14 elements in the sixth period (atomic no. = 90 to 103) that are filling $4f$ sub level.

14 elements in the seventh period (atomic no. = 90 to 103) that are filling $5f\;$ sub level.

14 elements in the seventh period (atomic no. = 58 to 71) that are filling $4f$ sub-level.

$VO_2^ + < C{r_2}O_7^{ - 2} < MnO_4^ - $ – oxidizing power

Co> Mn >Fe – Electrode potential $\left( {{M^{ + 3}}/{M^{ + 2}}} \right)$

The colour of ${\rm{CuC}}{{\rm{r}}_2}{{\rm{O}}_7}$ solution in water is green because:

${\rm{C}}{{\rm{u}}^{2 + }}$ ion is blue and ${\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }$ ion is yellow

${\rm{C}}{{\rm{u}}^{2 + }}$ ions is green

${\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }$ ion is green

${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}\mathop \to \limits^\Delta {{\rm{K}}_2}{\rm{Cr}}{{\rm{O}}_4} + {{\rm{O}}_2} + X{\rm{\;}}$ In the above reaction $X\;$ is

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

${\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$

In the extraction of Ag, ${\rm{A}}{{\rm{g}}_2}{\rm{S}}$ is dissolved in:

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

d & f- BLOCK ELEMENTS MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

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Remembering +4 | -1

Ruthenium carbonyl is:

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because Ruthenium pentacarbonyl has the formula

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

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}$ . The formula for Ruthenium carbonyl is $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}$ .

Step-by-Step Solution:

Step 1: Recall the formula for Ruthenium carbonyl.

Step 2: Ruthenium pentacarbonyl has the formula $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}$ .

Why other options are wrong:
- Option A: $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_4}$ is not the correct formula for Ruthenium carbonyl.
- Option C: $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_8}$ is not the correct formula for Ruthenium carbonyl.
- Option D: $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_6}$ is not the correct formula for Ruthenium carbonyl.

Key Concepts:
- Coordination compounds

Memory Aids:
- Remember the formula for common metal carbonyls.

Quick Tips:
- Know the formulas for common coordination compounds.

32 Medium

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Lanthanides are

View answer and explanation

Correct option: C

14 elements in the sixth period (atomic no. = 58 to 71) that are filling

4f

sub-level.

Option C (14 elements in the sixth period (atomic no. = 58 to 71) that are filling

4f

sub-level) is correct because lanthanides are defined as the 14 elements following lanthanum in the sixth period, filling the 4f sub-level.

Solution approach

Correct Answer + Core Reason: 14 elements in the sixth period (atomic no. = 58 to 71) that are filling $4f$ sub-level. This is the correct definition of lanthanides.

Step-by-Step Solution:

Step 1: Recall the definition of lanthanides.

Step 2: Lanthanides are the 14 elements from Cerium (Ce, atomic number 58) to Lutetium (Lu, atomic number 71) in the sixth period, filling the 4f sub-level.

Why other options are wrong:
- Option A: Atomic numbers 90 to 103 belong to the actinides, not lanthanides.
- Option B: Atomic numbers 90 to 103 belong to the actinides, and they are in the seventh period, filling the 5f sub-level.
- Option D: Atomic numbers 58 to 71 belong to the lanthanides, but they are in the sixth period, not the seventh.

Key Concepts:
- Definition of lanthanides
- Electronic configuration
- Periodic table

Memory Aids:
- Lanthanides: 4f, 6th period, atomic numbers 58-71.

Quick Tips:
- Remember the definition and position of lanthanides in the periodic table.

33 Medium

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The trivalent ion having largest size in lanthanide series is

View answer and explanation

Correct option: D

La

Option D (La) is correct because La (Lanthanum) is the first element in the lanthanide series and has the largest ionic size among the trivalent lanthanide ions.

Solution approach

Correct Answer + Core Reason: La. La has the largest size in the lanthanide series.

Step-by-Step Solution:

Step 1: Recall the trend in ionic size across the lanthanide series.

Step 2: Due to lanthanide contraction, the ionic size decreases from La to Lu.

Step 3: Therefore, La has the largest size.

Why other options are wrong:
- Option A: Ti (Titanium) is not a lanthanide.
- Option B: Zr (Zirconium) is not a lanthanide.
- Option C: Hf (Hafnium) is not a lanthanide.

Key Concepts:
- Lanthanide contraction
- Ionic size trends

Memory Aids:
- Remember that ionic size decreases across the lanthanide series.

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

34 Medium

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Which form of iron is least ductile?

View answer and explanation

Correct option: B

Cast iron

Option B (Cast iron) is correct because cast iron has a high carbon content, making it brittle and less ductile compared to other forms of iron.

Solution approach

Correct Answer + Core Reason: Cast iron. Cast iron is the least ductile form of iron due to its high carbon content.

Step-by-Step Solution:

Step 1: Define ductility.

Step 2: Ductility is the ability of a material to be drawn into wires.

Step 3: Consider the composition and properties of each form of iron.

Step 4: Hard steel has a higher carbon content than mild steel and wrought steel, making it less ductile than wrought steel and mild steel but more ductile than cast iron.

Step 5: Cast iron has the highest carbon content (2-4%), making it brittle and the least ductile.

Step 6: Mild steel has a low carbon content (0.05-0.25%), making it more ductile than hard steel and cast iron.

Step 7: Wrought steel has a very low carbon content (<0.08%), making it the most ductile.

Step 8: Therefore, cast iron is the least ductile.

Why other options are wrong:
- Option A: Hard steel is more ductile than cast iron.
- Option C: Mild steel is more ductile than cast iron.
- Option D: Wrought steel is the most ductile form of iron.

Key Concepts:
- Ductility
- Types of Iron
- Carbon Content

Memory Aids:
- Cast iron = high carbon = low ductility.

Quick Tips:
- Remember the relationship between carbon content and ductility in iron.

35 Medium

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Which one of the following pairs of elements is called ‘chemical twins’ because of their very similar chemical properties?

View answer and explanation

Correct option: D

Hf and Zr

Option D (Hf and Zr) is correct because Hafnium and Zirconium have very similar ionic radii and chemical properties due to the lanthanide contraction.

