Oxygen is divalent, but sulphur exhibits variable valency of $2,{\rm{ }}4{\rm{ }} and {\rm{ }}6$, because:

Sulphur is less electronegative than oxygen

Sulphur is bigger atom than oxygen

Ionisation potential of sulphur is more than oxygen

The formation of the oxide ion ${{\rm{O}}^{2 - }}\left( g \right)$ requires first an exothermic and then an endothermic step as shown below: ${\rm{O}}\left( {\rm{g}} \right) + {e^ - } = {{\rm{O}}^ - }\left( {\rm{g}} \right);{\rm{\;}}\Delta H^\circ = - 142{\rm{\;kJmo}}{1^{ - 1}}$ , ${\rm{O}}{\left( {\rm{g}} \right)^ - } + {e^ - } = {{\rm{O}}^{2 - }}\left( {\rm{g}} \right);\Delta H^\circ = 844{\rm{\;kJmo}}{1^{ - 1}}$ This is because

Oxygen has high electron affinity

${\text{O}^ - }$ has comparatively larger size than oxygen atom

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

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Remembering +4 | -1

Mendeleev corrected the atomic mass of:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because Mendeleev corrected the atomic masses of Beryllium, Indium, and Osmium to better fit them into his periodic table based on their properties.

Solution approach

Correct Answer + Core Reason: All of these. Mendeleev corrected the atomic masses of Be, In, and Os to better fit them in his periodic table based on their properties.

Step-by-Step Solution:

Step 1: Recall Mendeleev's work on the periodic table. Mendeleev arranged elements in his periodic table based on their atomic weights and chemical properties.

Step 2: Understand that Mendeleev sometimes corrected atomic masses. In some cases, the accepted atomic masses of elements did not align with their observed chemical properties, so Mendeleev corrected these values.

Step 3: Consider the given options:
- $Be$ : Mendeleev corrected the atomic mass of Beryllium.
- $In$ : Mendeleev corrected the atomic mass of Indium.
- $Os$ : Mendeleev corrected the atomic mass of Osmium.

Step 4: Conclude that all of the given elements had their atomic masses corrected by Mendeleev.

Why other options are wrong:
- **Option A ( $Be$ ): While Mendeleev corrected the atomic mass of Be, it's not the only one.
- Option B ( $In$ ): While Mendeleev corrected the atomic mass of In, it's not the only one.
- Option C ( $Os$ ): While Mendeleev corrected the atomic mass of Os, it's not the only one.

Key Concepts:
- Mendeleev's periodic table
- Correction of atomic masses

Memory Aids:
- Remember that Mendeleev made adjustments to atomic masses to better fit elements into his periodic table.

Quick Tips:**
- Know that Mendeleev's periodic table was based on both atomic weight and chemical properties, and he made corrections when necessary.

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Out of the four blocks in which periodic table is divided, helium belongs to which block?

View answer and explanation

Correct option: B

p

-block

Option B is correct. Helium (He) belongs to the p-block because its position is in group 18, even though it has an s² configuration.

Solution approach

Correct Answer + Core Reason: $p$ -block. Although Helium has an $s^2$ configuration, its properties and position in the periodic table place it in the p-block.

Step-by-Step Solution:

Step 1: Consider Helium's position in the periodic table: Helium is placed in Group 18 (Noble Gases), which is part of the p-block.

Step 2: Consider Helium's properties: Helium is a noble gas, and noble gases are located in the p-block.

Step 3: Conclusion: Despite having an $s^2$ electronic configuration, Helium belongs to the p-block due to its properties and position in the periodic table.

Why other options are wrong:
- Option A: While Helium has an $s^2$ configuration, it's not considered an s-block element due to its properties.
- Option C: Helium does not have any d electrons, so it's not a d-block element.
- Option D: Helium does not have any f electrons, so it's not an f-block element.

Key Concepts:
- Periodic Table Blocks (s, p, d, f)
- Electronic Configuration
- Noble Gases

Memory Aids:
- Remember that Helium is a noble gas and is located in Group 18 (p-block).

Quick Tips:
- Consider both electronic configuration and properties when determining the block of an element.

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The common oxidation state exhibited by inner transition elements usually in their compound is

View answer and explanation

Correct option: B

+ 3

Option B is correct because the most common oxidation state exhibited by inner transition elements in their compounds is +3, due to the ease of losing the two s-electrons and one f-electron.

Solution approach

Correct Answer + Core Reason: $+ 3$ . The most stable oxidation state for inner transition elements is +3.

Step-by-Step Solution:

Step 1: Define inner transition elements.\nInner transition elements are the lanthanides and actinides.

Step 2: Determine the common oxidation states of inner transition elements.\nThe most common oxidation state exhibited by inner transition elements is +3. This is because they readily lose the two s-electrons and one f-electron to achieve a stable electronic configuration.

Step 3: Conclusion.\nThe common oxidation state exhibited by inner transition elements is +3.

Why other options are wrong:
- Option A: Incorrect because +2 is not the most common oxidation state for inner transition elements.
- Option B: Correct because +3 is the most common oxidation state for inner transition elements.
- Option C: Incorrect because +5 is not the most common oxidation state for inner transition elements.
- Option D: Incorrect because Zero is not the common oxidation state for inner transition elements in their compounds.

Key Concepts:
- Oxidation state
- Inner transition elements

Memory Aids:
- Remember that +3 is the most common oxidation state for lanthanides and actinides.

Quick Tips:
- Review the electronic configurations and oxidation states of lanthanides and actinides.

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The set that represents the pair of neutral oxides nitrogen is

View answer and explanation

Correct option: A

NO

and

{N_2}O

Option A (

NO

and

{N_2}O

) is correct because both nitric oxide (

NO

) and nitrous oxide (

{N_2}O

) are neutral oxides.

Solution approach

Correct Answer + Core Reason: $NO$ and ${N_2}O$ are neutral oxides.

Step-by-Step Solution:

Step 1: Define neutral oxides: Neutral oxides are oxides that do not react with either acids or bases to form salts.

Step 2: Consider the given options:
- $NO$ (Nitric oxide) is a neutral oxide.
- ${N_2}O$ (Nitrous oxide) is a neutral oxide.
- ${N_2}{O_3}$ is an acidic oxide.
- $N{O_2}$ is an acidic oxide.

Step 3: $NO$ and ${N_2}O$ do not react with acids or bases.

Step 4: Conclusion: $NO$ and ${N_2}O$ are the neutral oxides.

Why other options are wrong:
- Option A: $NO$ and ${N_2}O$ is the correct answer as both are neutral oxides.
- Option B: ${N_2}O$ is neutral, but ${N_2}{O_3}$ is acidic.
- Option C: ${N_2}O$ is neutral, but $N{O_2}$ is acidic.
- Option D: $NO$ is neutral, but $N{O_2}$ is acidic.

Key Concepts:
- Neutral Oxides
- Acidic Oxides
- Basic Oxides

Memory Aids:
- Remember common neutral oxides: $CO$ , $NO$ , ${N_2}O$ .

