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

1 Easy

Remembering +4 | -1

In the first transition series, the differentiating electron enters:

View answer and explanation

Correct option: C

3

d

-orbital

Option C (3

d

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

d

orbitals.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the electronic configuration of transition metals.

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

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

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

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

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

2 Easy

2 of 80Question permalink

Understanding +4 | -1

Philosopher’s wool on heating with BaO at 1100

^\circ

C produce

View answer and explanation

Correct option: C

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

Option C (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

Memory Aids:
- ZnO + BaO -> BaZnO2

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

3 Easy

3 of 80Question permalink

Remembering +4 | -1

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

is used as ……. Indicator for

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

titrations.

View answer and explanation

Correct option: B

External

Option B (External) is correct because

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

is used as an external indicator in

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

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

Solution approach

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

Step-by-Step Solution:

Step 1: Understand the concept of indicators in titrations.

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

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

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

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

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

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

4 Easy

4 of 80Question permalink

Analyzing +4 | -1

Given below are two statements:

Statement I : Aqueous solution of

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

is preferred as a primary standard in volumetric analysis over

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

aqueous solution.

Statement II :

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

has a higher solubility in water than

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

.

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

View answer and explanation

Correct option: C

Statement I is true but Statement II is false

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

5 Easy

5 of 80Question permalink

Analyzing +4 | -1

Which of the following statements is not correct about

{Cu}^{2+}

and

{Cu}^{+}:

View answer and explanation

Correct option: D

Both

{Cu}^{+}

and

{Cu}^{2+}

are stable with

{I}^{-}

Option D is correct because

{Cu}^{+}

is not stable with

{I}^{-}

due to the formation of

CuI

precipitate.

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

6 Easy

6 of 80Question permalink

Application +4 | -1

Maximum oxidation state is presented by:

View answer and explanation

Correct option: A

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

Key Concepts:
- Oxidation state
- Transition metal compounds

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

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

7 Easy

7 of 80Question permalink

Analyzing +4 | -1

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

View answer and explanation

Correct option: A

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

Option A (

MnO_4^{2-}

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

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

8 Easy

8 of 80Question permalink

Application +4 | -1

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

View answer and explanation

Correct option: A

\mathrm{Mn}^{+2}

Option A (

\mathrm{Mn}^{+2}

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

Solution approach

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

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each ion.

Step 2: Calculate the number of unpaired electrons for each ion:
- $\mathrm{Mn}^{+2}$ : [Ar] 3d⁵ (5 unpaired electrons)
- $\mathrm{Fe}^{+2}$ : [Ar] 3d⁶ (4 unpaired electrons)
- $\mathrm{Cr}^{+3}$ : [Ar] 3d³ (3 unpaired electrons)
- $\mathrm{Ni}^{+2}$ : [Ar] 3d⁸ (2 unpaired electrons)

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

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

Why other options are wrong:
- Option B: $\mathrm{Fe}^{+2}$ has 4 unpaired electrons, less than $\mathrm{Mn}^{+2}$ .
- Option C: $\mathrm{Cr}^{+3}$ has 3 unpaired electrons, less than $\mathrm{Mn}^{+2}$ .
- Option D: $\mathrm{Ni}^{+2}$ has 2 unpaired electrons, less than $\mathrm{Mn}^{+2}$ .

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

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

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

9 Easy

9 of 80Question permalink

Remembering +4 | -1

Mercury is purified by:

View answer and explanation

Correct option: B

Distillation in vacuum

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

Solution approach

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

Step-by-Step Solution:

Step 1: Consider the properties of mercury.

Step 2: Mercury has a relatively low boiling point.

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

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

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

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

Step 7: Solidifying would not purify mercury.

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

Key Concepts:
- Distillation
- Vacuum Distillation
- Purification Techniques

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

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

10 Easy

10 of 80Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: C

Oxidation state of iron is +3

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

FeSO_4.(NH_4)_2SO_4.6H_2O

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

11 Easy

11 of 80Question permalink

Understanding +4 | -1

The colour of an electrolyte solution depends on:

View answer and explanation

Correct option: C

The nature of both the ions

The nature of both the ions determines the color of an electrolyte solution because the electronic transitions within both cations and anions contribute to light absorption.

Solution approach

Correct Answer + Core Reason: The nature of both the ions. Both cation and anion can contribute to the color.

Step-by-Step Solution:

Step 1: Consider the factors that contribute to the color of a solution.

Step 2: Color arises from the absorption of specific wavelengths of light, which corresponds to electronic transitions within the ions.

Step 3: Both the cation and the anion can have electronic transitions that absorb light in the visible region, thus contributing to the color of the solution.

Step 4: Therefore, the color depends on the nature of both ions.