Solution approach

Correct Answer + Core Reason: Hf and Zr. Hf and Zr have similar chemical properties.

Step-by-Step Solution:

Step 1: Recall the concept of chemical twins. Chemical twins are elements with very similar chemical properties due to similar ionic radii and electronic configurations.

Step 2: Consider the electronic configurations and ionic radii of the given pairs of elements.

Step 3: Hf and Zr have very similar ionic radii due to the lanthanide contraction, which causes their chemical properties to be remarkably similar. This makes them difficult to separate.

Why other options are wrong:
- Option A: Mn and W have different chemical properties and oxidation states.
- Option B: Mo and Tc have different chemical properties.
- Option C: Fe and Re have different chemical properties.
- Option D: Hf and Zr have very similar chemical properties due to lanthanide contraction.

Key Concepts:
- Lanthanide contraction
- Chemical properties of elements
- Ionic radii

Memory Aids:
- Remember that Hf and Zr are known as chemical twins.

Quick Tips:
- The lanthanide contraction significantly affects the properties of elements in the 5d series, making them similar to their 4d counterparts.

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Which of the following is not a characteristic of transition elements?

View answer and explanation

Correct option: D

Natural radioactivity

Option D (Natural radioactivity) is correct because transition elements are not characterized by natural radioactivity; this is a property of certain heavy elements.

Solution approach

Correct Answer + Core Reason: Natural radioactivity. Transition elements are not typically radioactive.

Step-by-Step Solution:

Step 1: Recall the characteristics of transition elements.

Step 2: Transition elements exhibit variable oxidation states, form colored compounds, and form interstitial compounds.

Step 3: Transition elements are generally not naturally radioactive.

Why other options are wrong:
- Option A: Variable oxidation states: Transition elements are known for exhibiting multiple oxidation states due to the involvement of d-electrons in bonding.
- Option B: Formation of coloured compounds: The partially filled d-orbitals in transition metal ions allow for d-d transitions, which result in the absorption of light and the appearance of color.
- Option C: Formation of interstitial compounds: Transition metals can trap smaller atoms like hydrogen, carbon, and nitrogen in the interstitial spaces of their crystal lattices, forming interstitial compounds.
- Option D: Natural radioactivity: This is incorrect because radioactivity is not a general characteristic of transition elements. It is a property of certain unstable nuclei, typically found in heavier elements.

Key Concepts:
- Characteristics of transition elements
- Variable oxidation states
- Colored compounds
- Interstitial compounds
- Radioactivity

Memory Aids:
- Transition elements: Variable oxidation, colored ions, complex formation.

Quick Tips:
- Remember the general properties of transition metals.

37 Medium

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Understanding +4 | -1

Consider the following statements.

(I)

\mathrm{La}{\left( {\mathrm{OH}} \right)_3}

is the least basic among hydroxides of lanthanides

(II)

\mathrm{Z}{\mathrm{r}^{4 + }}

and

\mathrm{H}{\mathrm{f}^{4 + }}

possess almost the same ionic radii

(III)

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

can act as an oxidizing agent

Which of the above is/are true?

View answer and explanation

Correct option: D

(I) only

Option A ((I) and (III)) is incorrect. Option B ((II) and (III)) is correct because

\mathrm{Z}{\mathrm{r}^{4 + }}

and

\mathrm{H}{\mathrm{f}^{4 + }}

possess almost the same ionic radii due to lanthanoid contraction, and

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

can act as an oxidizing agent.

Solution approach

Correct Answer + Core Reason: (II) and (III). Statements II and III are true.

Step-by-Step Solution:

Step 1: Analyze statement I: $\mathrm{La}{\left( {\mathrm{OH}} \right)_3}$ is the most basic among hydroxides of lanthanides, not the least.

Step 2: Analyze statement II: $\mathrm{Z}{\mathrm{r}^{4 + }}$ and $\mathrm{H}{\mathrm{f}^{4 + }}$ possess almost the same ionic radii due to lanthanoid contraction. This is true.

Step 3: Analyze statement III: $\mathrm{C}{\mathrm{e}^{4 + }}$ can act as an oxidizing agent because it can be reduced to $\mathrm{C}{\mathrm{e}^{3 + }}$ . This is true.

Why other options are wrong:
- Option A: (I) and (III) is incorrect because statement I is false.
- Option C: (II) only is incorrect because statement III is also true.
- Option D: (I) only is incorrect because statement I is false.

Key Concepts:
- Lanthanide properties
- Lanthanoid contraction
- Oxidizing agents

Memory Aids:
- Remember the trends in basicity and oxidizing power in lanthanides.

Quick Tips:
- Know the properties of lanthanides and their compounds.

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Analyzing +4 | -1

Wrong match is

View answer and explanation

Correct option: B

Cr (+6) >Mo (+6) >W (+6) - stability of oxidation state

Option B is the wrong match because the stability of the +6 oxidation state decreases down the group due to the inert pair effect.

Solution approach

Correct Answer + Core Reason: Cr (+6) >Mo (+6) >W (+6) - stability of oxidation state. The stability of higher oxidation states decreases down the group.

Step-by-Step Solution:

Step 1: Analyze the trend of stability of oxidation states in group 6.

Step 2: The stability of higher oxidation states generally decreases down a group due to the inert pair effect.

Step 3: Therefore, W(+6) should be less stable than Mo(+6) and Cr(+6).

Step 4: The correct order of stability should be Cr(+6) > Mo(+6) > W(+6) is incorrect. The correct order is W(+6) > Mo(+6) > Cr(+6).

Why other options are wrong:
- Option A: Cu > Au > Ag – Melting point. This is the correct order of melting points.
- Option C: $VO_2^ + < C{r_2}O_7^{ - 2} < MnO_4^ -$ – oxidizing power. This is the correct order of oxidizing power.
- Option D: Co> Mn >Fe – Electrode potential $\left( {{M^{ + 3}}/{M^{ + 2}}} \right)$ . This is the correct order of electrode potential.

Key Concepts:
- Stability of oxidation states
- Inert pair effect
- Melting point trends
- Oxidizing power trends
- Electrode potential trends

Memory Aids:
- Down the group, higher oxidation states become less stable.

