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

Oxygen has high electro affinity

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

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

1 Medium

1 of 175Question permalink

Remembering +4 | -1

The elements with zero electron affinity are

View answer and explanation

Correct option: B

Beryllium and Helium

Option B (Beryllium and Helium) is correct because both beryllium and helium have stable, filled electron configurations, making it energetically unfavorable to add an electron.

Solution approach

Correct Answer + Core Reason: Beryllium and Helium. These elements have stable, filled electron configurations.

Step-by-Step Solution:

Step 1: Understand electron affinity.

Step 2: Electron affinity is the energy change when an electron is added to a neutral gaseous atom.

Step 3: Elements with stable, filled or half-filled electron configurations tend to have low or zero electron affinity.

Step 4: Beryllium (Be) has the electronic configuration $1s^2 2s^2$ (filled s-orbital).

Step 5: Helium (He) has the electronic configuration $1s^2$ (filled s-orbital).

Step 6: Boron (B) and Carbon (C) have partially filled p-orbitals and have non-zero electron affinities.

Step 7: Lithium (Li) and Sodium (Na) have low ionization energies and positive electron affinities.

Step 8: Fluorine (F) and Chlorine (Cl) have high electron affinities.

Why other options are wrong:
- Option A: Boron and Carbon: These elements have non-zero electron affinities.
- Option C: Lithium and Sodium: These elements have positive electron affinities.
- Option D: Fluorine and Chlorine: These elements have high electron affinities.

Key Concepts:
- Electron affinity
- Electronic configuration
- Stability of filled orbitals

Memory Aids:
- Filled orbitals lead to low or zero electron affinity.

Quick Tips:
- Consider the electronic configuration of the elements when determining electron affinity.

2 Medium

2 of 175Question permalink

Analyzing +4 | -1

S-I:

\mathrm{Cu}^{+2}

paramagnetic.

S-II: All the orbitals present in

\mathrm{Cu}^{+2}

ion are doubly occupied.

View answer and explanation

Correct option: C

Only I is true

Option C is correct because S-I is true (Cu+2 is paramagnetic due to the presence of unpaired electrons), but S-II is false (not all orbitals in Cu+2 are doubly occupied).

Solution approach

Correct Answer + Core Reason: Only I is true. $\mathrm{Cu}^{+2}$ is paramagnetic, but not all its orbitals are doubly occupied.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Cu: [Ar] $3d^{10}4s^1$

Step 2: Determine the electronic configuration of $\mathrm{Cu}^{+2}$ : [Ar] $3d^9$

Step 3: Analyze Statement I: $\mathrm{Cu}^{+2}$ has 9 d-electrons, meaning there is one unpaired electron. Therefore, it is paramagnetic. Statement I is true.

Step 4: Analyze Statement II: Since $\mathrm{Cu}^{+2}$ has a $3d^9$ configuration, not all orbitals are doubly occupied. There is one singly occupied d-orbital. Statement II is false.

Why other options are wrong:
- Option A: Both statements are not true.
- Option B: Both statements are not false.
- Option D: Statement II is not true.

Key Concepts:
- Electronic configuration
- Paramagnetism and diamagnetism
- Hund's rule

Memory Aids:
- Paramagnetic: Unpaired electrons
- Diamagnetic: All electrons paired

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

3 Medium

3 of 175Question permalink

Remembering +4 | -1

The law of Octaves is not applicable to:

View answer and explanation

Correct option: D

Se,Te,As

Option D (Se,Te,As) is correct because the Law of Octaves, proposed by Newlands, states that every eighth element has similar properties. This law was applicable up to Calcium (Ca). Se, Te, and As are beyond Calcium, making the law inapplicable.

Solution approach

Correct Answer + Core Reason: $Se,Te,As$ . The law of Octaves is not applicable to these elements because they are beyond Calcium.

Step-by-Step Solution:

Step 1: Recall the Law of Octaves.\nThe Law of Octaves states that every eighth element has similar properties when arranged in order of increasing atomic weight. This law was applicable up to Calcium (Ca).

Step 2: Check the applicability of the Law of Octaves for each set.\nA. B, N, C: These elements are before Ca, but the Law of Octaves doesn't directly apply to such a small set.\nB. Be, Mg, Ca: These elements follow the Law of Octaves (alkaline earth metals).\nC. Ar, K, Ca: These elements are near the limit of the Law of Octaves.\nD. Se, Te, As: These elements are beyond Ca, so the Law of Octaves is not applicable.

Why other options are wrong:
- Option A: B, N, and C are lighter elements, but the Law of Octaves doesn't clearly apply to just these three.
- Option B: Be, Mg, and Ca follow the Law of Octaves as they are in the same group.
- Option C: Ar, K, and Ca are near the limit of the Law of Octaves' applicability.

Key Concepts:
- Law of Octaves
- Periodic table

Memory Aids:
- Remember that the Law of Octaves was applicable up to Calcium.

Quick Tips:
- Know the limitations of early attempts to classify elements.

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Understanding +4 | -1

Assertion (A): Mendeleev's arranged elements in horizontal rows and vertical columns.
Reason (R): Mendeleev's ignored the order of atomic weight thinking that the atomic measurements might be incorrect.

View answer and explanation

Correct option: B

Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion

Option B (Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion) is correct because Mendeleev did arrange elements in rows and columns and sometimes ignored atomic weight order, but the latter doesn't explain the former.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion. Both statements are true, but unrelated.

Step-by-Step Solution:

Step 1: Verify Assertion (A): Mendeleev did arrange elements in horizontal rows and vertical columns.

Step 2: Verify Reason (R): Mendeleev sometimes ignored the order of atomic weight.

Step 3: Determine if Reason (R) explains Assertion (A): Ignoring atomic weight order doesn't explain why he arranged elements in rows and columns.

Why other options are wrong:
- Option A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion): The reason does not explain the assertion.
- Option C (Assertion is correct; Reason is incorrect): Both are correct.
- Option D (Reason is correct; Assertion is incorrect): Both are correct.

Key Concepts:
- Mendeleev's Periodic Table
- Arrangement of Elements
- Atomic Weight

Memory Aids:
- Remember the key features of Mendeleev's periodic table.

Quick Tips:
- Distinguish between different aspects of Mendeleev's work.

5 Medium

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Analyzing +4 | -1

Which of the following electronic configuration does not belong to same block as others?