Quick Tips:
- Learn the common examples of neutral, acidic, and basic oxides.

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Analyzing +4 | -1

Which of the following is different from other three oxides

View answer and explanation

Correct option: A

MgO

Option A (MgO) is correct because it is a basic oxide, while SnO, ZnO, and PbO are amphoteric oxides.

Solution approach

Correct Answer + Core Reason: MgO. Magnesium oxide (MgO) is a basic oxide, while SnO, ZnO, and PbO are amphoteric oxides.

Step-by-Step Solution:

Step 1: Determine the nature of each oxide.\nMgO: Basic oxide (reacts with acids)\nSnO: Amphoteric oxide (reacts with both acids and bases)\nZnO: Amphoteric oxide (reacts with both acids and bases)\nPbO: Amphoteric oxide (reacts with both acids and bases)

Step 2: Identify the oxide that is different from the others.\nMgO is basic, while the others are amphoteric.

Step 3: Final Answer: MgO is different from the other three oxides.

Why other options are wrong:
- Option B: SnO: This is an amphoteric oxide, similar to ZnO and PbO.
- Option C: ZnO: This is an amphoteric oxide, similar to SnO and PbO.
- Option D: PbO: This is an amphoteric oxide, similar to SnO and ZnO.

Key Concepts:
- Basic oxides
- Amphoteric oxides
- Acid-base properties of oxides

Memory Aids:
- Oxides of highly electropositive metals are basic.
- Oxides of less electropositive metals and metalloids are amphoteric.

Quick Tips:
- Remember the common examples of basic and amphoteric oxides.

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The atomic number of elements A, B, C and D are Z-1, Z+1, and Z+2, respectively. If ‘B’ is a noble gas, choose the correct answer from the following statements.
V) ‘A’ has higher electron affinity
VI) ‘C’ exists in +2 oxidation state
VII) ‘D’ is an alkaline earth metal

View answer and explanation

Correct option: C

I and III

Option C (I and III) is correct because if B is a noble gas (Z+1), then A (Z-1) has a high electron affinity and D (Z+2) is an alkaline earth metal.

Solution approach

Correct Answer + Core Reason: I and III. If B is a noble gas, then A has higher electron affinity and D is an alkaline earth metal.

Step-by-Step Solution:

Step 1: Assume B is a noble gas. Noble gases are in group 18.

Step 2: Determine the positions of A, C, and D relative to B.
- A is Z-1, so it is one element before B in the periodic table. It is a halogen (group 17).
- C is Z, so it is two elements before B in the periodic table. It is a chalcogen (group 16).
- D is Z+2, so it is one element after B in the periodic table. It is an alkali metal (group 1).

Step 3: Evaluate the statements:
- I: A (halogen) has higher electron affinity. This is true because halogens have a strong tendency to gain an electron to achieve a stable noble gas configuration.
- II: C exists in +2 oxidation state. This is false. Chalcogens typically exhibit -2 oxidation state.
- III: D is an alkaline earth metal. This is false. D is an alkali metal (group 1).

Step 4: Conclude that statements I and III are correct.

Why other options are wrong:
- Option A: I and II - Statement II is incorrect.
- Option B: II and III - Statements II and III are incorrect.
- Option D: I, II and III - Statement II is incorrect.

Key Concepts:
- Periodic table groups: noble gases, halogens, chalcogens, alkali metals, alkaline earth metals
- Electron affinity
- Oxidation states

Memory Aids:
- Remember the groups and their properties.

Quick Tips:
- Knowing the group to which an element belongs helps determine its properties.

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Application +4 | -1

The atomic masses of

Li

and

K

are 7 and 39, respectively. According to law of triads the atomic mass of

Na

will be

View answer and explanation

Correct option: A

23

Option A (23) is correct because, according to the Law of Triads, the atomic mass of Na is approximately the average of the atomic masses of Li and K: (7+39)/2 = 23.

Solution approach

Correct Answer + Core Reason: 23. The Law of Triads states that the atomic mass of the middle element is the average of the other two.

Step-by-Step Solution:

Step 1: State the Law of Triads: The atomic mass of the middle element in a triad is approximately the arithmetic mean of the atomic masses of the other two elements.

Step 2: Given atomic masses: Li = 7, K = 39.

Step 3: Calculate the atomic mass of Na using the Law of Triads: $\frac{{Atomic \\ mass \\ of \\ Li + Atomic \\ mass \\ of \\ K}}{{2}} = \frac{{7 + 39}}{{2}}$

Step 4: $\frac{{46}}{{2}} = 23$

Step 5: Therefore, the atomic mass of Na is 23.

Why other options are wrong:
- Option B (32): This value is incorrect and doesn't follow from the Law of Triads.
- Option C (46): This is the sum of the atomic masses of Li and K, not the average.
- Option D (64): This value is also incorrect and doesn't relate to the Law of Triads.

Key Concepts:
- Law of Triads
- Atomic mass
- Periodic table history

Memory Aids:
- Remember that the Law of Triads is an early attempt to find relationships between atomic masses.

Quick Tips:
- The Law of Triads is a historical concept; modern periodic law is based on atomic number.

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Analyzing +4 | -1

Which of the following has the largest size

View answer and explanation

Correct option: A

{N^{ - 3}}

Option A (

{N^{ - 3}})

is correct because among isoelectronic species, the size increases with increasing negative charge.

Solution approach

Correct Answer + Core Reason: ${N^{ - 3}}$ . For isoelectronic species, size increases with increasing negative charge.

Step-by-Step Solution:

Step 1: Determine the number of electrons for each ion:
- ${N^{ - 3}}$ : 7 + 3 = 10 electrons
- ${O^{ - 2}}$ : 8 + 2 = 10 electrons
- $M{a^ + }$ : 11 - 1 = 10 electrons
- $M{g^{2 + }}$ : 12 - 2 = 10 electrons

Step 2: All the given ions are isoelectronic (have the same number of electrons).

Step 3: For isoelectronic species, the size decreases with increasing nuclear charge (number of protons). Therefore, the size increases with increasing negative charge.

Step 4: Arrange the ions in order of increasing nuclear charge:
- ${N^{ - 3}}$ (7 protons)
- ${O^{ - 2}}$ (8 protons)
- $M{a^ + }$ (11 protons)
- $M{g^{2 + }}$ (12 protons)

Step 5: The ion with the smallest nuclear charge ( ${N^{ - 3}}$ ) will have the largest size.

Why other options are wrong:
- Option B: ${O^{ - 2}}$ has a higher nuclear charge than ${N^{ - 3}}$ , so it's smaller.
- Option C: $M{a^ + }$ has a higher nuclear charge than ${N^{ - 3}}$ , so it's smaller.
- Option D: $M{g^{2 + }}$ has the highest nuclear charge, so it's the smallest.

Key Concepts:
- Isoelectronic species
- Ionic size
- Nuclear charge

Memory Aids:
- More negative charge = larger size (for isoelectronic species).
- More positive charge = smaller size (for isoelectronic species).