Quick Tips:
- Remember the general trends for stability of oxidation states in groups.

39 Medium

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Understanding +4 | -1

The colour of

\mathrm{CuC}{\mathrm{r}_2}{\mathrm{O}_7}

solution in water is green because:

View answer and explanation

Correct option: D

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

ion is blue and

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

ion is yellow

The green color of

\mathrm{CuC}{\mathrm{r}_2}{\mathrm{O}_7}

solution is due to the combination of blue

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

ions and yellow

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{u}^{2 + }}$ ion is blue and $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$ ion is yellow. The combination of blue and yellow gives green.

Step-by-Step Solution:

Step 1: Identify the colors of the individual ions.

Step 2: $\mathrm{C}{\mathrm{u}^{2 + }}$ ions are blue in aqueous solution.

Step 3: $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$ ions are yellow in aqueous solution.

Step 4: When both ions are present in the same solution, the combination of blue and yellow results in a green color.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}^{2 + }}$ ions alone are blue, not green.
- Option B: $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$ ions alone are yellow, not green.
- Option C: Neither ion is green on its own; the green color arises from their combination.

Key Concepts:
- Color of ions in solution
- Complementary colors

Memory Aids:
- $\mathrm{C}{\mathrm{u}^{2 + }}$ is blue, $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$ is yellow.

Quick Tips:
- Remember the common colors of transition metal ions.

40 Medium

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Understanding +4 | -1

which of the following does not exist

View answer and explanation

Correct option: A

{\left[ {CC{l_6}} \right]^2}

{\left[ {CC{l_6}} \right]^2}

does not exist because carbon cannot expand its octet due to the absence of available d-orbitals.

Solution approach

Correct Answer + Core Reason: ${\left[ {CC{l_6}} \right]^2}$ . Carbon cannot expand its octet.

Step-by-Step Solution:

Step 1: Understand the octet rule and its limitations.

Step 2: Carbon, being a second-period element, does not have available d-orbitals and therefore cannot expand its octet. It can only accommodate a maximum of 8 electrons in its valence shell.

Step 3: Silicon, Germanium, and Tin can expand their octets because they have available d-orbitals.

Step 4: Therefore, ${\left[ {CC{l_6}} \right]^2}$ does not exist.

Why other options are wrong:
- Option B: ${\left[ {GeC{l_6}} \right]^2}^ -$ exists because Germanium can expand its octet.
- Option C: ${\left[ {SiC{l_6}} \right]^2}^ -$ exists because Silicon can expand its octet.
- Option D: ${\left[ {SnC{l_6}} \right]^2}^ -$ exists because Tin can expand its octet.

Key Concepts:
- Octet rule
- Expansion of octet
- Availability of d-orbitals

Memory Aids:
- Carbon cannot expand its octet.

Quick Tips:
- Elements in the second period generally cannot expand their octets.

41 Medium

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Remembering +4 | -1

The process of heating of steel to temperature much below redness and then slowly cooling is called:

View answer and explanation

Correct option: C

Tempering

Option C (Tempering) is correct because tempering involves heating steel to a temperature below its critical point and then slowly cooling it to reduce brittleness and increase toughness.

Solution approach

Correct Answer + Core Reason: Tempering. Tempering involves heating steel below redness and slowly cooling it.

Step-by-Step Solution:

Step 1: Define the given heat treatment processes.

Step 2: Annealing involves heating a material to a specific temperature and then slowly cooling it to reduce hardness and increase ductility.

Step 3: Hardening involves heating steel to a high temperature and then rapidly cooling it (quenching) to increase hardness.

Step 4: Tempering involves heating hardened steel to a temperature below its critical point and then slowly cooling it to reduce brittleness and increase toughness.

Step 5: Case hardening involves hardening the surface of steel while leaving the core soft.

Step 6: The process described (heating below redness and slowly cooling) is tempering.

Why other options are wrong:
- Option A: Annealing typically involves heating to a higher temperature than tempering.
- Option B: Hardening involves rapid cooling (quenching), not slow cooling.
- Option D: Case hardening is a surface treatment, not a bulk treatment.

Key Concepts:
- Heat Treatment of Steel
- Annealing
- Hardening
- Tempering
- Case Hardening

Memory Aids:
- Tempering = reduces brittleness.

Quick Tips:
- Remember the definitions of the different heat treatment processes for steel.

42 Medium

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Application +4 | -1

The formula mass of Mohr’s salt is 392. The iron present in it is oxidised by

KMn{O_4}

in acid medium. The equivalent mass of Mohr’s salt is

View answer and explanation

Correct option: A

392

The correct answer is 392 because Mohr's salt,

FeSO_4(NH_4)_2SO_4.6H_2O

, undergoes a one-electron change (

Fe^{2+} \to Fe^{3+}

) when oxidized by

KMnO_4

in acidic medium, so its equivalent mass equals its molar mass.

Solution approach

Correct Answer + Core Reason: 392. The equivalent mass of Mohr's salt is equal to its formula mass because it undergoes a one-electron change during oxidation.

Step-by-Step Solution:

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

Step 2: Identify the oxidation state change of iron during the reaction with $KMnO_4$ . Iron is oxidized from $Fe^{2+}$ to $Fe^{3+}$ . This is a one-electron change.

Step 3: Calculate the equivalent mass using the formula: Equivalent mass = Formula mass / n-factor, where n-factor is the number of electrons transferred per mole of the substance.

Step 4: In this case, the n-factor is 1 because iron undergoes a one-electron change.

Step 5: Equivalent mass = 392 / 1 = 392.

Why other options are wrong:
- Option B: 31.6 is incorrect. This value is close to the equivalent weight of KMnO4 when it changes by 5 electrons, but this is not related to Mohr's salt.
- Option C: 278 is incorrect. This value doesn't correspond to any relevant calculation in this context.
- Option D: 156 is incorrect. This value doesn't correspond to any relevant calculation in this context.

Key Concepts:
- Equivalent Mass
- Oxidation-Reduction Reactions
- n-factor

Memory Aids:
- Equivalent mass = Molar mass / n-factor

Quick Tips:
- Determine the oxidation state change to find the n-factor.
- If the substance undergoes a one-electron change, its equivalent mass equals its molar mass.

