The IUPAC name of ${{\rm{K}}_2}\left[ {{\rm{Cr}}{{\left( {{\rm{CN}}} \right)}_2}{{\rm{O}}_2}{\rm{\;}}{{\left( {\rm{O}} \right)}_2}\left( {{\rm{N}}{{\rm{H}}_3}} \right)} \right]{\rm{\;}}$ is

Potassium ammine cyano peroxodioxochromium(VI)

Potassium ammine cyano peroxodioxochromium(V)

Potassium ammine cyano peroxodioxochromatic (IV)

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

91 Easy

91 of 98Question permalink

Understanding +4 | -1

Which of the following statements is correct?

View answer and explanation

Correct option: C

In

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

the ligand has satisfied only the secondary valency of ferric ion

The correct answer is C. In

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

, the ligand (CN-) satisfies only the secondary valency (coordination number) of the ferric ion (Fe3+).

Solution approach

Correct Answer + Core Reason: In ${\mathrm{K}_3}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}]$ , the ligand has satisfied only the secondary valency of ferric ion. Ligands in coordination complexes primarily satisfy the secondary valency (coordination number) of the central metal ion.

Step-by-Step Solution:

Step 1: Define primary and secondary valencies. Primary valency refers to the oxidation state of the metal ion, while secondary valency refers to the coordination number (number of ligands directly bonded to the metal ion).

Step 2: Analyze ${\mathrm{K}_3}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}]$ . In this complex, the iron ion (Fe) has an oxidation state of +3 (primary valency). The cyanide ligands (CN-) are coordinated to the iron ion, and there are six CN- ligands, so the coordination number (secondary valency) is 6.

Step 3: Determine which valency the ligand satisfies. The cyanide ligands (CN-) are directly bonded to the iron ion, satisfying the coordination number (secondary valency). The charge of the complex is balanced by the potassium ions outside the coordination sphere.

Step 4: Analyze $(\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{S}{\mathrm{O}_4}$ . In this complex, the copper ion (Cu) has an oxidation state of +2. The ammonia ligands (NH3) are coordinated to the copper ion, and there are four NH3 ligands, so the coordination number is 4. The ammonia ligands satisfy the secondary valency of copper.

Step 5: Conclusion. Therefore, in ${\mathrm{K}_3}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}]$ , the ligand has satisfied only the secondary valency of the ferric ion.

Why other options are wrong:
- Option A: In ${\mathrm{K}_3}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}]$ , the ligand has satisfied both primary and secondary valencies of ferric ion: This is incorrect because ligands primarily satisfy the secondary valency (coordination number), not the primary valency (oxidation state).
- Option B: In $(\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{S}{\mathrm{O}_4}$ , the ligand has satisfied only the secondary valency of copper: This statement is correct, but the question asks for which *one* of the statements is correct. Option C is also correct, but it is the *most* correct in the context of the question.
- Option D: Both (2) and (3): Option B is correct, and Option C is correct. However, the question is asking for which *one* of the statements is correct. Option C is the most correct answer.

Key Concepts:
- Coordination complexes
- Primary valency (oxidation state)
- Secondary valency (coordination number)
- Ligands

Memory Aids:
- Primary = Oxidation State
- Secondary = Coordination Number

Quick Tips:
- Ligands directly bonded to the metal satisfy the secondary valency (coordination number).

92 Easy

92 of 98Question permalink

Analyzing +4 | -1

Which statement is not correct in the case of

{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}

complex?

View answer and explanation

Correct option: B

It involves

{d^2}s{p^2}

- hybridization

Option B (It involves

{d^2}s{p^2}

- hybridization) is not correct because

{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}

involves

d^2sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: It involves ${d^2}s{p^2}$ - hybridization. The complex involves $d^2sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Identify the complex: ${\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}$

Step 2: Determine the oxidation state of Co: Co is in +3 oxidation state.

Step 3: Electronic configuration of Co(III): $3d^6$

Step 4: Since NH3 is a strong field ligand, it causes pairing of electrons. The hybridization is $d^2sp^3$ and the geometry is octahedral.

