${\rm{KMn}}{{\rm{O}}_4}$ in basic medium is reduced to

${{\rm{K}}_2}{\rm{Mn}}{{\rm{O}}_4}$

${\rm{Mn}}{\left( {{\rm{OH}}} \right)_2}$

${\left( {{\rm{N}}{{\rm{H}}_4}} \right)_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ on heating gives a gas which is also given by

Heating ${\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_2}$

Heating ${\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}$

${\rm{M}}{{\rm{g}}_3}{{\rm{N}}_2} + {{\rm{H}}_2}{\rm{O}}$

Na(Comp.)+ ${{\rm{H}}_2}{{\rm{O}}_2}$

Is salt of dibasic acid ${H}_2 {O}_2$

Oxidizes ${Cr}^{3+}$ (green) to ${CrO}_4^{2-}$ (yellow)

When excess of ${\rm{SnC}}{{\rm{l}}_2}$ is added to a solution of ${\rm{HgC}}{{\rm{l}}_2}$, a white ppt. turning to grey is obtained. This grey colour is due to the formation of:

${\rm{H}}{{\rm{g}}_2}{\rm{C}}{{\rm{l}}_2}$

d & f- BLOCK ELEMENTS MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

61 of 85Question permalink

Remembering +4 | -1

\mathrm{KMn}{\mathrm{O}_4}

in basic medium is used as

View answer and explanation

Correct option: A

Strong oxidising agent

Option A (Strong oxidising agent) is correct because potassium permanganate (

\mathrm{KMn}{\mathrm{O}_4}

) acts as a strong oxidizing agent in both acidic and basic media.

Solution approach

Correct Answer + Core Reason: Strong oxidising agent. $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent in basic medium.

Step-by-Step Solution:

Step 1: Recall the properties of $\mathrm{KMn}{\mathrm{O}_4}$ .

Step 2: $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent in acidic, neutral, and basic media.

Step 3: In basic medium, $\mathrm{KMn}{\mathrm{O}_4}$ typically reduces to $MnO_2$ .

Step 4: Oxidizing agents accept electrons, causing oxidation in other substances.

Step 5: Reducing agents donate electrons, causing reduction in other substances.

Step 6: Hydrogenating agents add hydrogen to a substance.

Step 7: Therefore, $\mathrm{KMn}{\mathrm{O}_4}$ in basic medium is a strong oxidizing agent.

Why other options are wrong:
- Option B: $\mathrm{KMn}{\mathrm{O}_4}$ is an oxidizing agent, not a reducing agent.
- Option C: $\mathrm{KMn}{\mathrm{O}_4}$ is not a hydrogenating agent.
- Option D: $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent, not a poor reducing agent.

Key Concepts:
- Oxidizing Agents
- Reducing Agents
- Potassium Permanganate

Memory Aids:
- $\mathrm{KMn}{\mathrm{O}_4}$ = strong oxidizer.

Quick Tips:
- Remember that $\mathrm{KMn}{\mathrm{O}_4}$ is a powerful oxidizing agent in various media.

62 Medium

62 of 85Question permalink

Understanding +4 | -1

Lanthanide contraction is caused due to

View answer and explanation

Correct option: D

The imperfect shielding on outer electrons by

4f

-electrons from the nuclear charge.

Option D (The imperfect shielding on outer electrons by

4f

-electrons from the nuclear charge) is correct because the lanthanide contraction is caused by the imperfect shielding of the nuclear charge by the 4f electrons.

Solution approach

Correct Answer + Core Reason: The imperfect shielding on outer electrons by $4f$ -electrons from the nuclear charge. This is the cause of lanthanide contraction.

Step-by-Step Solution:

Step 1: Understand the concept of lanthanide contraction.

Step 2: The 4f electrons provide poor shielding to the outer electrons from the increasing nuclear charge.

Step 3: This results in an increased effective nuclear charge, causing the contraction.

Why other options are wrong:
- Option A: The shielding is imperfect, not appreciable.
- Option B: $5d$ -electrons are not the primary cause of lanthanide contraction.
- Option C: The effective nuclear charge increases, it is not the same.

Key Concepts:
- Lanthanide contraction
- Shielding effect
- Effective nuclear charge

Memory Aids:
- Remember that poor shielding by 4f electrons leads to increased effective nuclear charge and contraction.

Quick Tips:
- Understand the factors affecting atomic size.

63 Medium

63 of 85Question permalink

Remembering +4 | -1

Permanent magnets are generally made of alloys of

View answer and explanation

Correct option: B

Co

Option B (Co) is correct because Cobalt (Co) is a key component in alloys used for making permanent magnets.

Solution approach

Correct Answer + Core Reason: Co. Cobalt alloys are commonly used to make permanent magnets.

Step-by-Step Solution:

Step 1: Recall the materials used for making permanent magnets.

Step 2: Cobalt, along with other elements like iron and nickel, forms alloys with high magnetic properties.

Step 3: These alloys are used in the production of permanent magnets.

Why other options are wrong:
- Option A: Mn (Manganese) is used in some alloys, but not typically for permanent magnets.
- Option C: Pb (Lead) is not used in permanent magnets.
- Option D: Zn (Zinc) is not used in permanent magnets.

Key Concepts:
- Permanent magnets and their composition

Memory Aids:
- Remember that Cobalt is a ferromagnetic material.

Quick Tips:
- Know the common elements used in magnetic materials.

64 Medium

64 of 85Question permalink

Analyzing +4 | -1

The correct order of decreasing second ionization enthalpy of

Ti(22),\,\,V(23),\,\,Cr(24)

and

Mn(25)

is:-

View answer and explanation

Correct option: C

Cr > Mn > V > Ti

Option C (

Cr > Mn > V > Ti

) is correct because the second ionization enthalpy is affected by the stability of the resulting electronic configurations.

