${\rm{C}}{{\rm{N}}^ - }$ ion and ${{\rm{N}}_2}$ are isoelectronic but in contrast to ${\rm{C}}{{\rm{N}}^ - },{\rm{\;}}{{\rm{N}}_2}$ is chemically inert because of:

Unsymmetrical electron distribution

Presence of more number of electrons in bonding orbitals

P-BLOCK ELEMENTS MCQs for JEE CHEMISTRY — Mixed difficulty

1 Easy

1 of 680Question permalink

Remembering +4 | -1

The gas used for inflating the tyres of aeroplanes is:

View answer and explanation

Correct option: B

\mathrm{He}

Option B (He) is correct because Helium is the lightest non-inflammable gas after Hydrogen, making it suitable for inflating aircraft tires.

Solution approach

Correct Answer + Core Reason: He. Helium is the lightest non-inflammable gas after H2.

Step-by-Step Solution:

Step 1: Identify the requirements for the gas used in airplane tires: It should be light to reduce weight and non-flammable for safety.

Step 2: Consider the given options: H2, He, N2, Ar.

Step 3: Evaluate each option:
- H2 is the lightest gas but highly flammable.
- He is the second lightest and non-flammable.
- N2 is non-flammable but heavier than He.
- Ar is also non-flammable but heavier than He.

Step 4: Choose the best option based on the requirements: He is the most suitable gas.

Why other options are wrong:
- Option A (H2): Hydrogen is extremely flammable, posing a significant safety risk.
- Option C (N2): Nitrogen is heavier than Helium, increasing the weight of the aircraft.
- Option D (Ar): Argon is heavier than Helium, increasing the weight of the aircraft.

Key Concepts:
- Properties of Gases (flammability, density)
- Safety considerations in aviation

Memory Aids:
- Remember that Helium is inert and lighter than air (except for Hydrogen).

Quick Tips:
- Always consider safety when selecting materials for aviation applications.

2 Medium

2 of 680Question permalink

Remembering +4 | -1

Which of the following is correct stability order of oxides of

17^{\text {th }}

group elements is:

View answer and explanation

Correct option: C

\mathrm{I}_2>\mathrm{Cl}_2>\mathrm{Br}_2

Option D (

\mathrm{Cl}_2>\mathrm{I}_2>\mathrm{Br}_2

) is incorrect, the correct order is

\mathrm{Cl}_2O > \mathrm{Br}_2O > \mathrm{I}_2O

.

Solution approach

Correct Answer + Core Reason: $\mathrm{I}_2>\mathrm{Cl}_2>\mathrm{Br}_2$ . The stability of oxides decreases down the group due to decreasing bond strength.

Step-by-Step Solution:

Step 1: Understand the trend in bond strength for halogen oxides. The stability of oxides generally decreases as you move down the halogen group (Group 17).

Step 2: Consider the factors affecting bond strength: electronegativity and atomic size. Smaller, more electronegative halogens form stronger bonds with oxygen.

Step 3: Apply the trend to the given halogens: Chlorine (Cl) is smaller and more electronegative than Bromine (Br) and Iodine (I). Bromine is smaller and more electronegative than Iodine.

Step 4: Determine the stability order based on bond strength: $\mathrm{Cl}_2O > \mathrm{Br}_2O > \mathrm{I}_2O$

Step 5: The question is about the stability order of the elements themselves, not their oxides. The correct order is $\mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$ due to decreasing bond dissociation energy down the group.

Why other options are wrong:
- Option A: $\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : This is the correct order for the stability of the halogens themselves, but not their oxides.
- Option B: $\mathrm{I}_2>\mathrm{Br}_2>\mathrm{Cl}_2$ : Incorrect order for both the halogens and their oxides.
- Option C: $\mathrm{I}_2>\mathrm{Cl}_2>\mathrm{Br}_2$ : Incorrect order for both the halogens and their oxides.

Key Concepts:
- Halogens
- Oxides of halogens
- Electronegativity
- Bond strength
- Stability of compounds

Memory Aids:
- Remember that smaller, more electronegative atoms form stronger bonds.

Quick Tips:
- Consider both electronegativity and atomic size when comparing bond strengths.

3 Medium

3 of 680Question permalink

Analyzing +4 | -1

Assertion: Ozone is a powerful oxidising agent in comparison to

{O}_2

.

Reason: Ozone is diamagnetic but

{O}_2

is paramagnetic.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

Option B (If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion) is correct because ozone is a stronger oxidizing agent than oxygen, and ozone is diamagnetic while oxygen is paramagnetic, but the paramagnetism of oxygen is not the reason for ozone's stronger oxidizing power.

Solution approach

Correct Answer + Core Reason: Both the assertion and the reason are correct, but the reason is not the correct explanation for the assertion. Ozone's oxidizing power is due to its instability and ease of decomposition, not its diamagnetic nature.

Step-by-Step Solution:

Step 1: Ozone ( $O_3$ ) is a powerful oxidizing agent because it readily decomposes to form oxygen ( $O_2$ ) and nascent oxygen ( $[O]$ ), which is highly reactive: $O_3 \rightarrow O_2 + [O]$

Step 2: Oxygen ( $O_2$ ) is paramagnetic due to the presence of two unpaired electrons in its molecular orbital configuration.

Step 3: Ozone ( $O_3$ ) is diamagnetic because all its electrons are paired.

Step 4: The oxidizing power of ozone is related to its ability to release nascent oxygen, not its magnetic properties.

Why other options are wrong:
- Option A: The reason is not the correct explanation for the assertion.
- Option C: The reason is correct, so this option is incorrect.
- Option D: Both the assertion and the reason are correct, so this option is incorrect.

Key Concepts:
- Oxidizing agent
- Paramagnetism
- Diamagnetism
- Ozone decomposition

Memory Aids:
- Ozone readily releases nascent oxygen, making it a strong oxidizing agent.
- Paramagnetic substances have unpaired electrons.
- Diamagnetic substances have all electrons paired.

Quick Tips:
- Understand the difference between oxidizing power and magnetic properties.
- Remember that ozone's oxidizing power is due to its instability.

4 Easy

4 of 680Question permalink

Remembering +4 | -1

Which of the following is kept in water?

View answer and explanation

Correct option: A

White phosphorus

Option A (White phosphorus) is correct because white phosphorus is highly reactive and spontaneously ignites in air, so it's stored under water.

Solution approach

Correct Answer + Core Reason: White phosphorus. White phosphorus is highly reactive and stored under water to prevent ignition.

Step-by-Step Solution:

Step 1: Recall the properties of different elements and their storage conditions.

Step 2: White phosphorus is highly reactive and ignites spontaneously in air.