Why other options are wrong:
- Option A: The nature of the anion alone is insufficient because the cation can also contribute to the color.
- Option B: The nature of the cation alone is insufficient because the anion can also contribute to the color.
- Option D: The nature of the solvent usually does not directly determine the color of the solution. The solvent can affect the color by influencing the electronic transitions of the ions, but the primary source of color is the ions themselves.

Key Concepts:
- Color of solutions
- Electronic transitions
- Role of ions in solution color

Memory Aids:
- Remember that both cations and anions can contribute to the color of a solution.

Quick Tips:
- Consider both ions when determining the color of a solution.

12 Easy

12 of 80Question permalink

Remembering +4 | -1

Actinides

View answer and explanation

Correct option: A

Have variable valency

Option A (Have variable valency) is correct because actinides exhibit multiple oxidation states due to the small energy difference between the 5f, 6d, and 7s orbitals.

Solution approach

Correct Answer + Core Reason: Have variable valency. Actinides exhibit multiple oxidation states.

Step-by-Step Solution:

Step 1: Recall the properties of actinides.

Step 2: Actinides show variable valency due to the small energy difference between the 5f, 6d, and 7s orbitals, allowing them to lose different numbers of electrons.

Why other options are wrong:
- Option B (Include element 12): Element 12 is Magnesium, which is not an actinide.
- Option C (Are all synthetic elements): Not all actinides are synthetic. Thorium and Uranium are naturally occurring.
- Option D (Have only short lived isotopes): Some actinides have long-lived isotopes, such as Uranium-238.

Key Concepts:
- Properties of actinides
- Variable valency

Memory Aids:
- Actinides have variable oxidation states.

Quick Tips:
- Remember that actinides are generally radioactive and exhibit variable valency.

13 Easy

13 of 80Question permalink

Remembering +4 | -1

The meniscus of mercury in a glass tube is:

View answer and explanation

Correct option: A

Convex upwards

Option A (Convex upwards) is correct because mercury has a stronger cohesive force than adhesive force with glass, resulting in a convex meniscus.

Solution approach

Correct Answer + Core Reason: Convex upwards. Mercury's cohesive forces are stronger than its adhesive forces with glass.

Step-by-Step Solution:

Step 1: Understand the concepts of cohesive and adhesive forces.

Step 2: Cohesive forces are the attractive forces between molecules of the same substance.

Step 3: Adhesive forces are the attractive forces between molecules of different substances.

Step 4: When cohesive forces are stronger than adhesive forces, the liquid forms a convex meniscus (e.g., mercury in glass).

Step 5: When adhesive forces are stronger than cohesive forces, the liquid forms a concave meniscus (e.g., water in glass).

Step 6: Since mercury has stronger cohesive forces than adhesive forces with glass, it forms a convex meniscus.

Why other options are wrong:
- Option B: Concave meniscus forms when adhesive forces are stronger (e.g., water in glass).
- Option C: Plane meniscus forms when cohesive and adhesive forces are approximately equal.
- Option D: Convex inwards is not a standard term; the correct term is convex upwards.

Key Concepts:
- Cohesive Forces
- Adhesive Forces
- Meniscus Formation

Memory Aids:
- Mercury is \"bulging\" (convex) because it sticks more to itself than the glass.

Quick Tips:
- Remember that mercury forms a convex meniscus, while water forms a concave meniscus in glass.

14 Easy

14 of 80Question permalink

Understanding +4 | -1

Which of the following compounds volatises on heating?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because mercuric chloride is known to sublime upon heating, meaning it transitions directly from solid to gas phase.

Solution approach

Correct Answer + Core Reason: $\mathrm{HgC}{\mathrm{l}_2}$ . Mercuric chloride sublimes on heating.

Step-by-Step Solution:

Step 1: Recall the properties of the given compounds.

Step 2: $\mathrm{FeC}{\mathrm{l}_3}$ sublimes but at higher temperatures and also tends to decompose.
$\mathrm{HgC}{\mathrm{l}_2}$ readily sublimes on heating.
$\mathrm{CaC}{\mathrm{l}_2}$ and $\mathrm{MgC}{\mathrm{l}_2}$ are ionic compounds with high melting and boiling points and do not readily volatilize.

Step 3: Identify the compound that volatizes most readily.

Why other options are wrong:
- Option A: $\mathrm{FeC}{\mathrm{l}_3}$ : Ferric chloride can sublime, but it's not as volatile as mercuric chloride and may decompose upon heating.
- Option C: $\mathrm{CaC}{\mathrm{l}_2}$ : Calcium chloride is an ionic compound with a high melting point and does not readily volatilize.
- Option D: $\mathrm{MgC}{\mathrm{l}_2}$ : Magnesium chloride is an ionic compound with a high melting point and does not readily volatilize.