43 Medium

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Analyzing +4 | -1

In chromyl chloride, the number of d-electrons present on chromium is same as in (Given at. no. of

\text{Ti:} 22, \, \text{V:} 23, \, \text{Cr:}24, \, \text{Mn:} 25, \, \text{Fe:} 26

)

View answer and explanation

Correct option: B

\mathrm{Mn}

(VII)

Option B (

\mathrm{Mn}

(VII)) is correct because in chromyl chloride (

CrO_2Cl_2

), Cr has an oxidation state of +6, meaning it has lost 6 electrons, leaving it with the same number of d-electrons as

Mn^{7+}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mn}$ (VII). In chromyl chloride ( $CrO_2Cl_2$ ), Cr has an oxidation state of +6, meaning it has lost 6 electrons, leaving it with the same number of d-electrons as $Mn^{7+}$ .

Step-by-Step Solution:

Step 1: Determine the oxidation state of Cr in chromyl chloride ( $CrO_2Cl_2$ ).

Step 2: The oxidation state of Cr is +6.

Step 3: The electronic configuration of Cr is $[Ar] 3d^5 4s^1$ .

Step 4: The electronic configuration of $Cr^{6+}$ is $[Ar]$ . It has 0 d-electrons.

Step 5: Determine the oxidation state and electronic configuration of the other options.

Step 6: V(IV): $V^{4+}$ has 1 d-electron.\nMn(VII): $Mn^{7+}$ has 0 d-electrons.\nFe(III): $Fe^{3+}$ has 5 d-electrons.\nTi(III): $Ti^{3+}$ has 1 d-electron.

Step 7: $Mn^{7+}$ has the same number of d-electrons as $Cr^{6+}$ in chromyl chloride.

Why other options are wrong:
- Option A: V(IV) has 1 d-electron, which is not the same as Cr(VI) in chromyl chloride.
- Option C: Fe(III) has 5 d-electrons, which is not the same as Cr(VI) in chromyl chloride.
- Option D: Ti(III) has 1 d-electron, which is not the same as Cr(VI) in chromyl chloride.

Key Concepts:
- Oxidation state
- Electronic configuration
- d-electrons

Memory Aids:
- Remember how to calculate oxidation states in compounds.

Quick Tips:
- Determine the oxidation state and electronic configuration to compare the number of d-electrons.

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Application +4 | -1

The fraction of chlorine precipitated by

\mathrm{AgN}{\mathrm{O}_3}

solution from

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

is:

View answer and explanation

Correct option: B

2/3

Option B (2/3) is correct because only the chloride ions outside the coordination sphere will precipitate with

AgNO_3

.

Solution approach

Correct Answer + Core Reason: 2/3. Only the $Cl^-$ ions outside the coordination sphere precipitate.

Step-by-Step Solution:

Step 1: Identify the complex: $[Co(NH_3)_5Cl]Cl_2$

Step 2: Determine the number of chloride ions outside the coordination sphere. There are 2 $Cl^-$ ions outside the brackets.

Step 3: Determine the total number of chloride ions in the compound. There are 3 $Cl^-$ ions in total (1 inside and 2 outside).

Step 4: Calculate the fraction of chlorine precipitated: (Number of $Cl^-$ ions precipitated) / (Total number of $Cl^-$ ions) = 2/3

Why other options are wrong:
- Option A: 1/2 is incorrect because it doesn't account for the total number of chloride ions.
- Option C: 1/3 is incorrect because it only considers one chloride ion precipitating.
- Option D: 1/4 is incorrect and doesn't relate to the stoichiometry of the compound.

Key Concepts:
- Coordination complexes
- Precipitation reactions
- Stoichiometry

Memory Aids:
- Remember that only ions outside the coordination sphere will react with $AgNO_3$ .

Quick Tips:
- Carefully examine the coordination complex to determine which ions are outside the coordination sphere.

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Remembering +4 | -1

Which of the following lanthanide is commonly used?

View answer and explanation

Correct option: D

Cerium

Option D (Cerium) is correct because Cerium is a commonly used lanthanide element in various applications.

Solution approach

Correct Answer + Core Reason: Cerium. Cerium is a commonly used lanthanide.

Step-by-Step Solution:

Step 1: Identify the lanthanides from the given options.

Step 2: Recall the common uses of lanthanides.

Step 3: Cerium is used in various applications such as catalytic converters, lighter flints, and as a polishing compound.

Why other options are wrong:
- Option A: Lanthanum is a lanthanide, but not as commonly used as Cerium.
- Option B: Nobelium is a synthetic, radioactive element and not commonly used.
- Option C: Thorium is an actinide, not a lanthanide.

Key Concepts:
- Lanthanides and their uses

Memory Aids:
- Remember common lanthanide applications.

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

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Analyzing +4 | -1

Which of the following has the maximum number of unpaired

d

-elelments?

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it has the maximum number of unpaired d-electrons (4) compared to the other options.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}{\mathrm{e}^{2 + }}$ has the maximum number of unpaired d-electrons due to its electronic configuration.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each ion/atom.

Step 2: $\mathrm{F}{\mathrm{e}^{2 + }}$ : Fe has an electronic configuration of $[\mathrm{Ar}]3{d^6}4{s^2}$ . $\mathrm{F}{\mathrm{e}^{2 + }}$ loses 2 electrons, resulting in $[\mathrm{Ar}]3{d^6}$ . According to Hund's rule, the 6 d-electrons will be arranged as follows: t2g4 eg2, resulting in 4 unpaired electrons.

Step 3: $\mathrm{C}{\mathrm{u}^ + }$ : Cu has an electronic configuration of $[\mathrm{Ar}]3{d^{10}}4{s^1}$ . $\mathrm{C}{\mathrm{u}^ + }$ loses 1 electron, resulting in $[\mathrm{Ar}]3{d^{10}}$ . This has 0 unpaired electrons.

Step 4: $\mathrm{Zn}$ : Zn has an electronic configuration of $[\mathrm{Ar}]3{d^{10}}4{s^2}$ . This has 0 unpaired electrons.

Step 5: $\mathrm{N}{\mathrm{i}^{3 + }}$ : Ni has an electronic configuration of $[\mathrm{Ar}]3{d^8}4{s^2}$ . $\mathrm{N}{\mathrm{i}^{3 + }}$ loses 3 electrons, resulting in $[\mathrm{Ar}]3{d^7}$ . According to Hund's rule, the 7 d-electrons will be arranged as follows: t2g5 eg2, resulting in 3 unpaired electrons.

Step 6: Compare the number of unpaired electrons: $\mathrm{F}{\mathrm{e}^{2 + }}$ (4), $\mathrm{C}{\mathrm{u}^ + }$ (0), $\mathrm{Zn}$ (0), $\mathrm{N}{\mathrm{i}^{3 + }}$ (3).

Step 7: Conclusion: $\mathrm{F}{\mathrm{e}^{2 + }}$ has the maximum number of unpaired d-electrons.

Why other options are wrong:
- Option B: $\mathrm{C}{\mathrm{u}^ + }$ has a completely filled d-orbital ( $3d^{10}$ ), so it has no unpaired electrons.
- Option C: $\mathrm{Zn}$ has a completely filled d-orbital ( $3d^{10}$ ), so it has no unpaired electrons.
- Option D: $\mathrm{N}{\mathrm{i}^{3 + }}$ has an electronic configuration of $3d^7$ , which results in 3 unpaired electrons, less than $\mathrm{F}{\mathrm{e}^{2 + }}$ .

Key Concepts:
- Electronic configuration
- Hund's rule
- Unpaired electrons

Memory Aids:
- Hund's rule: Electrons fill orbitals and subshells singly before pairing up.

Quick Tips:
- Write out the electronic configurations to determine the number of unpaired electrons.

47 Medium

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Remembering +4 | -1

The catalyst used for the hydrogenation of vegetable oils for making margarine is:

View answer and explanation

Correct option: C

Ni

Option C (Ni) is correct because nickel is a common catalyst used in the hydrogenation of vegetable oils.

Solution approach

Correct Answer + Core Reason: Ni. Nickel is a common catalyst for hydrogenation of vegetable oils.

Step-by-Step Solution:

Step 1: Recall the process of hydrogenation of vegetable oils.

Step 2: Hydrogenation of vegetable oils involves adding hydrogen to unsaturated fats to convert them into saturated fats.

Step 3: Nickel (Ni) is a commonly used catalyst in this process.

Why other options are wrong:
- Option A: Cu: Copper is not typically used as a catalyst for the hydrogenation of vegetable oils.
- Option B: Na: Sodium is a strong reducing agent but not used as a catalyst for hydrogenation.
- Option D: Zn: Zinc is not typically used as a catalyst for the hydrogenation of vegetable oils.

Key Concepts:
- Hydrogenation of vegetable oils
- Catalysis

Memory Aids:
- Nickel for hydrogenation.

Quick Tips:
- Remember common catalysts used in industrial processes.

48 Medium

48 of 85Question permalink

Understanding +4 | -1

Which series of elements have nearly the same atomic radii?

View answer and explanation

Correct option: D

Fe, Co, Ni, Cu

Fe, Co, Ni, and Cu have nearly the same atomic radii due to the shielding effect and filling of d-orbitals.

Solution approach

Correct Answer + Core Reason: Fe, Co, Ni, Cu. Shielding effect and filling of d-orbitals cause similar radii.

Step-by-Step Solution:

Step 1: Understand the factors affecting atomic radii.

Step 2: Atomic radii generally decrease across a period due to increasing nuclear charge. However, in the d-block elements, the filling of d-orbitals provides shielding, which counteracts the effect of increasing nuclear charge.

Step 3: Fe, Co, Ni, and Cu are consecutive elements in the d-block, and the shielding effect causes their atomic radii to be nearly the same.

Why other options are wrong:
- Option A: F, Cl, Br, I are halogens and their atomic radii increase down the group.
- Option B: Na, K, Rb, Cs are alkali metals and their atomic radii increase down the group.
- Option C: Li, Be, B, C are elements in the second period and their atomic radii decrease across the period.

Key Concepts:
- Atomic radii trends
- Shielding effect
- d-block elements

Memory Aids:
- Remember that shielding effect causes similar atomic radii in the d-block.

Quick Tips:
- Consider the shielding effect when comparing atomic radii of d-block elements.

49 Medium

49 of 85Question permalink

Remembering +4 | -1

The actinoids showing +7 oxidation state are

View answer and explanation

Correct option: C

Np, Pu

Option C (Np, Pu) is correct because Neptunium and Plutonium are known to exhibit a +7 oxidation state.

Solution approach

Correct Answer + Core Reason: Np, Pu. Neptunium and Plutonium can exhibit a +7 oxidation state.

Step-by-Step Solution:

Step 1: Recall the oxidation states of actinides.

Step 2: Neptunium (Np) and Plutonium (Pu) are known to exhibit a +7 oxidation state, although it is not as common as +3, +4, +5, or +6.

Why other options are wrong:
- Option A (U, Np): Uranium does not commonly exhibit a +7 oxidation state.
- Option B (Pu, Am): Americium does not commonly exhibit a +7 oxidation state.
- Option D (Am, Cm): Americium and Curium do not commonly exhibit a +7 oxidation state.

Key Concepts:
- Oxidation states of actinides
- Neptunium and Plutonium

Memory Aids:
- Np and Pu can reach +7 oxidation state.

Quick Tips:
- Remember the common oxidation states of actinides.

50 Medium

50 of 85Question permalink

Analyzing +4 | -1

{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}\mathop \to \limits^\Delta {\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4} + {\mathrm{O}_2} + X\mathrm{\;}

In the above reaction

X\;

is

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}

Option C (Cr2O3) is correct because the decomposition of K2Cr2O7 upon heating yields K2CrO4, O2, and Cr2O3.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ . This is the missing product in the thermal decomposition of K2Cr2O7.

Step-by-Step Solution:

Step 1: Write the unbalanced equation:
${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}\mathop \to \limits^\Delta {\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4} + {\mathrm{O}_2} + X\mathrm{\;}$

Step 2: Balance the equation:
$4{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}\mathop \to \limits^\Delta 4{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4} + 2C{r_2}{O_3} + 3{O_2}$

Step 3: Identify X:\nFrom the balanced equation, $X = C{r_2}{O_3}$

Why other options are wrong:
- Option A: $\mathrm{Cr}{\mathrm{O}_3}$ is not a product of this reaction.
- Option B: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ is the reactant, not a product.
- Option D: $\mathrm{Cr}{\mathrm{O}_5}$ is not a product of this reaction.

Key Concepts:
- Thermal decomposition reactions
- Balancing chemical equations

Memory Aids:
- Remember the products of common decomposition reactions.

Quick Tips:
- Balance the equation to identify the missing product.

51 Medium

51 of 85Question permalink

Analyzing +4 | -1

Which one of the following does not correctly represent the correct order of the property indicated against it?

View answer and explanation

Correct option: C

Ti < V < Cr < Mn : increasing melting points

Option C is correct because the melting points of transition metals do not increase monotonically from Ti to Mn. Cr has a higher melting point than Mn due to its electronic structure.

Solution approach

Correct Answer + Core Reason: Ti < V < Cr < Mn : increasing melting points. The melting point of Cr is higher than Mn.

Step-by-Step Solution:

Step 1: Analyze the given options and compare the properties of Ti, V, Cr, and Mn.

Step 2: Consider the increasing number of oxidation states. The number of oxidation states increases from Ti to Mn.

Step 3: Consider the increasing magnetic moment. The magnetic moment increases from Ti3+ to Mn3+ due to the increasing number of unpaired electrons.

Step 4: Consider the increasing melting points. The melting points generally increase with the number of unpaired electrons, but Cr has a higher melting point than Mn due to its electronic structure and strong metallic bonding.

Step 5: Consider the increasing second ionization enthalpy. The second ionization enthalpy increases from Ti to Cr, but Mn has a lower value due to the stable d5 configuration.

Why other options are wrong:
- Option A: Ti < V < Cr < Mn : increasing number of oxidation states. This statement is correct.
- Option B: $\mathrm{T}{\mathrm{i}^{3 + }} < {\mathrm{V}^{3 + }} < C{\mathrm{r}^{3 + }} < M{\mathrm{n}^{3 + }}$ : increasing magnetic moment. This statement is correct.
- Option D: Ti < V < Mn < Cr : increasing ${2^{nd}}$ ionization enthalpy. This statement is correct.

Key Concepts:
- Properties of transition metals
- Oxidation states
- Magnetic moment
- Melting points
- Ionization enthalpy

Memory Aids:
- Remember the trends in properties of transition metals.

Quick Tips:
- Recall the properties of transition metals to identify the incorrect order.

52 Medium

52 of 85Question permalink

Application +4 | -1

An element is in

{M^{3 + }}

form. Its electronic configuration is

\left[ {\mathrm{Ar}} \right]3{d^1}

, the ion is

View answer and explanation

Correct option: D

\mathrm{T}{\mathrm{i}^{3 + }}

Option D is correct because if M3+ has the electronic configuration [Ar]3d1, then the neutral atom M would have the configuration [Ar]3d14s2, which corresponds to Titanium (Ti), and therefore M3+ is Ti3+.

Solution approach

Correct Answer + Core Reason: $\mathrm{T}{\mathrm{i}^{3 + }}$ . If M³⁺ has the electronic configuration [Ar]3d¹, then the neutral atom M would have the configuration [Ar]3d¹4s², which corresponds to Titanium (Ti), and therefore M³⁺ is Ti³⁺.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the neutral atom M.\nM³⁺ has the electronic configuration [Ar]3d¹. To find the electronic configuration of the neutral atom M, add 3 electrons back. These electrons will fill the 4s orbital first, then the 3d orbital.\nSo, M would have the electronic configuration [Ar]3d¹4s².

Step 2: Identify the element with the electronic configuration [Ar]3d¹4s².\nThe element with the electronic configuration [Ar]3d¹4s² is Titanium (Ti), which has an atomic number of 22.

Step 3: Determine the ion.\nSince the ion is M³⁺, and M is Titanium (Ti), the ion is Ti³⁺.

Step 4: Conclude the answer.\nTherefore, the correct answer is Ti³⁺.

Why other options are wrong:
- Option A: Ca²⁺ has the electronic configuration [Ar], which is not related to [Ar]3d¹.
- Option B: Sc⁺ has the electronic configuration [Ar]4s¹, which is not related to [Ar]3d¹.
- Option C: Ti⁴⁺ has the electronic configuration [Ar], which is not related to [Ar]3d¹.

Key Concepts:
- Electronic configuration
- Ion formation
- Transition metal ions

Memory Aids:
- Remember the order of filling orbitals: s, p, d, f.
- When forming ions, electrons are removed from the outermost shell first (4s before 3d).

Quick Tips:
- Write the electronic configuration of the ion.
- Add electrons back to the neutral atom, filling the orbitals in the correct order.
- Identify the element based on its electronic configuration.

53 Medium

53 of 85Question permalink

Analyzing +4 | -1

The incorrect statement(s) among (1)-(3) is (are)
1) W(VI) is more stable than Cr(VI).
2) In the presence of HCl, permanganate titrations provide satisfactory results.
3) Some lanthanoid oxides can be used as phosphors.

View answer and explanation

Correct option: B

2 only

Option B (2 only) is the correct answer because permanganate titrations in the presence of HCl are not satisfactory due to the oxidation of HCl to chlorine.

Solution approach

Correct Answer + Core Reason: 2 only. Permanganate titrations in the presence of HCl are not satisfactory.

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: Statement 1: W(VI) is more stable than Cr(VI). This is generally true due to the inert pair effect and better shielding of the nuclear charge in heavier elements.

Step 3: Statement 2: In the presence of HCl, permanganate titrations provide satisfactory results. This is incorrect. $MnO_4^-$ oxidizes $Cl^-$ to $Cl_2$ in the presence of HCl, leading to inaccurate results.

Step 4: Statement 3: Some lanthanoid oxides can be used as phosphors. This is true. Lanthanoid oxides are known for their luminescent properties and are used in phosphors.

Step 5: Identify the incorrect statement(s).

Why other options are wrong:
- Option A: 2 and 3 only. Statement 3 is correct.
- Option C: 1 only. Statement 1 is correct.
- Option D: 1 and 2 only. Statement 1 is correct.

Key Concepts:
- Stability of oxidation states
- Permanganate titrations
- Lanthanoid oxides as phosphors

Memory Aids:
- Remember that permanganate titrations should not be performed in the presence of HCl.

Quick Tips:
- Be aware of the limitations of using certain reagents in titrations.

54 Medium

54 of 85Question permalink

Understanding +4 | -1

Which among the following is smallest in size?

View answer and explanation

Correct option: A

Tm

Tm is the smallest in size because lanthanide contraction causes a decrease in size across the lanthanide series.

Solution approach

Correct Answer + Core Reason: Tm. Lanthanide contraction causes size to decrease across the series.

Step-by-Step Solution:

Step 1: Understand the concept of lanthanide contraction.

Step 2: Lanthanide contraction is the steady decrease in the size of the lanthanide ions with increasing atomic number.

Step 3: The order of the given elements in the lanthanide series is Ce, Eu, Tb, Tm. Therefore, Tm has the smallest size due to lanthanide contraction.

Why other options are wrong:
- Option B: Ce is larger than Tm because it is earlier in the lanthanide series.
- Option C: Eu is larger than Tm because it is earlier in the lanthanide series.
- Option D: Tb is larger than Tm because it is earlier in the lanthanide series.

Key Concepts:
- Lanthanide contraction
- Atomic size trends

Memory Aids:
- Remember that size decreases across the lanthanide series due to lanthanide contraction.

Quick Tips:
- Lanthanide contraction is important for understanding the properties of lanthanides.

55 Medium

55 of 85Question permalink

Understanding +4 | -1

Which of the following elements has the maximum first ionization potential?

View answer and explanation

Correct option: C

Mn

Option C (Mn) is correct because Mn has a stable half-filled

d^5

configuration, making it more difficult to remove an electron compared to its neighbors.

Solution approach

Correct Answer + Core Reason: Mn. Mn has a stable half-filled d-orbital configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the given elements.\nV: $[Ar]3d^34s^2$ \nTi: $[Ar]3d^24s^2$ \nMn: $[Ar]3d^54s^2$ \nCr: $[Ar]3d^54s^1$

Step 2: Consider the stability of the electronic configurations.\nHalf-filled and fully filled electronic configurations are more stable.

Step 3: Mn has a half-filled $d^5$ configuration, which is more stable than the configurations of V, Ti, and Cr.

Why other options are wrong:
- Option A: V does not have a particularly stable electronic configuration.
- Option B: Ti does not have a particularly stable electronic configuration.
- Option D: Cr has a $3d^54s^1$ configuration. Although the $d$ subshell is half-filled, the removal of an electron from the 4s orbital is easier than removing an electron from the half-filled $d$ subshell of Mn.

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

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

Quick Tips:
- Consider the electronic configurations and stability of orbitals when comparing ionization potentials.

56 Medium

56 of 85Question permalink

Remembering +4 | -1

In the extraction of Ag,

\mathrm{A}{\mathrm{g}_2}\mathrm{S}

is dissolved in:

View answer and explanation

Correct option: C

KCN

Option C (KCN) is correct because silver sulfide (

\mathrm{A}{\mathrm{g}_2}\mathrm{S}

) dissolves in potassium cyanide (KCN) solution to form a soluble complex.

Solution approach

Correct Answer + Core Reason: KCN. $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ is dissolved in KCN.

Step-by-Step Solution:

Step 1: Understand the process of silver extraction, specifically the cyanide process.

Step 2: Silver sulfide ( $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ ) reacts with potassium cyanide (KCN) in the presence of oxygen and water to form a soluble silver cyanide complex and potassium sulfide.

Step 3: The reaction is: $\mathrm{A}{\mathrm{g}_2}\mathrm{S} + 4KCN \rightleftharpoons 2K[Ag(CN)_2] + K_2S$

Why other options are wrong:
- Option A: HCl does not dissolve $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ effectively.
- Option B: $\mathrm{HN}{\mathrm{O}_3}$ can react with $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ , but it's not the primary method for extraction.
- Option C: KCN dissolves $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ to form a soluble complex.
- Option D: ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ does not effectively dissolve $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ .

Key Concepts:
- Extraction of silver
- Cyanide process
- Complex formation

Memory Aids:
- Remember that KCN is used to dissolve silver ores in the cyanide process.

Quick Tips:
- The cyanide process is a common method for extracting gold and silver from their ores.

57 Medium

57 of 85Question permalink

Application +4 | -1

In the equation,

4M\mathrm{\;} + \mathrm{\;}8\mathrm{C}{\mathrm{N}^ - }\mathrm{\;} + \mathrm{\;}2{\mathrm{H}_2}\mathrm{O\;} + \mathrm{\;}{\mathrm{O}_2}\mathrm{\;\;} \to \mathrm{\;}4{\left[ {M{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }\mathrm{\;} + \mathrm{\;}4\mathrm{O}{\mathrm{H}^ - }

Identify the metal

M

View answer and explanation

Correct option: C

Silver

Option C is correct because silver (Ag) reacts with cyanide in the presence of water and oxygen to form a complex ion [Ag(CN)2]- and hydroxide ions.

Solution approach

Correct Answer + Core Reason: Silver. Silver (Ag) reacts with cyanide in the presence of water and oxygen to form a complex ion and hydroxide ions.

Step-by-Step Solution:

Step 1: Analyze the given equation.\nThe equation is:
4M + 8CN⁻ + 2H₂O + O₂ → 4[M(CN)₂]⁻ + 4OH⁻\nThis reaction involves the formation of a complex ion [M(CN)₂]⁻.

Step 2: Consider the metals that form cyanide complexes.\nSilver (Ag) is known to form a stable cyanide complex, [Ag(CN)₂]⁻, which is used in the extraction of silver from its ore.

Step 3: Write the reaction for silver.\nThe reaction for silver is:
4Ag + 8CN⁻ + 2H₂O + O₂ → 4[Ag(CN)₂]⁻ + 4OH⁻

Step 4: Consider other metals.
- Copper can form cyanide complexes, but the reaction is not as favorable as with silver.
- Iron and zinc do not readily form cyanide complexes under these conditions.

Step 5: Conclude the answer.\nTherefore, the metal M is silver (Ag).

Why other options are wrong:
- Option A: Copper can form cyanide complexes, but the reaction is not as favorable as with silver.
- Option B: Iron does not readily form cyanide complexes under these conditions.
- Option D: Zinc does not readily form cyanide complexes under these conditions.

Key Concepts:
- Complex formation
- Cyanide complexes
- Extraction of silver

Memory Aids:
- Remember that silver forms a stable cyanide complex, [Ag(CN)₂]⁻.

Quick Tips:
- Consider the metals that are known to form stable cyanide complexes.
- Silver is often extracted using cyanide complexes.

58 Medium

58 of 85Question permalink

Remembering +4 | -1

Thermal decomposition of zinc nitrate gives:

View answer and explanation

Correct option: B

ZnO

Option B (ZnO) is correct because the thermal decomposition of zinc nitrate produces zinc oxide, nitrogen dioxide, and oxygen.

Solution approach

Correct Answer + Core Reason: ZnO. Thermal decomposition of zinc nitrate gives zinc oxide.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the thermal decomposition of zinc nitrate.

Step 2: $2Zn(NO_3)_2(s) \rightarrow 2ZnO(s) + 4NO_2(g) + O_2(g)$

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

Why other options are wrong:
- Option A: Zn is not formed during the thermal decomposition of zinc nitrate.
- Option B: ZnO is formed during the thermal decomposition of zinc nitrate.
- Option C: $Zn(NO_2)_2$ is not formed during the thermal decomposition of zinc nitrate.
- Option D: NO is not formed during the thermal decomposition of zinc nitrate; $NO_2$ is formed instead.

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

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

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

59 Medium

59 of 85Question permalink

Application +4 | -1

Among the following, the compound that is both paramagnetic and coloured is

View answer and explanation

Correct option: D

\mathrm{VOS}{\mathrm{O}_4}

Option D is correct because VOSO4 contains V4+ which has an unpaired electron (d1 configuration), making it both paramagnetic and colored due to d-d transitions.

Solution approach

Correct Answer + Core Reason: $\mathrm{VOS}{\mathrm{O}_4}$ . VOSO₄ contains V⁴⁺, which has an unpaired electron, making it paramagnetic and colored.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the metal in each compound.
- (NH₄)₂[TiCl₆]: TiCl₆²⁻ has Ti in the +4 oxidation state. Ti⁴⁺ has a d⁰ configuration, so it is diamagnetic and colorless.
- K₂Cr₂O₇: Cr₂O₇²⁻ has Cr in the +6 oxidation state. Cr⁶⁺ has a d⁰ configuration, so it is diamagnetic, but it is colored due to charge transfer.
- K₃[Cu(CN)₄]: [Cu(CN)₄]³⁻ has Cu in the +1 oxidation state. Cu⁺ has a d¹⁰ configuration, so it is diamagnetic and colorless.
- VOSO₄: VO²⁺ has V in the +4 oxidation state. V⁴⁺ has a d¹ configuration, so it is paramagnetic and colored due to d-d transitions.

Step 2: Identify the compound that is both paramagnetic and colored.\nVOSO₄ contains V⁴⁺, which has an unpaired electron (d¹ configuration), making it paramagnetic. The d-d transition also makes it colored.

Step 3: Conclude the answer.\nTherefore, the correct answer is VOSO₄.

Why other options are wrong:
- Option A: (NH₄)₂[TiCl₆] is diamagnetic and colorless because Ti⁴⁺ has a d⁰ configuration.
- Option B: K₂Cr₂O₇ is diamagnetic but colored due to charge transfer, not d-d transitions. Cr⁶⁺ has a d⁰ configuration.
- Option C: K₃[Cu(CN)₄] is diamagnetic and colorless because Cu⁺ has a d¹⁰ configuration.

Key Concepts:
- Paramagnetism
- Diamagnetism
- d-d transitions
- Charge transfer
- Oxidation state

Memory Aids:
- Paramagnetic substances have unpaired electrons.
- Colored compounds often have d-d transitions or charge transfer.

Quick Tips:
- Determine the oxidation state of the metal in the compound.
- Write the electronic configuration of the metal ion.
- If there are unpaired electrons, the compound is paramagnetic.
- If there are d-d transitions or charge transfer, the compound is colored.

60 Medium

60 of 85Question permalink

Analyzing +4 | -1

Which of the following is not correct for

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

?

View answer and explanation

Correct option: D

Its colour is yellow

Option D (Its colour is yellow) is correct because the colour of

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

is orange, not yellow.

Solution approach

Correct Answer + Core Reason: Its colour is yellow. ${K}_2 {Cr}_2 {O}_7$ is orange, not yellow.

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: $6(\mathrm{Cr}-\mathrm{O})$ bonds are equivalent. This is correct due to resonance.
- B: Equivalent mass is $\mathrm{M} / 6$ in acidic medium. This is correct because the oxidation state of Cr changes from +6 to +3, a change of 3 per Cr atom, and there are two Cr atoms, so the total change is 6.
- C: $\mathrm{Cr}-\mathrm{O}-\mathrm{Cr}$ bond angle is more than $100^{\circ}$ . This is correct; the angle is approximately 109°.
- D: Its colour is yellow. This is incorrect. ${K}_2 {Cr}_2 {O}_7$ is orange.

Step 2: Recall the structure and properties of ${K}_2 {Cr}_2 {O}_7$ .

Step 3: Identify the incorrect statement: The color of ${K}_2 {Cr}_2 {O}_7$ is orange, not yellow.

Why other options are wrong:
- Option A: The Cr-O bonds are indeed equivalent due to resonance.
- Option B: The equivalent mass is M/6 in acidic medium, as the oxidation state change of Cr is 6.
- Option C: The Cr-O-Cr bond angle is greater than 100 degrees.

Key Concepts:
- Structure of ${K}_2 {Cr}_2 {O}_7$
- Equivalent mass calculation
- Properties of ${K}_2 {Cr}_2 {O}_7$

Memory Aids:
- Remember the orange color of potassium dichromate.

Quick Tips:
- Know the properties of common oxidizing agents like potassium dichromate.

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