Step 3: To prevent ignition, white phosphorus is stored under water.

Why other options are wrong:
- Option B: Sodium is stored under kerosene to prevent reaction with moisture in the air.
- Option C: Potassium is stored under kerosene to prevent reaction with moisture in the air.
- Option D: Calcium reacts slowly with water but does not require storage under water.

Key Concepts:
- Reactivity of elements
- Storage of reactive elements

Memory Aids:
- Remember that white phosphorus is "white hot" and needs to be kept away from air.

Quick Tips:
- Memorize the storage conditions for common reactive elements.

5 Easy

5 of 680Question permalink

Understanding +4 | -1

\mathrm{C}{\mathrm{N}^ - }

ion and

{\mathrm{N}_2}

are isoelectronic but in contrast to

\mathrm{C}{\mathrm{N}^ - },\mathrm{\;}{\mathrm{N}_2}

is chemically inert because of:

View answer and explanation

Correct option: B

Absence of bond polarity

Option B (Absence of bond polarity) is correct because N2 is chemically inert due to its non-polar nature and high bond energy, unlike the polar CN- ion.

Solution approach

Correct Answer + Core Reason: Absence of bond polarity. N2 is non-polar and has a high bond energy, making it inert.

Step-by-Step Solution:

Step 1: Analyze CN- ion: CN- is a polar ion with a negative charge, making it reactive.

Step 2: Analyze N2 molecule: N2 is a non-polar molecule with a triple bond, resulting in high bond energy. The absence of bond polarity contributes to its inertness.

Step 3: Compare: Although both are isoelectronic, the difference in polarity and reactivity arises from the bonding characteristics.

Why other options are wrong:
- Option A: Low bond energy - Incorrect because N2 has a very high bond energy due to the triple bond.
- Option B: Absence of bond polarity - Correct because N2 is non-polar, making it less reactive.
- Option C: Unsymmetrical electron distribution - Incorrect because N2 has a symmetrical electron distribution.
- Option D: Presence of more number of electrons in bonding orbitals - Incorrect because the number of electrons in bonding orbitals contributes to bond strength, not inertness directly.

Key Concepts:
- Isoelectronic Species
- Bond Polarity
- Bond Energy
- Chemical Inertness

Memory Aids:
- Remember that non-polar molecules are generally less reactive than polar molecules.

Quick Tips:
- Consider the polarity and bond strength when comparing the reactivity of molecules.

6 Medium

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Understanding +4 | -1

Consider the following statements

(i) Reaction

2 {Fe}^{3+}+{SO}_2+2 {H}_2 {O} \rightarrow 2 {Fe}^{2+}+{SO}_4^{2-}+4 {H}^{+}

shows reducing character of sulphur dioxide

(ii)

{H}_2 {~S}_2 {O}_8

contains four

{S}={O}

, two

{S}-{OH}

and one

{O}-

{O}

bond

(iii)

{NH}_3

gas can be dried effectively by using conc.

{H}_2 {SO}_4

.

(iv) One of the major use of

{H}_2 {SO}_4

is in the manufacture of fertilizers.

Which of the following is the correct code for the statements above?

View answer and explanation

Correct option: B

TTFT

Option B (TTFT) is correct because statements (i), (ii), and (iv) are true, while statement (iii) is false due to the reaction between ammonia and sulfuric acid.

Solution approach

Correct Answer + Core Reason: TTFT. Statements (i), (ii), and (iv) are true, while (iii) is false.

Step-by-Step Solution:

Step 1: Analyze statement (i): $2 {Fe}^{3+}+{SO}_2+2 {H}_2 {O} \rightarrow 2 {Fe}^{2+}+{SO}_4^{2-}+4 {H}^{+}$ . Sulfur dioxide ( $SO_2$ ) is oxidized to sulfate ( $SO_4^{2-}$ ), indicating its reducing character. This statement is TRUE.

Step 2: Analyze statement (ii): Structure of ${H}_2 {~S}_2 {O}_8$ (peroxodisulfuric acid) contains two $S-OH$ bonds, two $S=O$ bonds on each S atom (total four $S=O$ bonds) and one $O-O$ bond. This statement is TRUE.

Step 3: Analyze statement (iii): ${NH}_3$ gas cannot be dried effectively by using conc. ${H}_2 {SO}_4$ . Ammonia ( $NH_3$ ) is a base and reacts with concentrated sulfuric acid ( $H_2SO_4$ ), which is an acid. Therefore, $H_2SO_4$ cannot be used to dry $NH_3$ . This statement is FALSE.

Step 4: Analyze statement (iv): One of the major uses of ${H}_2 {SO}_4$ is in the manufacture of fertilizers, such as ammonium sulfate and superphosphate. This statement is TRUE.

Step 5: Determine the correct code based on the truth values of the statements: TTFT.

Why other options are wrong:
- Option A (TTFF): Incorrect because statement (iv) is true.
- Option C (FTFT): Incorrect because statement (i) is true.
- Option D (TFFT): Incorrect because statement (ii) is true.

Key Concepts:
- Redox reactions
- Structure of peroxodisulfuric acid
- Acid-base reactions
- Uses of sulfuric acid

Memory Aids:
- Remember that acids and bases react with each other.

Quick Tips:
- Analyze each statement individually and then combine the results.

7 Easy

7 of 680Question permalink

Remembering +4 | -1

Superphosphate of lime is used in:

View answer and explanation

Correct option: C

Agriculture

Option C (Agriculture) is correct because superphosphate of lime is a fertilizer used to provide phosphorus to plants.

Solution approach

Correct Answer + Core Reason: Agriculture. Superphosphate of lime is a fertilizer used in agriculture.

Step-by-Step Solution:

Step 1: Recall the uses of superphosphate of lime.

Step 2: Superphosphate of lime is a fertilizer, primarily used in agriculture to supply phosphorus to plants.

Step 3: The chemical formula is $Ca(H_2PO_4)_2 \cdot H_2O + 2(CaSO_4 \cdot 2H_2O)$ .

Why other options are wrong:
- Option A (Cement industry): Cement industry uses different raw materials like limestone, silica, alumina, and iron oxide.
- Option B (Glass industry): Glass industry uses silica as a major component.
- Option D (metallurgy): Metallurgy involves extraction and purification of metals from their ores. Superphosphate of lime is not directly involved in this process.

Key Concepts:
- Fertilizers
- Superphosphate of lime
- Uses of chemical compounds

Memory Aids:
- Superphosphate = Phosphate = Agriculture (think of plant growth)

Quick Tips:
- Remember common uses of chemical compounds.