Key Concepts:
- Volatility
- Sublimation
- Properties of ionic and covalent compounds

Memory Aids:
- Mercuric chloride is volatile.

Quick Tips:
- Remember the volatility of common compounds.

15 Easy

15 of 80Question permalink

Analyzing +4 | -1

Which shows a jump in second ionization potential?

View answer and explanation

Correct option: D

Cu

Option D (Cu) shows a jump in the second ionization potential because removing the second electron disrupts a stable, fully-filled d-orbital (

d^{10}

).

Solution approach

Correct Answer + Core Reason: Cu shows a jump in the second ionization potential due to the removal of an electron from a stable $d^{10}$ configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the elements and their ions.
- Co: $[\mathrm{Ar}]3{d^7}4{s^2}$ , $\mathrm{C}{\mathrm{o}^ + }$ : $[\mathrm{Ar}]3{d^8}$ , $\mathrm{C}{\mathrm{o}^{2 + }}$ : $[\mathrm{Ar}]3{d^7}$
- Ni: $[\mathrm{Ar}]3{d^8}4{s^2}$ , $\mathrm{N}{\mathrm{i}^ + }$ : $[\mathrm{Ar}]3{d^9}$ , $\mathrm{N}{\mathrm{i}^{2 + }}$ : $[\mathrm{Ar}]3{d^8}$
- Zn: $[\mathrm{Ar}]3{d^{10}}4{s^2}$ , $\mathrm{Z}{\mathrm{n}^ + }$ : $[\mathrm{Ar}]3{d^{10}}4{s^1}$ , $\mathrm{Z}{\mathrm{n}^{2 + }}$ : $[\mathrm{Ar}]3{d^{10}}$
- Cu: $[\mathrm{Ar}]3{d^{10}}4{s^1}$ , $\mathrm{C}{\mathrm{u}^ + }$ : $[\mathrm{Ar}]3{d^{10}}$ , $\mathrm{C}{\mathrm{u}^{2 + }}$ : $[\mathrm{Ar}]3{d^9}$

Step 2: Analyze the electronic configurations after the first and second ionization.

Step 3: Identify the element where the second ionization involves removing an electron from a stable, fully-filled or half-filled configuration.

Step 4: For Cu, the first ionization removes the $4s^1$ electron, resulting in a stable $3d^{10}$ configuration. Removing the second electron from the stable $3d^{10}$ configuration requires a significantly higher amount of energy, leading to a jump in the second ionization potential.

Why other options are wrong:
- Option A (Co): The electronic configurations of Co, $\mathrm{C}{\mathrm{o}^ + }$ , and $\mathrm{C}{\mathrm{o}^{2 + }}$ do not involve removing an electron from a stable configuration.
- Option B (Ni): The electronic configurations of Ni, $\mathrm{N}{\mathrm{i}^ + }$ , and $\mathrm{N}{\mathrm{i}^{2 + }}$ do not involve removing an electron from a stable configuration.
- Option C (Zn): The electronic configurations of Zn, $\mathrm{Z}{\mathrm{n}^ + }$ , and $\mathrm{Z}{\mathrm{n}^{2 + }}$ do not involve removing an electron from a stable configuration.

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

Memory Aids:
- Remember that removing an electron from a stable configuration (fully filled or half-filled) requires a large amount of energy.

Quick Tips:
- Write out the electronic configurations to determine the stability of the resulting ions after each ionization.

16 Easy

16 of 80Question permalink

Analyzing +4 | -1

A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

View answer and explanation

Correct option: B

\mathrm{Hg}{\mathrm{I}_2}

Option B (

\mathrm{Hg}{\mathrm{I}_2}

) is correct because it matches the description of a red solid that dissolves in KI solution, and decomposes upon heating to release violet iodine fumes and metallic mercury.

Solution approach

Correct Answer + Core Reason: $\mathrm{Hg}{\mathrm{I}_2}$ . The red solid is mercury(II) iodide, which dissolves in KI solution to form a complex and decomposes upon heating to give mercury and iodine.

Step-by-Step Solution:

Step 1: Analyze the given information.
- Red solid insoluble in water.
- Soluble in KI solution.
- Heating gives violet fumes and metal droplets.

Step 2: Consider the options.
- ${\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ : Ammonium dichromate is orange, not red, and decomposes to give nitrogen gas, water, and chromium(III) oxide.
- $\mathrm{Hg}{\mathrm{I}_2}$ : Mercury(II) iodide is a red solid that is insoluble in water but dissolves in KI solution due to the formation of a complex ion, ${\left[ {\mathrm{Hg}{\mathrm{I}_4}} \right]^{2 - }}$ . Heating $\mathrm{Hg}{\mathrm{I}_2}$ decomposes it to mercury metal and iodine vapor (violet fumes).
- $\mathrm{HgO}$ : Mercury(II) oxide is a red or yellow solid that decomposes upon heating to give mercury and oxygen gas.
- $\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}$ : Lead(II,IV) oxide is a red solid that decomposes upon heating, but it does not produce violet fumes or metallic droplets directly.