Quick Tips:
- Always check if the species are isoelectronic before comparing their sizes based on charge.

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Remembering +4 | -1

Metal used as catalyst in the hydrogenation of vegetable oils

View answer and explanation

Correct option: C

Nickel

Option C (Nickel) is correct because Nickel is commonly used as a catalyst in the hydrogenation of vegetable oils.

Solution approach

Correct Answer + Core Reason: Nickel. Nickel is widely used as a catalyst in the hydrogenation of vegetable oils.

Step-by-Step Solution:

Step 1: Hydrogenation of vegetable oils involves adding hydrogen to unsaturated fatty acids to convert them into saturated fatty acids, thus solidifying the oil.

Step 2: Nickel acts as a heterogeneous catalyst, providing a surface for the reaction to occur more efficiently.

Step 3: The hydrogen molecules adsorb onto the nickel surface, weakening the H-H bond and facilitating the addition of hydrogen to the double bonds in the oil.

Why other options are wrong:
- Option A (Iron): Iron is used in some catalytic processes, but not typically for hydrogenation of vegetable oils.
- Option B (Molybdoneum): Molybdenum is used in various industrial catalysts, but not primarily for hydrogenation of vegetable oils.
- Option D (Sodium): Sodium is a highly reactive alkali metal and is not used as a catalyst in hydrogenation reactions. It would react violently with water present in the system.

Key Concepts:
- Catalysis
- Hydrogenation
- Heterogeneous Catalysis

Memory Aids:
- Remember that Nickel is a common catalyst for hydrogenation reactions.

Quick Tips:
- Know the common catalysts used in various industrial processes.

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Application +4 | -1

Among the following, pair of elements belonging to same
period is

View answer and explanation

Correct option: C

Z = 55 and Z = 77

Option C (Z = 55 and Z = 77) is correct because both elements with atomic numbers 55 (Cesium) and 77 (Iridium) belong to the 6th period.

Solution approach

Correct Answer + Core Reason: Z = 55 and Z = 77. Both elements belong to the 6th period.

Step-by-Step Solution:

Step 1: Identify the elements corresponding to the given atomic numbers.
- Z = 35: Bromine (Br)
- Z = 38: Strontium (Sr)
- Z = 28: Nickel (Ni)
- Z = 55: Cesium (Cs)
- Z = 77: Iridium (Ir)
- Z = 86: Radon (Rn)
- Z = 96: Curium (Cm)

Step 2: Determine the period to which each element belongs.
- Br: 4th period
- Sr: 5th period
- Ni: 4th period
- Cs: 6th period
- Ir: 6th period
- Rn: 6th period
- Cm: 7th period

Step 3: Check which pair of elements belongs to the same period.
- Z = 55 (Cs) and Z = 77 (Ir) both belong to the 6th period.

Why other options are wrong:
- Option A: Z = 35 (Br) is in the 4th period, and Z = 38 (Sr) is in the 5th period.
- Option B: Z = 28 (Ni) is in the 4th period, and Z = 38 (Sr) is in the 5th period.
- Option D: Z = 86 (Rn) is in the 6th period, and Z = 96 (Cm) is in the 7th period.

Key Concepts:
- Periodic table periods
- Atomic number and element identification

Memory Aids:
- Remember the periods and their corresponding elements.

Quick Tips:
- Knowing the element's position in the periodic table helps determine the period to which it belongs.

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Understanding +4 | -1

Which of the following sets represents the collection of isoelectronic species?

View answer and explanation

Correct option: D

{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }},\mathrm{C}{\mathrm{l}^ - }

Option D (

{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }},\mathrm{C}{\mathrm{l}^ - }

) is correct because all the species have the same number of electrons (18), making them isoelectronic.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }},\mathrm{C}{\mathrm{l}^ - }$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Define isoelectronic species. Isoelectronic species have the same number of electrons.

Step 2: Determine the number of electrons for each ion in each option:
- A: $\mathrm{N}{\mathrm{a}^ + }$ (10), $\mathrm{M}{\mathrm{g}^{2 + }}$ (10), $\mathrm{A}{\mathrm{l}^{3 + }}$ (10), $\mathrm{C}{\mathrm{l}^ - }$ (18) - Not isoelectronic.
- B: $\mathrm{N}{\mathrm{a}^ + }$ (10), $\mathrm{C}{\mathrm{a}^{2 + }}$ (18), $\mathrm{S}{\mathrm{c}^{3 + }}$ (18), ${\mathrm{F}^ - }$ (10) - Not isoelectronic.
- C: ${\mathrm{K}^ + }$ (18), $\mathrm{C}{\mathrm{l}^ - }$ (18), $\mathrm{M}{\mathrm{g}^{2 + }}$ (10), $\mathrm{S}{\mathrm{c}^{3 + }}$ (18) - Not isoelectronic.
- D: ${\mathrm{K}^ + }$ (18), $\mathrm{C}{\mathrm{a}^{2 + }}$ (18), $\mathrm{S}{\mathrm{c}^{3 + }}$ (18), $\mathrm{C}{\mathrm{l}^ - }$ (18) - Isoelectronic.

Step 3: Conclusion: Only option D contains isoelectronic species.

Why other options are wrong:
- Option A: The ions in this set do not have the same number of electrons.
- Option B: The ions in this set do not have the same number of electrons.
- Option C: The ions in this set do not have the same number of electrons.

Key Concepts:
- Isoelectronic species
- Number of electrons in ions

Memory Aids:
- Iso = same, electronic = electrons.

Quick Tips:
- Calculate the number of electrons for each ion carefully.

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Understanding +4 | -1

The ionization potential order for which set is correct?

View answer and explanation

Correct option: B

B>Li>K

Option B (B>Li>K) is correct because ionization potential generally increases across a period and decreases down a group.

Solution approach

Correct Answer + Core Reason: B>Li>K. Ionization potential increases across a period and decreases down a group.

Step-by-Step Solution:

Step 1: Recall the trends in ionization potential.

Step 2: Ionization potential generally increases across a period (from left to right) and decreases down a group (from top to bottom).

Step 3: Locate the elements Li, K, and B on the periodic table. Li and K are in Group 1, with Li above K. B is in Group 13 and is to the right of Li.

Step 4: Based on the trends, B should have a higher ionization potential than Li, and Li should have a higher ionization potential than K. Therefore, the order is B > Li > K.

Why other options are wrong:
- Option A: Li>K>Cs is incorrect because ionization potential decreases down a group. Cs should have the lowest ionization potential among these elements.
- Option C: Cs>Li>B is incorrect because ionization potential increases across a period and decreases down a group. B should have the highest ionization potential.
- Option D: Cs<Li<K is incorrect because ionization potential decreases down a group. K should have a lower ionization potential than Li.

Key Concepts:
- Periodic trends: Ionization potential
- Position of elements in the periodic table

Memory Aids:
- Ionization potential: Increase across, decrease down.

Quick Tips:
- Use the periodic table to determine the relative positions of the elements and apply the periodic trends.