8 Easy

8 of 680Question permalink

Analyzing +4 | -1

Statement I: The bond angle in

\mathrm{NH}_3

is more than that of

\mathrm{PH}_3

Statement II: Increasing size and decreasing electronegativity of central atom permit the bonding electrons to be drawn out further, lowering the repulsion between the bonding pairs

View answer and explanation

Correct option: A

Statement I and statement II are correct and statement II is the correct explanation of statement I

Option A (Statement I and statement II are correct and statement II is the correct explanation of statement I) is correct because the bond angle in NH3 is greater than PH3 due to the smaller size and higher electronegativity of Nitrogen.

Solution approach

Correct Answer + Core Reason: Statement I and statement II are correct and statement II is the correct explanation of statement I. The bond angle in NH3 is greater than PH3 due to the smaller size and higher electronegativity of Nitrogen.

Step-by-Step Solution:

Step 1: Analyze the bond angles in NH₃ and PH₃:
- NH₃ has a bond angle of approximately 107°.
- PH₃ has a bond angle of approximately 93.5°.
- Therefore, the bond angle in NH₃ is greater than that of PH₃.

Step 2: Consider the factors affecting bond angles:
- Electronegativity of the central atom: Higher electronegativity of the central atom leads to greater repulsion between bonding pairs and larger bond angles.
- Size of the central atom: Smaller size of the central atom leads to greater repulsion between bonding pairs and larger bond angles.
- Lone pair repulsion: Lone pair repulsion decreases the bond angle.

Step 3: Compare the electronegativity and size of N and P:
- Nitrogen is more electronegative than Phosphorus.
- Nitrogen is smaller in size than Phosphorus.

Step 4: Explain the difference in bond angles:
- Due to the higher electronegativity and smaller size of Nitrogen, the bonding pairs in NH₃ are closer to the central atom and experience greater repulsion, leading to a larger bond angle.
- In PH₃, the bonding pairs are further away from the central atom due to the larger size and lower electronegativity of Phosphorus, resulting in less repulsion and a smaller bond angle.

Step 5: Evaluate the given statements:
- Statement I: The bond angle in NH₃ is more than that of PH₃. This is correct.
- Statement II: Increasing size and decreasing electronegativity of central atom permit the bonding electrons to be drawn out further, lowering the repulsion between the bonding pairs. This is correct and explains why the bond angle in PH₃ is smaller than in NH₃.

Why other options are wrong:
- Option B (Statement I and statement II are correct but statement II is not the correct explanation of statement I): Statement II correctly explains Statement I.
- Option C (Statement I is correct but statement II is false): Statement II is true.
- Option D (Statement I if false but statement II is correct): Statement I is true.

Key Concepts:
- Bond angle
- Electronegativity
- Atomic size
- Lone pair repulsion

Memory Aids:
- Remember that bond angle increases with increasing electronegativity and decreasing size of the central atom.

Quick Tips:
- Consider the electronegativity and size of the central atom when comparing bond angles.
- Remember that lone pair repulsion decreases the bond angle.

9 Hard

9 of 680Question permalink

Understanding +4 | -1

Assertion (A):

{SO}_2

converts ferric ions to ferrous ions.

Reason (R):

{SO}_2

act as reducing agent.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

The correct answer is A (Both (A) and (R) are true and (R) is the correct explanation of (A)) because sulfur dioxide reduces ferric ions to ferrous ions by acting as a reducing agent.

Solution approach

Correct Answer + Core Reason: A (Both (A) and (R) are true and (R) is the correct explanation of (A)) because $SO_2$ reduces $Fe^{3+}$ to $Fe^{2+}$ by acting as a reducing agent.

Step-by-Step Solution:

Step 1: Recall the properties of $SO_2$ .

Step 2: $SO_2$ can act as a reducing agent.

Step 3: $SO_2$ reduces ferric ions ( $Fe^{3+}$ ) to ferrous ions ( $Fe^{2+}$ ).

Step 4: The reaction is: $2Fe^{3+} + SO_2 + 2H_2O \rightarrow 2Fe^{2+} + SO_4^{2-} + 4H^+$ .

Step 5: Assertion (A) is true because $SO_2$ converts ferric ions to ferrous ions.

Step 6: Reason (R) is true because $SO_2$ acts as a reducing agent.

Step 7: Reason (R) correctly explains why $SO_2$ converts ferric ions to ferrous ions.

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

Key Concepts:
- Redox reactions
- Reducing agent
- Sulfur dioxide
- Iron ions

Memory Aid:
- $SO_2$ is a reducing agent.

Quick Tip:
- Remember the common reducing and oxidizing agents.

10 Medium

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Analyzing +4 | -1

Assertion (A) : Each

\mathrm{N}

-atom is

s p^3

-hybridised in

\mathrm{N}_2 \mathrm{H}_4

.

Reason (R) : It has gauche conformation.

View answer and explanation

Correct option: B

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

.

Both A and R are true but R is not the correct explanation of A because while N_2H_4 does have sp³ hybridized N atoms and a gauche conformation, the conformation doesn't explain the hybridization.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ because the gauche conformation doesn't explain the sp³ hybridization.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): In N_2H_4 (hydrazine), each nitrogen atom is bonded to two hydrogen atoms and one nitrogen atom. This means each nitrogen atom has four electron domains (3 bonding pairs and 1 lone pair). According to VSEPR theory, four electron domains result in sp³ hybridization. Therefore, the assertion is true.

Step 2: Analyze Reason (R): N_2H_4 has a gauche conformation. This is also true. The gauche conformation is a staggered conformation where the dihedral angle between the two N-H bonds is approximately 60 degrees. This minimizes steric hindrance and lone pair repulsions.

Step 3: Relate Assertion and Reason: While both the assertion and the reason are true, the gauche conformation does not explain why the nitrogen atoms are sp³ hybridized. The hybridization is determined by the number of electron domains around the nitrogen atom, not the conformation of the molecule.

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

Key Concepts:
- Hybridization (sp³)
- VSEPR Theory
- Conformations (Gauche)
- Electron Domains

Memory Aids:
- sp³ hybridization: 4 electron domains
- Gauche conformation: Staggered with a 60-degree dihedral angle

Quick Tips:
- Hybridization is determined by the number of electron domains around an atom.
- Conformation refers to the spatial arrangement of atoms in a molecule due to rotation around single bonds.

11 Hard

11 of 680Question permalink

Analyzing +4 | -1

Assertion: Salts of

\mathrm{H}_3 \mathrm{PO}_3

and

\mathrm{H}_3 \mathrm{PO}_4

used as a fertilizers.

Reason: Formed

\mathrm{H}_3 \mathrm{PO}_3

increases solubility of water in soil on addition of phosphate fertilizers.