Step 3: Based on the analysis, $\mathrm{Hg}{\mathrm{I}_2}$ fits the description best.

Why other options are wrong:
- Option A: ${\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ is incorrect because it is orange, not red, and its decomposition products do not match the description.
- Option C: $\mathrm{HgO}$ is incorrect because it decomposes to give oxygen gas, not violet fumes of iodine.
- Option D: $\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}$ is incorrect because its decomposition products do not match the description.

Key Concepts:
- Solubility of ionic compounds
- Complex ion formation
- Thermal decomposition of compounds

Memory Aids:
- $\mathrm{Hg}{\mathrm{I}_2}$ is red and gives violet fumes of iodine upon heating.

Quick Tips:
- Consider the color of the solid and the products of its decomposition when identifying an unknown compound.

17 Easy

17 of 80Question permalink

Application +4 | -1

The magnetic moment of a transition metal of

3d

series is

6.92 \, \mathrm{BM}

. Its electronic configuration would be:

View answer and explanation

Correct option: D

3 d^5 4 s^1

The electronic configuration is

3 d^5 4 s^1

because a magnetic moment of 6.92 BM corresponds to approximately 6 unpaired electrons, which is achieved with this configuration.

Solution approach

Correct Answer + Core Reason: $3 d^5 4 s^1$ . This configuration gives 6 unpaired electrons, resulting in a magnetic moment close to 6.92 BM.

Step-by-Step Solution:

Step 1: Use the spin-only formula to estimate the number of unpaired electrons (n).
$\mu = \sqrt{n(n+2)}$ BM\nGiven $\mu = 6.92$ BM, we have:
$6.92 = \sqrt{n(n+2)}$ \nSquaring both sides:
$47.8864 = n(n+2)$ \nWe can approximate this by testing integer values of n. If n = 6:
$6(6+2) = 6(8) = 48$ \nThis is close to 47.8864, so we can assume n ≈ 6.

Step 2: Determine the electronic configuration that gives 6 unpaired electrons.
- Option A: $3 d^{16}$ . This is not a valid configuration.
- Option B: $3 d^5 4 s^0$ . This gives 5 unpaired electrons.
- Option C: $3 d^4 4 s^2$ . This gives 4 + 0 = 4 unpaired electrons.
- Option D: $3 d^5 4 s^1$ . This gives 5 + 1 = 6 unpaired electrons.

Step 3: Therefore, the electronic configuration is $3 d^5 4 s^1$ .

Why other options are wrong:
- Option A: $3 d^{16}$ . This is not a valid electronic configuration as it exceeds the maximum number of electrons that can be accommodated in the 3d orbitals.
- Option B: $3 d^5 4 s^0$ . This configuration has 5 unpaired electrons, leading to a magnetic moment of approximately 5.92 BM, which is not 6.92 BM.
- Option C: $3 d^4 4 s^2$ . This configuration has 4 unpaired electrons, leading to a magnetic moment of approximately 4.90 BM, which is not 6.92 BM.

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

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

Quick Tips:
- Use the spin-only formula to estimate the number of unpaired electrons. Then, determine the electronic configuration that corresponds to that number of unpaired electrons.

18 Easy

18 of 80Question permalink

Understanding +4 | -1

The electronic configurations of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:

View answer and explanation

Correct option: D

\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{6}{\mathrm{s}^\mathrm{2}},\,\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}\,\mathrm{and}\,\mathrm{[Xe]4}{\mathrm{f}^9}\mathrm{6}{\mathrm{s}^\mathrm{2}}

Option D (

\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{6}{\mathrm{s}^\mathrm{2}},\,\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}\,\mathrm{and}\,\mathrm{[Xe]4}{\mathrm{f}^9}\mathrm{6}{\mathrm{s}^\mathrm{2}}

) is correct because it represents the correct electronic configurations of Eu, Gd, and Tb, considering Hund's rule and stability of half-filled orbitals.

Solution approach

Correct Answer + Core Reason: $\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{6}{\mathrm{s}^\mathrm{2}},\,\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}\,\mathrm{and}\,\mathrm{[Xe]4}{\mathrm{f}^9}\mathrm{6}{\mathrm{s}^\mathrm{2}}$ . These are the correct electronic configurations for Eu, Gd, and Tb.