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Understanding +4 | -1

The set representing the correct order of first ionisation potential is:

View answer and explanation

Correct option: B

\mathrm{Be} > Mg > Ca

Option B (

\mathrm{Be} > Mg > Ca

) is correct because ionization energy decreases down a group, and Be, Mg, and Ca are in Group 2.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be} > Mg > Ca$ . Ionization energy decreases down a group.

Step-by-Step Solution:
Step 1: Recall the trend of ionization energy within a group. Ionization energy generally decreases as you move down a group in the periodic table.
Step 2: Identify the positions of the given elements in the periodic table.
- Be, Mg, and Ca are all alkaline earth metals (Group 2), with Be at the top, Mg in the middle, and Ca below Mg.
- K, Na, and Li are alkali metals (Group 1), with Li at the top, Na in the middle, and K below Na. The correct order is Li > Na > K.
- B, C, and N are in the same period (Period 2). The general trend is B < C < N, but there's an exception between N and O. The correct order is N > C > B (considering the question asks for first ionization potential, and not electron affinity).
- Ge, Si, and C are in Group 14, with C at the top, Si in the middle, and Ge below Si. The correct order is C > Si > Ge.
Step 3: Apply the trend to the given options. Since ionization energy decreases down the group, the correct order for Group 2 elements is Be > Mg > Ca.

Why other options are wrong:
- Option A: K > Na > Li: Incorrect, the correct order is Li > Na > K.
- Option C: B > C > N: Incorrect, the correct order is N > C > B.
- Option D: Ge > Si > C: Incorrect, the correct order is C > Si > Ge.

Key Concepts:
- Ionization energy
- Periodic trends
- Group 1 (alkali metals) and Group 2 (alkaline earth metals) elements

Memory Aids:
- Ionization energy decreases down a group.

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

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Understanding +4 | -1

The element with the electronic configuration as

\left[ {\mathrm{Ar}} \right]3{d^{10}}4{s^2}4{p^3}

represents a

View answer and explanation

Correct option: C

Metalloid

Option C (Metalloid) is correct because the electronic configuration

[Ar]3d^{10}4s^24p^3

corresponds to Arsenic (As), which is a metalloid.

Solution approach

Correct Answer + Core Reason: Metalloid. The electronic configuration corresponds to Arsenic, which is a metalloid.

Step-by-Step Solution:

Step 1: Determine the element represented by the electronic configuration $[Ar]3d^{10}4s^24p^3$ .

Step 2: The electronic configuration has 18 (Ar) + 10 + 2 + 3 = 33 electrons. The element with 33 electrons is Arsenic (As).

Step 3: Identify the type of element Arsenic is. Arsenic is a metalloid.

Step 4: Conclude that the element represents a metalloid.

Why other options are wrong:
- Option A: Metal - Metals typically have electronic configurations with few electrons in their outermost shell.
- Option B: Non-metal - Non-metals typically have electronic configurations with more electrons in their outermost shell, but Arsenic is a metalloid.
- Option D: Transition element - Transition elements have partially filled d-orbitals, but Arsenic is a p-block element.

Key Concepts:
- Electronic configuration
- Types of elements: metals, non-metals, metalloids, transition elements

Memory Aids:
- Metalloids are elements with properties intermediate between metals and non-metals.

Quick Tips:
- Knowing the electronic configuration helps determine the element and its type.

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Understanding +4 | -1

A sudden large jump between the value of first and second ionisation energies of elements would be associated with which of the following electronic configurations?

View answer and explanation

Correct option: A

1{s^2},2{s^2}2{p^6},3{s^1}

Option A (

1{s^2},2{s^2}2{p^6},3{s^1}

) is correct because removing the single electron from the 3s orbital results in a stable noble gas configuration, making the second ionization energy significantly higher.

Solution approach

Correct Answer + Core Reason: $1{s^2},2{s^2}2{p^6},3{s^1}$ . Removing the single electron from the 3s orbital results in a stable noble gas configuration.

Step-by-Step Solution:

Step 1: Analyze each electronic configuration.

Step 2: Consider the effect of removing one electron from each configuration.

Step 3: A sudden large jump in ionization energy indicates that removing the second electron requires breaking a stable electron configuration.

Step 4: In option A ( $1{s^2},2{s^2}2{p^6},3{s^1}$ ), removing the first electron results in the stable noble gas configuration $1{s^2},2{s^2}2{p^6}$ . Removing a second electron would require breaking this stable configuration, resulting in a large jump in ionization energy.

Why other options are wrong:
- Option B: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^1}$ - Removing the first electron is relatively easy, and removing the second is also not drastically more difficult.
- Option C: $1{s^2},2{s^2},2{p^6},3{s^1}3{p^2}$ - Removing the first electron is relatively easy, and removing the second is also not drastically more difficult.
- Option D: $1{s^2},2{s^2}2{p^6},3{s^2}$ - Removing the first electron is relatively easy, and removing the second is also not drastically more difficult.

Key Concepts:
- Ionization Energy
- Electronic Configuration
- Stability of Noble Gas Configurations

Memory Aids:
- Noble gas configurations are very stable.

Quick Tips:
- Look for electronic configurations that, upon losing one electron, achieve a noble gas configuration.

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Application +4 | -1

The pair, in which ions are isoelectronic with

A{l^{3 + }}

is:

View answer and explanation

Correct option: D

{O^{2 - }}\,\,{\text{and}}\,\,M{g^{2 + }}

Option D (

{O^{2 - }}\,{\text{and}}\,M{g^{2 + }})

is correct because both ions have the same number of electrons as

A{l^{3 + }}

.

Solution approach

Correct Answer + Core Reason: ${O^{2 - }}\,{\text{and}}\,M{g^{2 + }}$ . Both ions have 10 electrons, the same as $A{l^{3 + }}$ .

Step-by-Step Solution:

Step 1: Determine the number of electrons in $A{l^{3 + }}$ . Aluminum (Al) has an atomic number of 13, so a neutral Al atom has 13 electrons. $A{l^{3 + }}$ has lost 3 electrons, so it has 13 - 3 = 10 electrons.

Step 2: Check each option to see which pair of ions also has 10 electrons.

Step 3: Analyze Option A: $B{r^ - }$ has 36 electrons (35 + 1) and $B{e^{2 + }}$ has 2 electrons (4 - 2).

Step 4: Analyze Option B: $C{l^ - }$ has 18 electrons (17 + 1) and $L{i^ + }$ has 2 electrons (3 - 1).

Step 5: Analyze Option C: ${S^{2 - }}$ has 18 electrons (16 + 2) and ${K^ + }$ has 18 electrons (19 - 1).

Step 6: Analyze Option D: ${O^{2 - }}$ has 10 electrons (8 + 2) and $M{g^{2 + }}$ has 10 electrons (12 - 2).

Step 7: Conclusion: Only Option D has both ions with 10 electrons.

