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

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

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

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

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

31 Easy

Analyzing +4 | -1

Which of the following statements are INCORRECT?
A. All the transition metals except scandium form

\text{MO}

oxides which are ionic.
B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in

\mathrm{Sc}_2 \mathrm{O}_3

to

\mathrm{Mn}_2 \mathrm{O}_7

.
C. Basic character increases from

\mathrm{V}_2 \mathrm{O}_3

to

\mathrm{V}_2 \mathrm{O}_4

to

\mathrm{V}_2 \mathrm{O}_5

.
D.

\mathrm{V}_2 \mathrm{O}_4

dissolves in acids to give

\mathrm{VO}_4^{3-}

salts.
E.

\mathrm{CrO}

is basic but

\mathrm{Cr}_2 \mathrm{O}_3

is amphoteric.
Choose the correct answer from the options given below:

View answer and explanation

Correct option: A

C and D only

Option A (C and D only) is correct because statements C and D are incorrect regarding the properties of vanadium oxides.

Solution approach

Correct Answer + Core Reason: C and D only. Statements C and D are incorrect regarding the properties of vanadium oxides.

Step-by-Step Solution:

Step 1: Analyze each statement.\nStatement A: All the transition metals except scandium form $\text{MO}$ oxides which are ionic. This statement is generally correct. Most transition metal monoxides are ionic.\nStatement B: The highest oxidation number corresponding to the group number in transition metal oxides is attained in $\mathrm{Sc}_2 \mathrm{O}_3$ to $\mathrm{Mn}_2 \mathrm{O}_7$ . This statement is correct. Scandium (+3) to Manganese (+7) show the highest oxidation state equal to their group number.\nStatement C: Basic character increases from $\mathrm{V}_2 \mathrm{O}_3$ to $\mathrm{V}_2 \mathrm{O}_4$ to $\mathrm{V}_2 \mathrm{O}_5$ . This statement is incorrect. As the oxidation state increases, the acidic character increases. So, the basic character decreases from $\mathrm{V}_2 \mathrm{O}_3$ to $\mathrm{V}_2 \mathrm{O}_5$ .\nStatement D: $\mathrm{V}_2 \mathrm{O}_4$ dissolves in acids to give $\mathrm{VO}_4^{3-}$ salts. This statement is incorrect. $\mathrm{V}_2 \mathrm{O}_4$ dissolves in acids to give $V^{4+}$ salts, not $\mathrm{VO}_4^{3-}$ .\nStatement E: $\mathrm{CrO}$ is basic but $\mathrm{Cr}_2 \mathrm{O}_3$ is amphoteric. This statement is correct.

Step 2: Identify the incorrect statements.\nStatements C and D are incorrect.

Final Answer: C and D only

Why other options are wrong:
- Option B (B and C only): This is incorrect because statement B is correct, and D is incorrect.
- Option C (A and E only): This is incorrect because statements A and E are correct, and C and D are incorrect.
- Option D (B , D and E only): This is incorrect because statements B and E are correct, and only C and D are incorrect.

Key Concepts:
- Transition metal oxides
- Oxidation states
- Acidic and basic character of oxides

Memory Aids:
- As the oxidation state of a metal increases, the acidic character of its oxide increases.

Quick Tips:
- Remember the trends in acidic and basic character of oxides with increasing oxidation state.

32 Easy

32 of 80Question permalink

Remembering +4 | -1

Essential constituent of an amalgam is:

View answer and explanation

Correct option: D

Mercury

Option D (Mercury) is correct because an amalgam is defined as an alloy of mercury with another metal.

Solution approach

Correct Answer + Core Reason: Mercury. An amalgam is an alloy of mercury with another metal.

Step-by-Step Solution:

Step 1: Define an amalgam.

Step 2: An amalgam is an alloy of mercury with another metal. Mercury is the essential constituent.

Step 3: Consider the other options.

Step 4: Fe (Iron) is not a constituent of amalgams.

Step 5: Alkali metals can form amalgams, but mercury is still the essential constituent.

Step 6: Silver is a common metal used in dental amalgams, but mercury is the essential constituent.

Step 7: Therefore, Mercury is the essential constituent of an amalgam.

Why other options are wrong:
- Option A: Iron (Fe) is not a constituent of amalgams.
- Option B: Alkali metals can be part of amalgams, but mercury is essential.
- Option C: Silver (Silver) can be part of amalgams, but mercury is essential.

Key Concepts:
- Alloys
- Amalgams

Memory Aids:
- Amalgam = Mercury alloy

Quick Tips:
- Remember the definition of an amalgam.

33 Easy

33 of 80Question permalink

Understanding +4 | -1

Zr\mathrm{ }(Z = 40)\mathrm{ and }Hf(Z = 72)

have similar atomic and ionic radii because of:

View answer and explanation

Correct option: A

Lanthanoid contraction

Option A (Lanthanoid contraction) is correct because Lanthanoid contraction causes

Zr

and

Hf

to have similar atomic and ionic radii.

Solution approach

Correct Answer + Core Reason: Lanthanoid contraction. Lanthanoid contraction is responsible for the similar radii of Zr and Hf.

Step-by-Step Solution:

Step 1: Understand the concept of lanthanoid contraction.

Step 2: Lanthanoid contraction is the decrease in atomic and ionic radii of the lanthanides as the atomic number increases.

Step 3: This contraction affects the elements that follow the lanthanides, such as Hf, causing it to have a similar size to Zr.

Why other options are wrong:
- Option B: Having similar chemical properties is a consequence of similar size, not the cause.
- Option C: Belonging to the same group is not the primary reason for their similar radii.
- Option D: Diagonal relationship applies to elements in different groups and periods, not within the same group.

Key Concepts:
- Lanthanoid contraction

Memory Aids:
- Remember that lanthanoid contraction affects the size of subsequent elements.

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

34 Easy

34 of 80Question permalink

Analyzing +4 | -1

An aqueous solution of

\mathrm{FeS}{\mathrm{O}_4},\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}

and chrome alum is heated with excess of

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

and filtered. The materials obtained are:

View answer and explanation

Correct option: C

A yellow filtrate and a brown residue

Option C (A yellow filtrate and a brown residue) is correct because

\mathrm{FeS}{\mathrm{O}_4}

forms iron(III) oxide (brown residue) and chrome alum forms sodium chromate (yellow filtrate) in the presence of excess

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

.