Step-by-Step Solution:

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

Step 2: Determine the number of electrons in the 4f orbital for each element.
- Eu (Z=63): $[Xe]4f^76s^2$ (half-filled f orbital)
- Gd (Z=64): $[Xe]4f^75d^16s^2$ (one electron goes to 5d for extra stability)
- Tb (Z=65): $[Xe]4f^96s^2$

Why other options are wrong:
- Option A: Incorrect electronic configuration for Gd.
- Option B: Incorrect electronic configuration for Eu and Gd.
- Option C: Incorrect electronic configuration for Eu and Gd.

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.

19 Easy

19 of 80Question permalink

Analyzing +4 | -1

Identify the statement which is not correct regarding copper sulphate

View answer and explanation

Correct option: C

It reacts with

\mathrm{KCl\;}

to give

\mathrm{C}{\mathrm{u}_2}\mathrm{C}{\mathrm{l}_2}

Option C is correct because copper sulphate does not react with KCl to form Cu2Cl2. The reaction is not thermodynamically favorable.

Solution approach

Correct Answer + Core Reason: It reacts with $\mathrm{KCl\;}$ to give $\mathrm{C}{\mathrm{u}_2}\mathrm{C}{\mathrm{l}_2}$ . Copper sulphate does not react with KCl to form Cu2Cl2.

Step-by-Step Solution:

Step 1: Analyze the given options and identify the incorrect statement regarding the reactions of copper sulphate.

Step 2: Consider the reaction of copper sulphate with NaOH and glucose. This reaction produces Cu2O, which is a red precipitate.

Step 3: Consider the reaction of copper sulphate on strong heating in air. This reaction produces CuO.

Step 4: Consider the reaction of copper sulphate with KI. This reaction produces iodine (I2).

Step 5: Consider the reaction of copper sulphate with KCl. Copper sulphate does not react with KCl to form Cu2Cl2. A double displacement reaction would require the formation of an insoluble chloride, which is not the case here.

Why other options are wrong:
- Option A: Copper sulphate reacts with NaOH and glucose to give Cu2O. This statement is correct.
- Option B: Copper sulphate gives CuO on strong heating in air. This statement is correct.
- Option D: Copper sulphate reacts with KI to give iodine. This statement is correct.

Key Concepts:
- Reactions of copper sulphate
- Properties of transition metal compounds

Memory Aids:
- Remember the common reactions of copper sulphate with different reagents.

Quick Tips:
- Recall the characteristic reactions of copper sulphate to identify the incorrect statement.

20 Easy

20 of 80Question permalink

Understanding +4 | -1

Which of the following statement is wrong for the d-block elements

View answer and explanation

Correct option: D

They are placed from third period to sixth period in modern periodic table

Option D (They are placed from third period to sixth period in modern periodic table) is wrong because d-block elements are placed from fourth period to seventh period in the modern periodic table.

Solution approach

Correct Answer + Core Reason: They are placed from third period to sixth period in modern periodic table. d-block elements are placed from fourth period to seventh period.

Step-by-Step Solution:

Step 1: Recall the general electronic configuration of d-block elements.

Step 2: The general electronic configuration of d-block elements is $(n-1)d^{1-10}ns^{0-2}$ .

Step 3: The last electron enters the d subshell in d-block elements.

Step 4: d-block elements generally exhibit variable valency due to the involvement of d-electrons in bonding.

Step 5: d-block elements are placed from the 4th period to the 7th period in the modern periodic table. The 3d series starts in the 4th period (Sc to Zn), the 4d series in the 5th period (Y to Cd), the 5d series in the 6th period (La, Hf to Hg), and the 6d series in the 7th period (Ac, Rf to Cn).

Why other options are wrong:
- Option A: The general electronic configuration is correct.
- Option B: The last electron enters the d subshell is correct.
- Option C: Variable valency is a characteristic of d-block elements.

Key Concepts:
- d-Block Elements
- Electronic Configuration
- Variable Valency
- Periodic Table

Memory Aids:
- d-block starts from the 4th period.

Quick Tips:
- Remember the position of d-block elements in the periodic table.

21 Easy

21 of 80Question permalink

Remembering +4 | -1

Identity the ore not containing iron.

View answer and explanation

Correct option: C

Carnallite

Option C (Carnallite) is correct because Carnallite is a hydrated double chloride of potassium and magnesium, and does not contain iron.

Solution approach

Correct Answer + Core Reason: Carnallite. Carnallite does not contain iron.

Step-by-Step Solution:

Step 1: Identify the chemical formulas of the given ores.

Step 2: Limonite: $Fe_2O_3.nH_2O$ (contains iron)\nSiderite: $FeCO_3$ (contains iron)\nCarnallite: $KCl.MgCl_2.6H_2O$ (does not contain iron)\nChalcopyrites: $CuFeS_2$ (contains iron)

Step 3: Based on the chemical formulas, determine which ore does not contain iron.

