3 + AgCl(s)$

Step 5: Analyze Option D: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]$ \nThis complex has no chloride ions outside the coordination sphere. It will not produce any $AgCl$ precipitate.

Step 6: Identify the complex that does not give a white precipitate.\nOnly $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]$ does not give a white precipitate with silver nitrate solution.

Why other options are wrong:
- Option A: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}$ gives 3 moles of AgCl precipitate.
- Option B: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_2}$ gives 2 moles of AgCl precipitate.
- Option C: $\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]\mathrm{Cl}$ gives 1 mole of AgCl precipitate.

Key Concepts:
- Coordination Chemistry
- Ionization of Coordination Complexes
- Precipitation Reactions

Memory Aids:
- Only ions outside the coordination sphere will react with other ions in solution.

Quick Tips:
- Determine the number of ions outside the coordination sphere to predict the number of moles of precipitate formed.

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Understanding +4 | -1

The IUPAC name of the complex

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

is

View answer and explanation

Correct option: B

Tetraammine dichloro cobalt(III) chloride

Option B (Tetraammine dichloro cobalt(III) chloride) is correct because it follows the IUPAC naming conventions for coordination complexes, listing ligands alphabetically before the metal and indicating the oxidation state of the metal.

Solution approach

Correct Answer + Core Reason: Tetraammine dichloro cobalt(III) chloride. The IUPAC naming convention dictates the order and nomenclature of ligands and the central metal ion.

Step-by-Step Solution:

Step 1: Identify the ligands: There are four ammine ligands (NH3) and two chloro ligands (Cl).

Step 2: Determine the oxidation state of cobalt: The complex has a +1 charge outside the brackets, and the chloride ion has a -1 charge. The chloro ligands contribute -2, and ammine ligands are neutral. Therefore, Co must have a +3 oxidation state to balance the charges.

Step 3: Name the complex according to IUPAC rules: List ligands alphabetically (ammine before chloro), then the metal, and finally the counter ion.

Step 4: Combine the names: Tetraammine dichloro cobalt(III) chloride.

Why other options are wrong:
- Option A: Dichloro tetraammine cobalt (III) chloride: Incorrect ligand order (chloro before ammine).
- Option C: Tetraammine dichloro cobalt (II) chloride: Incorrect oxidation state of cobalt (should be III).
- Option D: Tetraammine dichloro cobalt (IV) chloride: Incorrect oxidation state of cobalt (should be III).

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

Memory Aids:
- Alphabetical order of ligands
- Roman numerals for oxidation state

Quick Tips:
- Always list ligands alphabetically.
- Determine the oxidation state of the metal accurately.

59 Easy

59 of 98Question permalink

Understanding +4 | -1

A ligand can also be regarded as

View answer and explanation

Correct option: C

Lewis base

Option C (Lewis base) is correct because ligands donate electron pairs to the central metal atom, which is the definition of a Lewis base.

Solution approach

Correct Answer + Core Reason: Lewis base. Ligands donate electron pairs to the central metal.

Step-by-Step Solution:

Step 1: Define a ligand. A ligand is an ion or molecule that binds to a central metal atom to form a coordination complex.

Step 2: Recall Lewis acid/base definitions. A Lewis base is an electron pair donor, and a Lewis acid is an electron pair acceptor.

Step 3: Consider the interaction between a ligand and a metal. The ligand donates a pair of electrons to the metal to form a coordinate covalent bond.

Step 4: Conclude that a ligand acts as a Lewis base.

Why other options are wrong:
- Option A: A Lewis acid accepts electron pairs, which is the role of the metal ion, not the ligand.
- Option B: A Bronsted base accepts protons, which is not the primary function of a ligand in complex formation.
- Option D: A Bronsted acid donates protons, which is not the role of a ligand.

Key Concepts:
- Lewis acid/base theory
- Coordination complexes
- Ligands

Memory Aids:
- LB = Ligand is a Base (Lewis)

Quick Tips:
- Remember that ligands donate electron pairs, thus acting as Lewis bases.

60 Easy

60 of 98Question permalink

Application +4 | -1

EAN of

\mathrm{Cr}

in

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

is:

View answer and explanation

Correct option: B

33

The correct answer is 33 because the EAN of Cr in

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

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

Solution approach

Correct Answer + Core Reason: 33. The EAN is calculated by subtracting the oxidation state from the atomic number and adding the number of electrons donated by the ligands.

Step-by-Step Solution:

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

Step 2: Determine the atomic number of Cr: 24

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

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

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

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

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

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

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

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

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