Solution approach

Correct Answer + Core Reason: $Cr > Mn > V > Ti$ . The second ionization enthalpy is affected by the stability of the resulting electronic configurations.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the elements.\nTi: $[Ar] 3d^2 4s^2$ \nV: $[Ar] 3d^3 4s^2$ \nCr: $[Ar] 3d^5 4s^1$ \nMn: $[Ar] 3d^5 4s^2$

Step 2: Write the electronic configurations after the first ionization.\nTi $^+$ : $[Ar] 3d^2 4s^1$ \nV $^+$ : $[Ar] 3d^3 4s^1$ \nCr $^+$ : $[Ar] 3d^5$ \nMn $^+$ : $[Ar] 3d^5 4s^1$

Step 3: Write the electronic configurations after the second ionization.\nTi $^{2+}$ : $[Ar] 3d^2$ \nV $^{2+}$ : $[Ar] 3d^3$ \nCr $^{2+}$ : $[Ar] 3d^4$ \nMn $^{2+}$ : $[Ar] 3d^5$

Step 4: Cr $^+$ loses an electron from a stable half-filled $3d^5$ configuration, requiring more energy. Mn $^+$ loses an electron from 4s, which is easier than removing from the 3d. V and Ti follow the general trend.

Step 5: Therefore, the correct order is $Cr > Mn > V > Ti$ .

Why other options are wrong:
- Option A: Incorrect because it doesn't account for the stability of the half-filled configuration of Cr after the first ionization.
- Option B: Incorrect because it doesn't account for the stability of the half-filled configuration of Cr after the first ionization.
- Option D: Incorrect because it doesn't account for the stability of the half-filled configuration of Cr after the first ionization.

Key Concepts:
- Ionization enthalpy
- Electronic configuration
- Stability of half-filled and fully-filled orbitals

Memory Aids:
- Remember that removing an electron from a stable configuration requires more energy.

Quick Tips:
- Consider the electronic configurations after each ionization when comparing ionization enthalpies.

65 Medium

65 of 85Question permalink

Understanding +4 | -1

\mathrm{KMn}{\mathrm{O}_4}

in basic medium is reduced to

View answer and explanation

Correct option: B

\mathrm{Mn}{\mathrm{O}_2}

Option B (

\mathrm{Mn}{\mathrm{O}_2}

) is correct because in a basic medium,

KMnO_4

is reduced to

MnO_2

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mn}{\mathrm{O}_2}$ . In basic medium, $KMnO_4$ is reduced to $MnO_2$ .

Step-by-Step Solution:

Step 1: Write the reduction half-reaction of $KMnO_4$ in basic medium.
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

Step 2: Identify the product formed.\nThe product is $MnO_2$ .

Why other options are wrong:
- Option A: ${\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4}$ is formed in a less basic or neutral medium.
- Option C: $\mathrm{Mn}{\left( {\mathrm{OH}} \right)_2}$ is not the direct product of reduction in basic medium.
- Option D: $\mathrm{M}{\mathrm{n}^{2 + }}$ is formed in acidic medium.

Key Concepts:
- Redox reactions
- Reduction of permanganate
- Effect of pH on redox reactions

Memory Aids:
- Remember the reduction products of permanganate in different media.

Quick Tips:
- Recall the reduction products of $KMnO_4$ in different pH conditions.

66 Medium

66 of 85Question permalink

Analyzing +4 | -1

In context with the transition elements, which of the following statements is incorrect?

View answer and explanation

Correct option: B

In the highest oxidation state, the transition metal shows basic character and form cationic complexes.

Option B is correct because in the highest oxidation state, transition metals tend to form acidic oxides and oxoanions, not basic oxides and cationic complexes.

Solution approach

Correct Answer + Core Reason: In the highest oxidation state, the transition metal shows basic character and form cationic complexes. Transition metals in high oxidation states form acidic oxides and oxoanions.

Step-by-Step Solution:

Step 1: Analyze the given statements regarding transition elements.

Step 2: Consider the statement about zero oxidation state. Transition elements can exhibit zero oxidation state in complexes (e.g., Ni(CO)4).

Step 3: Consider the statement about the highest oxidation state. In the highest oxidation state, transition metals form acidic oxides and oxoanions (e.g., Mn2O7, CrO3, VO5). They do not form basic oxides and cationic complexes.

Step 4: Consider the statement about the use of 4s and 3d electrons. In the highest oxidation state, the first five transition elements (Sc to Mn) can use all 4s and 3d electrons for bonding.

Step 5: Consider the statement about the tendency to involve 3d electrons in bonding. Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.

Why other options are wrong:
- Option A: In addition to the normal oxidation state, the zero oxidation state is also shown by these elements in complexes. This statement is correct.
- Option C: In the highest oxidation state of the first five transition elements (Sc to Mn), all the $4s$ and $4d$ electrons are used for bonding. This statement is correct.
- Option D: Once the ${d^5}$ configuration is exceeded, the tendency to involve all the $3d$ electrons in bonding decreases. This statement is correct.

Key Concepts:
- Oxidation states of transition elements
- Properties of transition metal oxides
- Electronic configuration and bonding

Memory Aids:
- Remember that transition metals in high oxidation states form acidic oxides.

Quick Tips:
- Recall the properties of transition metal oxides in different oxidation states.

67 Medium

67 of 85Question permalink

Analyzing +4 | -1

The correct order of decreasing Second ionisation enthalpy of

Ti,\,V,\,Cr

and

Mn

is

View answer and explanation

Correct option: C

Cr > Mn > V > Ti

Option C (Cr > Mn > V > Ti) is correct because the second ionization enthalpy depends on electronic configuration and stability of the resulting ion.

Solution approach

Correct Answer + Core Reason: $Cr > Mn > V > Ti$ . The second ionization enthalpy is affected by the stability of electronic configurations.

