${\rm{FeS}}{{\rm{O}}_4}.7{{\rm{H}}_2}{\rm{O}}$

${\rm{ZnS}}{{\rm{O}}_4}.7{{\rm{H}}_2}{\rm{O}}$

${\rm{CuS}}{{\rm{O}}_4}.5{{\rm{H}}_2}{\rm{O}}$

${\rm{CaS}}{{\rm{O}}_4}.\frac{1}{2}{\rm{\;}}{{\rm{H}}_2}{\rm{O}}$

The reaction of ${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ with NaCl and conc ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ gives

${\rm{Cr}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}$

${\rm{C}}{{\rm{r}}_2}{{\rm{O}}_3}$

d & f- BLOCK ELEMENTS MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Medium

1 of 206Question permalink

Application +4 | -1

A substance which is not paramagnetic is:

View answer and explanation

Correct option: B

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

The correct answer is

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

because it is diamagnetic, having no unpaired electrons due to the

Mn

being in +7 oxidation state with

d^0

configuration.

Solution approach

Correct Answer + Core Reason: $\mathrm{KMn}{\mathrm{O}_4}$ . Potassium permanganate is diamagnetic, not paramagnetic.

Step-by-Step Solution:

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

Step 2: Determine the electronic configuration of the metal ion.

Step 3: Identify if there are any unpaired electrons. Paramagnetic substances have unpaired electrons, while diamagnetic substances do not.

Step 4: Analyze each option:
- A. $\mathrm{Cr}{\left( {\mathrm{Cl}{\mathrm{O}_4}} \right)_3}$ : Cr is in +3 oxidation state, $d^3$ configuration, 3 unpaired electrons (paramagnetic).
- B. $\mathrm{KMn}{\mathrm{O}_4}$ : Mn is in +7 oxidation state, $d^0$ configuration, 0 unpaired electrons (diamagnetic).
- C. $\mathrm{TiC}{\mathrm{l}_3}$ : Ti is in +3 oxidation state, $d^1$ configuration, 1 unpaired electron (paramagnetic).
- D. $\mathrm{VOB}{\mathrm{r}_2}$ : V is in +4 oxidation state, $d^1$ configuration, 1 unpaired electron (paramagnetic).

Step 5: The substance without unpaired electrons is $\mathrm{KMn}{\mathrm{O}_4}$ .

Why other options are wrong:
- Option A: $\mathrm{Cr}{\left( {\mathrm{Cl}{\mathrm{O}_4}} \right)_3}$ is paramagnetic because $Cr^{3+}$ has a $d^3$ configuration with 3 unpaired electrons.
- Option C: $\mathrm{TiC}{\mathrm{l}_3}$ is paramagnetic because $Ti^{3+}$ has a $d^1$ configuration with 1 unpaired electron.
- Option D: $\mathrm{VOB}{\mathrm{r}_2}$ is paramagnetic because $V^{4+}$ has a $d^1$ configuration with 1 unpaired electron.

Key Concepts:
- Paramagnetism and Diamagnetism
- Electronic Configuration
- Oxidation State

Memory Aids:
- Paramagnetic: unpaired electrons, attracted to magnetic field.
- Diamagnetic: paired electrons, repelled by magnetic field.

Quick Tips:
- Determine the oxidation state of the metal ion first.
- Write the electronic configuration to identify unpaired electrons.

2 Medium

2 of 206Question permalink

Remembering +4 | -1

Green vitriol is

View answer and explanation

Correct option: A

\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}

Option A (

\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}

) is correct because green vitriol is the common name for iron(II) sulfate heptahydrate.

Solution approach

Correct Answer + Core Reason: $\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ . Green vitriol is iron(II) sulfate heptahydrate.

Step-by-Step Solution:

Step 1: Recall the definition of green vitriol. Green vitriol is the common name for hydrated iron(II) sulfate.

Step 2: Identify the chemical formula for iron(II) sulfate heptahydrate. The formula is $\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ .

Step 3: Conclude that the correct answer is $\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ .

Why other options are wrong:
- Option B: $\mathrm{ZnS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ : This is white vitriol (zinc sulfate heptahydrate).
- Option C: $\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}$ : This is blue vitriol (copper(II) sulfate pentahydrate).
- Option D: $\mathrm{CaS}{\mathrm{O}_4}.\frac{1}{2}\mathrm{\;}{\mathrm{H}_2}\mathrm{O}$ : This is calcium sulfate hemihydrate (plaster of Paris).

Key Concepts:
- Common names of chemical compounds
- Hydrated salts

Memory Aids:
- Green vitriol = Iron(II) sulfate heptahydrate
- Blue vitriol = Copper(II) sulfate pentahydrate
- White vitriol = Zinc sulfate heptahydrate

Quick Tips:
- Memorize the common names and formulas of important chemical compounds.

3 Medium

3 of 206Question permalink

Analyzing +4 | -1

Which is the correct order of ionic sizes?) At.no.∶Ce=58,Sn=50,Yb=70 and Lu=71)

View answer and explanation

Correct option: C

\mathrm{Sn} > Ce > Yb > Lu

Option C (

\mathrm{Sn} > Ce > Yb > Lu

) is correct because ionic size decreases across the lanthanide series due to lanthanide contraction, and Sn is not a lanthanide.

Solution approach

Correct Answer + Core Reason: $\mathrm{Sn} > Ce > Yb > Lu$ . Ionic size decreases across the lanthanide series due to lanthanide contraction, and Sn is not a lanthanide.

Step-by-Step Solution:

Step 1: Identify the elements and their positions in the periodic table.

Step 2: Ce, Yb, and Lu are lanthanides. Sn is in group 14.

Step 3: Lanthanide contraction: The ionic size decreases from Ce to Lu due to the poor shielding of the 4f electrons.

Step 4: Sn is not a lanthanide and has a larger ionic size than the lanthanides.

Step 5: Therefore, the correct order is $\mathrm{Sn} > Ce > Yb > Lu$ .

Why other options are wrong:
- Option A: Incorrect because it places Ce as larger than Sn, which is not generally true.
- Option B: Incorrect because it places Sn as smaller than Ce, which is not generally true.
- Option D: Incorrect because it shows the reverse order of lanthanide contraction and places Sn incorrectly.

Key Concepts:
- Ionic size
- Lanthanide contraction
- Periodic trends

Memory Aids:
- Remember that lanthanide contraction causes a decrease in ionic size across the lanthanide series.

Quick Tips:
- Consider the position of the elements in the periodic table and the effect of lanthanide contraction.

4 Easy

4 of 206Question permalink

Remembering +4 | -1

In the first transition series, the differentiating electron enters:

View answer and explanation

Correct option: C

3

d

-orbital

Option C (3

d

-orbital) is correct because the first transition series involves filling the 3

d

orbitals.

Solution approach

Correct Answer + Core Reason: 3 $d$ -orbital. The first transition series involves filling the 3 $d$ orbitals.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of transition metals.

Step 2: The first transition series consists of elements from Scandium (Sc) to Zinc (Zn). These elements have their differentiating electron entering the 3 $d$ orbitals.

Why other options are wrong:
- **Option A (5 $d$ -orbital):** The 5 $d$ orbitals are filled in the third transition series.
- **Option B (4 $d$ -orbital):** The 4 $d$ orbitals are filled in the second transition series.
- **Option D (2 $d$ -orbital):** 2 $d$ orbitals do not exist.