View answer and explanation

Correct option: C

[{Kr}] 5 \mathrm{s}^2

Option C ([Kr] 5s²) is correct because it represents an s-block element, while the others represent d-block elements.

Solution approach

Correct Answer + Core Reason: $[{Kr}] 5 \mathrm{s}^2$ . This configuration represents an s-block element, while the others represent d-block elements.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations:
- A: $[{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{d}^{10} 6 \mathrm{s}^2$ - The last electron enters the d-orbital (after filling the f-orbital), so it's a d-block element.
- B: $[{Kr}] 4 \mathrm{d}^{10} 5 \mathrm{s}^2$ - The last electron enters the d-orbital, so it's a d-block element.
- C: $[{Kr}] 5 \mathrm{s}^2$ - The last electron enters the s-orbital, so it's an s-block element.
- D: $[{Ar}] 3 \mathrm{d}^6 4 \mathrm{s}^2$ - The last electron enters the d-orbital, so it's a d-block element.

Step 2: Identify the outlier: Configurations A, B, and D represent d-block elements, while configuration C represents an s-block element.

Why other options are wrong:
- Option A: $[{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{d}^{10} 6 \mathrm{s}^2$ : This configuration belongs to the d-block.
- Option B: $[{Kr}] 4 \mathrm{d}^{10} 5 \mathrm{s}^2$ : This configuration belongs to the d-block.
- Option D: $[{Ar}] 3 \mathrm{d}^6 4 \mathrm{s}^2$ : This configuration belongs to the d-block.

Key Concepts:
- Electronic configuration
- Block identification in the periodic table

Memory Aids:
- s-block: Last electron enters the s-orbital.
- p-block: Last electron enters the p-orbital.
- d-block: Last electron enters the d-orbital.
- f-block: Last electron enters the f-orbital.

Quick Tips:
- Focus on the last electron's orbital to determine the block.

6 Medium

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Understanding +4 | -1

Identify the correct order in which the covalent radius of the following elements increases (I)

Ti

(II)

Ca

(III)

Sc

View answer and explanation

Correct option: D

(I), (III), (II)

Option D (I), (III), (II) is correct because covalent radius generally increases down a group and decreases across a period, with exceptions due to electronic configurations and effective nuclear charge.

Solution approach

Correct Answer + Core Reason: (I), (III), (II). Covalent radius increases down a group and decreases across a period.

Step-by-Step Solution:

Step 1: Identify the positions of the elements in the periodic table.
- Ti (Titanium) is in Group 4, Period 4.
- Ca (Calcium) is in Group 2, Period 4.
- Sc (Scandium) is in Group 3, Period 4.

Step 2: Analyze the trend across a period. Generally, covalent radius decreases from left to right across a period due to increasing effective nuclear charge. However, there are exceptions.

Step 3: Compare Ti, Sc, and Ca. Since they are all in the same period (Period 4), we expect Ca to have the largest radius because it is furthest to the left. Ti should have the smallest radius, and Sc should be in between.

Step 4: The order is Ti < Sc < Ca. Therefore, the correct order is (I), (III), (II).

Why other options are wrong:
- Option A: (I), (II), (III): Incorrect. It suggests Ti < Ca < Sc, which is not the correct trend.
- Option B: (III), (II), (I): Incorrect. It suggests Sc < Ca < Ti, which is not the correct trend.
- Option C: (II), (I), (III): Incorrect. It suggests Ca < Ti < Sc, which is not the correct trend.

Key Concepts:
- Periodic trends (covalent radius)
- Effective nuclear charge
- Position of elements in the periodic table

Memory Aids:
- Covalent radius decreases across a period (left to right).
- Covalent radius increases down a group.

Quick Tips:
- Know the general trends of atomic properties in the periodic table.
- Consider the position of the elements in the periodic table.

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Remembering +4 | -1

Which among the following is/are amphoteric oxides

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because

BeO

,

SnO

, and

ZnO

are all known to exhibit amphoteric behavior, reacting with both acids and bases.

Solution approach

Correct Answer + Core Reason: All of these. $BeO$ , $SnO$ , and $ZnO$ are amphoteric oxides.

Step-by-Step Solution:

Step 1: Define amphoteric oxides: Amphoteric oxides are those that can react with both acids and bases to form salts and water.

Step 2: Consider the given options:
- $BeO$ (Beryllium oxide) is an amphoteric oxide.
- $SnO$ (Tin(II) oxide) is an amphoteric oxide.
- $ZnO$ (Zinc oxide) is an amphoteric oxide.

Step 3: $BeO$ , $SnO$ , and $ZnO$ react with both acids and bases.

Step 4: Conclusion: All of the given oxides are amphoteric.

Why other options are wrong:
- Option A: $BeO$ is amphoteric, but the question asks for all amphoteric oxides among the given options.
- Option B: $SnO$ is amphoteric, but the question asks for all amphoteric oxides among the given options.
- Option C: $ZnO$ is amphoteric, but the question asks for all amphoteric oxides among the given options.
- Option D: All of these is the correct answer as all the given oxides are amphoteric.

Key Concepts:
- Amphoteric Oxides

Memory Aids:
- Remember common amphoteric oxides: $BeO$ , $ZnO$ , $A{l_2}{O_3}$ , $SnO$ , $PbO$ .

Quick Tips:
- Learn the common examples of amphoteric oxides.

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8 of 175Question permalink

Analyzing +4 | -1

Which one of the following is smallest in size

View answer and explanation

Correct option: C

Na

Option C (

\mathrm{N}{\mathrm{a}^ + }

) is correct because it has the highest nuclear charge among the isoelectronic species, leading to the smallest ionic radius.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}^ + }$ . It has the highest nuclear charge among the isoelectronic species.

Step-by-Step Solution:

Step 1: Determine the number of protons and electrons for each species:
- ${\mathrm{N}^{3 - }}$ : 7 protons, 10 electrons
- ${\mathrm{K}^ + }$ : 19 protons, 18 electrons
- $\mathrm{N}{\mathrm{a}^ + }}$ : 11 protons, 10 electrons
- ${\mathrm{F}^ - }$ : 9 protons, 10 electrons

Step 2: ${\mathrm{N}^{3 - }}$ , $\mathrm{N}{\mathrm{a}^ + }}$ , and ${\mathrm{F}^ - }$ are isoelectronic with 10 electrons. For isoelectronic species, the ionic radius decreases with increasing nuclear charge.