Step-by-Step Solution:

Step 1: Write the electronic configurations of Ti, V, Cr, and Mn.\nTi: [Ar] 3d2 4s2\nV: [Ar] 3d3 4s2\nCr: [Ar] 3d5 4s1\nMn: [Ar] 3d5 4s2

Step 2: Write the electronic configurations of the ions after the first ionization.\nTi+: [Ar] 3d2 4s1\nV+: [Ar] 3d3 4s1\nCr+: [Ar] 3d5\nMn+: [Ar] 3d5 4s1

Step 3: Write the electronic configurations of the ions after the second ionization.\nTi2+: [Ar] 3d2\nV2+: [Ar] 3d3\nCr2+: [Ar] 3d4\nMn2+: [Ar] 3d5

Step 4: Analyze the stability of the resulting ions. Mn2+ has a stable d5 configuration. Cr+ loses an electron from a stable d5 configuration, requiring more energy. V2+ and Ti2+ have less stable configurations.

Step 5: Determine the order of decreasing second ionization enthalpy.\nCr > Mn > V > Ti

Why other options are wrong:
- Option A: $Mn > Cr > Ti > V$ - Incorrect because Cr+ loses an electron from a stable d5 configuration, requiring more energy than Mn+.
- Option B: $Ti > V > Cr > Mn$ - Incorrect because Mn has a stable d5 configuration after the second ionization, and Cr requires more energy to remove an electron from d5.
- Option D: $V.Mn > Cr > Ti$ - Incorrect because Cr requires more energy to remove an electron from d5 than Mn.

Key Concepts:
- Electronic configuration
- Ionization enthalpy
- Stability of half-filled and fully-filled orbitals

Memory Aids:
- Remember that half-filled and fully-filled orbitals are more stable.

Quick Tips:
- Write the electronic configurations of the elements and their ions to determine the ionization enthalpy order.

68 Medium

68 of 85Question permalink

Application +4 | -1

Which of the following ion is diamagnetic?

View answer and explanation

Correct option: B

\mathrm{L}{\mathrm{a}^{3 + }}

Option B is correct because La3+ has the electronic configuration [Xe], which means it has no unpaired electrons and is therefore diamagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{L}{\mathrm{a}^{3 + }}$ . La³⁺ has the electronic configuration [Xe], meaning it has no unpaired electrons and is diamagnetic.

Step-by-Step Solution:

Step 1: Write the electronic configuration of the neutral atoms.
- Nd (Z = 60): [Xe]4f⁴6s²
- La (Z = 57): [Xe]5d¹6s²
- Tb (Z = 65): [Xe]4f⁹6s²
- Er (Z = 68): [Xe]4f¹²6s²

Step 2: Determine the electronic configuration of the +3 ions.
- Nd³⁺: [Xe]4f³ (loses 6s² and one 4f electron)
- La³⁺: [Xe] (loses 5d¹ and 6s² electrons)
- Tb³⁺: [Xe]4f⁸ (loses 6s² and one 4f electron)
- Er³⁺: [Xe]4f¹¹ (loses 6s² and one 4f electron)

Step 3: Determine if the ions are diamagnetic or paramagnetic.
- Nd³⁺: [Xe]4f³ has 3 unpaired electrons, so it is paramagnetic.
- La³⁺: [Xe] has no unpaired electrons, so it is diamagnetic.
- Tb³⁺: [Xe]4f⁸ has unpaired electrons, so it is paramagnetic.
- Er³⁺: [Xe]4f¹¹ has unpaired electrons, so it is paramagnetic.

Step 4: Conclude the answer.\nTherefore, the correct answer is La³⁺.

Why other options are wrong:
- Option A: Nd³⁺ has unpaired electrons and is paramagnetic.
- Option C: Tb³⁺ has unpaired electrons and is paramagnetic.
- Option D: Er³⁺ has unpaired electrons and is paramagnetic.

Key Concepts:
- Diamagnetism
- Paramagnetism
- Electronic configuration
- Lanthanide ions

Memory Aids:
- Diamagnetic substances have no unpaired electrons.
- Paramagnetic substances have unpaired electrons.

Quick Tips:
- Write the electronic configuration of the ion.
- If there are unpaired electrons, the ion is paramagnetic.
- If there are no unpaired electrons, the ion is diamagnetic.

69 Medium

69 of 85Question permalink

Application +4 | -1

Which of the following is paramagnetic?

View answer and explanation

Correct option: A

\mathrm{CuC}{\mathrm{l}_2}

Option A (

\mathrm{CuC}{\mathrm{l}_2}

) is correct because Cu(II) has an unpaired electron, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{CuC}{\mathrm{l}_2}$ . Paramagnetic substances have unpaired electrons.

Step-by-Step Solution:

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

Step 2: For $\mathrm{CuC}{\mathrm{l}_2}$ , copper is in the +2 oxidation state ( $\mathrm{Cu}^{2+}$ ). The electronic configuration of $\mathrm{Cu}$ is $[\mathrm{Ar}]3d^{10}4s^1$ . The electronic configuration of $\mathrm{Cu}^{2+}$ is $[\mathrm{Ar}]3d^9$ . This has one unpaired electron.

Step 3: For $\mathrm{CaC}{\mathrm{l}_2}$ , calcium is in the +2 oxidation state ( $\mathrm{Ca}^{2+}$ ). The electronic configuration of $\mathrm{Ca}$ is $[\mathrm{Ar}]4s^2$ . The electronic configuration of $\mathrm{Ca}^{2+}$ is $[\mathrm{Ar}]$ . This has no unpaired electrons.

Step 4: For $\mathrm{CdC}{\mathrm{l}_2}$ , cadmium is in the +2 oxidation state ( $\mathrm{Cd}^{2+}$ ). The electronic configuration of $\mathrm{Cd}$ is $[\mathrm{Kr}]4d^{10}5s^2$ . The electronic configuration of $\mathrm{Cd}^{2+}$ is $[\mathrm{Kr}]4d^{10}$ . This has no unpaired electrons.

Step 5: Since $\mathrm{CuC}{\mathrm{l}_2}$ has unpaired electrons, it is paramagnetic.

Why other options are wrong:
- Option B: $\mathrm{CaC}{\mathrm{l}_2}$ is wrong because $\mathrm{Ca}^{2+}$ has a noble gas configuration and no unpaired electrons, making it diamagnetic.
- Option C: $\mathrm{CdC}{\mathrm{l}_2}$ is wrong because $\mathrm{Cd}^{2+}$ has a completely filled d-orbital configuration and no unpaired electrons, making it diamagnetic.
- Option D: None of these is wrong because option A is paramagnetic.