Why other options are wrong:
- Option A: Limonite ( $Fe_2O_3.nH_2O$ ) contains iron.
- Option B: Siderite ( $FeCO_3$ ) contains iron.
- Option C: Carnallite ( $KCl.MgCl_2.6H_2O$ ) does not contain iron.
- Option D: Chalcopyrites ( $CuFeS_2$ ) contains iron.

Key Concepts:
- Ores of metals
- Chemical formulas
- Composition of minerals

Memory Aids:
- Remember the common ores and their compositions.

Quick Tips:
- Knowing the chemical formulas of common ores is essential for identifying their constituents.

22 Easy

22 of 80Question permalink

Remembering +4 | -1

The colour of zinc sulphide is:

View answer and explanation

Correct option: B

White

Option B (White) is correct because pure zinc sulfide (ZnS) is white in color.

Solution approach

Correct Answer + Core Reason: White. Pure zinc sulphide is white in color.

Step-by-Step Solution:

Step 1: Recall the color of zinc sulfide (ZnS).

Step 2: Zinc sulfide is a white solid in its pure form.

Step 3: Impurities can affect the color of ZnS, but the question asks for the color of zinc sulfide itself.

Why other options are wrong:
- Option A: Yellow is not the color of pure zinc sulfide.
- Option C: Brown is not the color of pure zinc sulfide.
- Option D: Black is not the color of pure zinc sulfide.

Key Concepts:
- Color of compounds
- Zinc sulfide

Memory Aids:
- Remember that zinc sulfide is white.

Quick Tips:
- Know the colors of common compounds.

23 Easy

23 of 80Question permalink

Understanding +4 | -1

Ferrous sulphate on heating gives:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because ferrous sulphate (

FeSO_4

) decomposes upon heating to produce

Fe_2O_3

,

SO_2

, and

SO_3

.

Solution approach

Correct Answer + Core Reason: All of these. Ferrous sulphate decomposes on heating to give ferric oxide, sulphur dioxide and sulphur trioxide.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the thermal decomposition of ferrous sulphate.
$2FeSO_4(s) \rightarrow Fe_2O_3(s) + SO_2(g) + SO_3(g)$

Step 2: Identify the products formed.

Step 3: The products are $Fe_2O_3$ , $SO_2$ , and $SO_3$ .

Why other options are wrong:
- Option A: $SO_3$ is only one of the products, not the only product.
- Option B: $SO_2$ is only one of the products, not the only product.
- Option C: $Fe_2O_3$ is only one of the products, not the only product.

Key Concepts:
- Thermal decomposition
- Chemical reactions of transition metal compounds

Memory Aids:
- Remember the decomposition reaction of ferrous sulphate.

Quick Tips:
- Recall common decomposition reactions to quickly answer such questions.

24 Easy

24 of 80Question permalink

Understanding +4 | -1

The aqueous solution containing which one of the following ions will be colourless?

View answer and explanation

Correct option: B

Z{n^{2 + }}

Option B (

Z{n^{2 + }}

) is correct because

Z{n^{2 + }}

has a

d^{10}

configuration, meaning all d-orbitals are fully filled, and thus no d-d transitions are possible, resulting in a colorless solution.

Solution approach

Correct Answer + Core Reason: $Z{n^{2 + }}$ . The aqueous solution containing $Z{n^{2 + }}$ will be colourless.

Step-by-Step Solution:

Step 1: Understand the concept of color in transition metal complexes. Color arises from d-d transitions, where electrons in d-orbitals absorb light and move to higher energy d-orbitals.

Step 2: Consider the electronic configurations of the given ions:
$F{e^{2 + }}$ : $[Ar]3d^6$ (colored)
$Z{n^{2 + }}$ : $[Ar]3d^{10}$ (colorless)
$N{i^{2 + }}$ : $[Ar]3d^8$ (colored)
${V^{3 + }}$ : $[Ar]3d^2$ (colored)

Step 3: Determine which ion has a completely filled d-orbital ( $d^{10}$ ). $Z{n^{2 + }}$ has a $d^{10}$ configuration, meaning no d-d transitions are possible.

Why other options are wrong:
- Option A: $F{e^{2 + }}$ has a $d^6$ configuration, allowing d-d transitions and resulting in a colored solution.
- Option B: $Z{n^{2 + }}$ has a $d^{10}$ configuration, preventing d-d transitions and resulting in a colorless solution.
- Option C: $N{i^{2 + }}$ has a $d^8$ configuration, allowing d-d transitions and resulting in a colored solution.
- Option D: ${V^{3 + }}$ has a $d^2$ configuration, allowing d-d transitions and resulting in a colored solution.

Key Concepts:
- d-d transitions
- Electronic configurations
- Color of transition metal complexes

Memory Aids:
- Remember that $d^{0}$ and $d^{10}$ configurations typically result in colorless complexes.