Why other options are wrong:
- Option A: $B{r^ - }$ and $B{e^{2 + }}$ do not have 10 electrons each. $B{r^ - }$ has 36 and $B{e^{2 + }}$ has 2.
- Option B: $C{l^ - }$ and $L{i^ + }$ do not have 10 electrons each. $C{l^ - }$ has 18 and $L{i^ + }$ has 2.
- Option C: ${S^{2 - }}$ and ${K^ + }$ do not have 10 electrons each. They both have 18 electrons.
- Option D: ${O^{2 - }}$ and $M{g^{2 + }}$ both have 10 electrons, making them isoelectronic with $A{l^{3 + }}$ .

Key Concepts:
- Isoelectronic species
- Number of electrons in ions
- Atomic number

Memory Aids:
- Anions gain electrons (negative charge), cations lose electrons (positive charge).

Quick Tips:
- Remember to add electrons for negative ions and subtract electrons for positive ions when determining the number of electrons.

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Understanding +4 | -1

Oxygen is divalent, but sulphur exhibits variable valency of

2,\mathrm{ }4\mathrm{ } and \mathrm{ }6

, because:

View answer and explanation

Correct option: D

Of the presence of

d

-orbitals in sulphur

Option D (Of the presence of

d

-orbitals in sulphur) is correct because sulphur has available d-orbitals in its valence shell, allowing it to expand its octet and exhibit variable valency.

Solution approach

Correct Answer + Core Reason: Of the presence of $d$ -orbitals in sulphur. Sulphur can expand its octet due to available $d$ -orbitals.

Step-by-Step Solution:

Step 1: Compare the electronic configurations of oxygen and sulphur.
- Oxygen (O): $1s^2 2s^2 2p^4$ . It has no available $d$ -orbitals in its valence shell.
- Sulphur (S): $1s^2 2s^2 2p^6 3s^2 3p^4$ . It has available $3d$ -orbitals in its valence shell.

Step 2: Explain how $d$ -orbitals allow for variable valency: The presence of $d$ -orbitals allows sulphur to expand its octet and form more than four bonds, leading to valencies of 4 and 6 in addition to 2.

Why other options are wrong:
- Option A: Sulphur is less electronegative than oxygen: Electronegativity differences do not directly explain variable valency.
- Option B: Sulphur is bigger atom than oxygen: Size difference does not directly explain variable valency.
- Option C: Ionisation potential of sulphur is more than oxygen: Ionization potential is related to the ease of losing electrons, not the ability to form more bonds.

Key Concepts:
- Electronic configuration and valence electrons.
- Expansion of octet.
- Availability of $d$ -orbitals.

Memory Aids:
- Elements in the third period and beyond can expand their octet due to the availability of $d$ -orbitals.

Quick Tips:
- Consider the electronic configuration and the availability of $d$ -orbitals when explaining variable valency.

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Remembering +4 | -1

The atomic orbitals used for filling in sixth period of periodic table are

View answer and explanation

Correct option: C

6 s, 4 f, 5 d, 6 p

Option C (

6 s, 4 f, 5 d, 6 p

) is correct because the filling of orbitals in the sixth period follows the order 6s, then 4f, then 5d, and finally 6p.

Solution approach

Correct Answer + Core Reason: $6 s, 4 f, 5 d, 6 p$ . The filling of orbitals in the sixth period follows the order 6s, 4f, 5d, and 6p according to the Aufbau principle.

Step-by-Step Solution:

Step 1: Recall the Aufbau principle. The Aufbau principle states that electrons first fill the lowest energy orbitals available.

Step 2: Determine the order of filling orbitals based on the (n+l) rule. The (n+l) rule states that orbitals with lower (n+l) values are filled first. If two orbitals have the same (n+l) value, the orbital with the lower n value is filled first.

Step 3: Apply the (n+l) rule to the orbitals in the sixth period:
- 6s: n+l = 6+0 = 6
- 4f: n+l = 4+3 = 7
- 5d: n+l = 5+2 = 7
- 6p: n+l = 6+1 = 7

Step 4: Determine the filling order. Based on the (n+l) rule, the filling order is 6s, then 4f, then 5d, and finally 6p.

Why other options are wrong:
- **Option A ( $6 s, 6 p, 6 d, 6 f$ ): This is incorrect because it does not follow the correct filling order based on the (n+l) rule.
- Option B ( $6 s, 6 p, 4 d, 5 f$ ): This is incorrect because it does not follow the correct filling order based on the (n+l) rule.
- Option D ( $6s,6p,6d,5f$ ): This is incorrect because it does not follow the correct filling order based on the (n+l) rule.

Key Concepts:
- Aufbau principle
- (n+l) rule
- Electronic configuration

Memory Aids:
- Remember the order of filling orbitals based on the (n+l) rule.

Quick Tips:**
- Know the Aufbau principle and the (n+l) rule to determine the correct filling order of orbitals.

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Remembering +4 | -1

Among the following, basic oxide is:

View answer and explanation

Correct option: C

CaO

Option C (

CaO

) is correct because oxides of alkaline earth metals are basic.

Solution approach

Correct Answer + Core Reason: $CaO$ is a basic oxide because it is an oxide of an alkaline earth metal.

Step-by-Step Solution:

Step 1: Recall that oxides of metals are generally basic, while oxides of non-metals are acidic or neutral.

Step 2: Identify the nature of the elements in each oxide.

Step 3: $S{O_3}$ is an oxide of a non-metal (Sulfur) and is acidic.

Step 4: $Si{O_2}$ is an oxide of a non-metal (Silicon) and is acidic.

Step 5: $CaO$ is an oxide of a metal (Calcium) and is basic.

Step 6: $A{l_2}{O_3}$ is an amphoteric oxide.

Step 7: Conclusion: $CaO$ is the basic oxide.

Why other options are wrong:
- Option A: $S{O_3}$ is an acidic oxide.
- Option B: $Si{O_2}$ is an acidic oxide.
- Option C: $CaO$ is the correct answer as it is a basic oxide.
- Option D: $A{l_2}{O_3}$ is an amphoteric oxide.

Key Concepts:
- Basic Oxides: Oxides of metals.
- Acidic Oxides: Oxides of non-metals.
- Amphoteric Oxides

Memory Aids:
- Metals form bases, non-metals form acids.

Quick Tips:
- Remember the position of elements in the periodic table to determine their metallic or non-metallic character.

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Understanding +4 | -1

Which pair of elements belongs to the same period in periodic table

View answer and explanation

Correct option: A

Ca

,

Ti

Option A (

Ca

,

Ti

) is correct because both Calcium (Ca) and Titanium (Ti) belong to the 4th period of the periodic table.

Solution approach

Correct Answer + Core Reason: $Ca$ , $Ti$ . Both Calcium and Titanium belong to the 4th period.

Step-by-Step Solution:

Step 1: Identify the period of each element:
- Ca (Calcium): Period 4
- Ti (Titanium): Period 4
- Mg (Magnesium): Period 3
- Sn (Tin): Period 5
- Li (Lithium): Period 2
- Ga (Gallium): Period 4
- Na (Sodium): Period 3
- La (Lanthanum): Period 6

Step 2: Determine which pair belongs to the same period.