Key Concepts:
- Paramagnetism
- Electronic configuration
- Oxidation states

Memory Aids:
- Paramagnetic: unpaired electrons
- Diamagnetic: all electrons paired

Quick Tips:
- Determine the number of unpaired electrons in the metal ion. If there are unpaired electrons, the substance is paramagnetic.

70 Medium

70 of 85Question permalink

Remembering +4 | -1

Copper nitrate on strongly heating gives:

View answer and explanation

Correct option: B

Cupric oxide

Option B (Cupric oxide) is correct because copper nitrate decomposes upon strong heating to form cupric oxide, nitrogen dioxide, and oxygen.

Solution approach

Correct Answer + Core Reason: Cupric oxide. Copper nitrate on strongly heating gives cupric oxide.

Step-by-Step Solution:

Step 1: Write the chemical formula for copper nitrate: $Cu(NO_3)_2$ .

Step 2: Write the balanced chemical equation for the thermal decomposition of copper nitrate:
$2Cu(NO_3)_2(s) \rightarrow 2CuO(s) + 4NO_2(g) + O_2(g)$

Step 3: Identify the products of the reaction. Copper nitrate decomposes to form cupric oxide (CuO), nitrogen dioxide ( $NO_2$ ), and oxygen ( $O_2$ ).

Why other options are wrong:
- Option A: Cu is not formed during the thermal decomposition of copper nitrate.
- Option B: Cupric oxide (CuO) is formed during the thermal decomposition of copper nitrate.
- Option C: Cuprous oxide ( $Cu_2O$ ) is not the primary product of the thermal decomposition of copper nitrate under strong heating.
- Option D: cupric nitrate is the reactant, not the product.

Key Concepts:
- Thermal decomposition
- Chemical reactions
- Products of decomposition

Memory Aids:
- Remember the products of thermal decomposition of metal nitrates.

Quick Tips:
- Metal nitrates typically decompose to form metal oxides, nitrogen dioxide, and oxygen.

71 Medium

71 of 85Question permalink

Remembering +4 | -1

Lanthanoids and actinoids resembles in:

View answer and explanation

Correct option: A

Electronic configuration

Option A (Electronic configuration) is correct because both lanthanoids and actinoids involve the filling of f-orbitals in their electronic configurations.

Solution approach

Correct Answer + Core Reason: Electronic configuration. Both involve filling f-orbitals.

Step-by-Step Solution:

Step 1: Recall the properties of lanthanoids and actinoids.

Step 2: Both lanthanoids and actinoids involve the filling of f-orbitals (4f for lanthanoids and 5f for actinoids) in their electronic configurations.

Why other options are wrong:
- Option B (Oxidation state): While both exhibit variable oxidation states, the common oxidation states and their stability differ.
- Option C (Ionisation energy): Ionization energies differ between lanthanoids and actinoids.
- Option D (Formation of complex): Complex formation tendencies differ between lanthanoids and actinoids.

Key Concepts:
- Electronic configuration of lanthanoids and actinoids
- Filling of f-orbitals

Memory Aids:
- Both lanthanoids and actinoids are f-block elements.

Quick Tips:
- Remember that both lanthanoids and actinoids involve f-orbital filling.

72 Medium

72 of 85Question permalink

Understanding +4 | -1

Which one of the following forms a complex of coordination number 2 with excess of

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

ions?

View answer and explanation

Correct option: B

\mathrm{A}{\mathrm{g}^ + }

Option B (

\mathrm{A}{\mathrm{g}^ + }

) is correct because

Ag^+

forms a linear complex

[Ag(CN)_2]^-

with a coordination number of 2.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{g}^ + }$ . $Ag^+$ forms a linear complex with a coordination number of 2.

Step-by-Step Solution:

Step 1: Consider the complex formation of each ion with excess $CN^-$ .

Step 2: Determine the coordination number of the complex formed.

Step 3: $Cu^+$ can form $[Cu(CN)_2]^-$ or $[Cu(CN)_4]^{3-}$ .
$Ag^+$ forms $[Ag(CN)_2]^-$ , which has a coordination number of 2.
$Ni^{2+}$ forms $[Ni(CN)_4]^{2-}$ , which has a coordination number of 4.
$Fe^{2+}$ forms $[Fe(CN)_6]^{4-}$ , which has a coordination number of 6.

Why other options are wrong:
- Option A: $Cu^+$ can form complexes with coordination number 2 or 4.
- Option C: $N{i^{2 + }}$ forms a complex with coordination number 4.
- Option D: $F{e^{2 + }}$ forms a complex with coordination number 6.

Key Concepts:
- Complex formation
- Coordination number
- Transition metal chemistry

Memory Aids:
- Remember the common complexes formed by these ions.

Quick Tips:
- Recall the coordination numbers of common complexes.

73 Medium

73 of 85Question permalink

Application +4 | -1

Ferrous ion changes to

X

ion, on reacting with acidified hydrogen peroxide. The number of

d

-electrons present in

X

and its magnetic moment (in BM) are, respectively

View answer and explanation

Correct option: B

5 and 5.92

Option B (5 and 5.92) is correct because ferrous ion (

Fe^{2+}

) is oxidized to ferric ion (

Fe^{3+}

), which has 5 d-electrons and a magnetic moment of approximately 5.92 BM.

Solution approach

Correct Answer + Core Reason: 5 and 5.92. $Fe^{2+}$ is oxidized to $Fe^{3+}$ , which has 5 d-electrons and a magnetic moment of approximately 5.92 BM.

Step-by-Step Solution:

Step 1: Identify the oxidation states of iron.

Step 2: Ferrous ion is $Fe^{2+}$ .

Step 3: Acidified hydrogen peroxide oxidizes $Fe^{2+}$ to $Fe^{3+}$ . Therefore, X is $Fe^{3+}$ .