Solution approach

Correct Answer + Core Reason: A yellow filtrate and a brown residue are obtained because iron forms a brown residue and chromium forms a yellow solution.

Step-by-Step Solution:

Step 1: Identify the compounds present in the aqueous solution: $\mathrm{FeS}{\mathrm{O}_4}$ , $\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}$ , and chrome alum $[\mathrm{KCr}{{\left( {\mathrm{S}{\mathrm{O}_4}} \right)}_2} \cdot 12{\mathrm{H}_2}\mathrm{O}]$ .

Step 2: $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ is a strong oxidizing agent and a source of $\mathrm{O}{\mathrm{H}^ - }$ ions in water.

Step 3: Reaction of $\mathrm{FeS}{\mathrm{O}_4}$ with $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ :
$\mathrm{FeS}{\mathrm{O}_4} + \mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{Fe}{\left( {\mathrm{OH}} \right)_3} + \mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}$
$\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}$ is not stable and gets converted to $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} \cdot x{\mathrm{H}_2}\mathrm{O}$ (brown precipitate).

Step 4: Reaction of $\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}$ with $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ :
$\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} + \mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{NaAl}{\mathrm{O}_2} + \mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{O}$
$\mathrm{NaAl}{\mathrm{O}_2}$ dissolves in water.

Step 5: Reaction of chrome alum with $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ :
$\mathrm{KCr}{{\left( {\mathrm{S}{\mathrm{O}_4}} \right)}_2} \cdot 12{\mathrm{H}_2}\mathrm{O} + \mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2} \to \mathrm{NaCr}{\mathrm{O}_4}$ (yellow solution)

Step 6: After filtration, the filtrate contains sodium chromate (yellow) and the residue contains iron(III) oxide (brown).

Why other options are wrong:
- Option A (A colourless filtrate and a green residue): The filtrate is yellow due to the formation of sodium chromate, and the residue is brown due to iron(III) oxide.
- Option B (A yellow filtrate and a green residue): The residue is brown, not green.
- Option D (A green filtrate and a brown residue): The filtrate is yellow, not green.

Key Concepts:
- Reactions of metal salts with strong oxidizing agents
- Formation of precipitates
- Color of transition metal compounds

Memory Aids:
- Remember that iron(III) oxide is brown and chromates are often yellow.

Quick Tips:
- Consider the oxidizing and basic nature of $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ when predicting the products of the reactions.

35 Easy

35 of 80Question permalink

Understanding +4 | -1

Gadolinium belongs of 4f series. Its atomic number is 64. Which of the following is the correct electric
Configuration of gadolinium?

View answer and explanation

Correct option: A

\,\left[ {\mathrm{Xe}} \right]\mathrm{4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}

Option A (

\,\left[ {\mathrm{Xe}} \right]\mathrm{4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}

) is correct because Gadolinium's electronic configuration is [Xe]4f⁷5d¹6s², due to the stability of the half-filled 4f orbital.

Solution approach

Correct Answer + Core Reason: $\,\left[ {\mathrm{Xe}} \right]\mathrm{4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}$ . This is the correct electronic configuration of Gadolinium.

Step-by-Step Solution:

Step 1: Write the general electronic configuration for lanthanides: $[Xe]4f^n6s^2$ .

Step 2: Gadolinium (Z=64) has 64 electrons. After accounting for Xenon (54 electrons), there are 10 electrons left.

Step 3: Due to the stability of half-filled orbitals, one electron from 5d orbital is occupied before pairing starts in 4f orbital, so the configuration becomes $[Xe]4f^75d^16s^2$ .

Why other options are wrong:
- Option B: Incorrect electronic configuration for Gadolinium.
- Option C: Incorrect electronic configuration for Gadolinium.
- Option D: Incorrect electronic configuration for Gadolinium.

Key Concepts:
- Electronic configuration of lanthanides
- Hund's rule
- Stability of half-filled orbitals

Memory Aids:
- Remember the exceptions to Hund's rule for half-filled and fully-filled orbitals.

Quick Tips:
- Practice writing electronic configurations for transition and inner transition elements.

36 Easy

36 of 80Question permalink

Analyzing +4 | -1

Assertion:

d^5

configuration is more stable than

d^4

Reason:

d^5

has more exchange energy as compared to

d^4

because 10 & 6 exchanges are possible in

d^5

&

d^4

respectively.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

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

d^5

configuration is indeed more stable than

d^4

due to the higher exchange energy resulting from the greater number of possible exchanges.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. The $d^5$ configuration is more stable due to higher exchange energy.

Step-by-Step Solution:

Step 1: Analyze the Assertion.\nAssertion: $d^5$ configuration is more stable than $d^4$ .\nThis is correct. Half-filled and fully-filled configurations are more stable due to exchange energy and symmetrical distribution of electrons.

Step 2: Analyze the Reason.\nReason: $d^5$ has more exchange energy as compared to $d^4$ because 10 & 6 exchanges are possible in $d^5$ & $d^4$ respectively.\nThis is also correct. Exchange energy is related to the number of possible exchanges of electrons with parallel spins. For $d^5$ , the number of exchanges is 10, and for $d^4$ , the number of exchanges is 6.

Step 3: Determine if the Reason is the correct explanation of the Assertion.\nThe Reason correctly explains why the Assertion is true. The higher exchange energy in $d^5$ contributes to its greater stability compared to $d^4$ .

Final Answer: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Why other options are wrong:
- Option B: This is incorrect because the Reason is the correct explanation of the Assertion.
- Option C: This is incorrect because both the Assertion and Reason are correct.
- Option D: This is incorrect because both the Assertion and Reason are correct.

Key Concepts:
- Electronic configuration
- Stability of half-filled and fully-filled orbitals
- Exchange energy

Memory Aids:
- Hund's rule states that electrons will individually occupy each orbital within a subshell before doubling up in any one orbital.

Quick Tips:
- Remember that half-filled and fully-filled configurations are more stable due to exchange energy and symmetrical distribution of electrons.

37 Easy

37 of 80Question permalink

Remembering +4 | -1

Which metal has the highest melting point?

View answer and explanation

Correct option: B

W

Option B (W) is correct because Tungsten (W) has the highest melting point among the given metals, making it suitable for applications like light bulb filaments.

Solution approach

Correct Answer + Core Reason: W. Tungsten has the highest melting point among the given metals.

Step-by-Step Solution:

Step 1: Recall the melting points of the given metals.

