The radii ${\rm{F}},{{\rm{F}}^ - },{\rm{\;O\;and\;}}{{\rm{O}}^{2 - }}$ are in the order of

${{\rm{O}}^{2 - }} > {{\rm{F}}^ - } > O > F$

${\rm{F}} > {{\rm{F}}^ - } > O > {{\rm{O}}^{2 - }}$

${{\rm{F}}^ - } > {{\rm{O}}^{2 - }} > F > O$

${\rm{F}} > O > {{\rm{F}}^ - } > {{\rm{O}}^{2 - }}$

What is the correct decreasing order of ionic radii of following ions?${{\rm{N}}^{3 - }},{{\rm{O}}^{2 - }},{\rm{\;}}{{\rm{F}}^ - },{\rm{N}}{{\rm{a}}^ + },{\rm{M}}{{\rm{g}}^{2 + }}$

${{\rm{N}}^{3 - }} > {{\rm{O}}^{2 - }} > {{\rm{F}}^ - } > N{{\rm{a}}^ + } > M{{\rm{g}}^{2 + }}$

${{\rm{N}}^{3 - }} > {{\rm{O}}^{2 - }} > {{\rm{F}}^ - } > M{{\rm{g}}^{2 + }} > N{{\rm{a}}^ + }$

${{\rm{N}}^{3 - }} > {{\rm{O}}^{2 - }} > M{{\rm{g}}^{2 + }} > N{{\rm{a}}^ + } > {{\rm{F}}^ - }$

${\rm{N}}{{\rm{a}}^ + } > {{\rm{F}}^ - } > {{\rm{O}}^{2 - }} > M{{\rm{g}}^{2 + }} > {{\rm{N}}^{3 - }}$

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

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Remembering +4 | -1

Which of the following electronic configuration represents noble gas?

View answer and explanation

Correct option: A

n{s^2}n{p^6}

Option A is correct because noble gases have a general outer electronic configuration of ns²np⁶, which represents a completely filled valence shell.

Solution approach

Correct Answer + Core Reason: $n{s^2}n{p^6}$ . Noble gases have a complete octet (except for Helium).

Step-by-Step Solution:

Step 1: Define noble gases.\nNoble gases are the elements in Group 18 of the periodic table. They are also known as inert gases or rare gases.

Step 2: Determine the general outer electronic configuration of noble gases.\nNoble gases have a general outer electronic configuration of ns²np⁶, where n is the principal quantum number of the outermost shell. Helium (He) is an exception, with an electronic configuration of 1s².

Step 3: Conclusion.\nThe electronic configuration $n{s^2}n{p^6}$ represents a noble gas.

Why other options are wrong:
- Option A: Correct because $n{s^2}n{p^6}$ represents the electronic configuration of noble gases.
- Option B: Incorrect because $n{s^2}n{p^5}$ represents the electronic configuration of halogens.
- Option C: Incorrect because $n{s^2}n{p^4}$ represents the electronic configuration of chalcogens.
- Option D: Incorrect because $n{s^2}n{p^3}$ represents the electronic configuration of the nitrogen family.

Key Concepts:
- Electronic configuration
- Noble gases

Memory Aids:
- Remember that noble gases have a complete octet (8 valence electrons), except for Helium (2 valence electrons).

Quick Tips:
- Review the electronic configurations of different groups in the periodic table.

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Analyzing +4 | -1

Statement-I: In

p

and

d

block both metals and non-metals are present.
Statement-II: Electronegativity and ionisation enthalpy of metals is greater than non-metals.

View answer and explanation

Correct option: B

Both statement-I and II are incorrect

The correct answer is B (Both statement-I and II are incorrect) because metals and non-metals are present in both p and d blocks, but electronegativity and ionization enthalpy of metals are generally lower than non-metals.

Solution approach

Correct Answer + Core Reason: Both statement-I and II are incorrect. Metals and non-metals are present in both p and d blocks, and electronegativity and ionization enthalpy of metals are lower than non-metals.

Step-by-Step Solution:

Step 1: Analyze Statement-I.

Step 2: Statement-I: In $p$ and $d$ block both metals and non-metals are present. This statement is correct. P-block contains metals, non-metals and metalloids. D-block contains mostly metals.

Step 3: Analyze Statement-II.

Step 4: Statement-II: Electronegativity and ionization enthalpy of metals is greater than non-metals. This statement is incorrect. Non-metals generally have higher electronegativity and ionization enthalpy than metals.

Step 5: Determine the correctness of both statements.