Step 4: Determine the electronic configuration of $Fe^{3+}$ .

Step 5: Fe (Z=26): $[\mathrm{Ar}] 3d^6 4s^2$

Step 6: $Fe^{3+}$ : $[\mathrm{Ar}] 3d^5$ . The number of d-electrons is 5.

Step 7: Calculate the spin-only magnetic moment using the formula: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

Step 8: Since $Fe^{3+}$ has 5 unpaired electrons, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM

Why other options are wrong:
- Option A: 6 and 6.93 is incorrect. $Fe^{3+}$ has 5 d-electrons, not 6.
- Option C: 5 and 4.9 is incorrect. The magnetic moment is approximately 5.92 BM, not 4.9 BM.
- Option D: 4 and 5.92 is incorrect. $Fe^{3+}$ has 5 d-electrons, not 4.

Key Concepts:
- Oxidation states
- Electronic configuration
- Spin-only magnetic moment

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

Quick Tips:
- Determine the number of unpaired electrons to calculate the magnetic moment.

74 Medium

74 of 85Question permalink

Analyzing +4 | -1

Identify the incorrect statement among the following

View answer and explanation

Correct option: D

4f

and

5f

-orbitals are equally shielded.

Option D is correct because 4f orbitals are shielded more effectively than 5f orbitals due to their spatial distribution.

Solution approach

Correct Answer + Core Reason: $4f$ and $5f$ -orbitals are equally shielded. $4f$ orbitals are shielded more effectively than $5f$ orbitals.

Step-by-Step Solution:

Step 1: Analyze the given statements regarding d-block and f-block elements.

Step 2: Consider the statement about irregular chemical properties of d-block elements. d-block elements show irregular and erratic chemical properties among themselves due to variable oxidation states and complex formation.

Step 3: Consider the statement about La and Lu. La and Lu have partially filled d-orbitals and no other partially filled orbital.

Step 4: Consider the statement about the chemistry of lanthanoids. The chemistry of various lanthanoids is very similar due to the lanthanoid contraction and similar electronic configurations.

Step 5: Consider the statement about the shielding of 4f and 5f orbitals. 4f orbitals are shielded more effectively than 5f orbitals due to their spatial distribution. The 5f orbitals extend further into space and are less shielded from the nuclear charge.

Why other options are wrong:
- Option A: $d$ -block elements show irregular and erratic chemical properties among themselves. This statement is correct.
- Option B: La and Lu have partially filled $d$ -orbitals and no other partially filled orbital. This statement is correct.
- Option C: The chemistry of various lanthanoids is very similar. This statement is correct.

Key Concepts:
- Properties of d-block and f-block elements
- Lanthanoid contraction
- Shielding effect

Memory Aids:
- Remember that 4f orbitals are more shielded than 5f orbitals.

Quick Tips:
- Recall the properties of d-block and f-block elements to identify the incorrect statement.

75 Medium

75 of 85Question permalink

Application +4 | -1

KMn{O_4}

reacts with oxalic acid according to the equation

2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O

Here 20 ml of 0.1 M

KMn{O_4}

will react with

View answer and explanation

Correct option: A

50 ml of 0.1 M

{H_2}{C_2}{O_4}

The correct answer is 50 ml of 0.1 M

{H_2}{C_2}{O_4}

because the stoichiometry of the reaction dictates that 2 moles of

KMnO_4

react with 5 moles of oxalic acid.

Solution approach

Correct Answer + Core Reason: 50 ml of 0.1 M ${H_2}{C_2}{O_4}$ . The stoichiometry of the reaction requires 2 moles of $KMnO_4$ to react with 5 moles of ${H_2}{C_2}{O_4}$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$

Step 2: Calculate the moles of $KMnO_4$ used: moles = Molarity x Volume (in L) = 0.1 M x (20 ml / 1000 ml/L) = 0.002 moles.

Step 3: Determine the moles of ${H_2}{C_2}{O_4}$ required using the stoichiometry of the reaction: (5 moles ${H_2}{C_2}{O_4}$ / 2 moles $KMnO_4$ ) x 0.002 moles $KMnO_4$ = 0.005 moles ${H_2}{C_2}{O_4}$ .

Step 4: Calculate the volume of 0.1 M ${H_2}{C_2}{O_4}$ required: Volume (in L) = moles / Molarity = 0.005 moles / 0.1 M = 0.05 L = 50 ml.

Why other options are wrong:
- Option A: 50 ml of 0.5 M ${H_2}{C_2}{O_4}$ would be an incorrect volume and concentration.
- Option B: 20 ml of 0.5 M ${H_2}{C_2}{O_4}$ would be an incorrect volume and concentration.
- Option D: 20 ml of 0.1 M ${H_2}{C_2}{O_4}$ would be an incorrect volume and concentration.

Key Concepts:
- Stoichiometry
- Molarity
- Redox Reactions

Memory Aids:
- Molarity = moles / Volume (in L)

Quick Tips:
- Always balance the chemical equation first.
- Convert volumes to liters before calculations.

76 Medium

76 of 85Question permalink

Understanding +4 | -1

{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

on heating gives a gas which is also given by

View answer and explanation

Correct option: A

Heating

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_2}

Option A (Heating

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_2}

) is correct because it also produces nitrogen gas upon heating, similar to the given reaction.

Solution approach

Correct Answer + Core Reason: Heating $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_2}$ produces the same gas, nitrogen, as the given reaction.

Step-by-Step Solution:

Step 1: The given reaction is the thermal decomposition of ammonium dichromate: ${\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}(s) \to {\mathrm{N}_2}(g) + 4{\mathrm{H}_2}\mathrm{O}(g) + \mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}(s)$ .

Step 2: Option A: Heating ammonium nitrite: $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_2}(s) \to {\mathrm{N}_2}(g) + 2{\mathrm{H}_2}\mathrm{O}(g)$ . This reaction also produces nitrogen gas.