Step 2: Pt (Platinum) has a high melting point, but not the highest.

Step 3: W (Tungsten) has the highest melting point among common metals (3422 °C).

Step 4: Pd (Palladium) has a lower melting point than Platinum and Tungsten.

Step 5: Au (Gold) has a relatively low melting point compared to the others.

Step 6: Therefore, Tungsten (W) has the highest melting point.

Why other options are wrong:
- Option A: Platinum (Pt) has a high melting point, but not the highest.
- Option C: Palladium (Pd) has a lower melting point than Tungsten.
- Option D: Gold (Au) has a relatively low melting point compared to Tungsten.

Key Concepts:
- Melting Point
- Properties of Metals

Memory Aids:
- Tungsten = highest melting point (think of light bulb filaments).

Quick Tips:
- Remember the general trends in melting points of common metals.

38 Easy

38 of 80Question permalink

Remembering +4 | -1

An element which is highly toxic for plants and animals is:

View answer and explanation

Correct option: C

Hg

Option C (Hg) is correct because mercury is a well-known heavy metal toxin that is highly detrimental to both plants and animals.

Solution approach

Correct Answer + Core Reason: Hg. Mercury is highly toxic to plants and animals.

Step-by-Step Solution:

Step 1: Consider the toxicity of each element.

Step 2: Au (Gold) is generally inert and not highly toxic.

Step 3: Mn (Manganese) is an essential nutrient in small amounts but can be toxic in high concentrations.

Step 4: Hg (Mercury) is a heavy metal and a well-known toxin that can cause severe health problems in plants and animals.

Step 5: Ca (Calcium) is an essential nutrient and not highly toxic.

Step 6: Therefore, Mercury (Hg) is the most toxic element among the given options.

Why other options are wrong:
- Option A: Gold (Au) is generally inert and not toxic.
- Option B: Manganese (Mn) is an essential nutrient in small amounts, not highly toxic in general.
- Option D: Calcium (Ca) is an essential nutrient and not toxic.

Key Concepts:
- Heavy Metal Toxicity
- Environmental Pollution

Memory Aids:
- Remember that mercury is a neurotoxin.

Quick Tips:
- Familiarize yourself with the common toxic elements and their effects.

39 Easy

39 of 80Question permalink

Remembering +4 | -1

Useful lanthanoid member is:

View answer and explanation

Correct option: A

Cerium

Option A (Cerium) is correct because Cerium is widely used in various applications, including as a catalyst and in polishing compounds.

Solution approach

Correct Answer + Core Reason: Cerium. Cerium has many useful applications.

Step-by-Step Solution:

Step 1: Recall the uses of lanthanoids.

Step 2: Cerium is used in:
- Alloys (e.g., mischmetal)
- As a catalyst in catalytic converters
- Polishing compounds (cerium oxide)

Why other options are wrong:
- Option B (Lanthanum): Lanthanum is used in some alloys and optical glasses, but cerium has more widespread uses.
- Option C (Neodymium): Neodymium is used in magnets and lasers, but cerium has more widespread uses.
- Option D (Lutetium): Lutetium has limited applications compared to cerium.

Key Concepts:
- Uses of lanthanoids
- Applications of cerium

Memory Aids:
- Cerium is a versatile lanthanoid with many applications.

Quick Tips:
- Remember the common uses of lanthanoids, especially cerium.

40 Easy

40 of 80Question permalink

Application +4 | -1

The EAN of metal atoms in

M{n_2}{\left( {CO} \right)_{10}}

and

C{o_2}{\left( {CO} \right)_8}

respectively are:

View answer and explanation

Correct option: C

36, 36

The EAN of metal atoms in

M{n_2}{\left( {CO} \right)_{10}}

and

C{o_2}{\left( {CO} \right)_8}

are both 36 because they follow the 18-electron rule after considering metal-metal bonds and CO ligands.

Solution approach

Correct Answer + Core Reason: 36, 36. Both complexes follow the 18-electron rule.

Step-by-Step Solution:

Step 1: Calculate the EAN for Mn in $M{n_2}{\left( {CO} \right)_{10}}$ :
- Mn has 7 valence electrons.
- There are 5 CO ligands attached to each Mn, contributing 2 electrons each: 5 x 2 = 10 electrons.
- There is one Mn-Mn bond, contributing 1 electron to each Mn.
- Total electrons = 7 + 10 + 1 = 18
- EAN = Atomic number - Oxidation state + Electrons from ligands + Electrons from metal-metal bond
- EAN = 25 - 0 + 10 + 1 = 36 (after accounting for the Mn-Mn bond)

Step 2: Calculate the EAN for Co in $C{o_2}{\left( {CO} \right)_8}$ :
- Co has 9 valence electrons.
- There are 4 CO ligands attached to each Co, contributing 2 electrons each: 4 x 2 = 8 electrons.
- There is one Co-Co bond, contributing 1 electron to each Co.
- Total electrons = 9 + 8 + 1 = 18
- EAN = Atomic number - Oxidation state + Electrons from ligands + Electrons from metal-metal bond
- EAN = 27 - 0 + 8 + 1 = 36 (after accounting for the Co-Co bond)

Why other options are wrong:
- Option A: 34, 35. These values are incorrect because they don't account for the metal-metal bond and the 18-electron rule.
- Option B: 34, 36. The first value is incorrect for Mn.
- Option D: 36, 35. The second value is incorrect for Co.

Key Concepts:
- Effective Atomic Number (EAN)
- 18-electron rule
- Metal-metal bonds

Memory Aids:
- Remember the 18-electron rule: transition metals tend to gain, lose, or share electrons to achieve a noble gas configuration.

Quick Tips:
- When calculating EAN for complexes with metal-metal bonds, remember to account for the electrons contributed by the metal-metal bond.

41 Easy

41 of 80Question permalink

Analyzing +4 | -1

Which of the following has more unpaired d-electrons?

View answer and explanation

Correct option: B

{Fe}^{2+}

Option B (

{Fe}^{2+}

) is correct because it has 4 unpaired d-electrons, which is more than the other options.

Solution approach

Correct Answer + Core Reason: ${Fe}^{2+}$ has more unpaired d-electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configurations of each ion.