Why other options are wrong:
- Option B: $Mg$ , $Sn$ : Magnesium is in period 3, and Tin is in period 5.
- Option C: $Li$ , $Ga$ : Lithium is in period 2, and Gallium is in period 4.
- Option D: $Na$ , $La$ : Sodium is in period 3, and Lanthanum is in period 6.

Key Concepts:
- Periodic table and periods.

Memory Aids:
- Periods are horizontal rows in the periodic table.

Quick Tips:
- Refer to a periodic table to quickly identify the period of elements.

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Application +4 | -1

The formation of

{O}^{2-}(\mathrm{g})

starting from

{O}(\mathrm{g})

is endothermic by

603 \mathrm{~kJ} \mathrm{~mol}^{-1}

. If electron affinity of

{O}(\mathrm{g})

is

-141 \mathrm{~kJ} \mathrm{~mol}^{-1}

, the second electron affinity of oxygen would be:

View answer and explanation

Correct option: A

+744 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option A (+744 kJ/mol) is correct because it correctly calculates the second electron affinity by considering the overall endothermic process and the first electron affinity.

Solution approach

Correct Answer + Core Reason: $+744 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . This is the second electron affinity of oxygen.

Step-by-Step Solution:

Step 1: Define the process.
$O(g) + e^- \rightarrow O^-(g) \quad \Delta H_1 = -141 \text{ kJ/mol}$ (First electron affinity)
$O^-(g) + e^- \rightarrow O^{2-}(g) \quad \Delta H_2 = ?$ (Second electron affinity)
$O(g) + 2e^- \rightarrow O^{2-}(g) \quad \Delta H_{total} = +603 \text{ kJ/mol}$

Step 2: Apply Hess's Law.
$\Delta H_{total} = \Delta H_1 + \Delta H_2$
$603 \text{ kJ/mol} = -141 \text{ kJ/mol} + \Delta H_2$

Step 3: Solve for the second electron affinity.
$\Delta H_2 = 603 \text{ kJ/mol} + 141 \text{ kJ/mol} = 744 \text{ kJ/mol}$

Why other options are wrong:
- Option B: $-744 \mathrm{~kJ} \mathrm{~mol}^{-1}$ : This has the wrong sign. The second electron affinity is endothermic.
- Option C: $+462 \mathrm{~kJ} \mathrm{~mol}^{-1}$ : This is likely a calculation error, possibly subtracting 141 from 603 instead of adding.
- Option D: $-462 \mathrm{~kJ} \mathrm{~mol}^{-1}$ : This has the wrong sign and is likely a calculation error.

Key Concepts:
- Electron affinity
- Hess's Law

Memory Aids:
- Remember that the second electron affinity is usually positive (endothermic).

Quick Tips:
- Always consider the sign of electron affinity values.
- Use Hess's Law to relate the overall enthalpy change to individual steps.

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Understanding +4 | -1

The formation of the oxide ion

{\mathrm{O}^{2 - }}\left( g \right)

requires first an exothermic and then an endothermic step as shown below:

\mathrm{O}\left( \mathrm{g} \right) + {e^ - } = {\mathrm{O}^ - }\left( \mathrm{g} \right);\mathrm{\;}\Delta H^\circ = - 142\mathrm{\;kJmo}{1^{ - 1}}

,

\mathrm{O}{\left( \mathrm{g} \right)^ - } + {e^ - } = {\mathrm{O}^{2 - }}\left( \mathrm{g} \right);\Delta H^\circ = 844\mathrm{\;kJmo}{1^{ - 1}}

This is because

View answer and explanation

Correct option: C

{\text{O}^ - }

ion will tend to resist the addition of another electron

Option C (

{\text{O}^ - }

ion will tend to resist the addition of another electron) is correct because adding an electron to a negatively charged ion (

{\text{O}^ - }

) requires energy to overcome the electrostatic repulsion, making the process endothermic.

Solution approach

Correct Answer + Core Reason: ${\text{O}^ - }$ ion will tend to resist the addition of another electron. Adding an electron to a negatively charged ion requires energy to overcome electrostatic repulsion.

Step-by-Step Solution:

Step 1: Analyze the given reactions:
$\mathrm{O}\left( \mathrm{g} \right) + {e^ - } = {\mathrm{O}^ - }\left( \mathrm{g} \right);\mathrm{\;}\Delta H^\circ = - 142\mathrm{\;kJmo}{1^{ - 1}}$ (Exothermic)
$\mathrm{O}{\left( \mathrm{g} \right)^ - } + {e^ - } = {\mathrm{O}^{2 - }}\left( \mathrm{g} \right);\Delta H^\circ = 844\mathrm{\;kJmo}{1^{ - 1}}$ (Endothermic)

Step 2: Explain the exothermic step: The first step is exothermic because the neutral oxygen atom has an affinity for an electron, releasing energy upon gaining it.

Step 3: Explain the endothermic step: The second step is endothermic because adding an electron to the already negatively charged $\mathrm{O}^-$ ion requires energy to overcome the electrostatic repulsion between the negative ion and the electron.

Step 4: Conclusion: The ${\text{O}^ - }$ ion resists the addition of another electron due to electrostatic repulsion.

Why other options are wrong:
- Option A: While oxygen is electronegative, this doesn't directly explain why the second electron addition is endothermic. Electronegativity relates to the attraction of electrons in a bond, not necessarily in forming an ion.
- Option B: Oxygen having high electron affinity only explains the first exothermic step, not the second endothermic step.
- Option D: While the ${\text{O}^ - }$ ion is larger than the oxygen atom, this size difference doesn't directly explain the endothermic nature of the second electron addition. The primary reason is electrostatic repulsion.

Key Concepts:
- Electron affinity
- Electrostatic repulsion
- Endothermic and exothermic reactions

Memory Aids:
- Adding an electron to a neutral atom is usually exothermic, but adding an electron to a negative ion is usually endothermic due to repulsion.

Quick Tips:
- Consider the charge of the ion when determining whether electron addition will be endothermic or exothermic.

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Remembering +4 | -1

3 d

-transition series of elements starts with

\mathrm{Sc}

and ends with

\mathrm{Zn}

. The electronic configuration of

\mathrm{Sc}

and

\mathrm{Zn}

are respectively.

View answer and explanation

Correct option: B

[\mathrm{Ar}] 3 d^1 4 s^2

and

[\mathrm{Ar}] 3 d^{10} 4 s^2

Option B is correct because Scandium (Sc) has the electronic configuration [Ar]3d¹4s², and Zinc (Zn) has the electronic configuration [Ar]3d¹⁰4s².

Solution approach

Correct Answer + Core Reason: $[\mathrm{Ar}] 3 d^1 4 s^2$ and $[\mathrm{Ar}] 3 d^{10} 4 s^2$ . These are the correct electronic configurations for Sc and Zn respectively.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Scandium (Sc).\nScandium (Sc) has an atomic number of 21. Its electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d¹, which can be abbreviated as [Ar]3d¹4s².