Step 3: Compare the nuclear charges of ${\mathrm{N}^{3 - }}$ , $\mathrm{N}{\mathrm{a}^ + }}$ , and ${\mathrm{F}^ - }}$ : Na (11) > F (9) > N (7).

Step 4: Therefore, $\mathrm{N}{\mathrm{a}^ + }}$ has the smallest ionic radius among these isoelectronic species.

Step 5: ${\mathrm{K}^ + }$ has 18 electrons, which is more than the other species (10 electrons). Therefore, ${\mathrm{K}^ + }$ will be larger than the other species.

Why other options are wrong:
- Option A: ${\mathrm{N}^{3 - }}$ is incorrect because it has the lowest nuclear charge among the isoelectronic species, making it the largest.
- Option B: ${\mathrm{K}^ + }$ is incorrect because it has more electrons than the other species, making it larger.
- Option D: ${\mathrm{F}^ - }$ is incorrect because it has a lower nuclear charge than $\mathrm{N}{\mathrm{a}^ + }$ , making it larger.

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

Memory Aids:
- For isoelectronic species, higher nuclear charge = smaller ionic radius.

Quick Tips:
- When comparing ionic radii, first check if the species are isoelectronic. If so, compare their nuclear charges.

9 Medium

9 of 175Question permalink

Understanding +4 | -1

Chemical properties of

Li

and

Mg

are similar because:

View answer and explanation

Correct option: B

Shows diagonal relationship

The correct answer is B (Shows diagonal relationship) because Lithium (Li) and Magnesium (Mg) exhibit similar chemical properties due to their diagonal relationship in the periodic table.

Solution approach

Correct Answer + Core Reason: Shows diagonal relationship. Li and Mg exhibit similar properties due to their diagonal relationship in the periodic table.

Step-by-Step Solution:

Step 1: Recall the concept of diagonal relationships in the periodic table.

Step 2: Diagonal relationships occur between elements in adjacent groups and periods (e.g., Li and Mg, Be and Al, B and Si).

Step 3: The similarity in charge/size ratio leads to similar polarizing power and thus similar properties.

Why other options are wrong:
- Option A: These belong to same group is incorrect. Li and Mg belong to different groups (Group 1 and Group 2, respectively).
- Option C: Both has same ionization potential is incorrect. Their ionization potentials are different.
- Option D: Both has same electron affinity is incorrect. Their electron affinities are different.

Key Concepts:
- Diagonal Relationship
- Periodic Trends
- Charge/Size Ratio

Memory Aids:
- Remember the diagonal relationships: Li-Mg, Be-Al, B-Si.

Quick Tips:
- Understand the factors that contribute to diagonal relationships.

10 Medium

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Analyzing +4 | -1

The ionic radius of ‘Cr’ is minimum in which of the following compounds?

View answer and explanation

Correct option: B

{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}

Option B (

{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}

) is correct because Cr has the highest oxidation state (+6) in this compound, leading to the smallest ionic radius.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ . Higher oxidation state leads to smaller ionic radius due to increased effective nuclear charge.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Cr in each compound.

Step 2: $\mathrm{Cr}{\mathrm{O}_2}$ : Cr has an oxidation state of +4.

Step 3: ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ : Cr has an oxidation state of +6.

Step 4: $\mathrm{Cr}{\mathrm{F}_3}$ : Cr has an oxidation state of +3.

Step 5: $\mathrm{CrC}{\mathrm{l}_3}$ : Cr has an oxidation state of +3.

Step 6: The higher the oxidation state, the smaller the ionic radius because the effective nuclear charge increases, pulling the electrons closer to the nucleus.

Step 7: Therefore, ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ has the smallest ionic radius of Cr.

Why other options are wrong:
- Option A: $\mathrm{Cr}{\mathrm{O}_2}$ has a lower oxidation state (+4) than ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ (+6), hence a larger ionic radius.
- Option C: $\mathrm{Cr}{\mathrm{F}_3}$ has a lower oxidation state (+3) than ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ (+6), hence a larger ionic radius.
- Option D: $\mathrm{CrC}{\mathrm{l}_3}$ has a lower oxidation state (+3) than ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ (+6), hence a larger ionic radius.

Key Concepts:
- Oxidation state
- Ionic radius
- Effective nuclear charge

Memory Aids:
- Higher positive charge, smaller ionic radius.

Quick Tips:
- Determine the oxidation state of the element in each compound. The higher the oxidation state, the smaller the ionic radius.

11 Medium

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Application +4 | -1

If you are given Avogadro's number of atoms of a gas '

X

'. If half of the atoms are converted into

X_{(g)}^{+}

by energy

\Delta {H}

. The IE of

{X}

is

View answer and explanation

Correct option: A

\frac{2 \Delta {H}}{\mathrm{N}_{\mathrm{A}}}

Option A (

\frac{2 \Delta {H}}{\mathrm{N}_{\mathrm{A}}}

) is correct because ionization energy is the energy required to remove an electron from one atom, and the formula correctly scales the total energy by the number of ionized atoms.

Solution approach

Correct Answer + Core Reason: $\frac{2 \Delta {H}}{\mathrm{N}_{\mathrm{A}}}$ . Ionization energy is the energy required per atom, so we divide the total energy by the number of atoms ionized.

Step-by-Step Solution:

Step 1: Define the variables: $\Delta H$ is the total energy, $N_A$ is Avogadro's number, and $IE$ is the ionization energy.

Step 2: Determine the number of atoms ionized. Half of Avogadro's number of atoms are ionized, so the number of atoms ionized is $\frac{N_A}{2}$ .

Step 3: Calculate the ionization energy per atom: $IE = \frac{\Delta H}{\text{number of atoms ionized}} = \frac{\Delta H}{\frac{N_A}{2}} = \frac{2 \Delta H}{N_A}$ .

Why other options are wrong:
- **Option B ( $\frac{2 \mathrm{~N}_{\mathrm{A}}}{\Delta{H}}$ ): This is incorrect because it inverts the relationship between energy and the number of atoms.
- Option C ( $\frac{\Delta {H}}{2 \mathrm{~N}_{\mathrm{A}}}$ ): This is incorrect because it divides by twice Avogadro's number instead of half.
- Option D ( $\frac{N_A}{\Delta H}$ ): This is incorrect because it inverts the relationship between energy and the number of atoms.