Step 2: ${Zn}^{+}$ : $[Ar] 3d^{10}4s^1$ - 1 unpaired electron

Step 3: ${Fe}^{2+}$ : $[Ar] 3d^6$ - 4 unpaired electrons

Step 4: ${Ni}^{+}$ : $[Ar] 3d^9$ - 1 unpaired electron

Step 5: ${Cu}^{+}$ : $[Ar] 3d^{10}$ - 0 unpaired electrons

Step 6: Compare the number of unpaired electrons.

Why other options are wrong:
- Option A: ${Zn}^{+}$ has only 1 unpaired electron.
- Option C: ${Ni}^{+}$ has only 1 unpaired electron.
- Option D: ${Cu}^{+}$ has 0 unpaired electrons.

Key Concepts:
- Electronic configuration of transition metal ions
- Hund's rule
- Unpaired electrons and magnetic properties

Memory Aids:
- Remember the filling order of d-orbitals and Hund's rule for maximizing unpaired electrons.

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

42 Easy

42 of 80Question permalink

Application +4 | -1

Number of electrons present in

4 {f}

orbital of

{Ho}^{3+}

ion is .......
(Given Atomic No. of

{Ho}=67

)

View answer and explanation

Correct option: A

10

Option A (10) is correct because

{Ho}^{3+}

has 10 electrons in the 4f orbital.

Solution approach

Correct Answer + Core Reason: 10. ${Ho}^{3+}$ has 10 electrons in the 4f orbital.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Ho.

Step 2: Ho (Z=67): $[\mathrm{Xe}] 4f^{11} 6s^2$

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

Step 4: ${Ho}^{3+}$ : $[\mathrm{Xe}] 4f^{10}$

Step 5: The number of electrons in the 4f orbital of ${Ho}^{3+}$ is 10.

Why other options are wrong:
- Option B: 0 is incorrect. ${Ho}^{3+}$ has 10 electrons in the 4f orbital.
- Option C: 0 is incorrect. ${Ho}^{3+}$ has 10 electrons in the 4f orbital.
- Option D: 0 is incorrect. ${Ho}^{3+}$ has 10 electrons in the 4f orbital.

Key Concepts:
- Electronic configuration
- Lanthanide series

Memory Aids:
- Remember the filling order of electrons.

Quick Tips:
- Write the electronic configuration to determine the number of electrons in a specific orbital.

43 Easy

43 of 80Question permalink

Remembering +4 | -1

Blue vitriol is:

View answer and explanation

Correct option: C

\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}

Option C (

\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}

) is the correct formula for blue vitriol, which is copper(II) sulfate pentahydrate.

Solution approach

Correct Answer + Core Reason: $\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}$ . Blue vitriol is copper(II) sulfate pentahydrate.

Step-by-Step Solution:

Step 1: Recall the chemical formula for blue vitriol.

Step 2: Blue vitriol is the common name for copper(II) sulfate pentahydrate.

Step 3: The chemical formula for copper(II) sulfate pentahydrate is $\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}$ .

Why other options are wrong:
- Option A: $\mathrm{CuS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ is not the correct formula for blue vitriol.
- Option B: $\mathrm{ZnS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ is zinc sulfate heptahydrate (white vitriol).
- Option D: $\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}$ is iron(II) sulfate heptahydrate (green vitriol).

Key Concepts:
- Chemical formulas
- Hydrated salts

Memory Aids:
- Remember the common names and formulas of important chemical compounds.

Quick Tips:
- Memorize the formulas of common hydrated salts.

44 Easy

44 of 80Question permalink

Understanding +4 | -1

\mathrm{CuS}{\mathrm{O}_4}aq

. Absorbs

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because copper sulfate (

\mathrm{CuS}{\mathrm{O}_4}

) can form complexes with ammonia (

\mathrm{N}{\mathrm{H}_3}

), hydrogen sulfide (

{\mathrm{H}_2}\mathrm{S}

), and phosphine (

\mathrm{P}{\mathrm{H}_3}

).

Solution approach

Correct Answer + Core Reason: All of these. $\mathrm{CuS}{\mathrm{O}_4}$ can absorb $\mathrm{N}{\mathrm{H}_3}$ , ${\mathrm{H}_2}\mathrm{S}$ , and $\mathrm{P}{\mathrm{H}_3}$ .

Step-by-Step Solution:

Step 1: Consider the properties of $\mathrm{CuS}{\mathrm{O}_4}$ .

Step 2: $\mathrm{CuS}{\mathrm{O}_4}$ can form complexes with ligands such as $\mathrm{N}{\mathrm{H}_3}$ , ${\mathrm{H}_2}\mathrm{S}$ , and $\mathrm{P}{\mathrm{H}_3}$ .

Step 3: $\mathrm{CuS}{\mathrm{O}_4}$ reacts with $\mathrm{N}{\mathrm{H}_3}$ to form a complex ion, $[Cu(NH_3)_4]^{2+}$ .

Step 4: $\mathrm{CuS}{\mathrm{O}_4}$ reacts with ${\mathrm{H}_2}\mathrm{S}$ to form CuS precipitate.

Step 5: $\mathrm{CuS}{\mathrm{O}_4}$ can react with $\mathrm{P}{\mathrm{H}_3}$ to form copper phosphide.

Step 6: Therefore, $\mathrm{CuS}{\mathrm{O}_4}$ absorbs all of these gases.

Why other options are wrong:
- Option A: $\mathrm{CuS}{\mathrm{O}_4}$ also absorbs ${\mathrm{H}_2}\mathrm{S}$ and $\mathrm{P}{\mathrm{H}_3}$ .
- Option B: $\mathrm{CuS}{\mathrm{O}_4}$ also absorbs $\mathrm{N}{\mathrm{H}_3}$ and $\mathrm{P}{\mathrm{H}_3}$ .
- Option C: $\mathrm{CuS}{\mathrm{O}_4}$ also absorbs $\mathrm{N}{\mathrm{H}_3}$ and ${\mathrm{H}_2}\mathrm{S}$ .

Key Concepts:
- Complex Formation
- Copper Sulfate

Memory Aids:
- $\mathrm{CuS}{\mathrm{O}_4}$ is versatile and reacts with many gases.

Quick Tips:
- Remember that transition metal ions can form complexes with various ligands.

45 Easy

45 of 80Question permalink

Remembering +4 | -1

Chromium has most stable oxidation state of:

View answer and explanation

Correct option: B

+ 3

Option B (+3) is correct because chromium exhibits a stable +3 oxidation state due to the stable

d^3

electronic configuration.