Key Concepts:
- Electronic configuration of transition metals
- Transition series

Memory Aids:
- First transition series = 3d orbitals.

Quick Tips:
- Remember the order of filling orbitals: 3d, 4d, 5d for the transition series.

5 Easy

5 of 206Question permalink

Understanding +4 | -1

Philosopher’s wool on heating with BaO at 1100

^\circ

C produce

View answer and explanation

Correct option: C

\mathrm{BaZn}{\mathrm{O}_2}

Option C (

\mathrm{BaZn}{\mathrm{O}_2}

) is correct because philosopher's wool (ZnO) reacts with BaO at high temperatures to form barium zincate.

Solution approach

Correct Answer + Core Reason: $\mathrm{BaZn}{\mathrm{O}_2}$ . Philosopher's wool (ZnO) reacts with BaO to form barium zincate.

Step-by-Step Solution:

Step 1: Recall the composition of philosopher’s wool.

Step 2: Philosopher’s wool is zinc oxide (ZnO).

Step 3: The reaction between ZnO and BaO at 1100 $^\circ$ C is:
$\mathrm{ZnO} + \mathrm{BaO} \to \mathrm{BaZn}{\mathrm{O}_2}$

Why other options are wrong:
- Option A: $\mathrm{Ba} + \mathrm{ZnC}{\mathrm{l}_2}$ : This is incorrect. The reaction does not produce barium and zinc chloride.
- Option B: $\mathrm{BaCd}{\mathrm{O}_2}$ : This is incorrect. Cadmium is not involved in the reaction.
- Option D: $\mathrm{Ba}{\mathrm{O}_2} + \mathrm{Zn}$ : This is incorrect. The reaction does not produce barium peroxide and zinc.

Key Concepts:
- Reactions of metal oxides
- Formation of complex oxides

Memory Aids:
- ZnO + BaO -> BaZnO2

Quick Tips:
- Remember the reactions of metal oxides at high temperatures.

6 Hard

6 of 206Question permalink

Application +4 | -1

A compound in which a metal ion

{\mathrm{M}^{x + }}

(Z=25) has a spin only magnetic moment of

\sqrt {24}

BM. The number of unpaired electrons in the compound and the oxidation state of the metal ion are respectively.

View answer and explanation

Correct option: D

4 and 3

Option D (4 and 3) is correct because a magnetic moment of

\sqrt{24}

BM corresponds to 4 unpaired electrons, and with Z=25, an oxidation state of +3 leaves 4 d-electrons unpaired.

Solution approach

Correct Answer + Core Reason: 4 and 3. The magnetic moment indicates 4 unpaired electrons, and the oxidation state of +3 matches the electronic configuration.

Step-by-Step Solution:

Step 1: Calculate the number of unpaired electrons using the spin-only magnetic moment formula:
$\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.\nGiven $\mu = \sqrt{24}$ BM, we have:
$\sqrt{24} = \sqrt{n(n+2)}$
$24 = n(n+2)$
$n^2 + 2n - 24 = 0$
$(n+6)(n-4) = 0$ \nSince n must be positive, n = 4. Therefore, there are 4 unpaired electrons.

Step 2: Determine the electronic configuration of the metal ion ${\mathrm{M}^{x + }}$ (Z=25):\nThe electronic configuration of M (Z=25) is $[Ar]3d^5 4s^2$ .

Step 3: Determine the oxidation state of the metal ion:\nWe need to find the oxidation state (x) such that ${\mathrm{M}^{x + }}$ has 4 unpaired electrons.\nIf x = 2, the electronic configuration of ${\mathrm{M}^{2 + }}$ is $[Ar]3d^5$ , which has 5 unpaired electrons. This doesn't match.\nIf x = 3, the electronic configuration of ${\mathrm{M}^{3 + }}$ is $[Ar]3d^4$ , which has 4 unpaired electrons. This matches.

Step 4: The number of unpaired electrons is 4, and the oxidation state of the metal ion is +3.

Why other options are wrong:
- Option A: Incorrect number of unpaired electrons and oxidation state.
- Option B: Incorrect number of unpaired electrons and oxidation state.
- Option C: Incorrect number of unpaired electrons and oxidation state.

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

Memory Aids:
- $\mu = \sqrt{n(n+2)}$ BM

Quick Tips:
- Remember the electronic configurations of common transition metals.
- Use the spin-only formula to calculate the number of unpaired electrons.

7 Medium

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Understanding +4 | -1

The reaction of

{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

with NaCl and conc

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

gives

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the reaction of potassium dichromate with NaCl and concentrated sulfuric acid produces chromyl chloride.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ . The reaction produces chromyl chloride.

Step-by-Step Solution:

Step 1: Recall the reaction between ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ , NaCl, and conc ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ .

Step 2: The reaction produces chromyl chloride ( $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ ).
${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} + 4\mathrm{NaCl} + 6{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to 2\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2} + 2{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + 4\mathrm{NaHSO}_4 + 3{\mathrm{H}_2}\mathrm{O}$

Step 3: Identify the product.

Why other options are wrong:
- Option B: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ : Chromium(III) oxide is not formed in this reaction.
- Option C: $\mathrm{CrC}{\mathrm{l}_3}$ : Chromium(III) chloride is not formed in this reaction.
- Option D: $\mathrm{CroC}{\mathrm{l}_2}$ : This is not a known compound.

Key Concepts:
- Chromyl chloride test
- Reactions of potassium dichromate

Memory Aids:
- Chromyl chloride is formed from dichromate, chloride, and sulfuric acid.

Quick Tips:
- Remember the chromyl chloride test for chloride ions.

8 Hard

8 of 206Question permalink

Analyzing +4 | -1

Assertion :

{\text{C}}{{\text{e}}^{ + 4}}

acts as oxidising agent in aqueous medium

Reason :

+4

is common oxidation state of lanthanides

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C (If Assertion is True but the Reason is False) is correct because

{\text{C}}{{\text{e}}^{ + 4}}

acts as an oxidizing agent, but +4 is not the common oxidation state of lanthanides.

Solution approach

Correct Answer + Core Reason: If Assertion is True but the Reason is False. ${\text{C}}{{\text{e}}^{ + 4}}$ acts as an oxidizing agent, but +4 is not the common oxidation state of lanthanides.

Step-by-Step Solution:

Step 1: Analyze the assertion.

Step 2: ${\text{C}}{{\text{e}}^{ + 4}}$ has a strong tendency to gain electrons and convert to ${\text{C}}{{\text{e}}^{ + 3}}$ , thus acting as an oxidizing agent. The assertion is true.

Step 3: Analyze the reason.

Step 4: The most common oxidation state of lanthanides is +3, not +4. The reason is false.

Step 5: Therefore, the correct answer is that the assertion is true, but the reason is false.

Why other options are wrong:
- Option A: The reason is false, so this option is incorrect.
- Option B: The reason is false, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Oxidizing agent
- Oxidation state
- Lanthanides

Memory Aids:
- Remember that ${\text{C}}{{\text{e}}^{ + 4}}$ is a good oxidizing agent.

Quick Tips:
- Consider the stability of different oxidation states when determining oxidizing or reducing properties.

9 Easy

9 of 206Question permalink

Remembering +4 | -1

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

is used as ……. Indicator for

\mathrm{FeS}{\mathrm{O}_4}vs.{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

titrations.