Key Concepts:
- Ionization Energy
- Avogadro's Number

Memory Aids:
- Remember that ionization energy is energy per atom.

Quick Tips:**
- Pay attention to the units and the definition of ionization energy.

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Application +4 | -1

Total no. of elements in 6th period and 3rd group of modern periodic table.

View answer and explanation

Correct option: D

15

Option D (15) is correct. The 6th period and 3rd group contains Lanthanum (La) and the 14 Lanthanides, totaling 15 elements.

Solution approach

Correct Answer + Core Reason: 15. The 6th period and 3rd group includes Lanthanum and the 14 Lanthanides.

Step-by-Step Solution:

Step 1: Identify the 6th period and 3rd group in the periodic table.

Step 2: The 3rd group starts with Scandium (Sc) in the 4th period, Yttrium (Y) in the 5th period, and Lanthanum (La) in the 6th period.

Step 3: The 6th period, group 3 contains Lanthanum (La). Following Lanthanum are the 14 Lanthanides (Ce to Lu), which are placed separately at the bottom of the periodic table but belong to the 6th period and are chemically similar to Lanthanum.

Step 4: Therefore, the total number of elements in the 6th period and 3rd group is 1 (La) + 14 (Lanthanides) = 15.

Why other options are wrong:
- Option A (18): 18 is the number of elements in the 4th period, not the 6th period and 3rd group.
- Option B (20): This number doesn't correspond to any specific group and period combination.
- Option C (30): This number is also incorrect and doesn't relate to the number of elements in the specified group and period.

Key Concepts:
- Periodic table structure
- Group and period trends
- Lanthanides

Memory Aids:
- Remember the position of Lanthanides and Actinides in the periodic table.

Quick Tips:
- Understanding the layout of the periodic table is crucial for answering such questions.

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Application +4 | -1

An element having electronic configuration

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

forms

View answer and explanation

Correct option: B

Basic oxide

Option B (Basic oxide) is correct because the electronic configuration indicates the element is an alkali metal, which forms basic oxides.

Solution approach

Correct Answer + Core Reason: Basic oxide. The electronic configuration indicates an alkali metal, which forms basic oxides.

Step-by-Step Solution:

Step 1: Analyze the electronic configuration: $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1$ .

Step 2: Identify the element based on the electronic configuration. The outermost electron is in the 4s orbital, indicating that the element is in the 4th period and group 1 (alkali metal).

Step 3: Recall that alkali metals form basic oxides.

Step 4: Conclude that the element forms a basic oxide.

Why other options are wrong:
- Option A: Acidic oxide: Alkali metals do not form acidic oxides.
- Option C: Amphoteric oxide: Amphoteric oxides are formed by elements like Aluminum and Zinc, not alkali metals.
- Option D: Neutral oxide: Neutral oxides are formed by elements like Carbon and Nitrogen in some cases, not alkali metals.

Key Concepts:
- Electronic configuration and group number
- Nature of oxides of alkali metals
- Basic oxides

Memory Aids:
- Alkali metals form basic oxides

Quick Tips:
- Relate electronic configuration to group number to predict the nature of the oxide.

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Understanding +4 | -1

The ionic radii of

L{i^ + },\,B{e^{ + 2}}

and

B{e^{ + 3}}

follow the order

View answer and explanation

Correct option: D

L{i^ + } > B{e^{ + 2}} > {B^{3 + }}

Option D (

L{i^ + } > B{e^{ + 2}} > {B^{3 + }}

) is correct because for isoelectronic ions, the ionic radius decreases with increasing nuclear charge.

Solution approach

Correct Answer + Core Reason: $L{i^ + } > B{e^{ + 2}} > {B^{3 + }}$ . For isoelectronic ions, the ionic radius decreases with increasing nuclear charge.

Step-by-Step Solution:

Step 1: Identify the species: We have $L{i^ + },\,B{e^{ + 2}}$ and $B{^{3 + }}$ .

Step 2: Determine the number of electrons: All three ions are isoelectronic, meaning they have the same number of electrons. Li (3) loses 1 electron to become Li⁺ (2 electrons). Be (4) loses 2 electrons to become Be²⁺ (2 electrons). B (5) loses 3 electrons to become B³⁺ (2 electrons).

Step 3: Relate ionic radii to nuclear charge: For isoelectronic species, the ionic radius decreases as the nuclear charge (number of protons) increases. This is because the increased nuclear charge pulls the electrons closer to the nucleus, reducing the ionic size.

Step 4: Determine the order of nuclear charge: The nuclear charges are: Li (3), Be (4), B (5). Therefore, the order of increasing nuclear charge is Li < Be < B.

Step 5: Determine the order of ionic radii: Since ionic radii decrease with increasing nuclear charge, the order of ionic radii is $L{i^ + } > B{e^{ + 2}} > {B^{3 + }}$ .

Why other options are wrong:
- Option A: This option suggests that Be²⁺ has the largest ionic radius, which is incorrect because it has a higher nuclear charge than Li⁺.
- Option B: This option incorrectly places B³⁺ as having a larger ionic radius than Be²⁺, which is not true.
- Option C: This option incorrectly places both Be²⁺ and B³⁺ as having larger ionic radii than Li⁺, which is not true.

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

Memory Aids:
- For isoelectronic species, higher nuclear charge = smaller ionic radius.

Quick Tips:
- When comparing isoelectronic species, focus on the number of protons (nuclear charge).

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Understanding +4 | -1

The correct order of ionic radii is represented in

View answer and explanation

Correct option: B

A{l^ + } > A{l^{2 + }} > A{l^{3 + }}

Option B (

A{l^ + } > A{l^{2 + }} > A{l^{3 + }}

) is correct because as the positive charge increases, the ionic radii decreases due to increased effective nuclear charge.

Solution approach

Correct Answer + Core Reason: $A{l^ + } > A{l^{2 + }} > A{l^{3 + }}$ . Increasing positive charge decreases ionic size.

Step-by-Step Solution:

Step 1: Consider the effect of removing electrons from an atom.

Step 2: Removing electrons decreases electron-electron repulsion and increases the effective nuclear charge, which causes the electron cloud to contract.

Step 3: Therefore, the more positive the ion, the smaller its size.

Step 4: Comparing Al⁺, Al²⁺, and Al³⁺, the order of decreasing size is Al⁺ > Al²⁺ > Al³⁺.