View answer and explanation

Correct option: A

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

Option A is correct. Salts of both

H_3PO_3

and

H_3PO_4

are used as fertilizers, and

H_3PO_3

increases water solubility in soil when phosphate fertilizers are added.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is correct explanation of the Assertion. Both statements are correct and related.

Step-by-Step Solution:

Step 1: Analyze the assertion: Salts of $H_3PO_3$ and $H_3PO_4$ are indeed used as fertilizers. This is because phosphorus is an essential nutrient for plant growth.

Step 2: Analyze the reason: $H_3PO_3$ can increase the solubility of water in soil when phosphate fertilizers are added. This increased solubility helps in the uptake of nutrients by plants.

Step 3: Determine the relationship: The reason correctly explains why the assertion is true. The increased solubility due to $H_3PO_3$ enhances the effectiveness of phosphate fertilizers.

Why other options are wrong:
- Option B: If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion: Incorrect because the reason directly explains the assertion.
- Option C: If Assertion is True but the Reason is False: Incorrect because both assertion and reason are true.
- Option D: If both Assertion & Reason are false: Incorrect because both assertion and reason are true.

Key Concepts:
- Fertilizers and their role in plant growth
- Phosphorus as a plant nutrient
- Solubility and its effect on nutrient uptake

Memory Aids:
- Phosphorus is essential for plant growth (remember "P" for phosphorus and plants).

Quick Tips:
- Consider the role of each component in fertilizers and their impact on soil conditions.

12 Medium

12 of 680Question permalink

Understanding +4 | -1

Which is correct statement?

View answer and explanation

Correct option: C

Nitric oxide is an endothermic compound

Option C is correct. Nitric oxide (NO) is an endothermic compound, meaning its formation requires energy input (positive enthalpy change).

Solution approach

Correct Answer + Core Reason: Nitric oxide is an endothermic compound. Its formation requires energy input.

Step-by-Step Solution:

Step 1: Consider the formation reaction of nitric oxide: ${N_2} + {O_2} \to 2NO$

Step 2: The reaction is endothermic, meaning it absorbs heat from the surroundings. This is represented by a positive enthalpy change (ΔH > 0).

Step 3: Therefore, nitric oxide is an endothermic compound.

Why other options are wrong:
- Option A: Nitric oxide (NO) has 11+8 = 15 electrons. CO2 has 6+2(8) = 22 electrons. They are not isoelectronic.
- Option B: Nitric oxide is paramagnetic due to the presence of an unpaired electron.
- Option D: Nitrous oxide (N2O) is used as a general anesthetic, not nitric oxide (NO).

Key Concepts:
- Endothermic reactions
- Enthalpy change
- Nitric oxide properties

Memory Aids:
- Endothermic = absorbs heat.

Quick Tips:
- Remember the formation reaction of nitric oxide and its endothermic nature.

13 Easy

13 of 680Question permalink

Remembering +4 | -1

Freons are used as:

View answer and explanation

Correct option: A

Refrigerant

Option A (Refrigerant) is correct because freons are commonly used as refrigerants due to their properties.

Solution approach

Correct Answer + Core Reason: Refrigerant. Freons are widely used as refrigerants.

Step-by-Step Solution:

Step 1: Recall the uses of freons. They are known for their use in refrigeration systems.

Step 2: Consider the properties that make them suitable for this purpose (e.g., low boiling point, non-toxicity, non-flammability).

Why other options are wrong:
- Option B: Freons are not primarily used as catalysts.
- Option C: Freons are not primarily used as oxidants.
- Option D: This is incorrect because freons have a specific use.

Key Concepts:
- Freons
- Refrigerants
- Applications of chemical compounds

Memory Aids:
- Freons = Refrigerants

Quick Tips:
- Remember the common applications of freons.

14 Medium

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Analyzing +4 | -1

Which of the following will not form disproportionated product?

View answer and explanation

Correct option: D

{XeF}_2+{H}_2 {O} \longrightarrow

Option D is correct because XeF2 reacts with water to form Xe, HF, and O2, without disproportionation.

Solution approach

Correct Answer + Core Reason: ${XeF}_2+{H}_2 {O} \longrightarrow$ . XeF2 undergoes a redox reaction with water but does not form disproportionated products.

Step-by-Step Solution:\nStep 1: Define disproportionation. Disproportionation is a redox reaction where a single element is simultaneously oxidized and reduced.\nStep 2: Analyze the given reactions.
- A: $C{l_2} + NaOH \rightarrow NaCl + NaOCl + H_2O$ . Chlorine is both oxidized (to $OCl^-$ ) and reduced (to $Cl^-$ ).
- B: $B{r_2} + NaOH \rightarrow NaBr + NaBrO_3 + H_2O$ . Bromine is both oxidized (to $BrO_3^-$ ) and reduced (to $Br^-$ ).
- C: ${XeF}_4 + {H}_2 {O} \rightarrow {Xe} + {XeO}_3 + HF + {O}_2$ . Xenon is both oxidized (to $XeO_3$ ) and reduced (to $Xe$ ).
- D: ${XeF}_2 + {H}_2 {O} \rightarrow {Xe} + HF + {O}_2$ . Xenon is reduced to Xe, but oxygen is oxidized, not xenon itself. Thus, no disproportionation.\nStep 3: Conclude. Only reaction D does not involve disproportionation.

Why other options are wrong:
- Option A: Chlorine undergoes disproportionation.
- Option B: Bromine undergoes disproportionation.
- Option C: Xenon undergoes disproportionation.

Key Concepts:
- Disproportionation Reactions
- Redox Reactions

Memory Aids:
- Disproportionation: Same element oxidized and reduced.

Quick Tips:
- Identify the oxidation states of the elements involved in the reaction to determine if disproportionation occurs.

15 Easy

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Analyzing +4 | -1

Which of the following pair have same hybridisation for central atom?

View answer and explanation

Correct option: D

\mathrm{XeF}_2

and

\mathrm{I}_3^{-}

Option D (

\mathrm{XeF}_2

and

\mathrm{I}_3^{-}

) is correct because both have

sp^3d

hybridization for the central atom.

Solution approach

Correct Answer + Core Reason: $\mathrm{XeF}_2$ and $\mathrm{I}_3^{-}$ . Both have $sp^3d$ hybridization.

Step-by-Step Solution:

Step 1: Determine the hybridization of the central atom in each molecule/ion.

Step 2: For $\mathrm{XeF}_2$ : Xe has 8 valence electrons. Two are used for bonding with F atoms, leaving 3 lone pairs. Steric number = 2 + 3 = 5. Hybridization = $sp^3d$ .