View answer and explanation

Correct option: B

External

Option B (External) is correct because

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

is used as an external indicator in

\mathrm{FeS}{\mathrm{O}_4}vs.{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

titrations, meaning it's used outside the titration flask.

Solution approach

Correct Answer + Core Reason: External. ${\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ is used as an external indicator.

Step-by-Step Solution:

Step 1: Understand the concept of indicators in titrations.

Step 2: Differentiate between self, external, and internal indicators.

Step 3: ${\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ is used as an external indicator in the titration of $\mathrm{FeS}{\mathrm{O}_4}$ with ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ . A drop of the titration mixture is removed and mixed with ${\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ on a spot plate. The endpoint is reached when a blue color is no longer observed.

Why other options are wrong:
- Option A: Self. A self-indicator is a reactant that changes color at the endpoint.
- Option C: Internal. An internal indicator is added directly to the titration flask.
- Option D: Not an. ${\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ is indeed an indicator.

Key Concepts:
- Indicators in titrations
- External indicators
- Redox titrations

Memory Aids:
- Remember that external indicators are used outside the titration flask.

Quick Tips:
- Know the common indicators used in different types of titrations.

10 Hard

10 of 206Question permalink

Analyzing +4 | -1

Assertion :

{\text{E}}{{\text{u}}^{2 + }}\,\& \,{\text{Y}}{{\text{b}}^{{\text{2 + }}}}

are reducing agents for their ions.

Reason: Both ions have stable half - filled configuration.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C is correct because while Eu2+ and Yb2+ are indeed reducing agents, the reason given about stable half-filled configuration is not entirely accurate or complete.

Solution approach

Correct Answer + Core Reason: If Assertion is True but the Reason is False. The assertion is correct, but the reason is not the primary explanation.

Step-by-Step Solution:

Step 1: Analyze the assertion: Eu2+ and Yb2+ act as reducing agents, meaning they readily lose electrons.

Step 2: Consider the electronic configurations: Eu (Z=63) is [Xe]4f76s2. Eu2+ is [Xe]4f7, which is half-filled and relatively stable. Yb (Z=70) is [Xe]4f146s2. Yb2+ is [Xe]4f14, which is fully filled and very stable.

Step 3: Evaluate the reason: While Eu2+ has a half-filled f-orbital, Yb2+ has a fully filled f-orbital. The stability of these configurations contributes to their reducing power, but it's not the sole reason. The high negative reduction potentials are also important.

Step 4: Conclusion: The assertion is true because Eu2+ and Yb2+ are good reducing agents. However, the reason is not entirely correct as it only partially explains the reducing ability and doesn't fully apply to Yb2+ which has a fully filled configuration.

Why other options are wrong:
- Option A: The reason is not a correct explanation.
- Option B: The reason is not a correct explanation.
- Option D: The assertion is true.

Key Concepts:
- Reducing agents
- Electronic configuration
- Stability of half-filled and fully filled orbitals
- Lanthanide chemistry

Memory Aids:
- Reducing agent: Loses electrons, gets oxidized.
- Oxidation state stability is related to electronic configuration.

Quick Tips:
- Remember the electronic configurations of lanthanides and their common oxidation states.
- Consider both electronic configuration and reduction potentials when analyzing redox behavior.

11 Medium

11 of 206Question permalink

Understanding +4 | -1

Which pair consists only acidic oxides?

View answer and explanation

Correct option: A

\mathrm{Cr}{\mathrm{O}_3},\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

Option A (

\mathrm{Cr}{\mathrm{O}_3},\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

) is correct because both chromium trioxide and manganese heptoxide are acidic oxides.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\mathrm{O}_3},\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}$ . Both oxides react with water to form acids.

Step-by-Step Solution:

Step 1: Acidic oxides are oxides that react with water to form acids or react with bases to form salts.

Step 2: $\mathrm{Cr}{\mathrm{O}_3}$ reacts with water to form chromic acid: $\mathrm{Cr}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{H}_2}\mathrm{Cr}{\mathrm{O}_4}$ .

Step 3: $\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}$ reacts with water to form permanganic acid: $\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7} + {\mathrm{H}_2}\mathrm{O} \to 2\mathrm{HMn}{\mathrm{O}_4}$ .

Step 4: $\mathrm{Zn}{\mathrm{O}_2}$ and $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ are amphoteric oxides, meaning they can act as both acids and bases.

Step 5: $\mathrm{CaO}$ and $\mathrm{N}{\mathrm{a}_2}\mathrm{O}$ are basic oxides, meaning they react with water to form bases.

Why other options are wrong:
- **Option B ( $\mathrm{Zn}{\mathrm{O}_2},\mathrm{\;A}{\mathrm{l}_2}{\mathrm{O}_3}$ ): Both zinc oxide and aluminum oxide are amphoteric, not acidic.
- Option C ( $\mathrm{CaO},\mathrm{\;ZnO}$ ): Calcium oxide is a basic oxide, and zinc oxide is amphoteric.
- Option D ( $\mathrm{N}{\mathrm{a}_2}\mathrm{O},\mathrm{\;A}{\mathrm{l}_2}{\mathrm{O}_3}$ ): Sodium oxide is a basic oxide, and aluminum oxide is amphoteric.

Key Concepts:
- Acidic oxides
- Basic oxides
- Amphoteric oxides

Memory Aids:
- Remember that oxides of highly electronegative elements tend to be acidic.

Quick Tips:**
- Know the difference between acidic, basic, and amphoteric oxides.

12 Medium

12 of 206Question permalink

Application +4 | -1

Which of the following ions has a magnetic moment of 5.93 BM?

(At. no. V=23, Cr=24, Mn=25, Fe=26)

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because

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

has 5 unpaired electrons, resulting in a magnetic moment close to 5.93 BM.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}{\mathrm{n}^{2 + }}$ . $\mathrm{M}{\mathrm{n}^{2 + }}$ has 5 unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each ion.

Step 2: $\mathrm{V}^{3+}$ (Z=23): $[\mathrm{Ar}] 3d^2$ . 2 unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ BM

Step 3: $\mathrm{Cr}^{2+}$ (Z=24): $[\mathrm{Ar}] 3d^4$ . 4 unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM

Step 4: $\mathrm{Mn}^{2+}$ (Z=25): $[\mathrm{Ar}] 3d^5$ . 5 unpaired electrons. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM

Step 5: $\mathrm{Fe}^{2+}$ (Z=26): $[\mathrm{Ar}] 3d^6$ . 4 unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM

Step 6: $\mathrm{Mn}^{2+}$ has a magnetic moment closest to 5.93 BM.

Why other options are wrong:
- Option B: $\mathrm{Fe}^{2+}$ has a magnetic moment of approximately 4.90 BM.
- Option C: $\mathrm{C}{\mathrm{r}^{2 + }}$ has a magnetic moment of approximately 4.90 BM.
- Option D: ${\mathrm{V}^{3 + }}$ has a magnetic moment of approximately 2.83 BM.

Key Concepts:
- Electronic configuration
- Spin-only magnetic moment
- Hund's rule

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.

13 Easy

13 of 206Question permalink

Analyzing +4 | -1

Given below are two statements:

Statement I : Aqueous solution of

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

is preferred as a primary standard in volumetric analysis over

\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7

aqueous solution.

Statement II :

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

has a higher solubility in water than

\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7

.