Why other options are wrong:
- Option A: This order is incorrect because it suggests that the neutral atom is larger than its negative ions, which is not relevant here as we are dealing with positive ions.
- Option C: This order compares different elements, not different ions of the same element. It's not relevant to the question.
- Option D: This order compares different elements, not different ions of the same element. It's not relevant to the question.

Key Concepts:
- Ionic size
- Effective nuclear charge
- Effect of charge on size

Memory Aids:
- Positive charge = smaller size.

Quick Tips:
- Remember that adding electrons increases size, and removing electrons decreases size.

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Remembering +4 | -1

Which type of element occupy peak position in Lother mayer's curve :

View answer and explanation

Correct option: A

Alkali metal

Option A (Alkali metal) is correct because alkali metals have the largest atomic radii in their respective periods, thus occupying the peak positions in Lothar Meyer's curve.

Solution approach

Correct Answer + Core Reason: Alkali metal. Alkali metals have the largest atomic radii in their respective periods and thus occupy peak positions in Lothar Meyer's curve.

Step-by-Step Solution:

Step 1: Recall Lothar Meyer's curve. Lothar Meyer plotted the atomic volume of elements against their atomic weight and observed periodic trends.

Step 2: Understand the relationship between atomic volume and atomic radii. Atomic volume is related to atomic radii; larger atomic radii correspond to larger atomic volumes.

Step 3: Identify the elements with the largest atomic radii in each period. Alkali metals have the largest atomic radii in their respective periods due to their low effective nuclear charge.

Step 4: Relate the atomic radii to the peak positions in Lothar Meyer's curve. Elements with the largest atomic radii (and thus the largest atomic volumes) occupy the peak positions in Lothar Meyer's curve.

Why other options are wrong:
- Option B (Alkaline earthmetal): Alkaline earth metals have smaller atomic radii than alkali metals.
- Option C (Halogen): Halogens have relatively small atomic radii.
- Option D (Transition elements): Transition elements generally have smaller atomic radii than alkali metals.

Key Concepts:
- Lothar Meyer's curve
- Atomic volume
- Atomic radii
- Alkali metals

Memory Aids:
- Remember that alkali metals are the largest atoms in their periods and thus occupy the peaks in Lothar Meyer's curve.

Quick Tips:
- Know the trends in atomic radii across the periodic table.

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The corerct order of atomic /ionic sizes is

View answer and explanation

Correct option: B

{F^ - } < {O^{2 - }} < {N^{3 - }}

Option B (

{F^ - } < {O^{2 - }} < {N^{3 - }}

) is correct because these are isoelectronic species, and size increases with increasing negative charge (decreasing effective nuclear charge).

Solution approach

Correct Answer + Core Reason: ${F^ - } < {O^{2 - }} < {N^{3 - }}$ . For isoelectronic species, size increases with increasing negative charge.

Step-by-Step Solution:

Step 1: Identify that ${F^-}$ , ${O^{2-}}$ , and ${N^{3-}}$ are isoelectronic (they have the same number of electrons, 10).

Step 2: Recall that for isoelectronic species, ionic size decreases with increasing nuclear charge (more protons pulling on the same number of electrons).

Step 3: Since ${N^{3-}}$ has the fewest protons (7), it has the largest size. ${O^{2-}}$ has 8 protons, and ${F^-}$ has 9 protons, making ${F^-}$ the smallest.

Step 4: Therefore, the correct order is ${F^ - } < {O^{2 - }} < {N^{3 - }}$ .

Why other options are wrong:
- **Option A ( $N < Li < B$ ): This is incorrect. Generally, size decreases across a period (Li > B > N).
- Option C ( $C{a^{2 + }} < {S^{2 - }} < C{l^ - }$ ):** This is incorrect. These are not isoelectronic. $Ca^{2+}$ is smaller than $Cl^-$ and $S^{2-}$ .
- **Option D ( $N{a^ + } < M{g^{2 + }} < C{l^ - }$ ):** This is incorrect. $Na^+$ and $Mg^{2+}$ are isoelectronic, and $Mg^{2+}$ is smaller due to higher nuclear charge. $Cl^-$ has more electrons and is larger.

Key Concepts:
- Isoelectronic Species
- Ionic Size
- Effective Nuclear Charge
- Periodic Trends

Memory Aids:
- For isoelectronic species: more negative charge = larger size.

Quick Tips:
- When comparing ionic sizes, first check if the species are isoelectronic. If so, compare nuclear charges.

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According to IUPAC, total number of groups and periods in the periodic table respectively are

View answer and explanation

Correct option: B

18, 7

Option B (18, 7) is correct because IUPAC recognizes 18 groups and 7 periods in the periodic table.

Solution approach

Correct Answer + Core Reason: 18, 7. The modern periodic table has 18 groups and 7 periods according to IUPAC.

Step-by-Step Solution:

Step 1: The periodic table is organized into vertical columns called groups and horizontal rows called periods.

Step 2: IUPAC (International Union of Pure and Applied Chemistry) recognizes 18 groups, numbered 1 to 18.

Step 3: There are 7 periods in the periodic table, corresponding to the principal quantum numbers 1 to 7.

Why other options are wrong:
- Option A (16, 9): This is incorrect. There are not 16 groups or 9 periods.
- Option C (18, 9): This is incorrect. There are 18 groups but only 7 periods.
- Option D (13, 7): This is incorrect. There are 7 periods, but not 13 groups.

Key Concepts:
- Periodic Table
- Groups
- Periods
- IUPAC Nomenclature

Memory Aids:
- Remember the basic structure of the periodic table: 18 groups and 7 periods.

Quick Tips:
- Familiarize yourself with the organization of the periodic table.

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Diagonal relationship shown by

View answer and explanation

Correct option: D

Be,\,\,Al

Option D (

Be,\,Al

) is correct because Beryllium (Be) and Aluminum (Al) exhibit a diagonal relationship due to similarities in charge/size ratio.

Solution approach

Correct Answer + Core Reason: $Be,\,Al$ . Diagonal relationship arises due to similar polarizing power.

Step-by-Step Solution:

Step 1: Understand the concept of diagonal relationship.

Step 2: Diagonal relationship occurs between elements diagonally adjacent in the periodic table, primarily due to similar charge/size ratios and polarizing power.

Step 3: Consider the elements given in the options.

Step 4: Beryllium (Be) and Aluminum (Al) show a diagonal relationship. Be is in the second period and second group, while Al is in the third period and thirteenth group. Their similar electronegativity and polarizing power lead to similar properties.