Step 3: For $\mathrm{I}_3^{-}$ : The central I atom has 7 valence electrons + 1 from the negative charge. Two are used for bonding with I atoms, leaving 3 lone pairs. Steric number = 2 + 3 = 5. Hybridization = $sp^3d$ .

Step 4: Compare the hybridizations of the given pairs and identify the pair with the same hybridization.

Why other options are wrong:
- Option A: $\mathrm{CO}_2$ has $sp$ hybridization, while $\mathrm{SO}_2$ has $sp^2$ hybridization.
- Option B: $\mathrm{CH}_3^+$ has $sp^2$ hybridization, while $\mathrm{XeF}_4$ has $sp^3d^2$ hybridization.
- Option C: $\mathrm{I}_3^{+}$ has $sp^3$ hybridization, while $\stackrel{+}{\mathrm{N}} \mathrm{O}_2$ has $sp^2$ hybridization.

Key Concepts:
- Hybridization
- Steric Number
- VSEPR Theory

Memory Aids:
- Steric Number = Number of bonded atoms + Number of lone pairs

Quick Tips:
- Calculate the steric number to determine the hybridization.

16 Medium

16 of 680Question permalink

Application +4 | -1

Iodine reacts with concentrated

HN{O_3}

to yield Y along with other products. The oxidation state of iodine in

Y,

is

View answer and explanation

Correct option: D

5

Option D (5) is correct because iodine reacts with concentrated HNO₃ to form iodic acid (HIO₃), where the oxidation state of iodine is +5.

Solution approach

Correct Answer + Core Reason: 5. The oxidation state of iodine in $HIO_3$ is +5.

Step-by-Step Solution:

Step 1: Write the reaction of iodine with concentrated $HNO_3$ :
$I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O$

Step 2: Identify the product Y as $HIO_3$ (iodic acid).

Step 3: Determine the oxidation state of iodine in $HIO_3$ :\nLet the oxidation state of I be x.
$1 + x + 3(-2) = 0$ \nx - 5 = 0\nx = +5

Step 4: Final Answer: 5

Why other options are wrong:
- Option A (1): This is incorrect because iodine does not have an oxidation state of +1 in the product formed.
- Option B (3): This is incorrect because iodine does not have an oxidation state of +3 in the product formed.
- Option C (7): This is incorrect because iodine does not have an oxidation state of +7 in the product formed.

Key Concepts:
- Oxidation states
- Redox reactions

Memory Aids:
- Remember the rules for assigning oxidation states.

Quick Tips:
- Always write the correct chemical formula for the product before determining the oxidation state.

17 Easy

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Understanding +4 | -1

Xenon best reacts with:

View answer and explanation

Correct option: B

The most electronegative element

Option B (The most electronegative element) is correct because xenon, being a noble gas, reacts best with highly electronegative elements like fluorine and oxygen.

Solution approach

Correct Answer + Core Reason: The most electronegative element. Xenon reacts with highly electronegative elements.

Step-by-Step Solution:

Step 1: Understand the reactivity of Xenon.\nXenon is a noble gas and is generally unreactive. However, it can react with highly electronegative elements like fluorine and oxygen.

Step 2: Analyze the options:
- A: The most electropositive element (typically alkali metals) would not react well with Xenon.
- B: The most electronegative element (fluorine, oxygen) would react well with Xenon.
- C: The hydrogen halides (HF, HCl, HBr, HI) can react with some substances, but Xenon reacts better with highly electronegative elements directly.
- D: Non-metals is a broad category, but Xenon reacts best with the most electronegative non-metals.

Why other options are wrong:
- Option A: Xenon does not react well with electropositive elements.
- Option C: Xenon reacts better with highly electronegative elements directly than with hydrogen halides.
- Option D: Xenon reacts specifically with the most electronegative non-metals, not all non-metals.

Key Concepts:
- Noble Gas Reactivity
- Electronegativity
- Chemical Reactions

Memory Aids:
- Xenon reacts with \"FON\" (Fluorine, Oxygen, Nitrogen - in decreasing order of reactivity).

Quick Tips:
- Remember that Xenon's reactivity is limited to highly electronegative elements.

18 Medium

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Understanding +4 | -1

Which would quickly absorb oxygen?

View answer and explanation

Correct option: A

Alkaline solution of pyrogallic acid

Option A (Alkaline solution of pyrogallic acid) is correct because alkaline pyrogallol readily absorbs oxygen, making it suitable for removing oxygen from gas mixtures.

Solution approach

Correct Answer + Core Reason: Alkaline solution of pyrogallic acid. Alkaline pyrogallol readily absorbs oxygen.

Step-by-Step Solution:

Step 1: Understand the properties of pyrogallic acid.\nPyrogallic acid (1,2,3-trihydroxybenzene) is a reducing agent that reacts with oxygen.

Step 2: Understand the effect of alkaline conditions.\nIn alkaline conditions, pyrogallic acid is deprotonated, making it a stronger reducing agent and enhancing its ability to absorb oxygen.

Step 3: Analyze the other options.\nConcentrated sulfuric acid is a dehydrating agent and does not readily absorb oxygen.\nLime water (calcium hydroxide solution) absorbs carbon dioxide, not oxygen.\nAlkaline solution of copper sulfate does not readily absorb oxygen.

Step 4: Identify the best option for oxygen absorption.\nAlkaline solution of pyrogallic acid is the best option for quickly absorbing oxygen.

Why other options are wrong:
- Option B: Concentrated sulfuric acid is a dehydrating agent, not an oxygen absorber.
- Option C: Lime water absorbs carbon dioxide, not oxygen.
- Option D: Alkaline solution of copper sulfate does not readily absorb oxygen.

Key Concepts:
- Reducing Agents
- Oxygen Absorption
- Pyrogallic Acid

Memory Aids:
- Alkaline pyrogallol absorbs oxygen.

Quick Tips:
- Remember the properties of common reagents.

19 Medium

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Analyzing +4 | -1

The true statement for the acids of phosphorus,

\mathrm{H}_3 \mathrm{PO}_2, \mathrm{H}_3 \mathrm{PO}_3

and

\mathrm{H}_3 \mathrm{PO}_4

is

View answer and explanation

Correct option: D

The geometry of phosphorus is tetrahedral in all the three

Option D is correct because

\mathrm{H}_3 \mathrm{PO}_2, \mathrm{H}_3 \mathrm{PO}_3

and

\mathrm{H}_3 \mathrm{PO}_4

all have phosphorus with

sp^3

hybridization, resulting in a tetrahedral geometry.

Solution approach

Correct Answer + Core Reason: The geometry of phosphorus is tetrahedral in all the three. All three acids have phosphorus with $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Draw the structures of $\mathrm{H}_3 \mathrm{PO}_2, \mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{H}_3 \mathrm{PO}_4$ .

Step 2: Determine the hybridization of phosphorus in each acid.\nIn $\mathrm{H}_3 \mathrm{PO}_2$ , phosphorus is bonded to one oxygen with a double bond, one OH group, and two hydrogen atoms. The steric number is 4, so the hybridization is $sp^3$ .\nIn $\mathrm{H}_3 \mathrm{PO}_3$ , phosphorus is bonded to one oxygen with a double bond, two OH groups, and one hydrogen atom. The steric number is 4, so the hybridization is $sp^3$ .\nIn $\mathrm{H}_3 \mathrm{PO}_4$ , phosphorus is bonded to one oxygen with a double bond, and three OH groups. The steric number is 4, so the hybridization is $sp^3$ .

Step 3: Determine the geometry based on the hybridization.\nSince the hybridization is $sp^3$ in all three acids, the geometry around phosphorus is tetrahedral.

Why other options are wrong:
- Option A: The order of acidity is not $\mathrm{H}_3 \mathrm{PO}_2 > \mathrm{H}_3 \mathrm{PO}_3 > \mathrm{H}_3 \mathrm{PO}_4$ . The acidity depends on the number of ionizable hydrogen atoms, but other factors also influence it.
- Option B: Not all of them are reducing in nature. $\mathrm{H}_3 \mathrm{PO}_4$ is not a reducing agent.
- Option C: Not all of them are tribasic acids. $\mathrm{H}_3 \mathrm{PO}_2$ is monobasic, and $\mathrm{H}_3 \mathrm{PO}_3$ is dibasic.

Key Concepts:
- Hybridization
- Molecular geometry
- Acid strength
- Reducing agents
- Basicity of acids

Memory Aids:
- Steric number = number of sigma bonds + number of lone pairs
- $sp^3$ hybridization corresponds to tetrahedral geometry

Quick Tips:
- Draw the Lewis structures to determine the hybridization and geometry.
- Remember that the number of OH groups bonded to the central atom determines the basicity of the acid.

20 Medium

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Understanding +4 | -1

Assertion (A):

\mathrm{N}_2 \mathrm{O}_5

is not possible due to

Reason (R): incapability of 'N' to show pentavalency Max. covalency of 'N' can be 4.

View answer and explanation

Correct option: D

Both (A) and (R) are false

Option D is correct because N2O5 is a known compound, and nitrogen can exhibit a covalency of more than 4 using d-orbitals, though it is less common.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are false. $\mathrm{N}_2 \mathrm{O}_5$ exists, and nitrogen can have a covalency of 4.

Step-by-Step Solution:

Step 1: Analyze the assertion (A): $\mathrm{N}_2 \mathrm{O}_5$ is not possible.
- This is false. $\mathrm{N}_2 \mathrm{O}_5$ is a known compound.

Step 2: Analyze the reason (R): incapability of 'N' to show pentavalency Max. covalency of 'N' can be 4.
- While nitrogen cannot form 5 sigma bonds, it can have a covalency of 4 (e.g., in $NH_4^+$ ). The statement that nitrogen *cannot* show pentavalency is misleading, as it can form more than 4 bonds using coordinate bonds.

Step 3: Since both the assertion and the reason are incorrect, the correct answer is that both (A) and (R) are false.

Why other options are wrong:
- Option A: Both (A) & (R) are true and the (R) is the correct explanation of the (A): Both are false.
- Option B: Both (A) & (R) are true but the (R) is not the correct explanation of the (A): Both are false.
- Option C: (A) is true but (R) is false: Both are false.

Key Concepts:
- Covalency
- Nitrogen compounds
- Structure of $\mathrm{N}_2 \mathrm{O}_5$

Memory Aids:
- Remember that $\mathrm{N}_2 \mathrm{O}_5$ exists.

Quick Tips:
- Be careful about absolute statements. Nitrogen can form more than 4 bonds in some cases.

21 Medium

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Analyzing +4 | -1

When

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

water is added to an aqueous solution of potassium halide in presence of chloroform a violet colour is obtained. On adding more of

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

water, the violet colour disappears and a colourless solution is obtained. This test confirms the presence of the following in aqueous solution:

View answer and explanation

Correct option: A

Iodide

Option A (Iodide) is correct because the violet color in chloroform indicates the presence of iodine, which is formed by the reaction of iodide with chlorine water. The disappearance of the color with excess chlorine water indicates further oxidation of iodine.

Solution approach

Correct Answer + Core Reason: Iodide is confirmed because it reacts with chlorine water to form iodine, which dissolves in chloroform to give a violet color. Further addition of chlorine water oxidizes iodine to a colorless compound.

Step-by-Step Solution:

Step 1: Write the reaction of chlorine water with potassium iodide: $2KI + Cl_2 \rightarrow I_2 + 2KCl$

Step 2: Iodine ( $I_2$ ) dissolves in chloroform ( $CHCl_3$ ) to give a violet color.

Step 3: Write the reaction of excess chlorine water with iodine: $I_2 + Cl_2 + H_2O \rightarrow 2HIO_3 + 2HCl$ (colorless)

Step 4: The violet color disappears because iodine is oxidized to iodic acid ( $HIO_3$ ), which is colorless.

Why other options are wrong:
- Option B: Bromide reacts with chlorine water to form bromine, which dissolves in chloroform to give an orange-brown color, not violet.
- Option C: Chloride does not react with chlorine water.
- Option D: If both iodide and bromide were present, the initial color would be a mixture of violet and orange-brown, and the final solution would still not be colorless until all iodine and bromine are oxidized.

Key Concepts:
- Halide reactions with chlorine water
- Solubility of halogens in chloroform
- Oxidation of halogens

Memory Aids:
- Iodine gives a violet color in chloroform.
- Chlorine water oxidizes halides to halogens.

Quick Tips:
- Remember the colors of halogens in organic solvents.
- Understand the oxidizing power of chlorine water.

22 Easy

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Remembering +4 | -1

Ammonia is:

View answer and explanation

Correct option: A

Polar solvent

Option A (Polar solvent) is correct because ammonia (

N{H_3}

) is a polar molecule and acts as a polar solvent.

Solution approach

Correct Answer + Core Reason: Polar solvent. Ammonia is a polar molecule due to the lone pair of electrons on the nitrogen atom and the difference in electronegativity between nitrogen and hydrogen.

Step-by-Step Solution:

Step 1: Identify the molecule: Ammonia ( $N{H_3}$ ).

Step 2: Determine the molecular geometry: Ammonia has a trigonal pyramidal shape due to the lone pair on the nitrogen atom.

Step 3: Analyze bond polarity: The N-H bonds are polar because nitrogen is more electronegative than hydrogen.

Step 4: Determine overall molecular polarity: Due to the trigonal pyramidal shape and the polar N-H bonds, the bond dipoles do not cancel out, resulting in a net dipole moment. Therefore, ammonia is a polar molecule.

Step 5: Conclude that ammonia acts as a polar solvent because polar solvents dissolve polar solutes.

Why other options are wrong:
- Option B: Non-polar: This is incorrect because ammonia has a net dipole moment and is therefore polar.
- Option C: Paramagnetic: This is incorrect because all electrons in ammonia are paired, making it diamagnetic, not paramagnetic. Paramagnetic substances have unpaired electrons.
- Option D: None of these: This is incorrect because ammonia is a polar solvent.

Key Concepts:
- Molecular polarity
- Electronegativity
- Molecular geometry (VSEPR theory)
- Polar solvents dissolve polar solutes

Memory Aids:
- "Like dissolves Like" - Polar solvents dissolve polar solutes, and non-polar solvents dissolve non-polar solutes.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict polarity.
- Remember that lone pairs on the central atom often lead to polar molecules.

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Understanding +4 | -1

Assertion (A):

{Cr},{Al}, {Fe}

and

{Ni}

are passive with concentrated Nitric acid.

Reason (R): Thin film of oxide layer is formed which prevents further action of concentrated Nitric acid on metal.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

The correct answer is A (Both (A) and (R) are true and (R) is the correct explanation of (A)) because Cr, Al, Fe, and Ni become passive due to the formation of a protective oxide layer.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). Oxide layer prevents further reaction.

Step-by-Step Solution:

Step 1: Chromium (Cr), Aluminum (Al), Iron (Fe), and Nickel (Ni) exhibit passivity when treated with concentrated nitric acid ( $HNO_3$ ).

Step 2: This passivity is due to the formation of a thin, adherent, and impervious oxide layer on the metal surface.

Step 3: This oxide layer protects the underlying metal from further reaction with the nitric acid.

Why other options are wrong:
- Option B: Both (A) and (R) are true but (R) is not the correct explanation of (A) - The reason is the correct explanation.
- Option C: (A) is true but (R) is false - The reason is true.
- Option D: (A) is false but (R) is true - The assertion is true.

Key Concepts:
- Passivity
- Oxide Layer Formation
- Reaction with Nitric Acid

Memory Aids:
- Oxide layer protects the metal.

Quick Tips:
- Remember the metals that exhibit passivity with nitric acid.

24 Easy

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Application +4 | -1

In which of the following compound central atom has

+4

oxidation state?

View answer and explanation

Correct option: B

\mathrm{H}_2 \mathrm{SO}_3

The correct answer is B (

\mathrm{H}_2 \mathrm{SO}_3

) because the oxidation state of sulfur in this compound is +4.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_2 \mathrm{SO}_3$ . The oxidation state of S in $\mathrm{H}_2 \mathrm{SO}_3$ is +4.

Step-by-Step Solution:

Step 1: Assign oxidation numbers to known elements: H = +1, O = -2.

Step 2: Let the oxidation state of S be x.

Step 3: For $\mathrm{H}_2 \mathrm{SO}_3$ , the sum of oxidation numbers must be zero: $2(+1) + x + 3(-2) = 0$

Step 4: Solve for x: $2 + x - 6 = 0 \Rightarrow x = +4$

Why other options are wrong:
- **Option A ( $\mathrm{SO}_3$ ): Oxidation state of S is +6.
- Option C ( $\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7$ ): Oxidation state of S is +6.
- Option D ( $\mathrm{BaSO}_4$ ): Oxidation state of S is +6.

Key Concepts:
- Oxidation State
- Rules for Assigning Oxidation Numbers

Memory Aids:
- Remember the common oxidation states of H and O.

Quick Tips:**
- Always check that the sum of oxidation numbers in a neutral compound is zero.

25 Medium

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Analyzing +4 | -1

Given below are two statements:
Statement I: Noble gases have very high boiling
points
Statement II: Noble gases are monoatomic gases.
They are held together by strong dispersion forces.
Because of this they are liquefied at very low
temperature. Hence, they have very high boiling
points.
In the light of the above statements, choose the
correct answer from the options given below.

View answer and explanation

Correct option: B

Both statement I and statement II are false

The correct answer is B (Both statement I and statement II are false) because noble gases have very *low* boiling points due to *weak* dispersion forces.

Solution approach

Correct Answer + Core Reason: Both statement I and statement II are false. Noble gases have low boiling points due to weak dispersion forces.

Step-by-Step Solution:

Step 1: Analyze Statement I: Noble gases have very high boiling points.
This statement is incorrect. Noble gases have very low boiling points.

Step 2: Analyze Statement II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points.
This statement is also incorrect. Noble gases are held together by *weak* dispersion forces, which leads to *low* boiling points.

Step 3: Conclusion: Both statements are false.

Why other options are wrong:
- Option A: Both statement I and statement II are true - Incorrect because both statements are false.
- Option C: Statement I is false but statement II is true - Incorrect because both statements are false.
- Option D: Statement I is true but statement II is false - Incorrect because both statements are false.

Key Concepts:
- Intermolecular forces
- Dispersion forces (London forces)
- Boiling point trends
- Properties of noble gases

Memory Aids:
- Noble gases are "noble" because they are aloof and don't interact strongly (weak dispersion forces).

Quick Tips:
- Remember that weak intermolecular forces lead to low boiling points and vice versa.

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Understanding +4 | -1

The correct order of reducing abilities of hydrides of V group elements is

View answer and explanation

Correct option: A

N{H_3} < P{H_3} < As{H_3} < Sb{H_3} < Bi{H_3}

Option A (

N{H_3} < P{H_3} < As{H_3} < Sb{H_3} < Bi{H_3}

) is correct because the reducing ability of hydrides increases down the group due to decreasing bond strength and stability.

Solution approach

Correct Answer + Core Reason: $N{H_3} < P{H_3} < As{H_3} < Sb{H_3} < Bi{H_3}$ . Reducing ability increases down the group.

Step-by-Step Solution:

Step 1: Understand the trend in reducing ability of hydrides down a group.
- The reducing ability of hydrides increases down a group because the bond strength between the central atom and hydrogen decreases.
- This makes it easier for the hydride to lose hydrogen and act as a reducing agent.

Step 2: Apply the trend to the hydrides of Group 15 elements.
- $N{H_3}$ is the most stable and least reducing.
- $Bi{H_3}$ is the least stable and most reducing.
- The order of reducing ability is: $N{H_3} < P{H_3} < As{H_3} < Sb{H_3} < Bi{H_3}$

Why other options are wrong:
- Option B: This order is the reverse of the correct order.
- Option C: This order is the same as the correct order, but it is repeated. This is likely a typo.
- Option D: This order is incorrect and does not follow the trend of increasing reducing ability down the group.

Key Concepts:
- Reducing ability
- Bond strength
- Stability of hydrides
- Periodic trends

Memory Aids:
- Reducing ability increases down a group for hydrides.

Quick Tips:
- Remember the trends in properties of hydrides down a group.

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Analyzing +4 | -1

Assertion (A):

C l F

is more reactive than

F_2

.

Reason (R): The

F-F

bond is weaker than

ClF

bond.

View answer and explanation

Correct option: D

(A) is false and (R) is true

Option D is correct because the F-F bond is indeed weaker than the Cl-F bond, but this makes ClF more reactive, not less. Therefore, the assertion is false, and the reason is true.

Solution approach

Correct Answer + Core Reason: (A) is false and (R) is true. ClF is more reactive than F2 because the Cl-F bond is weaker, not stronger.

Step-by-Step Solution:
Step 1: Analyze the assertion: ClF is more reactive than F2. This is generally true. Interhalogen compounds are often more reactive than halogens due to the polarity of the bond.
Step 2: Analyze the reason: The F-F bond is weaker than the ClF bond. This is true. The F-F bond has a low bond dissociation energy due to lone pair repulsions.
Step 3: Determine if the reason explains the assertion. The weaker F-F bond contributes to the higher reactivity of F2, while the polarity of the Cl-F bond contributes to the higher reactivity of ClF. Since the assertion states that ClF is *less* reactive, it is false.

Why other options are wrong:
- Option A: The assertion is false.
- Option B: The assertion is false.
- Option C: The assertion is false.

Key Concepts:
- Reactivity of halogens and interhalogens
- Bond strength
- Bond polarity

Memory Aids:
- Remember that weaker bonds are generally more reactive.

Quick Tips:
- Consider both bond strength and bond polarity when assessing reactivity.

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Remembering +4 | -1

The only element in VIA group or group 16 elements, which is definitely a metal, is:

View answer and explanation

Correct option: D

Polonium

Option D (Polonium) is correct because metallic character increases down the group, and Polonium is the last element in Group 16 that exhibits metallic properties.

Solution approach

Correct Answer + Core Reason: Polonium. Metallic character increases down the group, making Polonium the most metallic element in Group 16.

Step-by-Step Solution:

Step 1: Recall the periodic trends of metallic character.

Step 2: Metallic character increases as you move down a group.

Step 3: Identify the position of each element in Group 16: Oxygen, Sulfur, Selenium, Tellurium, Polonium.

Step 4: Polonium is the last element in the group and therefore has the most metallic character.

Why other options are wrong:
- Option A: Tellurium is a metalloid, exhibiting properties of both metals and nonmetals, but it is not definitively a metal.
- Option B: Selenium is a nonmetal.
- Option C: Sulphur is a nonmetal.

Key Concepts:
- Periodic trends
- Metallic character
- Group 16 elements

Memory Aids:
- Metallic character increases down and to the left on the periodic table.

Quick Tips:
- Remember the general trends in the periodic table.

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Understanding +4 | -1

Among the halogens, the one which is oxidised by nitric acid is

View answer and explanation

Correct option: B

Iodine

Option B (Iodine) is correct because nitric acid oxidizes iodine to iodic acid.

Solution approach

Correct Answer + Core Reason: Iodine. Nitric acid oxidizes iodine to iodic acid.

Step-by-Step Solution:

Step 1: Recall the oxidizing properties of nitric acid: Nitric acid is a strong oxidizing agent.

Step 2: Consider the reactions of halogens with nitric acid:
- Fluorine: Fluorine is the strongest oxidizing agent among the halogens and is not oxidized by nitric acid.
- Chlorine, Bromine, and Iodine: These halogens can be oxidized by nitric acid, but iodine is the most readily oxidized among them.

Step 3: Write the reaction of iodine with nitric acid:
$10HNO_3 + I_2 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O$

Step 4: Conclude that iodine is oxidized by nitric acid.

Why other options are wrong:
- Option A: Fluorine is incorrect because fluorine is the strongest oxidizing agent and cannot be oxidized further by nitric acid.
- Option C: Chlorine is incorrect because while chlorine can be oxidized by nitric acid, iodine is more readily oxidized.
- Option D: Bromine is incorrect because while bromine can be oxidized by nitric acid, iodine is more readily oxidized.

Key Concepts:
- Oxidizing Agents
- Nitric Acid
- Halogens
- Redox Reactions

Memory Aids:
- Remember that iodine is easily oxidized due to its lower electronegativity compared to other halogens.

Quick Tips:
- Know the relative oxidizing and reducing strengths of common elements and compounds.

30 Medium

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Analyzing +4 | -1

The electronic configurations of four elements are given below. Which element does not belong to the same family as others?

View answer and explanation

Correct option: C

\left[ {\mathrm{Ne}} \right]3{s^2},\mathrm{\;}3{p^5}

Option C is correct because the element with electronic configuration

\left[ {\mathrm{Ne}} \right]3{s^2},\mathrm{\;}3{p^5}

is Chlorine, which is a halogen and belongs to a different family than the others, which are transition metals.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{Ne}} \right]3{s^2},\mathrm{\;}3{p^5}$ . This configuration represents a halogen, while the others are transition metals.

Step-by-Step Solution:
Step 1: Identify the element represented by each electronic configuration.
Step 2: A: $\left[ {\mathrm{Xe}} \right]4\mathrm{\;}{f^{10}},\mathrm{\;}5{d^{10}},\mathrm{\;}6{s^2}$ - A transition metal.
Step 3: B: $\left[ {\mathrm{Kr}} \right]4{d^{10}},\mathrm{\;}5{s^2}$ - A transition metal.
Step 4: C: $\left[ {\mathrm{Ne}} \right]3{s^2},\mathrm{\;}3{p^5}$ - Chlorine (a halogen).
Step 5: D: $\left[ {\mathrm{Ar}} \right]3{d^{10}},\mathrm{\;}4{s^2}$ - A transition metal.
Step 6: Determine which element does not belong to the same family as the others. Chlorine does not belong to the same family as the transition metals.

Why other options are wrong:
- Option A: This is a transition metal, belonging to the same family as B and D.
- Option B: This is a transition metal, belonging to the same family as A and D.
- Option D: This is a transition metal, belonging to the same family as A and B.

Key Concepts:
- Electronic configuration
- Transition metals
- Halogens

Memory Aids:
- Remember the general electronic configurations of different groups.

Quick Tips:
- Recognize the noble gas cores and focus on the valence electrons.

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