Solution approach

Correct Answer + Core Reason: + 3. Chromium has a stable +3 oxidation state due to the stable $d^3$ configuration.

Step-by-Step Solution:

Step 1: Consider the electronic configuration of chromium.

Step 2: The electronic configuration of Cr is [Ar] 3d⁵ 4s¹.

Step 3: Chromium can exhibit various oxidation states, but +3 is the most stable.

Step 4: In the +3 oxidation state, the electronic configuration is [Ar] 3d³, which is relatively stable due to the half-filled $t_{2g}$ orbitals in an octahedral field.

Why other options are wrong:
- Option A: + 5. +5 is not a common or stable oxidation state for chromium.
- Option C: + 2. +2 is less stable than +3 for chromium.
- Option D: + 4. +4 is less stable than +3 for chromium.

Key Concepts:
- Oxidation states
- Electronic configuration
- Stability of oxidation states

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

Quick Tips:
- Consider the electronic configuration when determining the stability of oxidation states.

46 Easy

46 of 80Question permalink

Understanding +4 | -1

In which of the following pairs are both the ions coloured in aqueous solution

View answer and explanation

Correct option: B

N{i^{2 + }},T{i^{3 + }}

Option B (

N{i^{2 + }},T{i^{3 + }}

) is correct because both ions have unpaired d-electrons which allow for d-d transitions, resulting in color.

Solution approach

Correct Answer + Core Reason: $N{i^{2 + }},T{i^{3 + }}$ . Colored ions have unpaired d-electrons that allow for d-d transitions.

Step-by-Step Solution:

Step 1: Identify the electronic configurations of the given ions.

Step 2: Determine if each ion has unpaired d-electrons.

Step 3: Ions with unpaired d-electrons in aqueous solution are colored due to d-d transitions.

$N{i^{2 + }}$ : $[Ar]3d^8$ (2 unpaired electrons, colored)
$C{u^ + }$ : $[Ar]3d^{10}$ (no unpaired electrons, colorless)
$T{i^{3 + }}$ : $[Ar]3d^1$ (1 unpaired electron, colored)
$S{c^{3 + }}$ : $[Ar]3d^0$ (no unpaired electrons, colorless)
$C{o^{2 + }}$ : $[Ar]3d^7$ (3 unpaired electrons, colored)

Step 4: Based on the above analysis, the pair with both colored ions is $N{i^{2 + }},T{i^{3 + }}$ .

Why other options are wrong:
- Option A: $C{u^ + }$ is colorless because it has a completely filled d-orbital ( $d^{10}$ ).
- Option C: $S{c^{3 + }}$ is colorless because it has an empty d-orbital ( $d^0$ ).
- Option D: $S{c^{3 + }}$ is colorless because it has an empty d-orbital ( $d^0$ ).

Key Concepts:
- d-d transitions
- Electronic configuration
- Color of transition metal ions

Memory Aids:
- Remember that filled or empty d-orbitals generally result in colorless ions.

Quick Tips:
- Quickly determine the d-electron count for each ion to assess its potential for color.

47 Easy

47 of 80Question permalink

Application +4 | -1

The reduction potential of hydrogen electrode when placed in a buffer solution is found to be –0.591 V. The pH of the buffer is:

View answer and explanation

Correct option: A

10

The pH of the buffer is 10 because the reduction potential of the hydrogen electrode is related to the hydrogen ion concentration, which can be used to calculate the pH.

Solution approach

Correct Answer + Core Reason: 10. The pH is calculated using the Nernst equation for the hydrogen electrode.

Step-by-Step Solution:

Step 1: Write the Nernst equation for the hydrogen electrode:
$E = E^0 - \frac{0.0591}{n} \log \frac{[H_2]}{[H^+]^2}$ \nSince $E^0 = 0$ for the standard hydrogen electrode and assuming the pressure of $H_2$ is 1 atm, the equation simplifies to:
$E = - \frac{0.0591}{2} \log \frac{1}{[H^+]^2} = -0.0591 \log \frac{1}{[H^+]} = 0.0591 \log [H^+]$

Step 2: Given that E = -0.591 V, substitute this value into the equation:
$-0.591 = -0.0591 \times pH$

Step 3: Solve for pH:
$pH = \frac{0.591}{0.0591} = 10$

Why other options are wrong:
- Option B: 12. This would result in a different reduction potential.
- Option C: 4. This would result in a different reduction potential.
- Option D: 7. This would result in a different reduction potential.

Key Concepts:
- Nernst equation
- Standard hydrogen electrode (SHE)
- pH calculation

Memory Aids:
- Remember the Nernst equation and its application to the hydrogen electrode.

Quick Tips:
- The Nernst equation relates the reduction potential of an electrode to the concentrations of the oxidized and reduced species. For the hydrogen electrode, the pH can be directly calculated from the reduction potential.

48 Easy

48 of 80Question permalink

Remembering +4 | -1

Which metal is not used for making coins?

View answer and explanation

Correct option: D

Tungsten

Option D (Tungsten) is correct because tungsten is primarily used in light bulb filaments and high-temperature applications, not for making coins.

Solution approach

Correct Answer + Core Reason: Tungsten. Tungsten is not used for making coins.

Step-by-Step Solution:

Step 1: Consider the properties required for metals used in coins.

Step 2: Coins need to be durable, resistant to corrosion, and relatively inexpensive.

Step 3: Gold, silver, and nickel have been used for making coins.

Step 4: Tungsten is a hard, brittle metal with a very high melting point. It is expensive and difficult to work with, making it unsuitable for coinage.

Why other options are wrong:
- Option A: Gold has been used for making coins.
- Option B: Silver has been used for making coins.
- Option C: Nickel has been used for making coins.

Key Concepts:
- Properties of metals
- Uses of metals

Memory Aids:
- Remember the common metals used for coinage.

Quick Tips:
- Consider the properties required for different applications of metals.

49 Easy

49 of 80Question permalink

Understanding +4 | -1

Zr (z=40) and Hf (z=72) have similar atomic radii because of

View answer and explanation

Correct option: D

Lanthanoid contraction

Option D (Lanthanoid contraction) is correct because the lanthanoid contraction is responsible for the similar atomic radii of Zr and Hf.

Solution approach

Correct Answer + Core Reason: Lanthanoid contraction. Lanthanoid contraction causes Zr and Hf to have similar atomic radii.

Step-by-Step Solution:

Step 1: Understand the concept of lanthanoid contraction.

Step 2: Lanthanoid contraction is the gradual decrease in the size of atoms and ions with increasing atomic number within the lanthanide series.

Step 3: This effect causes the atomic radii of Zr and Hf to be similar.

Why other options are wrong:
- Option A: Similar chemical properties are a result of similar size, not the cause.
- Option B: Belonging to the same group doesn't solely explain the similar radii.
- Option C: Diagonal relationship is not relevant in this case.

Key Concepts:
- Lanthanoid contraction

Memory Aids:
- Remember that lanthanoid contraction affects the size of elements following the lanthanides.

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

50 Easy

50 of 80Question permalink

Understanding +4 | -1

Copper sulphate is commercially made from copper scrap by:

View answer and explanation

Correct option: B

Action of dilute sulphuric acid and air

Copper sulphate is commercially made from copper scrap by the action of dilute sulphuric acid and air, where air acts as an oxidizing agent.

Solution approach

Correct Answer + Core Reason: Action of dilute sulphuric acid and air. Air acts as an oxidizing agent.

Step-by-Step Solution:

Step 1: Understand the process of making copper sulphate from copper scrap.

Step 2: Copper scrap reacts with dilute sulphuric acid in the presence of air (oxygen) to form copper sulphate.

Step 3: The reaction is: $2Cu + 2H_2SO_4 + O_2 \rightarrow 2CuSO_4 + 2H_2O$

Step 4: Therefore, the correct method is the action of dilute sulphuric acid and air.

Why other options are wrong:
- Option A: Dissolving in hot concentrated sulphuric acid is a possible method but not the commercially preferred one due to the formation of sulfur dioxide ( $SO_2$ ) as a byproduct, which is environmentally unfriendly.
- Option C: Heating with sodium sulphate does not directly produce copper sulphate.
- Option D: Heating with sulphur does not directly produce copper sulphate.

Key Concepts:
- Production of copper sulphate
- Role of air as an oxidizing agent

Memory Aids:
- Remember that air (oxygen) is needed to oxidize copper to copper sulphate.

Quick Tips:
- Air is often used as an oxidizing agent in industrial processes.

51 Easy

51 of 80Question permalink

Understanding +4 | -1

Powdered silver ore is treated with NaCN solution and air is bubbled through the mixture to give:

View answer and explanation

Correct option: D

\mathrm{Na}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

Option D (

\mathrm{Na}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

) is correct because silver ore reacts with NaCN in the presence of air to form a soluble complex, sodium dicyanoargentate(I).

Solution approach

Correct Answer + Core Reason: $\mathrm{Na}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]$ . Silver ore reacts with NaCN to form a soluble complex.

Step-by-Step Solution:

Step 1: Write the chemical reaction between silver ore ( $Ag_2S$ ) and NaCN in the presence of air.
$Ag_2S + 4NaCN \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$
$4Au + 8CN^- + O_2 + 2H_2O \rightarrow 4[Au(CN)_2]^- + 4OH^-$

Step 2: Identify the product formed.

Step 3: The product is $Na[Ag(CN)_2]$ .

Why other options are wrong:
- Option A: AgCN is an intermediate but reacts further to form the soluble complex.
- Option B: Ag is the final product after reduction, not the immediate product of the reaction with NaCN.
- Option C: $\mathrm{Ag}{\left( {\mathrm{CN}} \right)_2}$ is the complex ion, but it exists as the sodium salt in the solution.

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

Memory Aids:
- Remember the cyanide process for silver extraction.

Quick Tips:
- Recall the reactions involved in the extraction of metals.

52 Easy

52 of 80Question permalink

Remembering +4 | -1

The formula of mercurous ion is:

View answer and explanation

Correct option: C

\mathrm{Hg}_2^{2 + }

Option C (

\mathrm{Hg}_2^{2 + }

) is correct because the mercurous ion exists as a diatomic species with a +2 charge overall.

Solution approach

Correct Answer + Core Reason: $\mathrm{Hg}_2^{2 + }$ . Mercurous ion exists as a diatomic species with a +2 charge.

Step-by-Step Solution:

Step 1: Recall the common ions of mercury.

Step 2: Mercurous ion is a diatomic ion with the formula $\mathrm{Hg}_2^{2 + }$ .

Step 3: Mercuric ion is $\mathrm{Hg}^{2 + }$ .

Step 4: Identify the correct formula.

Why other options are wrong:
- Option A: $\mathrm{H}{\mathrm{g}^ + }$ : This represents a single mercury atom with a +1 charge, which is not the correct formula for the mercurous ion.
- Option B: $\mathrm{Hg}_2^ +$ : This represents a diatomic mercury ion with a +1 charge, which is incorrect for mercurous ion.
- Option D: None of these: This is incorrect because option C is the correct answer.

Key Concepts:
- Nomenclature of ions
- Mercury ions (mercurous and mercuric)

Memory Aids:
- Remember that "mercurous" implies a lower oxidation state and diatomic form.

Quick Tips:
- Memorize the common ions and their formulas.

53 Easy

53 of 80Question permalink

Remembering +4 | -1

Mercury forms amalgams with all except:

View answer and explanation

Correct option: D

Fe

Option D (Fe) is correct because iron does not form amalgams with mercury due to its high melting point and strong metallic bonding.

Solution approach

Correct Answer + Core Reason: Fe. Iron does not form an amalgam with mercury.

Step-by-Step Solution:

Step 1: Understand what an amalgam is. An amalgam is an alloy of mercury with another metal.

Step 2: Consider the properties of the given metals: Al, Zn, Ni, and Fe.

Step 3: Mercury readily forms amalgams with most metals, but iron (Fe) is an exception due to its strong metallic bonding and high melting point, which hinders the diffusion of mercury into its lattice.

Why other options are wrong:
- Option A: Al forms an amalgam with mercury.
- Option B: Zn forms an amalgam with mercury.
- Option C: Ni forms an amalgam with mercury.
- Option D: Fe does not form an amalgam with mercury.

Key Concepts:
- Amalgams
- Properties of metals
- Mercury alloys

Memory Aids:
- Remember that iron is an exception and does not form amalgams with mercury.

Quick Tips:
- Consider the metallic bonding strength and melting points of metals when determining whether they form amalgams with mercury.

54 Easy

54 of 80Question permalink

Application +4 | -1

The one which has lowest ox. no. of Hg:

View answer and explanation

Correct option: D

\mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2}

Option D is correct because in Hg2Cl2, mercury exists as Hg2²⁺, giving each Hg atom an oxidation state of +1, which is the lowest among the options.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2}$ . In Hg₂Cl₂, mercury exists as Hg₂²⁺, giving each Hg atom an oxidation state of +1.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Hg in each compound.
- Hg(NO₂)₂: Let the oxidation state of Hg be x. Then, x + 2(-1) = 0 => x = +2
- HgCl₂: Let the oxidation state of Hg be x. Then, x + 2(-1) = 0 => x = +2
- Hg(NO₃)₂: Let the oxidation state of Hg be x. Then, x + 2(-1) = 0 => x = +2
- Hg₂Cl₂: Let the oxidation state of Hg be x. Since it exists as Hg₂²⁺, 2x + 2(-1) = 0 => 2x = 2 => x = +1

Step 2: Compare the oxidation states of Hg in each compound.\nThe oxidation state of Hg is +2 in Hg(NO₂)₂, HgCl₂, and Hg(NO₃)₂. The oxidation state of Hg is +1 in Hg₂Cl₂.

Step 3: Conclude the answer.\nTherefore, the compound with the lowest oxidation number of Hg is Hg₂Cl₂.

Why other options are wrong:
- Option A: Hg(NO₂)₂ has Hg in the +2 oxidation state, which is higher than +1.
- Option B: HgCl₂ has Hg in the +2 oxidation state, which is higher than +1.
- Option C: Hg(NO₃)₂ has Hg in the +2 oxidation state, which is higher than +1.

Key Concepts:
- Oxidation state
- Mercury compounds

Memory Aids:
- Remember that Hg₂Cl₂ contains Hg₂²⁺ ions, where each Hg atom has a +1 oxidation state.

Quick Tips:
- Use the rules for assigning oxidation states to determine the oxidation state of each element in the compound.
- Be careful when dealing with compounds containing polyatomic ions.

55 Easy

55 of 80Question permalink

Remembering +4 | -1

Which ion is blue in colour in aqueous medium?

View answer and explanation

Correct option: B

C{u^{2 + }}

Option B (

C{u^{2 + }}

) is correct because copper(II) ions in aqueous solution form a complex with water molecules, resulting in a blue color.

Solution approach

Correct Answer + Core Reason: $C{u^{2 + }}$ . Copper(II) ions form a blue-colored complex in aqueous solution.

Step-by-Step Solution:

Step 1: Recall the colors of common transition metal ions in aqueous solution:
- $Z{n^{2 + }}$ : Colorless
- $C{u^{2 + }}$ : Blue
- $S{c^{3 + }}$ : Colorless
- $T{i^{4 + }}$ : Colorless

Step 2: Identify the ion with a blue color: $C{u^{2 + }}$ is blue due to the formation of the complex ion $[Cu(H_2O)_6]^{2+}$ .

Why other options are wrong:
- **Option A ( $Z{n^{2 + }}$ ): Zinc(II) ions are colorless.
- Option C ( $S{c^{3 + }}$ ): Scandium(III) ions are colorless.
- Option D ( $T{i^{4 + }}$ ): Titanium(IV) ions are colorless.

Key Concepts:
- Colors of transition metal ions
- Complex formation

Memory Aids:
- Copper(II) ions are blue in solution.

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

56 Easy

56 of 80Question permalink

Understanding +4 | -1

Silver nitrate produces a black stain on skin due to:

View answer and explanation

Correct option: B

Its reduction to metallic silver

Option B (Its reduction to metallic silver) is correct because silver nitrate reacts with organic matter in the skin and is reduced to metallic silver, which appears as a black stain.

Solution approach

Correct Answer + Core Reason: Its reduction to metallic silver. Silver nitrate is reduced to metallic silver on the skin.

Step-by-Step Solution:

Step 1: Recall the reaction of silver nitrate with organic matter.

Step 2: Silver nitrate ( $\mathrm{AgNO}_3$ ) reacts with organic matter in the skin and is reduced to metallic silver (Ag).

Step 3: Metallic silver appears as a black stain.

Why other options are wrong:
- Option A: Its corrosive action: While silver nitrate can be corrosive at high concentrations, the black stain is primarily due to the formation of metallic silver, not direct corrosion.
- Option C: Its strong reducing action: Silver nitrate is an oxidizing agent, not a reducing agent. It gets reduced itself.
- Option D: The formation of a complex compound: While silver can form complexes, the black stain is mainly due to the deposition of metallic silver.

Key Concepts:
- Redox reactions
- Silver nitrate
- Formation of metallic silver

Memory Aids:
- Silver nitrate reduces to black metallic silver.

Quick Tips:
- Remember the reactions of silver salts.

57 Easy

57 of 80Question permalink

Understanding +4 | -1

Lanthanide contraction is due to increase in

View answer and explanation

Correct option: C

Effective nuclear charge

Option C (Effective nuclear charge) is correct because lanthanide contraction is due to the increase in effective nuclear charge experienced by the outer electrons.

Solution approach

Correct Answer + Core Reason: Effective nuclear charge. Lanthanide contraction is due to the increase in effective nuclear charge.

Step-by-Step Solution:

Step 1: Understand the concept of lanthanide contraction.

Step 2: The increase in nuclear charge is not effectively shielded by the added 4f electrons.

Step 3: This leads to an increase in the effective nuclear charge experienced by the outer electrons, causing the contraction.

Why other options are wrong:
- Option A: Shielding by 4f-electrons is imperfect, not the cause of contraction.
- Option B: Atomic number increases, but it's the effective nuclear charge that matters.
- Option D: Size of 4f-orbitals is not the direct cause of contraction.

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

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

Quick Tips:
- Understand the factors affecting atomic size.

58 Easy

58 of 80Question permalink

Application +4 | -1

Which of following ions has electronic configuration

\left[ {Ar} \right]3{d^6}

View answer and explanation

Correct option: D

C{O^{3 + }}

Option D is correct because Co3+ has the electronic configuration [Ar]3d⁶ after losing 3 electrons from its neutral state.

Solution approach

Correct Answer + Core Reason: $C{O^{3 + }}$ . Cobalt (Co) loses 3 electrons to achieve the [Ar]3d⁶ configuration.