Why other options are wrong:
- Option A: $O,\,Cl$ - Oxygen and Chlorine are in the same period (2nd and 3rd respectively) but not diagonally related in terms of properties.
- Option B: $Na,\,Ca$ - Sodium and Calcium are in the same group (1st and 2nd respectively) but not diagonally related in terms of properties.
- Option C: $N,\,P$ - Nitrogen and Phosphorus are in the same group (15th) but not diagonally related in terms of properties.

Key Concepts:
- Diagonal Relationship in Periodic Table
- Polarizing Power
- Charge/Size Ratio

Memory Aids:
- Remember the common diagonal relationships: Li-Mg, Be-Al, B-Si

Quick Tips:
- Look for elements in the second and third periods that are diagonally adjacent.

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Element with which of the following atomic number was named by American Society as Rutherfordium, while by Soviet Society it was named as Kurchatovium?

View answer and explanation

Correct option: B

104

Option B is correct because element 104 was named Rutherfordium by the American Society and Kurchatovium by the Soviet Society, leading to a naming dispute.

Solution approach

Correct Answer + Core Reason: 104. Element 104 was named Rutherfordium by American Society and Kurchatovium by Soviet Society.

Step-by-Step Solution:

Step 1: Recall the naming controversy of element 104.\nElement 104 was synthesized independently by research groups in the United States and the Soviet Union. The American group named it Rutherfordium (Rf), while the Soviet group named it Kurchatovium (Ku).

Step 2: Conclusion.\nThe element with atomic number 104 was named Rutherfordium by the American Society and Kurchatovium by the Soviet Society.

Why other options are wrong:
- Option A: Incorrect because element 108 is Hassium (Hs).
- Option B: Correct because element 104 was named Rutherfordium by American Society and Kurchatovium by Soviet Society.
- Option C: Incorrect because element 114 is Flerovium (Fl).
- Option D: Incorrect because element 110 is Darmstadtium (Ds).

Key Concepts:
- Naming of elements
- Transuranic elements

Memory Aids:
- Remember the naming controversy surrounding element 104.

Quick Tips:
- Review the names and symbols of transuranic elements.

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Which of the following pairs of elements is/are correctly matched?
a) Eka Silicon-

Ge

b) Eka Aluminium-

Ga

c) Eka- Mangnese-

Te

View answer and explanation

Correct option: B

a, b

Option B is correct. Eka-silicon corresponds to Germanium (Ge) and Eka-aluminium corresponds to Gallium (Ga). Eka-Manganese corresponds to Technetium (Tc), not Tellurium (Te).

Solution approach

Correct Answer + Core Reason: a, b. Eka Silicon is Germanium (Ge) and Eka Aluminium is Gallium (Ga). Eka-Manganese is Technetium (Tc).

Step-by-Step Solution:

Step 1: Recall Mendeleev's predictions:
- Eka Silicon: Germanium (Ge)
- Eka Aluminium: Gallium (Ga)
- Eka Manganese: Technetium (Tc)

Step 2: Compare with the given options:\na) Eka Silicon - Ge: Correct\nb) Eka Aluminium - Ga: Correct\nc) Eka- Mangnese - Te: Incorrect (should be Tc)

Step 3: Conclusion: Only a and b are correctly matched.

Why other options are wrong:
- Option A: c is incorrect.
- Option C: c is incorrect.
- Option D: c is incorrect.

Key Concepts:
- Mendeleev's Predictions
- Eka-Silicon, Eka-Aluminium, Eka-Manganese

Memory Aids:
- Remember the elements predicted by Mendeleev.

Quick Tips:
- Know the historical significance of Mendeleev's predictions.

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The period containing max. number of element is:

View answer and explanation

Correct option: D

6^{\text {th }}

The correct answer is D (

6^{\text {th }}

) because the 6th period contains 32 elements, including lanthanides.

Solution approach

Correct Answer + Core Reason: $6^{\text {th }}$ . The 6th period contains the maximum number of elements (32) due to the inclusion of lanthanides.

Step-by-Step Solution:

Step 1: Recall the structure of the periodic table and the number of elements in each period.

Step 2: The first period has 2 elements, the second and third have 8, the fourth and fifth have 18, and the sixth has 32 (including lanthanides).

Step 3: The seventh period is incomplete but would also have 32 elements if complete (including actinides).

Why other options are wrong:
- Option A: The $1^{\text {st }}$ period contains only 2 elements (Hydrogen and Helium).
- Option B: The $18^{\text {th }}$ is not a period number. It refers to group 18 (noble gases).
- Option C: The $3^{\text {rd }}$ period contains 8 elements.

Key Concepts:
- Periodic Table Structure
- Number of Elements in Each Period

Memory Aids:
- Remember the sequence of element numbers in periods: 2, 8, 8, 18, 18, 32.

Quick Tips:
- Memorize the number of elements in each period to quickly answer this type of question.

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Outermost electronic configuration of a group-13 element

E

is

4{s^2}4{p^1}

. The electronic configuration of an element of

p

-block period-five placed diagonally to element,

E

is

View answer and explanation

Correct option: D

[Kr]4{d^{10}}5{s^2}5{p^2}

The correct answer is

[Kr]4{d^{10}}5{s^2}5{p^2}

(D) because the element E is Gallium (Ga) in period 4, group 13. The element diagonally placed to Ga is Germanium (Ge) in period 5, group 14, with the electronic configuration

[Kr]4{d^{10}}5{s^2}5{p^2}

.

Solution approach

Correct Answer + Core Reason: $[Kr]4{d^{10}}5{s^2}5{p^2}$ . The element diagonally placed to E (Ga) is Germanium (Ge).

Step-by-Step Solution:

Step 1: Identify the element E based on its electronic configuration.

Step 2: The element E has the electronic configuration $4{s^2}4{p^1}$ , which corresponds to Gallium (Ga) in group 13 and period 4.

Step 3: Find the element diagonally placed to Ga in the periodic table. This element is Germanium (Ge), which is in group 14 and period 5.

Step 4: Determine the electronic configuration of Germanium (Ge). Ge has an atomic number of 32. Its electronic configuration is $[Ar]3d^{10}4s^24p^2$ . However, the question asks for the element in period 5 diagonally to E. The element diagonally to Ga is Germanium (Ge) which is in period 5 with electronic configuration $[Kr]4d^{10}5s^25p^2$ .