In the light of the above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: C

Statement I is true but Statement II is false

Option C (Statement I is true but Statement II is false) is correct because K2Cr2O7 is preferred as a primary standard due to its lower hygroscopicity and higher molar mass, but it has lower solubility than Na2Cr2O7.

Solution approach

Correct Answer + Core Reason: Statement I is true but Statement II is false. K2Cr2O7 is preferred as a primary standard, but it has lower solubility.

Step-by-Step Solution:

Step 1: Analyze Statement I:\nAqueous solution of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is preferred as a primary standard in volumetric analysis over $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ aqueous solution. This statement is true. $K_2Cr_2O_7$ is preferred because it is less hygroscopic and has a higher molar mass, making it easier to weigh accurately.

Step 2: Analyze Statement II:
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ has a higher solubility in water than $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ . This statement is false. Sodium salts are generally more soluble than potassium salts due to the smaller size and higher charge density of $Na^+$ .

Step 3: Determine the correct answer:\nStatement I is true, and Statement II is false.

Why other options are wrong:
- Option A: Statement I is true, not false.
- Option B: Statement II is false, not true.
- Option D: Statement I is true, and Statement II is false.

Key Concepts:
- Primary standards in volumetric analysis
- Solubility of ionic compounds
- Hygroscopic nature of salts

Memory Aids:
- Sodium salts are generally more soluble than potassium salts.

Quick Tips:
- Know the properties of common laboratory reagents.

14 Easy

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Analyzing +4 | -1

Which of the following statements is not correct about

{Cu}^{2+}

and

{Cu}^{+}:

View answer and explanation

Correct option: D

Both

{Cu}^{+}

and

{Cu}^{2+}

are stable with

{I}^{-}

Option D is correct because

{Cu}^{+}

is not stable with

{I}^{-}

due to the formation of

CuI

precipitate.

Solution approach

Correct Answer + Core Reason: Both ${Cu}^{+}$ and ${Cu}^{2+}$ are stable with ${I}^{-}$ . ${Cu}^{+}$ is unstable with $I^-$ due to the formation of $CuI$ precipitate.

Step-by-Step Solution:

Step 1: Analyze the stability of copper ions with iodide ions.

Step 2: ${Cu}^{2+}$ is reduced to ${Cu}^{+}$ by $I^-$ , and ${Cu}^{+}$ immediately precipitates as $CuI$ .
${2Cu}^{2+} + 4I^- \rightarrow 2CuI(s) + I_2$

Step 3: ${Cu}^{2+}$ is more stable than ${Cu}^{+}$ in aqueous solution due to higher hydration energy.

Step 4: ${Cu}^{2+}$ (e.g., $CuSO_4$ ) is paramagnetic due to one unpaired electron, while ${Cu}^{+}$ is diamagnetic as it has a completely filled d-orbital.

Step 5: ${Cu}^{2+}$ solutions are generally blue, while ${Cu}^{+}$ compounds are usually colorless.

Why other options are wrong:
- Option A: ${Cu}^{2+}$ is indeed more stable than ${Cu}^{+}$ in aqueous solution due to its higher hydration energy.
- Option B: ${Cu}^{2+}$ is paramagnetic, and ${Cu}^{+}$ is diamagnetic because of the electronic configurations.
- Option C: ${Cu}^{2+}$ is generally blue in solution, while ${Cu}^{+}$ is colorless.

Key Concepts:
- Stability of ions in solution
- Redox reactions
- Paramagnetism and Diamagnetism
- Color of transition metal ions

Memory Aids:
- Remember that $CuI$ is a stable precipitate, driving the reaction forward.

Quick Tips:
- Consider the standard reduction potentials and the formation of precipitates when assessing the stability of ions in solution.

15 Medium

15 of 206Question permalink

Remembering +4 | -1

Which does not possess allotropic forms?

View answer and explanation

Correct option: C

Fe

Option C (Fe) is correct because Iron (Fe) exhibits allotropy, but the question asks which does NOT possess allotropic forms. Carbon, Tin, and Phosphorus all exhibit allotropy.

Solution approach

Correct Answer + Core Reason: Fe. Iron does not possess allotropic forms is incorrect. The question is asking which element does NOT possess allotropic forms, and the correct answer should be an element that does not exhibit allotropy. However, Iron DOES exhibit allotropy.

Step-by-Step Solution:

Step 1: Define allotropy. Allotropy is the property of some chemical elements to exist in two or more different forms, in the same physical state.

Step 2: Consider the allotropic forms of the given elements:\nCarbon (C): Diamond, graphite, fullerenes\nTin (Sn): Grey tin, white tin\nIron (Fe): Alpha, beta, gamma, delta\nPhosphorus (P): White phosphorus, red phosphorus, black phosphorus

Step 3: Determine which element does not exhibit allotropy. The question is flawed as all the options exhibit allotropy.

Why other options are wrong:
- Option A: C possesses allotropic forms (diamond, graphite, fullerenes).
- Option B: Sn possesses allotropic forms (grey tin, white tin).
- Option C: Fe possesses allotropic forms (alpha, beta, gamma, delta).
- Option D: P possesses allotropic forms (white phosphorus, red phosphorus, black phosphorus).

Key Concepts:
- Allotropy
- Allotropic forms of elements

Memory Aids:
- Remember the common allotropes of carbon, tin, iron, and phosphorus.

Quick Tips:
- Allotropy is related to the different ways atoms can bond together in the same physical state.

16 Easy

16 of 206Question permalink

Application +4 | -1

Maximum oxidation state is presented by:

View answer and explanation

Correct option: A

\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}\mathrm{\;and\;MnO}_4^ -

Option A is correct because CrO2Cl2 has Cr in +6 oxidation state and MnO4- has Mn in +7 oxidation state, which is the maximum possible for Mn.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}\mathrm{\;and\;MnO}_4^ -$ . Cr in CrO₂Cl₂ has +6 oxidation state, and Mn in MnO₄⁻ has +7 oxidation state, which is the maximum for Mn.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the central metal atom in each compound.
- CrO₂Cl₂: Let the oxidation state of Cr be x. Then, x + 2(-2) + 2(-1) = 0 => x - 4 - 2 = 0 => x = +6
- MnO₄⁻: Let the oxidation state of Mn be x. Then, x + 4(-2) = -1 => x - 8 = -1 => x = +7
- MnO₂: Let the oxidation state of Mn be x. Then, x + 2(-2) = 0 => x - 4 = 0 => x = +4
- [Fe(CN)₆]³⁻: Let the oxidation state of Fe be x. Then, x + 6(-1) = -3 => x - 6 = -3 => x = +3
- [Co(CN)₆]³⁻: Let the oxidation state of Co be x. Then, x + 6(-1) = -3 => x - 6 = -3 => x = +3
- MnO: Let the oxidation state of Mn be x. Then, x + (-2) = 0 => x = +2

Step 2: Identify the compounds with the maximum oxidation state.\nCrO₂Cl₂ has Cr in +6 oxidation state, and MnO₄⁻ has Mn in +7 oxidation state.

Step 3: Conclude the answer.\nTherefore, the correct answer is CrO₂Cl₂ and MnO₄⁻.