Step 2: Determine the electronic configuration of Zinc (Zn).\nZinc (Zn) has an atomic number of 30. Its electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰, which can be abbreviated as [Ar]3d¹⁰4s².

Step 3: Conclusion.\nThe electronic configurations of Sc and Zn are [Ar]3d¹4s² and [Ar]3d¹⁰4s² respectively.

Why other options are wrong:
- Option A: Incorrect because the electronic configuration of Sc is not [Ar]3d¹4s¹ and the electronic configuration of Zn is not [Ar]3d¹⁰4s¹.
- Option B: Correct because the electronic configuration of Sc is [Ar]3d¹4s² and the electronic configuration of Zn is [Ar]3d¹⁰4s².
- Option C: Incorrect because the electronic configuration of Sc is not [Ar]3d²4s¹ and the electronic configuration of Zn is not [Ar]3d⁹4s³.
- Option D: Incorrect because the electronic configuration of Sc is not [Ar]3d³4s² and the electronic configuration of Zn is not [Ar]3d⁸4s².

Key Concepts:
- Electronic configuration
- Transition elements

Memory Aids:
- Remember the order of filling of orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.

Quick Tips:
- Review the electronic configurations of the first-row transition elements.

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Analyzing +4 | -1

Which is mismatched regarding the position of the element as given below?

View answer and explanation

Correct option: D

Both (1) & (3)

Option D (Both (1) & (3)) is correct because element X (Z=89) belongs to d-block and 7th period and element Z(Z=115) belongs to p-block and 7th period.

Solution approach

Correct Answer + Core Reason: Both (1) & (3). X(Z=89) is in the d-block, and Z(Z=115) is in the p-block.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each element.

Step 2: Identify the block and period of each element based on its electronic configuration.

Step 3: X(Z=89): Actinium (Ac). Electronic configuration: $[Rn] 6d^1 7s^2$ . It belongs to the d-block and 7th period.

Step 4: Y(Z=100): Fermium (Fm). Electronic configuration: $[Rn] 5f^{12} 7s^2$ . It belongs to the f-block and 7th period.

Step 5: Z(Z=115): Moscovium (Mc). Electronic configuration: $[Rn] 5f^{14} 6d^{10} 7s^2 7p^3$ . It belongs to the p-block and 7th period.

Step 6: Therefore, X(Z=89) is mismatched as it is given as f-block, $6^{th}$ period, and Z(Z=115) is mismatched as it is given as d-block, $7^{th}$ period.

Why other options are wrong:
- Option A: $X(Z=89)-f$ block, $6^{\text {th }}$ period: Incorrect, X is in d-block, 7th period.
- Option B: $Y(Z=100)-f$ block, $7^{\text {th }}$ period: Correct.
- Option C: $Z(Z=115)-d$ block, $7^{\text {th }}$ period: Incorrect, Z is in p-block, 7th period.

Key Concepts:
- Electronic Configuration
- Block and Period in Periodic Table

Memory Aids:
- s-block: Groups 1 and 2
- p-block: Groups 13 to 18
- d-block: Groups 3 to 12
- f-block: Lanthanides and Actinides

Quick Tips:
- Knowing the electronic configuration helps determine the block and period.

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Understanding +4 | -1

Which of the following acts sometimes as a metal and sometimes as a non-metal?

View answer and explanation

Correct option: D

\mathrm{At}

The correct answer is D (At) because Astatine (At) is a metalloid, exhibiting properties of both metals and non-metals.

Solution approach

Correct Answer + Core Reason: $\mathrm{At}$ . Astatine is a metalloid, exhibiting properties intermediate between metals and non-metals.

Step-by-Step Solution:

Step 1: Recall the properties of metals, non-metals, and metalloids.

Step 2: Metals are typically shiny, conductive, and malleable. Non-metals are generally brittle and poor conductors.

Step 3: Metalloids (also known as semi-metals) exhibit properties of both metals and non-metals. Astatine is a radioactive metalloid.

Why other options are wrong:
- Option A: $\mathrm{Hg}$ (Mercury) is a metal (though it's a liquid at room temperature).
- Option B: $\mathrm{Cl}$ (Chlorine) is a non-metal.
- Option C: $\mathrm{K}$ (Potassium) is a metal.

Key Concepts:
- Metals, Non-metals, and Metalloids
- Properties of Elements

Memory Aids:
- Metalloids are located along the staircase line in the periodic table.

Quick Tips:
- Know the general properties and locations of metals, non-metals, and metalloids on the periodic table.

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Analyzing +4 | -1

The

\mathrm{I}{\mathrm{P}_1},\mathrm{I}{\mathrm{P}_2},\mathrm{I}{\mathrm{P}_3},\mathrm{I}{\mathrm{P}_4},\mathrm{ and } \, {\mathrm{IP}_5}

of an element are

7.1,\mathrm{ }14.3,\mathrm{ }34.5

,

46.8

,

162.2 \, \, \text{eV}

respectively. The element is likely to be:

View answer and explanation

Correct option: B

\mathrm{Si}

The element is most likely Silicon (Si) because there is a large jump in ionization energy between IP4 and IP5, indicating 4 valence electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{Si}$ . The large jump between IP4 and IP5 indicates 4 valence electrons.

Step-by-Step Solution:

Step 1: Analyze the given ionization potential (IP) values.

Step 2: Look for a significant jump in IP values. A large jump indicates the removal of an electron from a core shell, which requires much more energy than removing a valence electron.

Step 3: Observe that the largest jump occurs between IP4 (46.8 eV) and IP5 (162.2 eV). This suggests that the element has 4 valence electrons.

Step 4: Consider the electronic configurations of the given elements:
- Na: 1 valence electron
- Si: 4 valence electrons
- F: 7 valence electrons
- Ca: 2 valence electrons

Step 5: Since the element has 4 valence electrons, it is most likely Silicon (Si).

Why other options are wrong:
- Option A: Na has only one valence electron. The ionization energy would increase significantly after removing that one electron, but we see a large jump after the 4th ionization.
- Option B: This is the correct answer. Silicon has 4 valence electrons.
- Option C: F has 7 valence electrons. We would expect a large jump after the 7th ionization, not the 4th.
- Option D: Ca has 2 valence electrons. We would expect a large jump after the 2nd ionization, not the 4th.

Key Concepts:
- Ionization Potential (IP)
- Valence Electrons
- Electronic Configuration

Memory Aids:
- Large jump in IP indicates the number of valence electrons.

Quick Tips:
- Look for the largest jump in ionization energies to determine the number of valence electrons.

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Analyzing +4 | -1

Which is correct regarding the position of the element as given below?

View answer and explanation

Correct option: A

X(Z=60) - f

-block,

6^{\text {th}}

period

Option A (

X(Z=60) - f

-block,

6^{\text {th}}

period) is correct because element X with Z=60 (Neodymium) belongs to f-block and 6th period.