Step 3: Option B: Heating ammonium nitrate: $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}(s) \to {\mathrm{N}_2}\mathrm{O}(g) + 2{\mathrm{H}_2}\mathrm{O}(g)$ . This reaction produces nitrous oxide, not nitrogen.

Step 4: Option C: $\mathrm{M}{\mathrm{g}_3}{\mathrm{N}_2} + {\mathrm{H}_2}\mathrm{O} \to 3\mathrm{Mg(OH}{\mathrm{)}_2} + 2\mathrm{N}{\mathrm{H}_3}$ . This reaction produces ammonia gas.

Step 5: Option D: Na(Comp.)+ ${\mathrm{H}_2}{\mathrm{O}_2}$ is not a standard reaction and doesn't directly produce nitrogen gas.

Why other options are wrong:
- Option B: Heating ammonium nitrate produces nitrous oxide ( $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3} \to {\mathrm{N}_2}\mathrm{O} + 2{\mathrm{H}_2}\mathrm{O}$ ), not nitrogen.
- Option C: Magnesium nitride reacts with water to produce ammonia ( $\mathrm{M}{\mathrm{g}_3}{\mathrm{N}_2} + 3{\mathrm{H}_2}\mathrm{O} \to 3\mathrm{Mg(OH}{\mathrm{)}_2} + 2\mathrm{N}{\mathrm{H}_3}$ ), not nitrogen.
- Option D: This option is vague and doesn't represent a standard reaction producing nitrogen gas.

Key Concepts:
- Thermal decomposition reactions
- Reactions of ammonium salts
- Production of gases

Memory Aids:
- Remember the decomposition products of ammonium salts.

Quick Tips:
- Know the common reactions of ammonium salts and their decomposition products.

77 Medium

77 of 85Question permalink

Analyzing +4 | -1

{Na}_2 {O}_2

View answer and explanation

Correct option: D

All are correct properties of

{Na}_2 {O}_2

Option D (All are correct properties of

{Na}_2 {O}_2

) is correct because

{Na}_2 {O}_2

is diamagnetic, it's a salt of dibasic acid

{H}_2 {O}_2

, and it oxidizes

{Cr}^{3+}

to

{CrO}_4^{2-}

.

Solution approach

Correct Answer + Core Reason: All are correct properties of ${Na}_2 {O}_2$ . It is diamagnetic, a salt of ${H}_2 {O}_2$ , and oxidizes ${Cr}^{3+}$ to ${CrO}_4^{2-}$ .

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: Is diamagnetic in nature. ${Na}_2 {O}_2$ has the structure ${Na}^+ [{O}-{O}]^{2-} {Na}^+$ . The peroxide ion ( ${O}_2^{2-}$ ) has all its electrons paired, making it diamagnetic.
- B: Is salt of dibasic acid ${H}_2 {O}_2$ . ${Na}_2 {O}_2$ is formed by replacing both acidic hydrogens of ${H}_2 {O}_2$ with sodium ions.
- C: Oxidizes ${Cr}^{3+}$ (green) to ${CrO}_4^{2-}$ (yellow). ${Na}_2 {O}_2$ is a strong oxidizing agent and can oxidize ${Cr}^{3+}$ to ${CrO}_4^{2-}$ .

Step 2: Recall the properties of ${Na}_2 {O}_2$ .

Step 3: Determine the correctness of each statement: All three statements are correct.

Why other options are wrong:
- Option A: This is only one of the correct properties.
- Option B: This is only one of the correct properties.
- Option C: This is only one of the correct properties.

Key Concepts:
- Properties of ${Na}_2 {O}_2$
- Diamagnetism
- Oxidation reactions

Memory Aids:
- Remember that ${Na}_2 {O}_2$ is a strong oxidizing agent.

Quick Tips:
- Know the properties of common peroxides.

78 Medium

78 of 85Question permalink

Application +4 | -1

Which of the following ions has the highest magnetic moment?

View answer and explanation

Correct option: C

\mathrm{M}{\mathrm{n}^{2 + }}

Option C (

\mathrm{Mn}^{2+}

) is correct because it has the highest number of unpaired electrons, leading to the highest magnetic moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}{\mathrm{n}^{2 + }}$ . $\mathrm{M}{\mathrm{n}^{2 + }}$ has the most unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each ion.

Step 2: Calculate the number of unpaired electrons for each ion:
- $\mathrm{Ti}^{3 + }$ : [Ar] 3d¹ (1 unpaired electron)
- $\mathrm{Sc}^{3 + }$ : [Ar] (0 unpaired electrons)
- $\mathrm{Mn}^{2 + }$ : [Ar] 3d⁵ (5 unpaired electrons)
- $\mathrm{Zn}^{2 + }$ : [Ar] 3d¹⁰ (0 unpaired electrons)

Step 3: The magnetic moment is directly proportional to the number of unpaired electrons. The ion with the most unpaired electrons will have the highest magnetic moment.

Step 4: $\mathrm{Mn}^{2 + }$ has 5 unpaired electrons, which is the maximum among the given options.

Why other options are wrong:
- Option A: $\mathrm{Ti}^{3 + }$ has only 1 unpaired electron.
- Option B: $\mathrm{Sc}^{3 + }$ has 0 unpaired electrons.
- Option D: $\mathrm{Zn}^{2 + }$ has 0 unpaired electrons.

Key Concepts:
- Electronic configuration
- Hund's rule
- Magnetic moment and unpaired electrons

Memory Aids:
- Remember Hund's rule for filling orbitals.

Quick Tips:
- The more unpaired electrons, the higher the magnetic moment.

79 Medium

79 of 85Question permalink

Application +4 | -1

Which of the following electronic configuration represents the maximum magnetic moment?

View answer and explanation

Correct option: D

{d^6}

The correct answer is

{d^6}

because it has the maximum number of unpaired electrons (4) when considering Hund's rule, leading to the highest magnetic moment.

Solution approach

Correct Answer + Core Reason: ${d^6}$ . The $d^6$ configuration has the maximum number of unpaired electrons among the given options.

Step-by-Step Solution:

Step 1: Determine the number of unpaired electrons for each electronic configuration using Hund's rule.