Why other options are wrong:
- Option B: MnO₂ has Mn in the +4 oxidation state, which is not the maximum.
- Option C: [Fe(CN)₆]³⁻ and [Co(CN)₆]³⁻ have Fe and Co in the +3 oxidation state, which are not the maximum.
- Option D: MnO has Mn in the +2 oxidation state, which is not the maximum.

Key Concepts:
- Oxidation state
- Transition metal compounds

Memory Aids:
- Remember the common oxidation states of oxygen (-2) and chlorine (-1).

Quick Tips:
- Use the rules for assigning oxidation states to determine the oxidation state of the central metal atom.
- The maximum oxidation state of a transition metal is often equal to its group number.

17 Hard

17 of 206Question permalink

Understanding +4 | -1

{NaCl}

reacts with conc.

{H}_2 {SO}_4

and

{K}_2 {Cr}_2 {O}_7

to give reddish fumes (B), which react with

{NaOH}

to give yellow solution (C). (B) and (C) respectively are:

View answer and explanation

Correct option: A

{CrO}_2 {Cl}_2, {Na}_2 {CrO}_4

The correct answer is

{CrO}_2 {Cl}_2, {Na}_2 {CrO}_4

because the reddish fumes (B) are chromyl chloride, and it reacts with NaOH to form sodium chromate (yellow solution).

Solution approach

Correct Answer + Core Reason: ${CrO}_2 {Cl}_2, {Na}_2 {CrO}_4$ . Chromyl chloride is formed first, then it reacts with NaOH to form sodium chromate.

Step-by-Step Solution:

Step 1: Recognize the reaction as the chromyl chloride test.

Step 2: The reaction of NaCl with conc. H2SO4 and K2Cr2O7 produces chromyl chloride (CrO2Cl2), which are reddish fumes.
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 + 4 \mathrm{NaCl} + 6 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{CrO}_2 \mathrm{Cl}_2 + 2 \mathrm{KHSO}_4 + 4 \mathrm{NaHSO}_4 + 3 \mathrm{H}_2 \mathrm{O}$

Step 3: Chromyl chloride reacts with NaOH to form sodium chromate (Na2CrO4), which is a yellow solution.
$\mathrm{CrO}_2 \mathrm{Cl}_2 + 4 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{CrO}_4 + 2 \mathrm{NaCl} + 2 \mathrm{H}_2 \mathrm{O}$

Step 4: Therefore, B is CrO2Cl2 and C is Na2CrO4.

Why other options are wrong:
- Option B: ${Na}_2 {CrO}_4,{CrO}_2 {Cl}_2$ . This is incorrect because the reddish fumes are chromyl chloride, not sodium chromate.
- Option C: ${CrO}_2 {Cl}_2, {Na}_2 {Cr}_2{O}_7$ . Sodium dichromate is orange, not yellow, and is not directly formed in the reaction of chromyl chloride with NaOH.
- Option D: ${CrO}_2 {Cl}_2,{KHSO}_4$ . Potassium bisulfate is a byproduct of the first reaction but is not the yellow solution formed when the reddish fumes react with NaOH.

Key Concepts:
- Chromyl chloride test
- Reaction of chromyl chloride with NaOH

Memory Aids:
- Remember that chromyl chloride is reddish and reacts with NaOH to form yellow sodium chromate.

Quick Tips:
- The chromyl chloride test is a specific test for chloride ions. The yellow solution formed after reaction with NaOH is due to the formation of chromate ions.

18 Medium

18 of 206Question permalink

Remembering +4 | -1

Which metal gives hydrogen gas on heating with hot concentrated alkali?

View answer and explanation

Correct option: C

Zn

Option C (Zn) is correct because Zinc reacts with hot concentrated alkali to produce hydrogen gas and a zincate salt.

Solution approach

Correct Answer + Core Reason: Zn. Zinc reacts with hot concentrated alkali to produce hydrogen gas.

Step-by-Step Solution:

Step 1: Consider the reactivity of each metal with hot concentrated alkali.

Step 2: Ag (Silver) is a noble metal and does not react with alkali.

Step 3: Ni (Nickel) does not readily react with alkali.

Step 4: Zn (Zinc) reacts with hot concentrated alkali (e.g., NaOH) to produce hydrogen gas and a zincate salt. The reaction is: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$

Step 5: Cu (Copper) does not react with alkali.

Step 6: Therefore, Zinc (Zn) gives hydrogen gas on heating with hot concentrated alkali.

Why other options are wrong:
- Option A: Silver (Ag) does not react with alkali.
- Option B: Nickel (Ni) does not readily react with alkali.
- Option D: Copper (Cu) does not react with alkali.

Key Concepts:
- Reactivity of Metals
- Reactions with Alkalis

Memory Aids:
- Zinc is amphoteric and reacts with both acids and bases.

Quick Tips:
- Remember the metals that react with both acids and bases (amphoteric metals).

19 Easy

19 of 206Question permalink

Analyzing +4 | -1

Which one of the following ions exhibits d-d transition and paramagnetism as well ?

View answer and explanation

Correct option: A

\,\,\mathrm{Mn}\,\mathrm{O}_\mathrm{4}^{\mathrm{2 - }}

Option A (

MnO_4^{2-}

) is correct because it exhibits both d-d transition (due to the presence of d electrons) and paramagnetism (due to the presence of unpaired electrons).

Solution approach

Correct Answer + Core Reason: $\,\mathrm{Mn}\,\mathrm{O}_{\mathrm{4}}^{\mathrm{2 - }}$ . This ion exhibits both d-d transition and paramagnetism due to the presence of unpaired electrons in its d orbitals.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the central metal ion in each complex.
$MnO_4^{2-}$ : Mn is in +6 oxidation state.
$CrO_4^{2-}$ : Cr is in +6 oxidation state.
$MnO_4^-$ : Mn is in +7 oxidation state.
$Cr_2O_7^{2-}$ : Cr is in +6 oxidation state.

Step 2: Determine the electronic configuration of the central metal ion in each complex.
$Mn^{6+}$ : Electronic configuration is [Ar] $3d^1$
$Cr^{6+}$ : Electronic configuration is [Ar] $3d^0$
$Mn^{7+}$ : Electronic configuration is [Ar] $3d^0$
$Cr^{6+}$ : Electronic configuration is [Ar] $3d^0$

Step 3: Determine if d-d transitions are possible.\nd-d transitions are possible only if the metal ion has d electrons (i.e., not $d^0$ or $d^{10}$ configuration).
$Mn^{6+}$ ( $d^1$ ) can show d-d transitions.
$Cr^{6+}$ ( $d^0$ ) cannot show d-d transitions.
$Mn^{7+}$ ( $d^0$ ) cannot show d-d transitions.

Step 4: Determine if the complex is paramagnetic.\nA complex is paramagnetic if it has unpaired electrons.
$Mn^{6+}$ ( $d^1$ ) has one unpaired electron, so it is paramagnetic.
$Cr^{6+}$ ( $d^0$ ) has no unpaired electrons, so it is diamagnetic.
$Mn^{7+}$ ( $d^0$ ) has no unpaired electrons, so it is diamagnetic.

Step 5: Identify the complex that exhibits both d-d transition and paramagnetism.\nOnly $MnO_4^{2-}$ exhibits both d-d transition and paramagnetism.