Why other options are wrong:
- Option A: $[Kr]3{d^{10}}4{s^2}4{p^2}$ is incorrect because the element is in period 5, not period 4.
- Option B: $[Ar]3{d^{10}}4{s^2}4{p^2}$ is the electronic configuration of Germanium, but it is not written with Krypton core.
- Option C: $[Xe]5{d^{10}}6{s^2}6{p^2}$ is incorrect because it corresponds to an element in period 6, not period 5.

Key Concepts:
- Electronic configuration
- Periodic table trends
- Diagonal relationship

Memory Aids:
- Remember the positions of elements in the periodic table.

Quick Tips:
- Know the electronic configurations of common elements.

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Remembering +4 | -1

Half the internuclear distance separating the metal cores in the metallic crystal is termed as

View answer and explanation

Correct option: A

Metallic radius

Option A is correct because the metallic radius is defined as half the internuclear distance separating the metal cores in the metallic crystal.

Solution approach

Correct Answer + Core Reason: Metallic radius. Metallic radius is defined as half the distance between the nuclei of two adjacent metal atoms in a metallic crystal.

Step-by-Step Solution:

Step 1: Define metallic radius.\nMetallic radius is defined as one-half of the distance between the nuclei of two adjacent metal atoms in a metallic crystal lattice.

Step 2: Conclusion.\nHalf the internuclear distance separating the metal cores in the metallic crystal is termed as metallic radius.

Why other options are wrong:
- Option A: Correct because metallic radius is half the internuclear distance in a metallic crystal.
- Option B: Incorrect because atomic radius is a general term and can refer to different types of bonding (covalent, metallic, van der Waals).
- Option C: Incorrect because covalent radius is half the distance between two atoms bonded together by a single covalent bond.
- Option D: Incorrect because ionic radius is the radius of an ion in an ionic crystal.

Key Concepts:
- Metallic radius
- Atomic radius
- Covalent radius
- Ionic radius

Memory Aids:
- Remember the definitions of different types of atomic radii.

Quick Tips:
- Understand the differences between metallic, covalent, and ionic radii.

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According to modern periodic law, variations in the properties of elements is related to their

View answer and explanation

Correct option: C

Atomic numbers

Option C (Atomic numbers) is correct because the modern periodic law states that properties vary with atomic numbers.

Solution approach

Correct Answer + Core Reason: Atomic numbers. The modern periodic law states that the properties of elements are periodic functions of their atomic numbers.

Step-by-Step Solution:

Step 1: Recall the modern periodic law: The properties of elements are determined by their electronic configuration, which is related to the atomic number.

Step 2: Consider the options:
- Atomic weight: Used in Mendeleev's periodic table but replaced by atomic number.
- Nuclear weight: Similar to atomic weight.
- Atomic numbers: The basis of the modern periodic table.
- Neutron-proton ratios: Affect nuclear stability but not directly related to periodic properties.

Step 3: Conclusion: The variations in properties are related to atomic numbers.

Why other options are wrong:
- Option A (Atomic weight): Atomic weight was used in the earlier periodic table but is not the basis of the modern periodic law.
- Option B (Nuclear weight): Similar to atomic weight, not the basis of the modern periodic law.
- Option D (Neutron-proton ratios): This relates to nuclear stability, not the periodic properties of elements.

Key Concepts:
- Modern Periodic Law
- Atomic Number
- Periodic Trends

Memory Aids:
- Atomic Number determines the electronic configuration and thus the properties.

Quick Tips:
- Remember the definition of the modern periodic law.

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Application +4 | -1

The element with outer shell electronic configuration

(n-2) f^7(n-1) d^1 n s^2

for

n=6

is ... .

View answer and explanation

Correct option: B

gadolinium

The correct answer is B (gadolinium) because the electronic configuration

(n-2) f^7(n-1) d^1 n s^2

for

n=6

is

4f^7 5d^1 6s^2

, which corresponds to Gadolinium.

Solution approach

Correct Answer + Core Reason: gadolinium. The electronic configuration $4f^7 5d^1 6s^2$ corresponds to Gadolinium.

Step-by-Step Solution:

Step 1: Substitute n = 6 into the given electronic configuration.

Step 2: The electronic configuration becomes $(6-2)f^7 (6-1)d^1 6s^2 = 4f^7 5d^1 6s^2$ .

Step 3: Identify the element with the electronic configuration $4f^7 5d^1 6s^2$ .

Step 4: Gadolinium (Gd) has the atomic number 64 and the electronic configuration $[Xe] 4f^7 5d^1 6s^2$ .

Why other options are wrong:
- Option A: Uranium (U) has an electronic configuration of $[Rn] 5f^3 6d^1 7s^2$ , which does not match the given configuration.
- Option C: Lead (Pb) has an electronic configuration of $[Xe] 4f^{14} 5d^{10} 6s^2 6p^2$ , which does not match the given configuration.
- Option D: Niobium (Nb) has an electronic configuration of $[Kr] 4d^4 5s^1$ , which does not match the given configuration.

Key Concepts:
- Electronic configuration
- f-block elements
- Lanthanides

Memory Aids:
- Remember the electronic configurations of common lanthanides.

Quick Tips:
- Know the electronic configurations of common elements.

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Analyzing +4 | -1

{M^{ + 3}}

ion has the electronic configuration as

\left[ {Xe} \right]\,\,5{d^{10}}4{f^{14}}6{s^2},

hence the element

M

belongs to

View answer and explanation

Correct option: B

p-block

The correct answer is p-block (B) because the electronic configuration of

M^{+3}

is

[Xe] 4f^{14} 5d^{10} 6s^2

, adding back the 3 electrons gives a configuration ending in

6s^2 6p^1

, indicating a p-block element.

Solution approach

Correct Answer + Core Reason: p-block. Adding back the 3 electrons to $M^{+3}$ gives a configuration ending in $6s^2 6p^1$ , indicating a p-block element.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the neutral atom M.

Step 2: The ion $M^{+3}$ has the electronic configuration $[Xe] 4f^{14} 5d^{10} 6s^2$ .

Step 3: Add 3 electrons to the ion to obtain the electronic configuration of the neutral atom M. The electrons will be added to the outermost orbitals.

Step 4: The electronic configuration of M is $[Xe] 4f^{14} 5d^{10} 6s^2 6p^1$ .