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

Step 3: Compare the magnetic moments.

Step 4: Analyze each option:
- A. ${d^3}$ : 3 unpaired electrons, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM
- B. ${d^2}$ : 2 unpaired electrons, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ BM
- C. ${d^8}$ : 2 unpaired electrons, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ BM
- D. ${d^6}$ : 4 unpaired electrons (assuming weak field ligand), $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM

Step 5: The ${d^6}$ configuration has the maximum magnetic moment.

Why other options are wrong:
- Option A: ${d^3}$ has fewer unpaired electrons (3) than ${d^6}$ (4), resulting in a lower magnetic moment.
- Option B: ${d^2}$ has fewer unpaired electrons (2) than ${d^6}$ (4), resulting in a lower magnetic moment.
- Option C: ${d^8}$ has fewer unpaired electrons (2) than ${d^6}$ (4), resulting in a lower magnetic moment.

Key Concepts:
- Hund's Rule
- Spin-Only Magnetic Moment
- Electronic Configuration

Memory Aids:
- Hund's Rule: Maximize unpaired electrons before pairing.
- Magnetic moment formula: $\mu = \sqrt{n(n+2)}$

Quick Tips:
- Remember Hund's rule for filling orbitals.
- Consider the possibility of strong field ligands causing pairing.

80 Medium

80 of 85Question permalink

Application +4 | -1

Correct outer electronic configuration of trivalent cationic form of

{Np}

(Neptunium, atomic number of 93 ) is

View answer and explanation

Correct option: C

[R n] 5 f^4

Option C (

[R n] 5 f^4

) is correct because Neptunium (Np) has an atomic number of 93. After losing 3 electrons to form

Np^{3+}

, the resulting electronic configuration is

[Rn]5f^4

.

Solution approach

Correct Answer + Core Reason: $[R n] 5 f^4$ . After losing 3 electrons, the resulting electronic configuration is $[Rn]5f^4$ .

Step-by-Step Solution:

Step 1: Write the electronic configuration of neutral Neptunium (Np).\nNp (Z=93): $[Rn] 5f^4 6d^1 7s^2$

Step 2: Determine the electronic configuration of $Np^{3+}$ .\nWhen Neptunium loses 3 electrons, it loses the two 7s electrons and one 6d electron.
$Np^{3+}$ : $[Rn] 5f^4$

Why other options are wrong:
- **Option A ( $[R n] 5 f^2$ ): This is incorrect because it would correspond to losing 5 electrons from the neutral atom, not 3.
- Option B ( $[R n] 5 f^3$ ): This is incorrect because it would correspond to losing 4 electrons from the neutral atom, not 3.
- Option D ( $[R n] 5 f^3 6 d^1$ ): This is incorrect because the 6d electron is removed before the 5f electrons when forming the ion.

Key Concepts:
- Electronic configuration of actinides
- Ion formation
- Hund's rule

Memory Aids:
- Remember the order of filling orbitals: (n+l) rule.

Quick Tips:**
- Write out the electronic configuration of the neutral atom first, then remove electrons from the outermost shells (highest n value) first.

81 Medium

81 of 85Question permalink

Analyzing +4 | -1

In which of the following cases, the stability of two oxidation states is correctly represented

View answer and explanation

Correct option: B

M{n^{2 + }} > M{n^{3 + }}

Option B (

M{n^{2 + }} > M{n^{3 + }}

) is correct because

Mn^{2+}

has a stable half-filled

d^5

configuration, making it more stable than

Mn^{3+}

.

Solution approach

Correct Answer + Core Reason: $M{n^{2 + }} > M{n^{3 + }}$ . $Mn^{2+}$ has a stable half-filled $d^5$ configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the ions.

Step 2: $Ti^{3+}$ : $[Ar] 3d^1$
$Ti^{4+}$ : $[Ar]$

Step 3: $Mn^{2+}$ : $[Ar] 3d^5$ (half-filled, stable)
$Mn^{3+}$ : $[Ar] 3d^4$

Step 4: $Fe^{2+}$ : $[Ar] 3d^6$
$Fe^{3+}$ : $[Ar] 3d^5$ (half-filled, stable)

Step 5: $Cu^{+}$ : $[Ar] 3d^{10}$ (fully-filled, stable)
$Cu^{2+}$ : $[Ar] 3d^9$

Step 6: Compare the stability of the oxidation states.

Step 7: $Mn^{2+}$ is more stable than $Mn^{3+}$ due to the half-filled $d^5$ configuration.

Why other options are wrong:
- Option A: $Ti^{4+}$ is more stable than $Ti^{3+}$ because it has a noble gas configuration.
- Option C: $Fe^{3+}$ is more stable than $Fe^{2+}$ because it has a half-filled $d^5$ configuration.
- Option D: $Cu^{+}$ is more stable than $Cu^{2+}$ because it has a fully-filled $d^{10}$ configuration.

Key Concepts:
- Stability of ions
- Electronic configuration
- Half-filled and fully-filled orbitals

Memory Aids:
- Remember that half-filled and fully-filled orbitals are more stable.

Quick Tips:
- Consider the electronic configurations when comparing the stability of different oxidation states.

82 Medium

82 of 85Question permalink

Understanding +4 | -1

When excess of

\mathrm{SnC}{\mathrm{l}_2}

is added to a solution of

\mathrm{HgC}{\mathrm{l}_2}

, a white ppt. turning to grey is obtained. This grey colour is due to the formation of:

View answer and explanation

Correct option: D

\mathrm{H}{\mathrm{g}_2}

The grey color is due to the formation of mercury (

\mathrm{H}{\mathrm{g}_2}

) because

\mathrm{SnC}{\mathrm{l}_2}

reduces

\mathrm{HgC}{\mathrm{l}_2}

to metallic mercury.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}{\mathrm{g}_2}$ . $\mathrm{SnC}{\mathrm{l}_2}$ reduces $\mathrm{HgC}{\mathrm{l}_2}$ to Hg.