Step 5: All electrons are paired, hence it is diamagnetic.

Why other options are wrong:
- Option A: It is octahedral in shape, which is correct.
- Option C: It has diamagnetic nature, which is correct.
- Option D: None of the above is incorrect since option B is incorrect.

Key Concepts:
- Coordination complexes
- Oxidation state
- Hybridization
- Geometry of complexes
- Magnetic properties
- Strong field ligands

Memory Aids:
- NH3 is a strong field ligand and causes pairing.

Quick Tips:
- Determine the oxidation state of the metal ion first.
- Identify the ligands as strong or weak field ligands.
- Predict the hybridization and geometry based on the ligands.

93 Easy

93 of 98Question permalink

Remembering +4 | -1

The hardness of water is estimated by:

View answer and explanation

Correct option: B

EDTA method

Option B (EDTA method) is correct because the EDTA method is a complexometric titration used to determine the total hardness of water by measuring the concentration of calcium and magnesium ions.

Solution approach

Correct Answer + Core Reason: EDTA method. EDTA titration is used to estimate the hardness of water by complexing with $Ca^{2+}$ and $Mg^{2+}$ ions.

Step-by-Step Solution:

Step 1: Understand the concept of water hardness.

Step 2: Water hardness is primarily due to the presence of $Ca^{2+}$ and $Mg^{2+}$ ions.

Step 3: EDTA (ethylenediaminetetraacetic acid) forms stable complexes with these ions.

Step 4: The EDTA method involves titrating a water sample with EDTA solution, using an indicator to detect the endpoint.

Why other options are wrong:
- Option A: Conductivity method: Measures the ability of water to conduct electricity, which is related to the total ion concentration, but not specific to hardness.
- Option C: Titrimetric method: A general term for titration, but EDTA titration is the specific method used for hardness.
- Option D: Distillation method: Removes all ions, but does not estimate the original hardness.

Key Concepts:
- Water hardness
- EDTA complexometric titration

Memory Aids:
- EDTA: Remember "EDTA Eats Hardness" to associate it with water hardness estimation.

Quick Tips:
- EDTA is a common complexing agent used in analytical chemistry.

94 Easy

94 of 98Question permalink

Application +4 | -1

The compound having the lowest oxidation state of iron is

View answer and explanation

Correct option: D

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

Option D (Fe(CO)5) is correct because the oxidation state of Fe in this complex is 0, which is the lowest among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ . The oxidation state of Fe in this compound is 0, which is the lowest among the given options.

Step-by-Step Solution:

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

Step 2: Option A: ${\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ . K has +1 charge, and CN has -1 charge. Thus, 4(+1) + Fe + 6(-1) = 0, so Fe = +2.

Step 3: Option B: ${\mathrm{K}_2}\mathrm{Fe}{\mathrm{O}_4}$ . K has +1 charge, and O has -2 charge. Thus, 2(+1) + Fe + 4(-2) = 0, so Fe = +6.

Step 4: Option C: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ . O has -2 charge. Thus, 2Fe + 3(-2) = 0, so Fe = +3.

Step 5: Option D: $\mathrm{Fe}{\left( {\mathrm{CO}} \right)_5}$ . CO is a neutral ligand with 0 charge. Thus, Fe + 5(0) = 0, so Fe = 0.

Step 6: Comparing the oxidation states, Fe(CO)5 has the lowest oxidation state (0).

Why other options are wrong:
- Option A: The oxidation state of Fe is +2, which is not the lowest.
- Option B: The oxidation state of Fe is +6, which is not the lowest.
- Option C: The oxidation state of Fe is +3, which is not the lowest.

Key Concepts:
- Oxidation state determination
- Charge of common ions and ligands

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

Quick Tips:
- Assign known oxidation states to other elements/ligands first, then solve for the unknown oxidation state of the metal.

95 Easy

95 of 98Question permalink

Remembering +4 | -1

Among the ligands

N{H_3},en,C{N^ - }and\,CO,

the correct order of their increasing field strength, is

View answer and explanation

Correct option: B

N{H_3} < en < C{N^ - } < CO

Option B (

N{H_3} < en < C{N^ - } < CO

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

Solution approach

Correct Answer + Core Reason: $N{H_3} < en < C{N^ - } < CO$ . The 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: NH3 (ammonia), en (ethylenediamine), CN- (cyanide), and CO (carbon monoxide).
Step 3: Recall the relative positions of these ligands in the spectrochemical series. Generally, NH3 < en < CN- < CO.
Step 4: Arrange the ligands in increasing order of field strength based on their positions in the series.

Why other options are wrong:
- Option A: $\;CO < N{H_3} < en < C{N^ - }$ : This is incorrect because CO is a very strong field ligand and should be at the end, not the beginning.
- Option C: $C{N^ - } < N{H_3} < CO < en$ : This is incorrect because CN- is a stronger field ligand than NH3 and en is stronger than NH3.
- Option D: $en < C{N^ - } < N{H_3} < CO$ : This is incorrect because en is a stronger field ligand than NH3, but CN- is stronger than both, and CO is the strongest.

Key Concepts:
- Spectrochemical Series
- Ligand Field Strength

Memory Aids:
- Remember that CO and CN- are strong field ligands, while NH3 and en are intermediate.
- en (ethylenediamine) is a stronger field ligand than NH3 due to its chelate effect.

Quick Tips:
- Memorize the relative positions of common ligands in the spectrochemical series.
- Recognize that CO and CN- are almost always at the strong field end.

96 Easy

96 of 98Question permalink

Analyzing +4 | -1

Which among the following statements are true for the complex

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]

?
(a) It is anon electrolyte
(b) The magnitude of the charge on each complex ion is 3
(c) The complex will not conduct current.
(d) The complex will exhibit coordination isomerism
(e) The magnitude of the charge on each complex ion is 1

View answer and explanation

Correct option: C

b and d

Option C (b and d) is correct because the magnitude of the charge on each complex ion is 3, and the complex can exhibit coordination isomerism.

Solution approach

Correct Answer + Core Reason: b and d. The magnitude of the charge on each complex ion is 3, and the complex exhibits coordination isomerism.

Step-by-Step Solution:

Step 1: Determine the charge on each complex ion.
[Co(NH3)6][Cr(CN)6]\nNH3 is a neutral ligand, and Co is typically +3. Therefore, the charge on the [Co(NH3)6] complex is +3.\nCN is a -1 ligand, and Cr is typically +3. Therefore, the charge on the [Cr(CN)6] complex is -3.\nThus, the magnitude of the charge on each complex ion is 3.

Step 2: Determine if the complex is an electrolyte.\nThe complex dissociates into ions in solution, making it an electrolyte.

Step 3: Determine if the complex will conduct current.\nSince the complex is an electrolyte, it will conduct current in solution.

Step 4: Determine if the complex exhibits coordination isomerism.\nCoordination isomerism occurs when the ligands are exchanged between the cationic and anionic complex ions. In this case, the NH3 and CN ligands can be exchanged, resulting in [Cr(NH3)6][Co(CN)6]. Therefore, the complex exhibits coordination isomerism.

Step 5: Final Answer: Statements b and d are true.

Why other options are wrong:
- Option A (a and b): Statement a is incorrect because the complex is an electrolyte.
- Option B (a and c): Statements a and c are incorrect because the complex is an electrolyte and will conduct current.
- Option D (c and e): Statements c and e are incorrect because the complex will conduct current, and the magnitude of the charge on each complex ion is 3, not 1.

Key Concepts:
- Coordination complexes
- Electrolytes
- Coordination isomerism
- Charge of complex ions

Memory Aids:
- Coordination isomerism involves the exchange of ligands between complex ions.

Quick Tips:
- Determine the charge of each complex ion by considering the charges of the ligands and the metal ion.
- Remember that electrolytes conduct current in solution.

97 Easy

97 of 98Question permalink

Application +4 | -1

The IUPAC name of

{\mathrm{K}_2}\left[ {\mathrm{Cr}{{\left( {\mathrm{CN}} \right)}_2}{\mathrm{O}_2}\mathrm{\;}{{\left( \mathrm{O} \right)}_2}\left( {\mathrm{N}{\mathrm{H}_3}} \right)} \right]\mathrm{\;}

is

View answer and explanation

Correct option: A

Potassium ammine dicyano dioxoperoxochromate (VI)

Option A (Potassium ammine dicyano dioxoperoxochromate (VI)) is correct because it follows IUPAC naming conventions for coordination compounds.

Solution approach

Correct Answer + Core Reason: Potassium ammine dicyano dioxoperoxochromate (VI). This name correctly identifies the ligands, their order, and the oxidation state of chromium.

Step-by-Step Solution:

Step 1: Identify the cation and anion. The cation is Potassium (K+), and the anion is the complex ion [Cr(CN)2O2(O2)(NH3)]2-.

Step 2: Determine the ligands and their charges. Ammine (NH3) is neutral, Cyano (CN-) has a -1 charge, Dioxo (O2-) has a -2 charge each, and Peroxo (O2^2-) has a -2 charge.

Step 3: Determine the oxidation state of Cr. Let x be the oxidation state of Cr. The overall charge of the complex is -2. So, x + (0) + 2(-1) + 2(-1) + (-2) = -2 => x - 6 = -2 => x = +6. Therefore, the oxidation state of Cr is VI.

Step 4: Name the complex according to IUPAC rules: Potassium ammine dicyano dioxoperoxochromate (VI).

Why other options are wrong:
- Option B (Potassium ammine cyano peroxodioxochromium(VI)): Incorrect because it doesn't specify the number of cyano and oxo ligands correctly.
- Option C (Potassium ammine cyano peroxodioxochromium(V)): Incorrect because the oxidation state of Cr is VI, not V.
- Option D (Potassium ammine cyano peroxodioxochromatic (IV)): Incorrect because the oxidation state of Cr is VI, not IV, and the ending should be -ate.

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.

98 Easy

98 of 98Question permalink

Remembering +4 | -1

Lithium tetrahydridoaluminate is correctly represented as:

View answer and explanation

Correct option: C

\mathrm{Li}\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]

Option C (

\mathrm{Li}\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]

) is correct because it accurately represents the structure of lithium tetrahydridoaluminate, showing lithium as the cation and the tetrahydroaluminate as the anion.

Solution approach

Correct Answer + Core Reason: $\mathrm{Li}\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]$ . This is the correct representation of Lithium tetrahydridoaluminate.

Step-by-Step Solution:

Step 1: Identify the components of the compound: Lithium (Li), Aluminum (Al), and Hydrogen (H).

Step 2: Recognize that the compound is a complex hydride where aluminum is coordinated with four hydrogen atoms, forming the tetrahydroaluminate ion, $[\mathrm{AlH}_4]^−$ .

Step 3: Determine the correct formula by balancing the charges. Lithium has a +1 charge, and the tetrahydroaluminate ion has a -1 charge, so the formula is $\mathrm{Li}\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]$ .

Why other options are wrong:
- Option A: $\mathrm{Al}\left[ {\mathrm{Li}{\mathrm{H}_4}} \right]$ : This is incorrect because it places aluminum outside the complex ion and incorrectly associates lithium with four hydrogen atoms.
- Option B: $\mathrm{A}{\mathrm{l}_2}{\left[ {\mathrm{Li}{\mathrm{H}_4}} \right]_3}$ : This formula is incorrect as it suggests a different stoichiometry and incorrect association of atoms.
- Option D: $\mathrm{Li}{\left[ {\mathrm{Al}{\mathrm{H}_4}} \right]_2}$ : This formula is incorrect because it implies that two tetrahydroaluminate ions are associated with one lithium ion, which is not the case.

Key Concepts:
- Nomenclature of coordination compounds
- Complex hydrides
- Charge balancing in ionic compounds

Memory Aids:
- Remember that complex hydrides often involve a metal coordinated with hydrogen atoms.

Quick Tips:
- Always consider the charges of the ions involved when writing the formula of a compound.

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