Solution approach

Correct Answer + Core Reason: $X(Z=60) - f$ -block, $6^{\text {th}}$ period. Neodymium (Nd) with Z=60 is a lanthanide and belongs to the f-block and 6th period.

Step-by-Step Solution:

Step 1: Identify the element based on its atomic number.

Step 2: Determine the electronic configuration of the element.

Step 3: Identify the block and period based on the electronic configuration.

Step 4: X(Z=60): Neodymium (Nd). Electronic configuration: $[Xe] 4f^4 6s^2$ . It belongs to the f-block and 6th period.

Step 5: Y(Z=82): Lead (Pb). Electronic configuration: $[Xe] 4f^{14} 5d^{10} 6s^2 6p^2$ . It belongs to the p-block and 6th period.

Step 6: W(Z=101): Mendelevium (Md). Electronic configuration: $[Rn] 5f^{13} 7s^2$ . It belongs to the f-block and 7th period.

Step 7: Q(Z=48): Cadmium (Cd). Electronic configuration: $[Kr] 4d^{10} 5s^2$ . It belongs to the d-block and 5th period.

Why other options are wrong:
- Option B: $Y(Z=82)- p$ -block, $7^{\text {th}}$ period: Incorrect, Y is in p-block, 6th period.
- Option C: $W(Z=101)-d$ -block, $7^{\text {th}}$ period: Incorrect, W is in f-block, 7th period.
- Option D: $Q(Z=48) - s$ -block, $5^{\text {th}}$ period: Incorrect, Q is in d-block, 5th period.

Key Concepts:
- Electronic Configuration
- Block and Period in Periodic Table

Memory Aids:
- Lanthanides and Actinides belong to the f-block.

Quick Tips:
- Knowing the electronic configuration helps determine the block and period.

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Analyzing +4 | -1

The incorrect order of the radii of the species is

View answer and explanation

Correct option: D

C{l^ - } < {K^ + } < C{a^{2 + }}

Option D (

C{l^ - } < {K^ + } < C{a^{2 + }}

) is the incorrect order because isoelectronic species with greater nuclear charge have smaller radii.

Solution approach

Correct Answer + Core Reason: $C{l^ - } < {K^ + } < C{a^{2 + }}$ . The order is incorrect because for isoelectronic species, ionic size decreases with increasing nuclear charge.

Step-by-Step Solution:

Step 1: Identify the species: $C{l^-}$ , ${K^+}$ , and $C{a^{2+}}$ .

Step 2: Determine the number of electrons in each species: $C{l^-}$ (18 electrons), ${K^+}$ (18 electrons), and $C{a^{2+}}$ (18 electrons). They are isoelectronic.

Step 3: Determine the number of protons in each species: $C{l^-}$ (17 protons), ${K^+}$ (19 protons), and $C{a^{2+}}$ (20 protons).

Step 4: For isoelectronic species, the ionic radius decreases with increasing nuclear charge (number of protons). Therefore, the correct order should be ${C{l^-} > {K^+} > C{a^{2+}}}$ .

Step 5: Compare the given order with the correct order. The given order $C{l^ - } < {K^ + } < C{a^{2 + }}$ is incorrect.

Why other options are wrong:
- Option A: $O < {O^ - },{O^{2 - }}$ is correct. As more electrons are added, the electron-electron repulsion increases, and the ionic radius increases.
- Option B: $N < P < As$ is correct. As you go down a group, the atomic radius increases due to the addition of new electron shells.
- Option C: $Fe > F{e^{2 + }} > F{e^{3 + }}$ is correct. As more electrons are removed, the effective nuclear charge increases, and the ionic radius decreases.

Key Concepts:
- Isoelectronic species
- Ionic radius trend
- Effective nuclear charge

Memory Aids:
- For isoelectronic species: more protons = smaller radius.

Quick Tips:
- When comparing ionic radii of isoelectronic species, focus on the number of protons. The ion with more protons will have a smaller radius.

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Remembering +4 | -1

The enthalpy change accompanying in the given reaction,

X(g)+e^{-} \longrightarrow X^{-}(g)

is defined as

View answer and explanation

Correct option: A

Electron gain enthalpy

Option A (Electron gain enthalpy) is correct because the reaction describes the addition of an electron to a gaseous atom, which is the definition of electron gain enthalpy.

Solution approach

Correct Answer + Core Reason: Electron gain enthalpy. The reaction describes the addition of an electron to a gaseous atom.

Step-by-Step Solution:

Step 1: Understand the definition of electron gain enthalpy.

Step 2: Electron gain enthalpy is the enthalpy change when an electron is added to a neutral gaseous atom to form a negative ion.

Step 3: The given reaction is $X(g) + e^- \longrightarrow X^-(g)$ .

Step 4: This reaction represents the addition of an electron to a gaseous atom X.

Step 5: Therefore, the enthalpy change accompanying this reaction is the electron gain enthalpy.

Why other options are wrong:
- Option B: Electron loss enthalpy: This refers to the removal of an electron, not the addition.
- Option C: Ionisation enthalpy: This refers to the removal of an electron, not the addition.
- Option D: Both (1) and (3): Ionisation enthalpy is incorrect.

Key Concepts:
- Electron gain enthalpy
- Ionization enthalpy

Memory Aids:
- Electron gain = electron gain enthalpy.

Quick Tips:
- Remember the definitions of electron gain enthalpy and ionization enthalpy.

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Application +4 | -1

Predict the position of an element has the electronic configuration

1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1

.

View answer and explanation

Correct option: A

Period 4, group 6

Option A (Period 4, group 6) is correct. The highest principal quantum number is 4 (period 4), and the number of d electrons + s electrons is 5+1 = 6 (group 6).

Solution approach

Correct Answer + Core Reason: Period 4, group 6. The electronic configuration indicates the element is in the 4th period and group 6.

Step-by-Step Solution:

Step 1: Identify the electronic configuration: $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1$ .

Step 2: Determine the period: The highest principal quantum number (n) is 4, so the element is in the 4th period.

Step 3: Determine the block: The last electron enters the d-orbital, so the element is in the d-block.

Step 4: Determine the group number: For d-block elements, the group number is the sum of the number of electrons in the (n-1)d orbitals and the ns orbitals. In this case, it's 5 (from $3d^5$ ) + 1 (from $4s^1$ ) = 6.

Step 5: Therefore, the element is in Period 4 and Group 6.

Why other options are wrong:
- Option B (Period 6, group 4): The period and group are incorrect. The element is in the 4th period and group 6.
- Option C (Period 3, group 1): The period and group are incorrect. The element is in the 4th period and group 6.
- Option D (Period 4, group 5): The group number is incorrect. It should be 6, not 5.

Key Concepts:
- Electronic configuration
- Periodic table structure
- Group, period, and block identification

Memory Aids:
- Remember the rules for determining group numbers based on electronic configuration.

Quick Tips:
- The period number is the highest principal quantum number (n).
- For d-block elements, add the number of d and s electrons in the outermost shell to find the group number.

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