Step-by-Step Solution:

Step 1: Write the electronic configuration of the neutral atoms.
- Ni (Z = 28): [Ar]3d⁸4s²
- Mn (Z = 25): [Ar]3d⁵4s²
- Fe (Z = 26): [Ar]3d⁶4s²
- Co (Z = 27): [Ar]3d⁷4s²

Step 2: Determine the electronic configuration of the ions.
- Ni³⁺: [Ar]3d⁷ (loses 2 electrons from 4s and 1 from 3d)
- Mn³⁺: [Ar]3d⁴ (loses 2 electrons from 4s and 1 from 3d)
- Fe³⁺: [Ar]3d⁵ (loses 2 electrons from 4s and 1 from 3d)
- Co³⁺: [Ar]3d⁶ (loses 2 electrons from 4s and 1 from 3d)

Step 3: Compare the ion configurations with the target configuration [Ar]3d⁶.\nOnly Co³⁺ has the [Ar]3d⁶ configuration.

Step 4: Conclude the answer.\nTherefore, the correct answer is Co³⁺.

Why other options are wrong:
- Option A: Ni³⁺ has an electronic configuration of [Ar]3d⁷, not [Ar]3d⁶.
- Option B: Mn³⁺ has an electronic configuration of [Ar]3d⁴, not [Ar]3d⁶.
- Option C: Fe³⁺ has an electronic configuration of [Ar]3d⁵, not [Ar]3d⁶.

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 neutral atom first.
- Remove electrons from the outermost shell (highest principal quantum number) when forming cations.

59 Easy

59 of 80Question permalink

Application +4 | -1

In acidic medium

\mathrm{KMnO}_4

oxidises

\mathrm{FeSO}_4

solution. Which of the following statements is correct?

View answer and explanation

Correct option: B

10 \mathrm{~mL}

of

1 \mathrm{M} \mathrm{~KMnO}_4

solution oxidises

10 \mathrm{~mL}

of

5 \mathrm{~N} \mathrm{~FeSO}_4

solution

Option B (

10 \mathrm{~mL}

of

1 \mathrm{M} \mathrm{~KMnO}_4

solution oxidises

10 \mathrm{~mL}

of

5 \mathrm{~N} \mathrm{~FeSO}_4

solution) is correct because the n-factor for

KMnO_4

in acidic medium is 5, so 1 M

KMnO_4

is equivalent to 5 N

KMnO_4

.

Solution approach

Correct Answer + Core Reason: $10 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{~KMnO}_4$ solution oxidises $10 \mathrm{~mL}$ of $5 \mathrm{~N} \mathrm{~FeSO}_4$ solution. The n-factor for $KMnO_4$ in acidic medium is 5.

Step-by-Step Solution:

Step 1: Write the balanced equation for the reaction.
$MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

Step 2: Determine the n-factor for $KMnO_4$ and $FeSO_4$ .\nThe n-factor for $KMnO_4$ is 5 (change in oxidation state from +7 to +2).\nThe n-factor for $FeSO_4$ is 1 (change in oxidation state from +2 to +3).

Step 3: Use the normality equation: $N_1V_1 = N_2V_2$ .\nFor $KMnO_4$ : $1 \mathrm{M} = 5 \mathrm{N}$ \nSo, $5 \mathrm{N} \times 10 \mathrm{mL} = 50 \mathrm{meq}$ \nFor $FeSO_4$ : $5 \mathrm{N} \times 10 \mathrm{mL} = 50 \mathrm{meq}$

Why other options are wrong:
- Option A: $10 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{~KMnO}_4$ solution oxidises $10 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{~FeSO}_4$ solution, not $5 \mathrm{~N}$ .
- Option C: $10 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{~KMnO}_4$ solution oxidises $10 \mathrm{~mL}$ of $5 \mathrm{N}$ or $5 \mathrm{M} \mathrm{~FeSO}_4$ solution, not $1 \mathrm{M}$ .
- Option D: $10 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{~KMnO}_4$ solution oxidises $10 \mathrm{~mL}$ of $0.1 \mathrm{M} \times 5 = 0.5 \mathrm{~N} \mathrm{~FeSO}_4$ solution, not $0.1 \mathrm{M}$ .

Key Concepts:
- Redox titrations
- Normality
- Molarity
- n-factor

Memory Aids:
- Remember the n-factor for $KMnO_4$ in acidic medium is 5.

Quick Tips:
- Recall the n-factors for common oxidizing and reducing agents.

60 Easy

60 of 80Question permalink

Analyzing +4 | -1

Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?

View answer and explanation

Correct option: D

3{d^5}4{s^2}

Option D (

3{d^5}4{s^2}

) exhibits the largest number of oxidation states because it has the maximum number of electrons in the d and s orbitals available for bonding.

Solution approach

Correct Answer + Core Reason: $3{d^5}4{s^2}$ exhibits the largest number of oxidation states because it has the most electrons available for bonding.

Step-by-Step Solution:

Step 1: Determine the number of electrons in the $d$ and $s$ orbitals for each configuration.
- $3{d^2}4{s^2}$ : 2 + 2 = 4 electrons
- $3{d^3}4{s^2}$ : 3 + 2 = 5 electrons
- $3{d^5}4{s^1}$ : 5 + 1 = 6 electrons
- $3{d^5}4{s^2}$ : 5 + 2 = 7 electrons

Step 2: The number of oxidation states an element can exhibit is related to the number of electrons it can lose or share.

Step 3: The configuration with the largest number of electrons in the $d$ and $s$ orbitals will generally exhibit the largest number of oxidation states.

Step 4: The $3{d^5}4{s^2}$ configuration has 7 electrons, which is the largest number among the given options. Therefore, it is expected to exhibit the largest number of oxidation states.

Why other options are wrong:
- **Option A ( $3{d^2}4{s^2}$ ):** This configuration has only 4 electrons available for bonding, which is less than $3{d^5}4{s^2}$ .
- **Option B ( $3{d^3}4{s^2}$ ):** This configuration has only 5 electrons available for bonding, which is less than $3{d^5}4{s^2}$ .
- **Option C ( $3{d^5}4{s^1}$ ):** This configuration has only 6 electrons available for bonding, which is less than $3{d^5}4{s^2}$ .

Key Concepts:
- Oxidation states
- Electronic configuration
- Number of electrons available for bonding

Memory Aids:
- The more electrons available for bonding, the more oxidation states an element can exhibit.

Quick Tips:
- Count the number of electrons in the $d$ and $s$ orbitals to determine the potential number of oxidation states.

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