Step-by-Step Solution:

Step 1: Understand the redox reaction occurring.

Step 2: $\mathrm{SnC}{\mathrm{l}_2}$ is a reducing agent and reduces $\mathrm{HgC}{\mathrm{l}_2}$ to metallic mercury (Hg).

Step 3: The reaction is: $HgCl_2 + SnCl_2 \rightarrow Hg + SnCl_4$

Step 4: The metallic mercury is finely divided and appears grey.

Why other options are wrong:
- Option A: $\mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2}$ is white, not grey.
- Option B: $\mathrm{SnC}{\mathrm{l}_4}$ is a colorless solution.
- Option C: $\mathrm{Sn}$ is not formed in this reaction.

Key Concepts:
- Redox reactions
- Reducing agents
- Formation of metallic mercury

Memory Aids:
- $\mathrm{SnC}{\mathrm{l}_2}$ reduces $\mathrm{HgC}{\mathrm{l}_2}$ to Hg.

Quick Tips:
- Remember the common reducing agents and their reactions.

83 Medium

83 of 85Question permalink

Understanding +4 | -1

Transition elements exhibit positive oxidation states only. This is because of:

View answer and explanation

Correct option: B

Their electropositive nature

Option B (Their electropositive nature) is correct because transition elements readily lose electrons due to their relatively low ionization energies, leading to positive oxidation states.

Solution approach

Correct Answer + Core Reason: Their electropositive nature. Transition elements lose electrons to form positive oxidation states.

Step-by-Step Solution:

Step 1: Transition elements have a tendency to lose electrons due to their electronic configurations and relatively low ionization energies.

Step 2: Electropositivity is the tendency of an element to lose electrons and form positive ions.

Step 3: The electropositive nature of transition elements allows them to exhibit positive oxidation states.

Step 4: The large size of the atoms (Option A) is a contributing factor but not the primary reason.

Step 5: Electronegativity (Option C) is the opposite of electropositivity and is not relevant here.

Step 6: Paramagnetic nature (Option D) is related to the presence of unpaired electrons but does not directly explain the positive oxidation states.

Why other options are wrong:
- Option A (Their large size of the atoms): While size can influence ionization energy, it's not the primary reason for positive oxidation states.
- Option C (Their electronegative nature): Transition elements are not electronegative; they are electropositive.
- Option D (Their paramagnetic nature): Paramagnetism is a consequence of unpaired electrons, not the cause of positive oxidation states.

Key Concepts:
- Electropositivity
- Oxidation states
- Ionization energy

Memory Aids:
- Remember that electropositive elements tend to lose electrons and form positive ions.

Quick Tips:
- Understand the relationship between electropositivity and oxidation states.

84 Medium

84 of 85Question permalink

Understanding +4 | -1

The reaction

\mathrm{MnO}_4^ - + e \to \mathrm{MnO}_4^{2 - }

takes place in:

View answer and explanation

Correct option: A

Basic medium

Option A (Basic medium) is correct because the reduction of

MnO_4^-

to

MnO_4^{2-}

is favored in a basic medium.

Solution approach

Correct Answer + Core Reason: Basic medium. The reduction of $MnO_4^-$ to $MnO_4^{2-}$ is favored in basic medium.

Step-by-Step Solution:

Step 1: Write the half-reaction: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$

Step 2: Consider the effect of pH on the reaction.\nThis reaction is favored in a basic medium because it does not involve $H^+$ or $OH^-$ ions directly.

Why other options are wrong:
- Option B: Acidic medium would favor reactions that consume hydroxide ions or produce hydrogen ions, which is not the case here.
- Option C: Neutral medium is less favorable than basic medium for this specific reaction.
- Option D: While some reactions can occur in both acidic and basic media, this specific reduction is more favorable in basic conditions.

Key Concepts:
- Redox reactions
- Effect of pH on redox reactions

Memory Aids:
- Remember that the reduction of permanganate to manganate is favored in basic medium.

Quick Tips:
- Consider the effect of pH on redox reactions.

85 Medium

85 of 85Question permalink

Analyzing +4 | -1

For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?

View answer and explanation

Correct option: B

Mn > Fe > Cr > Co

Option B (Mn > Fe > Cr > Co) is correct because the stability of the +2 oxidation state is related to achieving stable electronic configurations.

Solution approach

Correct Answer + Core Reason: Mn > Fe > Cr > Co. The stability of +2 oxidation state is related to achieving stable electronic configurations.

Step-by-Step Solution:

Step 1: Write the electronic configurations of Cr, Mn, Fe, and Co.\nCr: [Ar] 3d5 4s1\nMn: [Ar] 3d5 4s2\nFe: [Ar] 3d6 4s2\nCo: [Ar] 3d7 4s2

Step 2: Write the electronic configurations of the M2+ ions.\nCr2+: [Ar] 3d4\nMn2+: [Ar] 3d5\nFe2+: [Ar] 3d6\nCo2+: [Ar] 3d7

Step 3: Analyze the stability of the M2+ ions. Mn2+ has a stable d5 configuration. Fe2+ is more stable than Co2+ because removing two electrons from Fe results in a less destabilized configuration than removing two electrons from Co. Cr2+ is less stable than Mn2+ because it does not have a stable d5 configuration.

Step 4: Determine the order of stability of the +2 oxidation state.\nMn > Fe > Cr > Co

Why other options are wrong:
- Option A: Cr > Mn > Co > Fe - Incorrect because Mn2+ has a stable d5 configuration.
- Option C: Fe > Mn > Co > Cr - Incorrect because Mn2+ has a stable d5 configuration.
- Option D: Co > Mn > Fe > Cr - Incorrect because Mn2+ has a stable d5 configuration.

Key Concepts:
- Electronic configuration
- Oxidation states
- Stability of half-filled orbitals

Memory Aids:
- Remember that half-filled orbitals are more stable.

Quick Tips:
- Write the electronic configurations of the elements and their ions to determine the stability of oxidation states.

Your Progress

D & f- BLOCK ELEMENTS practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice