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

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

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

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

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

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

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

Which of the following is not oxidized by ${{\rm{O}}_3}$?

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

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

1 Medium

1 of 85Question 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 85Question 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 85Question 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 Medium

4 of 85Question permalink

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.

5 Medium

5 of 85Question 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.

6 Medium

6 of 85Question 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.

7 Medium

7 of 85Question 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.

8 Medium

8 of 85Question 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).

9 Medium

9 of 85Question 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.

10 Medium

10 of 85Question permalink

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.

11 Medium

11 of 85Question permalink

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.

12 Medium

12 of 85Question permalink

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

13 Medium

13 of 85Question 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.

14 Medium

14 of 85Question permalink

Understanding +4 | -1

Acidified solution of chromic acid on treatment with hydrogen peroxide yields

View answer and explanation

Correct option: A

\mathrm{Cr}{\mathrm{O}_5} + {\mathrm{H}_2}\mathrm{O}

Option A (

\mathrm{Cr}{\mathrm{O}_5} + {\mathrm{H}_2}\mathrm{O}

) is the correct product because chromic acid reacts with hydrogen peroxide to form chromium pentoxide (

\mathrm{Cr}{\mathrm{O}_5}

), which has a butterfly structure.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\mathrm{O}_5} + {\mathrm{H}_2}\mathrm{O}$ . Acidified chromic acid reacts with hydrogen peroxide to form chromium pentoxide.

Step-by-Step Solution:

Step 1: Chromic acid is typically formed by dissolving chromium trioxide ( $\mathrm{Cr}{\mathrm{O}_3}$ ) in water.

Step 2: The reaction between acidified chromic acid and hydrogen peroxide forms chromium pentoxide ( $\mathrm{Cr}{\mathrm{O}_5}$ ) and water.

Step 3: The reaction is: ${\mathrm{H}_2}\mathrm{Cr}{\mathrm{O}_4} + 2{\mathrm{H}_2}{\mathrm{O}_2} \to \mathrm{Cr}{\mathrm{O}_5} + 3{\mathrm{H}_2}\mathrm{O}$ . Chromium pentoxide has a butterfly structure with two peroxide linkages.

Why other options are wrong:
- **Option B ( ${\mathrm{H}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} + {\mathrm{H}_2}\mathrm{O} + {\mathrm{O}_2}$ ): This option suggests the formation of dichromic acid, which is not the primary product of this reaction.
- Option C ( $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O} + {\mathrm{O}_2}$ ): This option suggests the formation of chromium(III) oxide, which is not the primary product of this reaction.
- Option D ( $\mathrm{Cr}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O} + {\mathrm{O}_2}$ ): This option suggests the formation of chromium trioxide, which is the starting material, not the product.

Key Concepts:
- Reactions of transition metals
- Chromium compounds
- Hydrogen peroxide reactions

Memory Aids:**
- Remember that chromium pentoxide ( $\mathrm{Cr}{\mathrm{O}_5}$ ) is formed in this reaction.

Quick Tips:
- Know the characteristic reactions of chromium compounds.

15 Medium

15 of 85Question permalink

Remembering +4 | -1

Which among the following has maximum density?

View answer and explanation

Correct option: C

Ni

Option C (Ni) is correct because Nickel has the highest density among the given options.

Solution approach

Correct Answer + Core Reason: Ni. Nickel has the highest density among the given options.

Step-by-Step Solution:

Step 1: Recall the approximate densities of the given metals.

Step 2: Fe (Iron) has a density of approximately 7.87 g/cm³.

Step 3: Ti (Titanium) has a density of approximately 4.51 g/cm³.

Step 4: Ni (Nickel) has a density of approximately 8.90 g/cm³.

Step 5: Cr (Chromium) has a density of approximately 7.19 g/cm³.

Step 6: Comparing the densities, Nickel (Ni) has the highest density.

Why other options are wrong:
- Option A: Iron (Fe) has a lower density than Nickel.
- Option B: Titanium (Ti) has a lower density than Nickel.
- Option D: Chromium (Cr) has a lower density than Nickel.

Key Concepts:
- Density of Metals

Memory Aids:
- Remember the relative densities of common metals.

Quick Tips:
- Familiarize yourself with the densities of common elements.

16 Medium

16 of 85Question permalink

Analyzing +4 | -1

Assertion: Manganate ion is paramagnetic while permanganate ion is diamagnetic.

Reason: Manganate ion has two unpaired electrons.

View answer and explanation

Correct option: C

Assertion is correct, but reason is wrong statement.

Option C (Assertion is correct, but reason is wrong statement) is correct because manganate ion (

MnO_4^{2-}

) has one unpaired electron and is paramagnetic, while permanganate ion (

MnO_4^{-}

) has no unpaired electrons and is diamagnetic. The reason states that manganate ion has two unpaired electrons, which is incorrect.

Solution approach

Correct Answer + Core Reason: Assertion is correct, but reason is wrong statement. Manganate ion has one unpaired electron, not two.

Step-by-Step Solution:

Step 1: Analyze the assertion.

Step 2: Manganate ion ( $MnO_4^{2-}$ ) has Mn in the +6 oxidation state. The electronic configuration of $Mn^{6+}$ is $d^1$ , so it has one unpaired electron and is paramagnetic.

Step 3: Permanganate ion ( $MnO_4^{-}$ ) has Mn in the +7 oxidation state. The electronic configuration of $Mn^{7+}$ is $d^0$ , so it has no unpaired electrons and is diamagnetic.

Step 4: Analyze the reason.

Step 5: The reason states that manganate ion has two unpaired electrons, which is incorrect. It has only one unpaired electron.

Step 6: Determine if the assertion and reason are true and if the reason is the correct explanation.

Why other options are wrong:
- Option A: Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. The reason is incorrect.
- Option B: Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. The reason is incorrect.
- Option D: Assertion is wrong, but reason is correct statement. The assertion is correct.

Key Concepts:
- Paramagnetism and diamagnetism
- Electronic configuration
- Oxidation states

Memory Aids:
- Remember that unpaired electrons lead to paramagnetism.

Quick Tips:
- Determine the number of unpaired electrons based on the electronic configuration.

17 Medium

17 of 85Question permalink

Understanding +4 | -1

Positive sol among the following is

View answer and explanation

Correct option: C

Ti{O_2}sol

Titanium dioxide (

Ti{O_2}sol

) is a positive sol because metal oxides generally form positive sols.

Solution approach

Correct Answer + Core Reason: $Ti{O_2}sol$ . Metal oxides generally form positive sols.

Step-by-Step Solution:

Step 1: Understand the concept of positive and negative sols.

Step 2: Positive sols are those in which the dispersed phase has a positive charge, while negative sols have a negatively charged dispersed phase.

Step 3: Metal oxides and basic dyes generally form positive sols, while metal sulfides and acidic dyes generally form negative sols.

Step 4: Therefore, $Ti{O_2}sol$ is a positive sol.

Why other options are wrong:
- Option A: Ag sol can be either positive or negative depending on the preparation method, but it's not always positive.
- Option B: $A{s_2}{S_3}sol$ is a negative sol because it is a metal sulfide.
- Option D: Eosin dye is an acidic dye and forms a negative sol.

Key Concepts:
- Positive and negative sols
- Metal oxides and sulfides
- Acidic and basic dyes

Memory Aids:
- Metal oxides are generally positive sols.
- Metal sulfides are generally negative sols.

Quick Tips:
- Remember the general rules for positive and negative sols.

18 Medium

18 of 85Question permalink

Analyzing +4 | -1

Which of the following is not oxidized by

{\mathrm{O}_3}

?

View answer and explanation

Correct option: B

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

Option B (

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

) is the correct answer because Mn is already in its highest oxidation state (+7) and cannot be further oxidized by

{\mathrm{O}_3}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{KMn}{\mathrm{O}_4}$ . Manganese in $\mathrm{KMn}{\mathrm{O}_4}$ is already in its highest oxidation state (+7) and cannot be further oxidized.

Step-by-Step Solution:

Step 1: Identify the oxidizing agent: ${\mathrm{O}_3}$ (ozone) is an oxidizing agent.

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

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

Step 4: $\mathrm{FeS}{\mathrm{O}_4}$ can be oxidized to $\mathrm{Fe}_2{(\mathrm{SO}_4})_3$ .

Step 5: $\mathrm{KMn}{\mathrm{O}_4}$ has Mn in +7 oxidation state, which is the highest possible, so it cannot be oxidized further.

Step 6: $\mathrm{KI}$ can be oxidized to ${\mathrm{I}_2}$ .

Step 7: ${\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4}$ can be oxidized to $\mathrm{KMn}{\mathrm{O}_4}$ .

Why other options are wrong:
- Option A: $\mathrm{FeS}{\mathrm{O}_4}$ can be oxidized by ${\mathrm{O}_3}$ .
- Option C: $\mathrm{KI}$ can be oxidized by ${\mathrm{O}_3}$ .
- Option D: ${\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4}$ can be oxidized by ${\mathrm{O}_3}$ .

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.

19 Medium

19 of 85Question permalink

Analyzing +4 | -1

Which of the following process does not involve oxidation of iron?

View answer and explanation

Correct option: A

Formation of

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

from

Fe

Option A (Formation of

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

from

Fe

) is correct because the oxidation state of iron remains 0 in this process.

Solution approach

Correct Answer + Core Reason: Formation of $Fe{\left( {CO} \right)_5}$ from $Fe$ . The oxidation state of iron remains 0.

Step-by-Step Solution:

Step 1: Determine the oxidation state of iron in each process.

Step 2: In $Fe{\left( {CO} \right)_5}$ , the oxidation state of Fe is 0 because CO is a neutral ligand.

Step 3: In the liberation of $H_2$ from steam by iron at high temperature, $Fe + H_2O \rightarrow Fe_3O_4 + H_2$ . The oxidation state of Fe changes from 0 to +8/3.

Step 4: In the rusting of iron sheets, $Fe \rightarrow Fe_2O_3$ . The oxidation state of Fe changes from 0 to +3.

Step 5: In the decolourisation of blue $CuSO_4$ solution by iron, $Fe + CuSO_4 \rightarrow FeSO_4 + Cu$ . The oxidation state of Fe changes from 0 to +2.

Step 6: Therefore, the formation of $Fe{\left( {CO} \right)_5}$ does not involve oxidation of iron.

Why other options are wrong:
- Option B: The liberation of $H_2$ from steam by iron involves the oxidation of iron.
- Option C: The rusting of iron sheets involves the oxidation of iron.
- Option D: The decolourisation of blue $CuSO_4$ solution by iron involves the oxidation of iron.

Key Concepts:
- Oxidation state
- Redox reactions
- Coordination compounds

Memory Aids:
- Remember that CO is a neutral ligand.

Quick Tips:
- Determine the oxidation states of the elements to identify oxidation processes.

20 Medium

20 of 85Question permalink

Application +4 | -1

Number of electrons transferred in each case when

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

acts as an oxidising agent to give

\mathrm{Mn}{\mathrm{O}_2},\mathrm{\;M}{\mathrm{n}^{2 + }},\mathrm{\;Mn}{\left( {\mathrm{OH}} \right)_3}\;

and

\mathrm{MnO}_4^{2 - }

, are respectively :

View answer and explanation

Correct option: A

3, 5, 4 and 1

Option A (3, 5, 4 and 1) is correct because the number of electrons transferred corresponds to the change in oxidation state of Mn in each reduction.

Solution approach

Correct Answer + Core Reason: 3, 5, 4 and 1. The number of electrons transferred corresponds to the change in oxidation state of Mn.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Mn in $\mathrm{KMn}{\mathrm{O}_4}$ .

Step 2: In $\mathrm{KMn}{\mathrm{O}_4}$ , the oxidation state of Mn is +7.

Step 3: Determine the oxidation state of Mn in each product:
- $\mathrm{Mn}{\mathrm{O}_2}$ : +4. Change in oxidation state: 7 - 4 = 3. Electrons transferred: 3
- $\mathrm{M}{\mathrm{n}^{2 + }}$ : +2. Change in oxidation state: 7 - 2 = 5. Electrons transferred: 5
- $\mathrm{Mn}{\left( {\mathrm{OH}} \right)_3}$ : +3. Change in oxidation state: 7 - 3 = 4. Electrons transferred: 4
- $\mathrm{MnO}_4^{2 - }$ : +6. Change in oxidation state: 7 - 6 = 1. Electrons transferred: 1

Step 4: The number of electrons transferred are 3, 5, 4, and 1, respectively.

Why other options are wrong:
- Option B: This order is incorrect. The correct order is 3, 5, 4, and 1.
- Option C: This order is incorrect. The correct order is 3, 5, 4, and 1.
- Option D: This order is incorrect. The correct order is 3, 5, 4, and 1.

Key Concepts:
- Oxidation states
- Redox reactions
- Electron transfer

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

Quick Tips:
- Determine the oxidation state of the element before and after the reaction to find the number of electrons transferred.

21 Medium

21 of 85Question permalink

Understanding +4 | -1

The lanthanoids contraction relates to

View answer and explanation

Correct option: B

Atomic as well as

{M^{3 + }}

radii

Option B (Atomic as well as

{M^{3 + }}

radii) is correct because lanthanoid contraction affects both the atomic and ionic radii (specifically

{M^{3 + }}

) of the lanthanides.

Solution approach

Correct Answer + Core Reason: Atomic as well as ${M^{3 + }}$ radii. Lanthanoid contraction affects both atomic and ionic radii.

Step-by-Step Solution:

Step 1: Understand the concept of lanthanide contraction.

Step 2: Lanthanide contraction results in a decrease in both atomic and ionic radii.

Step 3: The contraction is observed for both the atoms and their trivalent ions ( ${M^{3 + }}$ ).

Why other options are wrong:
- Option A: Atomic radii alone is not a complete answer.
- Option C: Valence electrons are affected, but the primary effect is on radii.
- Option D: Oxidation states are not directly related to lanthanide contraction.

Key Concepts:
- Lanthanide contraction
- Atomic and ionic radii

Memory Aids:
- Remember that lanthanide contraction affects both atomic and ionic sizes.

Quick Tips:
- Understand the trends in atomic and ionic radii.

22 Medium

22 of 85Question permalink

Understanding +4 | -1

Carbon in wrought iron is present as

View answer and explanation

Correct option: D

Partly iron carbide and partly as graphite

Option D (Partly iron carbide and partly as graphite) is correct because wrought iron contains a very small amount of carbon, which exists both as iron carbide (

Fe_3C

) and as free graphite.

Solution approach

Correct Answer + Core Reason: Partly iron carbide and partly as graphite. Wrought iron contains very little carbon, present in both forms.

Step-by-Step Solution:

Step 1: Define wrought iron.\nWrought iron is an iron alloy with a very low carbon content (less than 0.08%).

Step 2: Determine the form in which carbon is present in wrought iron.\nCarbon in wrought iron is present both as iron carbide ( $Fe_3C$ ) and as free graphite.

Why other options are wrong:
- Option A: Silicon carbide is not the primary form of carbon in wrought iron.
- Option B: While some carbon is present as iron carbide, it's not the only form.
- Option C: While some carbon is present as graphite, it's not the only form.

Key Concepts:
- Composition of iron alloys
- Forms of carbon in iron alloys

Memory Aids:
- Remember that wrought iron has very low carbon content in two forms.

Quick Tips:
- Recall the composition of different types of iron alloys.

23 Medium

23 of 85Question permalink

Remembering +4 | -1

\mathrm{C}{\mathrm{u}^{2 + }}

ions give precipitate with

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

. The colour of precipitate is:

View answer and explanation

Correct option: D

Brown

Option C (Red) is incorrect. Option D (Brown) is correct because the reaction between

\mathrm{C}{\mathrm{u}^{2 + }}

and

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

produces a brown precipitate of copper ferrocyanide.

Solution approach

Correct Answer + Core Reason: Brown. The reaction forms a brown precipitate of copper ferrocyanide.

Step-by-Step Solution:

Step 1: Recall the reaction between $\mathrm{C}{\mathrm{u}^{2 + }}$ ions and ${\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ .

Step 2: The reaction forms copper ferrocyanide, which is a brown precipitate.
$2\mathrm{C}{\mathrm{u}^{2 + }} + {\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6} \to \mathrm{C}{\mathrm{u}_2}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6} \downarrow + 4{\mathrm{K}^ + }$

Step 3: Identify the color of the precipitate.

Why other options are wrong:
- Option A: Blue: This is incorrect. Copper ferrocyanide is not blue.
- Option B: Green: This is incorrect. Copper ferrocyanide is not green.
- Option C: Red: This is incorrect. Copper ferrocyanide is not red.

Key Concepts:
- Qualitative analysis
- Reactions of metal ions
- Formation of precipitates

Memory Aids:
- Copper ferrocyanide is brown.

Quick Tips:
- Remember the colors of common precipitates in qualitative analysis.

24 Medium

24 of 85Question permalink

Application +4 | -1

A solution of

10 \mathrm{~mL} \frac{{M}}{10} {FeSO}_4

is titrated with

{KMnO}_4

solution in acidic medium the amount of

{KMnO}_4

used will be

View answer and explanation

Correct option: D

10 \mathrm{~mL}

of

0.02 \mathrm{M}

The correct answer is

10 \mathrm{~mL}

of

0.02 \mathrm{M}

because the stoichiometry of the reaction between

FeSO_4

and

KMnO_4

requires careful calculation of the molar ratios and volumes.

Solution approach

Correct Answer + Core Reason: $10 \mathrm{~mL}$ of $0.02 \mathrm{M}$ . The amount of $KMnO_4$ used is determined by the stoichiometry of the redox reaction with $FeSO_4$ .

Step-by-Step Solution:

Step 1: Write the balanced redox equation in acidic medium:
$MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

Step 2: Calculate the millimoles of $FeSO_4$ used:\nMoles = Molarity x Volume = (1/10) M x 10 mL = 1 millimole.

Step 3: From the balanced equation, 1 mole of $MnO_4^-$ reacts with 5 moles of $Fe^{2+}$ . Therefore, the millimoles of $KMnO_4$ required = (1/5) x millimoles of $FeSO_4$ = (1/5) x 1 millimole = 0.2 millimoles.

Step 4: Calculate the volume of $KMnO_4$ required for each option:
- A. $5 \mathrm{~mL}$ of $0.1 \mathrm{M}$ : millimoles = 5 mL x 0.1 M = 0.5 millimoles (incorrect)
- B. $10 \mathrm{~mL}$ of $0.1 \mathrm{M}$ : millimoles = 10 mL x 0.1 M = 1 millimole (incorrect)
- C. $10 \mathrm{~mL}$ of $0.5 \mathrm{M}$ : millimoles = 10 mL x 0.5 M = 5 millimoles (incorrect)
- D. $10 \mathrm{~mL}$ of $0.02 \mathrm{M}$ : millimoles = 10 mL x 0.02 M = 0.2 millimoles (correct)

Step 5: The correct option is $10 \mathrm{~mL}$ of $0.02 \mathrm{M}$ .

Why other options are wrong:
- Option A: $5 \mathrm{~mL}$ of $0.1 \mathrm{M}$ provides 0.5 millimoles of $KMnO_4$ , which is not the required amount.
- Option B: $10 \mathrm{~mL}$ of $0.1 \mathrm{M}$ provides 1 millimole of $KMnO_4$ , which is not the required amount.
- Option C: $10 \mathrm{~mL}$ of $0.5 \mathrm{M}$ provides 5 millimoles of $KMnO_4$ , which is not the required amount.

Key Concepts:
- Redox Titration
- Stoichiometry
- Molarity

Memory Aids:
- Molarity = moles / Volume

Quick Tips:
- Balance the redox equation to determine the molar ratio.
- Convert volumes to liters or use millimoles consistently.

25 Medium

25 of 85Question permalink

Application +4 | -1

1 mole of

\mathrm{Fe}\left(\mathrm{NO}_2\right)_2

is oxidised into

\mathrm{Fe}^{3+}

and

\mathrm{NO}_3^{-}

by

\mathrm{x}

mole of

\mathrm{Cr}_2 \mathrm{O}_7^{2-}

in acidic medium,

\mathrm{x}

is

View answer and explanation

Correct option: D

0.83

Option D (0.83) is correct because it is derived from balancing the oxidation and reduction half-reactions, showing that 1 mole of

Fe(NO_2)_2

requires 5/6 moles of

Cr_2O_7^{2-}

for complete oxidation.

Solution approach

Correct Answer + Core Reason: 0.83. The stoichiometry of the redox reaction requires 5/6 moles of $Cr_2O_7^{2-}$ to oxidize 1 mole of $Fe(NO_2)_2$ .

Step-by-Step Solution:

Step 1: Write the oxidation half-reaction.
$Fe(NO_2)_2 \rightarrow Fe^{3+} + 2NO_3^-$

Step 2: Balance the oxidation half-reaction.
$Fe(NO_2)_2 + 5H_2O \rightarrow Fe^{3+} + 2NO_3^- + 10H^+ + 11e^-$

Step 3: Write the reduction half-reaction.
$Cr_2O_7^{2-} \rightarrow Cr^{3+}$

Step 4: Balance the reduction half-reaction.
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Step 5: Balance the number of electrons in both half-reactions.\nMultiply the oxidation half-reaction by 6 and the reduction half-reaction by 11.
$6Fe(NO_2)_2 + 30H_2O \rightarrow 6Fe^{3+} + 12NO_3^- + 60H^+ + 66e^-$
$11Cr_2O_7^{2-} + 154H^+ + 66e^- \rightarrow 22Cr^{3+} + 77H_2O$

Step 6: Combine the balanced half-reactions.
$6Fe(NO_2)_2 + 11Cr_2O_7^{2-} + 154H^+ + 30H_2O \rightarrow 6Fe^{3+} + 12NO_3^- + 60H^+ + 22Cr^{3+} + 77H_2O$

Step 7: Simplify the equation.
$6Fe(NO_2)_2 + 11Cr_2O_7^{2-} + 94H^+ \rightarrow 6Fe^{3+} + 12NO_3^- + 22Cr^{3+} + 47H_2O$

Step 8: Determine the mole ratio of $Cr_2O_7^{2-}$ to $Fe(NO_2)_2$ .\nFrom the balanced equation, 6 moles of $Fe(NO_2)_2$ react with 11 moles of $Cr_2O_7^{2-}$ .\nTherefore, 1 mole of $Fe(NO_2)_2$ reacts with $\frac{{11}}{{6}}$ moles of $Cr_2O_7^{2-}$ .
$\frac{{11}}{{6}} \approx 1.83$ \nHowever, the oxidation state change for Fe is from +2 to +3 (change of 1), and for each $NO_2^-$ it is +3 to +5 (change of 2). So for one mole of $Fe(NO_2)_2$ , the total change is $1 + 2*2 = 5$ .\nFor $Cr_2O_7^{2-}$ , the change is from +6 to +3, so the total change is $2*(6-3) = 6$ .\nTherefore, $6x = 5$ , so $x = \frac{{5}}{{6}} = 0.833$

Why other options are wrong:
- Option A (0.167): This is incorrect as it doesn't account for the complete oxidation of both Fe and the nitrite ions.
- Option B (0.5): This is incorrect as it doesn't account for the complete oxidation of both Fe and the nitrite ions.
- Option C (1.2): This is incorrect as it doesn't account for the complete oxidation of both Fe and the nitrite ions.

Key Concepts:
- Balancing redox reactions
- Oxidation states
- Stoichiometry

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

Quick Tips:
- Always balance the half-reactions separately before combining them.
- Make sure the number of electrons lost in oxidation equals the number of electrons gained in reduction.

26 Medium

26 of 85Question permalink

Remembering +4 | -1

The extraction of which of the following metals involves bessemerization?

View answer and explanation

Correct option: D

Cu

Option D (Cu) is correct because bessemerization is a process used in the extraction of copper to remove iron impurities.

Solution approach

Correct Answer + Core Reason: Cu. Bessemerization is used in the extraction of copper.

Step-by-Step Solution:

Step 1: Recall the process of bessemerization.

Step 2: Bessemerization is a metallurgical process primarily used in the extraction of copper to remove iron impurities by oxidation.

Step 3: The process involves blowing air through molten matte (Cu2S and FeS) to oxidize iron and sulfur.

Why other options are wrong:
- Option A: Fe: While the Bessemer process was originally developed for steelmaking (iron), the question asks about extraction, not refining.
- Option B: Ag: Silver extraction does not typically involve bessemerization.
- Option C: Al: Aluminum extraction involves the Hall-Héroult process, not bessemerization.

Key Concepts:
- Bessemerization
- Extraction of copper

Memory Aids:
- Bessemerization is mainly for copper extraction.

Quick Tips:
- Remember the metallurgical processes associated with different metals.

27 Medium

27 of 85Question permalink

Remembering +4 | -1

Each transition series contains:

View answer and explanation

Correct option: B

10 elements

Option B (10 elements) is correct because each transition series corresponds to the filling of a d-orbital set, which can hold 10 electrons, hence 10 elements.

Solution approach

Correct Answer + Core Reason: 10 elements. Each transition series corresponds to the filling of a d-orbital.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of transition elements.

Step 2: Transition elements involve the filling of d-orbitals.

Step 3: Each d-orbital can hold 2 electrons, and there are 5 d-orbitals. Therefore, each transition series contains 10 elements (5 orbitals * 2 electrons/orbital).

Why other options are wrong:
- Option A: 12 elements: This is incorrect. 12 elements would correspond to filling both a d-orbital set and two electrons in an s-orbital, which doesn't define a transition series.
- Option C: 14 elements: This is the number of elements in a lanthanide or actinide series (f-block elements), not a transition series (d-block elements).
- Option D: 8 elements: This is incorrect. 8 elements would not fully utilize the d-orbitals.

Key Concepts:
- Transition elements
- d-orbitals
- Electronic configuration

Memory Aids:
- d-orbitals can hold 10 electrons, hence 10 transition elements.

Quick Tips:
- Remember the number of elements in each block of the periodic table.

28 Medium

28 of 85Question permalink

Application +4 | -1

Magnetic moment of

C{r^{2 + }}

ion is

View answer and explanation

Correct option: A

4.9 BM

Option A (4.9 BM) is correct because

Cr^{2+}

has 4 unpaired electrons, and the magnetic moment is calculated using the spin-only formula.

Solution approach

Correct Answer + Core Reason: 4.9 BM. $Cr^{2+}$ has 4 unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $Cr^{2+}$ .
- Cr: [Ar] 3d⁵ 4s¹
- $Cr^{2+}$ : [Ar] 3d⁴

Step 2: Determine the number of unpaired electrons (n).
- n = 4

Step 3: Calculate the magnetic moment using the spin-only formula: $\mu = \sqrt{n(n+2)}$ BM

Step 4: Substitute n = 4 into the formula: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM

Why other options are wrong:
- Option B: 2.82 BM corresponds to 2 unpaired electrons.
- Option C: 3.87 BM corresponds to 3 unpaired electrons.
- Option D: 5.9 BM corresponds to 5 unpaired electrons.

Key Concepts:
- Electronic configuration
- Spin-only magnetic moment formula

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

Quick Tips:
- Calculate the number of unpaired electrons accurately.

29 Medium

29 of 85Question permalink

Application +4 | -1

A magnetic moment of 1.73 BM will be shown by one among the following compounds:

View answer and explanation

Correct option: A

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

The correct answer is

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

because

Cu^{2+}

has a

d^9

configuration with one unpaired electron, resulting in a magnetic moment close to 1.73 BM.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ . This complex has one unpaired electron, resulting in a magnetic moment of approximately 1.73 BM.

Step-by-Step Solution:

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

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

Step 3: Calculate the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

Step 4: Analyze each option:
- A. ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ : Cu is in +2 oxidation state, $d^9$ configuration, 1 unpaired electron, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM
- B. ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}$ : Ni is in +2 oxidation state, $d^8$ configuration, but $CN^-$ is a strong field ligand, causing pairing of electrons. Therefore, it is diamagnetic, $\mu = 0$ BM
- C. $\mathrm{TiC}{\mathrm{l}_4}$ : Ti is in +4 oxidation state, $d^0$ configuration, 0 unpaired electrons, $\mu = 0$ BM
- D. ${\left[ {\mathrm{CoC}{\mathrm{l}_6}} \right]^{4 - }}$ : Co is in +2 oxidation state, $d^7$ configuration, 3 unpaired electrons (high spin), $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM

Step 5: The complex with a magnetic moment of 1.73 BM is ${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}$ .

Why other options are wrong:
- Option B: ${\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}$ is diamagnetic due to the strong field ligand causing pairing of electrons.
- Option C: $\mathrm{TiC}{\mathrm{l}_4}$ has a $d^0$ configuration and is therefore diamagnetic.
- Option D: ${\left[ {\mathrm{CoC}{\mathrm{l}_6}} \right]^{4 - }}$ has a magnetic moment of approximately 3.87 BM, not 1.73 BM.

Key Concepts:
- Spin-Only Magnetic Moment
- Electronic Configuration
- Strong and Weak Field Ligands

Memory Aids:
- Magnetic moment formula: $\mu = \sqrt{n(n+2)}$
- Strong field ligands cause pairing of electrons.

Quick Tips:
- Determine the number of unpaired electrons accurately.
- Consider the effect of strong field ligands.

30 Medium

30 of 85Question permalink

Remembering +4 | -1

When hydrogen peroxide is added to acidified potassium dichromate, a blue colour is produced due to formation of

View answer and explanation

Correct option: C

\mathrm{Cr}{\mathrm{O}_5}

Option C (

\mathrm{Cr}{\mathrm{O}_5}

) is correct because the reaction of hydrogen peroxide with acidified potassium dichromate forms chromium pentoxide, which has a blue color.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cr}{\mathrm{O}_5}$ . The blue color is due to the formation of Chromium pentoxide.

Step-by-Step Solution:

Step 1: Recall the reaction between hydrogen peroxide and acidified potassium dichromate.

Step 2: The reaction forms chromium pentoxide ( $\mathrm{Cr}{\mathrm{O}_5}$ ), which has a blue color.

Step 3: The reaction is:
${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + {\mathrm{H}_2}{\mathrm{O}_2} \to \mathrm{Cr}{\mathrm{O}_5} + {\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + {\mathrm{H}_2}{\mathrm{O}$

Why other options are wrong:
- Option A: $\mathrm{Cr}{\mathrm{O}_3}$ : Chromium trioxide is orange-red in color and is not formed in this reaction.
- Option B: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ : Chromium(III) oxide is green and is not directly formed in this reaction, although it can be a decomposition product of $\mathrm{Cr}{\mathrm{O}_5}$ .
- Option D: $\mathrm{CrO}_4^{2 - }$ : Chromate ion is yellow in solution and is not the blue species formed in this reaction.

Key Concepts:
- Reaction of hydrogen peroxide with potassium dichromate
- Formation of chromium pentoxide

Memory Aids:
- Remember that $\mathrm{Cr}{\mathrm{O}_5}$ gives a blue color.

Quick Tips:
- Know the characteristic colors of different chromium compounds.

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