Step 5: Determine the block to which the element M belongs.

Step 6: Since the last electron enters the p-orbital, the element M belongs to the p-block.

Why other options are wrong:
- Option A: s-block elements have their last electron entering the s-orbital, which is not the case here.
- Option C: d-block elements have their last electron entering the d-orbital, which is not the case here.
- Option D: f-block elements have their last electron entering the f-orbital, which is not the case here.

Key Concepts:
- Electronic configuration
- Periodic table blocks
- Ion formation

Memory Aids:
- Remember the electronic configurations of elements in different blocks.

Quick Tips:
- Know the electronic configurations of common elements.

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Which of the following is incorrect?

View answer and explanation

Correct option: D

Unununnium – Darmstadtium

Option A (Unnilunium – Mendelevium) is incorrect because Unnilunium corresponds to atomic number 101, which is indeed Mendelevium.

Solution approach

Correct Answer + Core Reason: Unununnium – Darmstadtium. Unununnium is not Darmstadtium.

Step-by-Step Solution:

Step 1: Determine the atomic number for Unnilunium: Unnilunium corresponds to 101, which is Mendelevium.

Step 2: Determine the atomic number for Unniltrium: Unniltrium corresponds to 103, which is Lawrencium.

Step 3: Determine the atomic number for Unnilhexium: Unnilhexium corresponds to 106, which is Seaborgium.

Step 4: Determine the atomic number for Unununnium: Unununnium corresponds to 111, which is Roentgenium, not Darmstadtium (110).

Step 5: Conclusion: The incorrect pairing is Unununnium – Darmstadtium.

Why other options are wrong:
- Option A: Unnilunium – Mendelevium: This is a correct pairing.
- Option B: Unniltrium – Lawrencium: This is a correct pairing.
- Option C: Unnilhexium – Seaborgium: This is a correct pairing.
- Option D: Unununnium – Darmstadtium: This is an incorrect pairing.

Key Concepts:
- IUPAC nomenclature
- Element names and symbols

Memory Aids:
- Remember the numerical roots used in IUPAC nomenclature.

Quick Tips:
- Practice converting atomic numbers to IUPAC names and element names.

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Understanding +4 | -1

The formation of the oxide ion

{\mathrm{O}^{2 - }}\left( \mathrm{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^o} = - 142\mathrm{\;kJ\;mo}{\mathrm{l}^{ - 1}}

,

{\mathrm{O}^ - }\left( \mathrm{g} \right) + {e^ - } = {\mathrm{O}^{2 - }}\left( \mathrm{g} \right),\mathrm{\;}\Delta {H^o} = 844\mathrm{\;kJ\;mo}{\mathrm{l}^{ - 1}}

This is because

View answer and explanation

Correct option: A

{\mathrm{O}^ - }

ion will tend to resist the addition of another electron

Option A (

{\mathrm{O}^ - }

ion will tend to resist the addition of another electron) is correct because the negative charge on the O- ion repels the incoming electron, making the process endothermic.

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}^ - }$ ion will tend to resist the addition of another electron. This explains why the second electron affinity is endothermic.

Step-by-Step Solution:

Step 1: Understand the first electron affinity.
$O(g) + e^- \rightarrow O^-(g) \quad \Delta H^o = -142 \text{ kJ/mol}$ . This is exothermic because the oxygen atom has an affinity for electrons.

Step 2: Understand the second electron affinity.
$O^-(g) + e^- \rightarrow O^{2-}(g) \quad \Delta H^o = 844 \text{ kJ/mol}$ . This is endothermic because the $O^-$ ion already has a negative charge, and adding another electron requires energy to overcome the electrostatic repulsion.

Step 3: Explain the resistance to adding another electron.\nThe $O^-$ ion has a net negative charge. Adding another electron increases the negative charge, leading to electron-electron repulsion. This repulsion requires energy to overcome, making the process endothermic.

Why other options are wrong:
- Option B: Oxygen has high electro affinity: While oxygen has a high first electron affinity, this doesn't explain why the *second* electron affinity is endothermic.
- Option C: Oxygen is more electronegative: Electronegativity describes the tendency to attract electrons in a chemical bond, not the energy change associated with adding an electron to a gaseous atom.
- Option D: ${\mathrm{O}^ - }$ ion has comparatively larger size than oxygen atom: While the ionic radius of $O^-$ is larger than O, this doesn't directly explain the endothermic nature of the second electron affinity. The primary reason is electron-electron repulsion.

Key Concepts:
- Electron affinity
- Electrostatic repulsion

Memory Aids:
- Remember that adding an electron to a negatively charged ion requires energy to overcome repulsion.

Quick Tips:
- Consider the charge of the ion when determining whether electron affinity is exothermic or endothermic.

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Remembering +4 | -1

The electronic configuration of inner transition elements is

View answer and explanation

Correct option: D

n s^2(n-1) d^{0-1}(n-2) f^{1-14}

Option D is correct. The general electronic configuration of inner transition elements is ns²(n-1)d⁰⁻¹(n-2)f¹⁻¹⁴.

Solution approach

Correct Answer + Core Reason: $n s^2(n-1) d^{0-1}(n-2) f^{1-14}$ . This is the general electronic configuration of inner transition elements, accounting for the filling of the f-orbitals.

Step-by-Step Solution:

Step 1: Define Inner Transition Elements: Inner transition elements are the f-block elements, consisting of the lanthanoids and actinoids.

Step 2: Recall the general electronic configuration: The general electronic configuration of inner transition elements is $n s^2(n-1) d^{0-1}(n-2) f^{1-14}$ , where n is the principal quantum number of the outermost shell.

Step 3: Conclusion: The correct electronic configuration reflects the filling of the (n-2)f orbitals.

Why other options are wrong:
- Option A: $n s^1$ is too simplistic and doesn't account for the f-orbitals.
- Option B: $n s^2 n p^5 n d^{10}$ represents a halogen with a filled d-orbital, not an inner transition element.
- Option C: $n s^{0-2}(n-1) d^{1-10}(n-2) f^{1-14}$ has incorrect d-orbital filling order and is not the standard representation.

Key Concepts:
- Inner Transition Elements (f-block)
- Electronic Configuration
- Lanthanoids and Actinoids

Memory Aids:
- Remember the (n-2)f filling for inner transition elements.

Quick Tips:
- Understand the filling order of orbitals in the periodic table.

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