Quick Tips:
- The presence of unpaired d-electrons is necessary for d-d transitions and color in transition metal complexes.

25 Easy

25 of 80Question permalink

Application +4 | -1

The spin only magnetic moment [in units of Bohr magneton] of

{Ni}^{2+}

in aqueous solution would be (At. No. 28) :

View answer and explanation

Correct option: A

2.84

Option A (2.84) is correct because

{Ni}^{2+}

has two unpaired electrons, resulting in a spin-only magnetic moment of 2.84 BM.

Solution approach

Correct Answer + Core Reason: 2.84. ${Ni}^{2+}$ has two unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Ni.

Step 2: Ni (Z=28): $[\mathrm{Ar}] 3d^8 4s^2$

Step 3: Determine the electronic configuration of ${Ni}^{2+}$ .

Step 4: ${Ni}^{2+}$ : $[\mathrm{Ar}] 3d^8$

Step 5: Determine the number of unpaired electrons in ${Ni}^{2+}$ . According to Hund's rule, there are 2 unpaired electrons.

Step 6: Calculate the spin-only magnetic moment using the formula: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

Step 7: $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.828 \approx 2.84$ BM

Why other options are wrong:
- Option B: 4.9 BM corresponds to 5 unpaired electrons, which is incorrect for ${Ni}^{2+}$ .
- Option C: 0 BM corresponds to no unpaired electrons, which is incorrect for ${Ni}^{2+}$ .
- Option D: 1.73 BM corresponds to 1 unpaired electron, which is incorrect for ${Ni}^{2+}$ .

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

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

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

26 Easy

26 of 80Question permalink

Remembering +4 | -1

The compound used in preservation of wood is:

View answer and explanation

Correct option: C

\mathrm{ZnC}{\mathrm{l}_2}

Option C (

\mathrm{ZnC}{\mathrm{l}_2}

) is correct because zinc chloride is a common wood preservative that protects wood from decay and insect damage.

Solution approach

Correct Answer + Core Reason: $\mathrm{ZnC}{\mathrm{l}_2}$ . Zinc chloride is used in the preservation of wood.

Step-by-Step Solution:

Step 1: Consider the properties required for wood preservatives.

Step 2: Wood preservatives need to be toxic to fungi and insects, and they should be able to penetrate the wood.

Step 3: Zinc chloride is a common wood preservative that protects wood from decay and insect damage.

Why other options are wrong:
- Option A: NaCl is common salt and is not very effective as a wood preservative.
- Option B: $\mathrm{HgC}{\mathrm{l}_2}$ is mercuric chloride, which is highly toxic and its use is restricted due to environmental concerns.
- Option D: $\mathrm{CaC}{\mathrm{l}_2}$ is calcium chloride and is not commonly used as a wood preservative.

Key Concepts:
- Wood preservation
- Properties of chemical compounds

Memory Aids:
- Remember the common uses of different chemical compounds.

Quick Tips:
- Know the common applications of different chemical compounds.

27 Easy

27 of 80Question permalink

Remembering +4 | -1

On strongly heating

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

we get:

View answer and explanation

Correct option: C

Ag

Option C (Ag) is correct because upon strong heating, silver nitrate decomposes to form silver metal, nitrogen dioxide, and oxygen.

Solution approach

Correct Answer + Core Reason: Ag. Silver nitrate decomposes to form silver metal upon strong heating.

Step-by-Step Solution:

Step 1: Consider the decomposition reaction of $AgNO_3$ upon strong heating.

Step 2: The reaction is: $2AgNO_3(s) \to 2Ag(s) + 2NO_2(g) + O_2(g)$ .

Step 3: The products are silver metal (Ag), nitrogen dioxide ( $NO_2$ ), and oxygen ( $O_2$ ).

Why other options are wrong:
- Option A: $AgNO_2$ is not formed in this reaction.
- Option B: Silver nitride is not formed in this reaction.
- Option D: $Ag_2O$ is not formed in this reaction.

Key Concepts:
- Decomposition reactions
- Thermal decomposition
- Reactions of nitrates

Memory Aids:
- Remember the products of thermal decomposition of silver nitrate.

Quick Tips:
- Know the general trends for the thermal decomposition of metal nitrates.

28 Easy

28 of 80Question permalink

Analyzing +4 | -1

Statement I:

\mathrm{Fe}^{+2}

is more stable than

\mathrm{Fe}^{+3}

ion

Statement II:

\mathrm{Fe}^{+3}

ion contain half-filled

3 \mathrm{d}

subshell

View answer and explanation

Correct option: D

Statement I if false but statement II is correct

Option D is correct because Statement I is false, and Statement II is correct.

{Fe}^{3+}

is more stable than

{Fe}^{2+}

due to the half-filled d-orbital configuration.

Solution approach

Correct Answer + Core Reason: Statement I if false but statement II is correct. ${Fe}^{3+}$ is more stable than ${Fe}^{2+}$ due to the half-filled $3d$ subshell.

Step-by-Step Solution:

Step 1: Analyze the stability of $Fe^{2+}$ and $Fe^{3+}$ ions.

Step 2: $Fe^{2+}$ has the electronic configuration $[Ar]3d^6$ .

Step 3: $Fe^{3+}$ has the electronic configuration $[Ar]3d^5$ , which is a half-filled, more stable configuration.

Step 4: Therefore, $Fe^{3+}$ is more stable than $Fe^{2+}$ .

Step 5: Statement I is false, and Statement II is correct.

Why other options are wrong:
- Option A: Statement I is false, so this option is incorrect.
- Option B: Statement I is false, so this option is incorrect.
- Option C: Statement I is false, so this option is incorrect.

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

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

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

29 Easy

29 of 80Question permalink

Understanding +4 | -1

Which of the following metal ions is not coloured?

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because zinc(II) ion has a completely filled d-orbital configuration (

d^{10}

), preventing d-d transitions and thus resulting in no color.

Solution approach

Correct Answer + Core Reason: $\mathrm{Z}{\mathrm{n}^{2 + }}$ . Zinc(II) ion has a completely filled d-orbital and is colorless.

Step-by-Step Solution:

Step 1: Recall the electronic configurations of the given metal ions.

Step 2: $\mathrm{T}{\mathrm{i}^{3 + }}$ has a $d^1$ configuration and is colored.
$\mathrm{F}{\mathrm{e}^{3 + }}$ has a $d^5$ configuration and is colored.
${\mathrm{V}^{2 + }}$ has a $d^3$ configuration and is colored.
$\mathrm{Z}{\mathrm{n}^{2 + }}$ has a $d^{10}$ configuration and is colorless.

Step 3: Color arises from d-d transitions, which require partially filled d-orbitals. A completely filled or empty d-orbital will not exhibit d-d transitions.

Why other options are wrong:
- Option A: $\mathrm{T}{\mathrm{i}^{3 + }}$ : Titanium(III) has a $d^1$ configuration and exhibits d-d transitions, making it colored.
- Option B: $\mathrm{F}{\mathrm{e}^{3 + }}$ : Iron(III) has a $d^5$ configuration and exhibits d-d transitions, making it colored.
- Option C: ${\mathrm{V}^{2 + }}$ : Vanadium(II) has a $d^3$ configuration and exhibits d-d transitions, making it colored.

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

Memory Aids:
- Filled or empty d-orbitals mean no color.

Quick Tips:
- Remember that color arises from d-d transitions in partially filled d-orbitals.

30 Easy

30 of 80Question permalink

Application +4 | -1

Magnetic moment in B.M. of

\mathrm{Ni}^{2+}

ion is

View answer and explanation

Correct option: B

\sqrt{8}

The magnetic moment of

\mathrm{Ni}^{2+}

ion is

\sqrt{8}

B.M. because

\mathrm{Ni}^{2+}

has two unpaired electrons, and the spin-only formula is used to calculate the magnetic moment.

Solution approach

Correct Answer + Core Reason: $\sqrt{8}$ . $\mathrm{Ni}^{2+}$ has two unpaired electrons, and the spin-only formula is used.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Ni.\nNi (atomic number 28): $[Ar] 3d^8 4s^2$

Step 2: Determine the electronic configuration of $\mathrm{Ni}^{2+}$ .
$\mathrm{Ni}^{2+}$ : $[Ar] 3d^8$

Step 3: Determine the number of unpaired electrons (n) in the $3d^8$ configuration.\nAccording to Hund's rule, the eight electrons will fill the five d-orbitals as follows: two orbitals will be filled with two electrons each, and three orbitals will have one electron each. This leaves two unpaired electrons (n = 2).

Step 4: Calculate the magnetic moment using the spin-only formula:
$\mu = \sqrt{n(n+2)}$ BM
$\mu = \sqrt{2(2+2)}$ BM
$\mu = \sqrt{2(4)}$ BM
$\mu = \sqrt{8}$ BM

Why other options are wrong:
- Option A: $\sqrt{3}$ . This would correspond to approximately 1.73 BM, which is the magnetic moment for one unpaired electron.
- Option C: $\sqrt{15}$ . This would correspond to approximately 3.87 BM, which is the magnetic moment for three unpaired electrons.
- Option D: $\sqrt{24}$ . This would correspond to approximately 4.90 BM, which is the magnetic moment for four unpaired electrons.

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

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

Quick Tips:
- Use Hund's rule to determine the number of unpaired electrons in a given electronic configuration.

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