Final Answer: $\,\mathrm{Mn}\,\mathrm{O}_{\mathrm{4}}^{\mathrm{2 - }}$

Why other options are wrong:
- **Option B ( $CrO_4^{2-}$ ):** This is incorrect because $Cr^{6+}$ has a $d^0$ configuration, so it cannot exhibit d-d transitions and is diamagnetic.
- **Option C ( $MnO_4^-$ ):** This is incorrect because $Mn^{7+}$ has a $d^0$ configuration, so it cannot exhibit d-d transitions and is diamagnetic.
- **Option D ( $Cr_2O_7^{2-}$ ):** This is incorrect because $Cr^{6+}$ has a $d^0$ configuration, so it cannot exhibit d-d transitions and is diamagnetic.

Key Concepts:
- d-d transitions
- Paramagnetism
- Electronic configuration
- Oxidation states

Memory Aids:
- Paramagnetic substances are attracted to magnetic fields due to unpaired electrons.
- d-d transitions require the presence of d electrons.

Quick Tips:
- Determine the oxidation state and electronic configuration of the metal ion first.
- Check for the presence of d electrons to determine if d-d transitions are possible.
- Check for unpaired electrons to determine if the complex is paramagnetic.

20 Easy

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Application +4 | -1

Which of the following d-block ions show the maximum value of magnetic moment?

View answer and explanation

Correct option: A

\mathrm{Mn}^{+2}

Option A (

\mathrm{Mn}^{+2}

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

Solution approach

Correct Answer + Core Reason: $\mathrm{Mn}^{+2}$ . $\mathrm{Mn}^{+2}$ 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:
- $\mathrm{Mn}^{+2}$ : [Ar] 3d⁵ (5 unpaired electrons)
- $\mathrm{Fe}^{+2}$ : [Ar] 3d⁶ (4 unpaired electrons)
- $\mathrm{Cr}^{+3}$ : [Ar] 3d³ (3 unpaired electrons)
- $\mathrm{Ni}^{+2}$ : [Ar] 3d⁸ (2 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 B: $\mathrm{Fe}^{+2}$ has 4 unpaired electrons, less than $\mathrm{Mn}^{+2}$ .
- Option C: $\mathrm{Cr}^{+3}$ has 3 unpaired electrons, less than $\mathrm{Mn}^{+2}$ .
- Option D: $\mathrm{Ni}^{+2}$ has 2 unpaired electrons, less than $\mathrm{Mn}^{+2}$ .

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.

21 Medium

21 of 206Question permalink

Application +4 | -1

1 mole of

{K}_2 {Cr}_2 {O}_7

cannot oxidise

View answer and explanation

Correct option: D

4 moles of

{Sn}^{2+}

to

{Sn}^{4+}

Option D (4 moles of Sn2+ to Sn4+) is correct because 1 mole of K2Cr2O7 can oxidize a maximum of 3 moles of Sn2+.

Solution approach

Correct Answer + Core Reason: 4 moles of ${Sn}^{2+}$ to ${Sn}^{4+}$ . 1 mole of K2Cr2O7 can oxidize a maximum of 3 moles of Sn2+.

Step-by-Step Solution:

Step 1: Write the balanced redox reaction between ${K_2}C{r_2}{O_7}$ and ${Sn}^{2+}$ :
$C{r_2}{O_7}^{2-} + 3S{n^{2+}} + 14{H^+} \rightarrow 2C{r^{3+}} + 3S{n^{4+}} + 7{H_2}O$

Step 2: Determine the mole ratio between ${K_2}C{r_2}{O_7}$ and ${Sn}^{2+}$ :\nFrom the balanced equation, 1 mole of ${K_2}C{r_2}{O_7}$ reacts with 3 moles of ${Sn}^{2+}$ .

Step 3: Analyze the given options:\nA. 3.6 moles of ${FeSO}_4$ to ${Fe}_2\left({SO}_4\right)_3$ : $Fe^{2+} \rightarrow Fe^{3+} + e^-$ . $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ . 1 mole of $K_2Cr_2O_7$ can oxidize 6 moles of $FeSO_4$ . So, it can oxidize 3.6 moles.\nB. 1 mole of ${FeSO}_4$ to ${Fe}_2\left({SO}_4\right)_3$ : As shown above, 1 mole of $K_2Cr_2O_7$ can oxidize 6 moles of $FeSO_4$ . So, it can oxidize 1 mole.\nC. 1.8 moles of ${Sn}^{2+}$ to ${Sn}^{4+}$ : As shown above, 1 mole of $K_2Cr_2O_7$ can oxidize 3 moles of $Sn^{2+}$ . So, it can oxidize 1.8 moles.\nD. 4 moles of ${Sn}^{2+}$ to ${Sn}^{4+}$ : 1 mole of $K_2Cr_2O_7$ can oxidize only 3 moles of $Sn^{2+}$ . Therefore, it cannot oxidize 4 moles.

Step 4: The option that ${K_2}C{r_2}{O_7}$ cannot oxidize is 4 moles of ${Sn}^{2+}$ to ${Sn}^{4+}$ .

Why other options are wrong:
- Option A: 1 mole of ${K_2}C{r_2}{O_7}$ can oxidize 3.6 moles of ${FeSO}_4$ .
- Option B: 1 mole of ${K_2}C{r_2}{O_7}$ can oxidize 1 mole of ${FeSO}_4$ .
- Option C: 1 mole of ${K_2}C{r_2}{O_7}$ can oxidize 1.8 moles of ${Sn}^{2+}$ .

Key Concepts:
- Redox reactions
- Balancing redox reactions
- Stoichiometry

Memory Aids:
- OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Quick Tips:
- Always balance the redox reaction before determining the mole ratios.

22 Medium

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Analyzing +4 | -1

Which one of the following does not decolourise an acidified

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

solution?

View answer and explanation

Correct option: B

\mathrm{FeC}{\mathrm{l}_3}

Option B (

\mathrm{FeC}{\mathrm{l}_3}

) is the correct answer because it is already in its highest oxidation state (+3) and cannot be further oxidized by acidified

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

.

Solution approach

Correct Answer + Core Reason: $\mathrm{FeC}{\mathrm{l}_3}$ . $\mathrm{FeC}{\mathrm{l}_3}$ is already in its highest oxidation state and cannot be oxidized further by acidified $\mathrm{KMn}{\mathrm{O}_4}$ .

Step-by-Step Solution:

Step 1: Identify the oxidizing agent: Acidified $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent.

Step 2: Determine the oxidation states of the species in each option.

Step 3: Check if the species can be oxidized further.

Step 4: $\mathrm{S}{\mathrm{O}_2}$ can be oxidized to $\mathrm{SO}_4^{2 - }$ .

Step 5: $\mathrm{FeC}{\mathrm{l}_3}$ is already in the +3 oxidation state and cannot be oxidized further.

Step 6: ${\mathrm{H}_2}{\mathrm{O}_2}$ can be oxidized to ${\mathrm{O}_2}$ .

Step 7: $\mathrm{FeS}{\mathrm{O}_4}$ can be oxidized to $\mathrm{Fe}_2{(\mathrm{SO}_4})_3$ .

Why other options are wrong:
- Option A: $\mathrm{S}{\mathrm{O}_2}$ can be oxidized by acidified $\mathrm{KMn}{\mathrm{O}_4}$ .
- Option C: ${\mathrm{H}_2}{\mathrm{O}_2}$ can be oxidized by acidified $\mathrm{KMn}{\mathrm{O}_4}$ .
- Option D: $\mathrm{FeS}{\mathrm{O}_4}$ can be oxidized by acidified $\mathrm{KMn}{\mathrm{O}_4}$ .

Key Concepts:
- Oxidation and reduction
- Oxidizing agents
- Oxidation states

Memory Aids:
- Remember that oxidizing agents oxidize other substances and get reduced themselves.

Quick Tips:
- Identify the oxidation states of the elements involved to determine if oxidation is possible.

23 Medium

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Analyzing +4 | -1

Assertion (A): Green solution of potassium manganate (VI),

\mathrm{K}_2 \mathrm{MnO}_4

turns purple and brown solid is precipitated when

\mathrm{CO}_2

is bubbled into it.

Reason (R):

\mathrm{K}_2 \mathrm{MnO}_4

changes to

\mathrm{K}_3 \mathrm{MnO}_8

by

\mathrm{CO}_2

.

View answer and explanation

Correct option: C

\mathbf{A}

is true but

\mathbf{R}

is false.

Option C (

\mathbf{A}

is true but

\mathbf{R}

is false) is correct because the green solution of potassium manganate(VI) turns purple due to disproportionation, not because it changes to

\mathrm{K}_3\mathrm{MnO}_8

.

Solution approach

Correct Answer + Core Reason: $\mathbf{A}$ is true but $\mathbf{R}$ is false. The reaction involves disproportionation, not formation of $\mathrm{K}_3\mathrm{MnO}_8$ .

Step-by-Step Solution:

Step 1: Analyze the assertion.

Step 2: The green solution of potassium manganate(VI), $\mathrm{K}_2\mathrm{MnO}_4$ , does turn purple when $\mathrm{CO}_2$ is bubbled into it. This is due to disproportionation.

Step 3: Analyze the reason.

Step 4: $\mathrm{K}_2\mathrm{MnO}_4$ does not change to $\mathrm{K}_3\mathrm{MnO}_8$ by $\mathrm{CO}_2$ . The reaction is a disproportionation reaction where $MnO_4^{2-}$ converts to $MnO_4^-$ and $MnO_2$ .

Step 5: Determine if the assertion and reason are true and if the reason is the correct explanation.

Why other options are wrong:
- Option A: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ . The reason is false.
- Option B: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ . The reason is false.
- Option D: $\mathbf{A}$ is false but $\mathbf{R}$ is true. The assertion is true.

Key Concepts:
- Disproportionation reactions
- Potassium manganate(VI) and potassium permanganate

Memory Aids:
- Remember that disproportionation involves a species being both oxidized and reduced.

Quick Tips:
- Recognize disproportionation reactions and their products.

24 Easy

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Remembering +4 | -1

Mercury is purified by:

View answer and explanation

Correct option: B

Distillation in vacuum

Option B (Distillation in vacuum) is correct because it allows mercury to be purified at lower temperatures, preventing decomposition or reaction with air.

Solution approach

Correct Answer + Core Reason: Distillation in vacuum. Vacuum distillation allows purification at lower temperatures.

Step-by-Step Solution:

Step 1: Consider the properties of mercury.

Step 2: Mercury has a relatively low boiling point.

Step 3: Distillation is a common method for purifying liquids.

Step 4: Distillation in vacuum lowers the boiling point of mercury, preventing decomposition or reaction with air during purification.

Step 5: Treatment with dilute $HNO_3$ might remove some impurities but can also react with mercury.

Step 6: Electrolytic methods are not typically used for mercury purification.

Step 7: Solidifying would not purify mercury.

Why other options are wrong:
- Option A: Solidifying does not purify mercury.
- Option C: Treatment with dil. $\mathrm{HN}{\mathrm{O}_3}$ can react with mercury.
- Option D: Electrolytic method is not a common method for mercury purification.

Key Concepts:
- Distillation
- Vacuum Distillation
- Purification Techniques

Memory Aids:
- Vacuum distillation = lower boiling point = less decomposition.

Quick Tips:
- Consider the boiling point and reactivity of the substance when choosing a purification method.

25 Medium

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Application +4 | -1

Number of electrons in 3

d

-orbital of

{\mathrm{V}^{2 + }},\mathrm{\;C}{\mathrm{r}^{2 + }},\mathrm{\;M}{\mathrm{n}^{2 + }}

, and

\mathrm{F}{\mathrm{e}^{2 + }}

are 3, 4, 5 and 6 respectively. Which of the following ions will have largest value of magnetic moment

\left( \mathrm{\mu } \right)

?

View answer and explanation

Correct option: C

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

Option C is correct because Mn2+ has 5 unpaired electrons in its 3d orbitals, resulting in the largest magnetic moment according to the formula μ = √n(n+2) BM, where n is the number of unpaired electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}{\mathrm{n}^{2 + }}$ . Mn²⁺ has 5 unpaired electrons, resulting in the largest magnetic moment.

Step-by-Step Solution:

Step 1: Determine the number of unpaired electrons in each ion.
- V²⁺: 3d³ has 3 unpaired electrons.
- Cr²⁺: 3d⁴ has 4 unpaired electrons.
- Mn²⁺: 3d⁵ has 5 unpaired electrons.
- Fe²⁺: 3d⁶ has 4 unpaired electrons (after pairing).

Step 2: Calculate the magnetic moment (μ) for each ion using the formula: $\nμ = \sqrt{{n(n+2)}}$ BM, where n is the number of unpaired electrons.
- V²⁺: μ = $\sqrt{{3(3+2)}}$ = $\sqrt{{15}}$ ≈ 3.87 BM
- Cr²⁺: μ = $\sqrt{{4(4+2)}}$ = $\sqrt{{24}}$ ≈ 4.90 BM
- Mn²⁺: μ = $\sqrt{{5(5+2)}}$ = $\sqrt{{35}}$ ≈ 5.92 BM
- Fe²⁺: μ = $\sqrt{{4(4+2)}}$ = $\sqrt{{24}}$ ≈ 4.90 BM

Step 3: Compare the magnetic moments.\nMn²⁺ has the largest magnetic moment (5.92 BM).

Step 4: Conclude the answer.\nTherefore, the correct answer is Mn²⁺.

Why other options are wrong:
- Option A: V²⁺ has a magnetic moment of approximately 3.87 BM, which is smaller than that of Mn²⁺.
- Option B: Cr²⁺ has a magnetic moment of approximately 4.90 BM, which is smaller than that of Mn²⁺.
- Option D: Fe²⁺ has a magnetic moment of approximately 4.90 BM, which is smaller than that of Mn²⁺.

Key Concepts:
- Magnetic moment
- Unpaired electrons
- Electronic configuration

Memory Aids:
- Remember the formula for calculating magnetic moment: μ = $\sqrt{{n(n+2)}}$ $BM.\n- The more unpaired electrons, the larger the magnetic moment.\n\n**Quick Tips:**\n- Determine the number of unpaired electrons in the ion.\n- Use the formula to calculate the magnetic moment.\n- Compare the magnetic moments to find the largest value.$

26 Hard

26 of 206Question permalink

Remembering +4 | -1

Oxygen is absorbed by molten Ag, which is evolved on cooling and the silver particles are scattered; the phenomenon is known as:

View answer and explanation

Correct option: B

Spitting of silver

Option B (Spitting of silver) is correct because the phenomenon of oxygen being absorbed by molten silver and then released upon cooling, causing the silver to scatter, is known as spitting of silver.

Solution approach

Correct Answer + Core Reason: Spitting of silver. The phenomenon is known as spitting of silver.

Step-by-Step Solution:

Step 1: Understand the phenomenon described in the question.

Step 2: Molten silver has the ability to absorb oxygen. As the silver cools, the dissolved oxygen is released, often violently, causing the silver to spit and scatter. This is known as the spitting of silver.

Step 3: Identify the correct term for this phenomenon.

Why other options are wrong:
- Option A: Silvering of mirror is a process of coating a surface with silver to create a mirror.
- Option B: Spitting of silver is the correct term for the described phenomenon.
- Option C: Frosting of silver is not a recognized term for this phenomenon.
- Option D: Hairing of silver is not a recognized term for this phenomenon.

Key Concepts:
- Metallurgy
- Properties of silver
- Spitting of silver

Memory Aids:
- Remember the term "spitting of silver" for this phenomenon.

Quick Tips:
- Spitting of silver is a characteristic behavior of molten silver due to its oxygen absorption properties.

27 Easy

27 of 206Question permalink

Understanding +4 | -1

Which of the following statements is not true for Mohr’s salt?

View answer and explanation

Correct option: C

Oxidation state of iron is +3

Option C (Oxidation state of iron is +3) is correct because Mohr's salt is

FeSO_4.(NH_4)_2SO_4.6H_2O

, where iron is in the +2 oxidation state, not +3.

Solution approach

Correct Answer + Core Reason: Oxidation state of iron is +3. In Mohr's salt, iron is in the +2 oxidation state.

Step-by-Step Solution:

Step 1: Write the formula of Mohr's salt: $FeSO_4.(NH_4)_2SO_4.6H_2O$ .

Step 2: Determine the oxidation state of iron in $FeSO_4$ .

Step 3: In $FeSO_4$ , the oxidation state of $SO_4^{2-}$ is -2. Therefore, the oxidation state of Fe is +2.

Step 4: Mohr's salt is a double salt that contains $Fe^{2+}$ ions.

Why other options are wrong:
- Option A: Mohr's salt decolourises $KMnO_4$ solution because $Fe^{2+}$ is oxidized to $Fe^{3+}$ .
- Option B: Mohr's salt is a double salt because it is formed by the combination of two salts ( $FeSO_4$ and $(NH_4)_2SO_4$ ).
- Option D: Mohr's salt is a primary standard because it can be obtained in a pure and stable form and its solution concentration can be accurately determined.

Key Concepts:
- Oxidation state
- Double salt
- Primary standard
- Redox titration

Memory Aids:
- Remember the formula of Mohr's salt and the oxidation state of iron in it.

Quick Tips:
- Recall the properties of Mohr's salt.

28 Hard

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Understanding +4 | -1

The reaction of

\mathrm{H}_2 \mathrm{O}_2

with potassium permanganate in acidic medium leads to the formation of mainly

View answer and explanation

Correct option: A

\mathrm{Mn}^{2+}

Option A (

\mathrm{Mn}^{2+}

) is correct because in acidic medium,

\mathrm{H}_2\mathrm{O}_2

reduces

\mathrm{KMnO}_4

to

\mathrm{Mn}^{2+}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mn}^{2+}$ . In acidic medium, $\mathrm{H}_2\mathrm{O}_2$ reduces $\mathrm{KMnO}_4$ to $\mathrm{Mn}^{2+}$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the reaction.
$2KMnO_4 + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 8H_2O + 5O_2 + 2K^+$

Step 2: Identify the product formed.\nThe product is $\mathrm{Mn}^{2+}$ .

Why other options are wrong:
- Option B: $\mathrm{Mn}^{4+}$ is not the product of this reaction in acidic medium.
- Option C: $\mathrm{Mn}^{3+}$ is not the product of this reaction in acidic medium.
- Option D: $\mathrm{Mn}^{6+}$ is not the product of this reaction in acidic medium.

Key Concepts:
- Redox reactions
- Oxidation and reduction
- Reaction of $H_2O_2$ with $KMnO_4$

Memory Aids:
- Remember the products of the reaction between $H_2O_2$ and $KMnO_4$ in acidic medium.

Quick Tips:
- Recall the common reactions of $H_2O_2$ .

29 Hard

29 of 206Question permalink

Understanding +4 | -1

Assertion :

{\text{KMn}}{{\text{O}}_{\text{4}}}

is dark pink coloured compound.

Reason : In the

{\text{KMn}}{{\text{O}}_{\text{4}}}

charge transfer spectrum occurs.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is correct explanation of the Assertion.

Option A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) is correct because

KMnO_4

is dark pink due to ligand to metal charge transfer.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is correct explanation of the Assertion. The color of $KMnO_4$ is due to charge transfer.

Step-by-Step Solution:

Step 1: Analyze the assertion.
$KMnO_4$ is indeed a dark pink colored compound.

Step 2: Analyze the reason.\nThe color of $KMnO_4$ arises from charge transfer transitions, where an electron is transferred from the oxygen ligand to the manganese metal center.

Step 3: Determine if the reason correctly explains the assertion.\nThe reason correctly explains why $KMnO_4$ is colored.

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

Key Concepts:
- Charge transfer spectra
- Color of coordination compounds

Memory Aids:
- Remember that charge transfer transitions can cause intense colors in coordination compounds.

Quick Tips:
- Recall the origin of color in coordination compounds.

30 Medium

30 of 206Question permalink

Understanding +4 | -1

The radius of

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

(Atomic number of La = 57) is 1.06 Å. Which one of the following given values will be closest to the radius of

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

?
(Atomic number of Lu=71)

View answer and explanation

Correct option: D

0.85 Å

Option D (0.85 Å) is correct because, due to lanthanide contraction, the ionic radii decrease across the lanthanide series.

Solution approach

Correct Answer + Core Reason: 0.85 Å. Lanthanide contraction causes a decrease in ionic radius across the lanthanide series.

Step-by-Step Solution:

Step 1: Identify the trend: As we move across the lanthanide series from La (Z=57) to Lu (Z=71), there is a gradual decrease in ionic radii due to the lanthanide contraction.

Step 2: Lanthanide contraction is caused by the poor shielding of the 4f electrons. The effective nuclear charge increases, pulling the electrons closer to the nucleus and decreasing the ionic size.

Step 3: Since Lu is further along the lanthanide series than La, its ionic radius will be significantly smaller than that of La $^{3+}$ .

Step 4: Given the radius of La $^{3+}$ is 1.06 Å, the radius of Lu $^{3+}$ must be smaller. Among the options, 0.85 Å is the only reasonable value.

Why other options are wrong:
- Option A (1.60 Å): This value is significantly larger than the radius of La $^{3+}$ , which is incorrect due to lanthanide contraction.
- Option B (1.40 Å): This value is also larger than the radius of La $^{3+}$ , which is incorrect due to lanthanide contraction.
- Option C (1.06 Å): This value is the same as the radius of La $^{3+}$ , which is incorrect because the radius should decrease across the lanthanide series.

Key Concepts:
- Lanthanide contraction
- Ionic radii trends
- Effective nuclear charge

Memory Aids:
- Remember that lanthanide contraction causes a decrease in size across the lanthanide series.

Quick Tips:
- Understand the concept of lanthanide contraction and its effect on ionic radii.

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D & f- BLOCK ELEMENTS practice · Mixed difficulty

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