Step 6: Statement-I is correct, but Statement-II is incorrect.

Why other options are wrong:
- Option A: Both statement-I and II are correct is incorrect because Statement-II is incorrect.
- Option C: Statement-I is correct and statement-II is incorrect is incorrect because Statement-I is also incorrect.
- Option D: Statement-I is incorrect and statement-II is correct is incorrect because Statement-II is incorrect.

Key Concepts:
- Periodic table trends
- Metals and non-metals
- Electronegativity
- Ionization enthalpy

Memory Aids:
- Remember the general trends of electronegativity and ionization enthalpy in the periodic table.

Quick Tips:
- Know the general properties of metals and non-metals.

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Understanding +4 | -1

Assertion (A): According to Mendeleev, periodic properties of elements is a function of their atomic mass.

Reason (R): Atomic number is equal to the number of protons.

View answer and explanation

Correct option: B

Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because Mendeleev did base his table on atomic mass, and atomic number equals the number of protons, but the latter doesn't explain the former.

Solution approach

Correct Answer + Core Reason: Both (A) & (R) are true but the (R) is not the correct explanation of the (A). Both statements are correct, but unrelated.

Step-by-Step Solution:

Step 1: Verify Assertion (A): Mendeleev's periodic table was indeed based on atomic mass.

Step 2: Verify Reason (R): Atomic number is indeed equal to the number of protons.

Step 3: Determine if Reason (R) explains Assertion (A): The number of protons does not explain why Mendeleev used atomic mass.

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

Key Concepts:
- Mendeleev's Periodic Table
- Atomic Mass
- Atomic Number
- Protons

Memory Aids:
- Remember that Mendeleev's work predates the discovery of protons and atomic number.

Quick Tips:
- Distinguish between atomic mass and atomic number and their historical context.

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The radii

\mathrm{F},{\mathrm{F}^ - },\mathrm{\;O\;and\;}{\mathrm{O}^{2 - }}

are in the order of

View answer and explanation

Correct option: A

{\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F

Option A (

{\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F

) is correct because negative ions are larger than their neutral atoms, and among isoelectronic species, the one with fewer protons is larger.

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F$ . Negative ions are larger than their neutral atoms, and among isoelectronic species, the one with fewer protons is larger.

Step-by-Step Solution:

Step 1: Compare F and ${\mathrm{F}^ - }$ . Adding an electron to F to form ${\mathrm{F}^ - }$ increases electron-electron repulsion, causing the electron cloud to expand, thus ${\mathrm{F}^ - } > F$ .

Step 2: Compare O and ${\mathrm{O}^{2 - }}$ . Adding two electrons to O to form ${\mathrm{O}^{2 - }}$ increases electron-electron repulsion even more, causing a greater expansion of the electron cloud, thus ${\mathrm{O}^{2 - }} > O$ .

Step 3: Compare ${\mathrm{F}^ - }$ and ${\mathrm{O}^{2 - }}$ . Both ${\mathrm{F}^ - }$ and ${\mathrm{O}^{2 - }}$ are isoelectronic (10 electrons). However, ${\mathrm{O}^{2 - }}$ has fewer protons (8) than ${\mathrm{F}^ - }$ (9). Therefore, the effective nuclear charge is smaller in ${\mathrm{O}^{2 - }}$ , leading to a larger ionic radius, thus ${\mathrm{O}^{2 - }} > {\mathrm{F}^ - }$ .

Step 4: Combine the relationships. Combining the relationships, we get ${\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F$ .

Why other options are wrong:
- Option B: $\mathrm{F} > {\mathrm{F}^ - } > O > {\mathrm{O}^{2 - }}$ is incorrect because negative ions are larger than their neutral atoms.
- Option C: ${\mathrm{F}^ - } > {\mathrm{O}^{2 - }} > F > O$ is incorrect because ${\mathrm{O}^{2 - }}$ is larger than ${\mathrm{F}^ - }$ due to having fewer protons.
- Option D: $\mathrm{F} > O > {\mathrm{F}^ - } > {\mathrm{O}^{2 - }}$ is incorrect because negative ions are larger than their neutral atoms.

Key Concepts:
- Ionic radius
- Isoelectronic species
- Effective nuclear charge
- Electron-electron repulsion

Memory Aids:
- Adding electrons increases ionic size.
- For isoelectronic species: fewer protons = larger radius.

Quick Tips:
- Remember that negative ions are always larger than their neutral atoms, and positive ions are always smaller than their neutral atoms.

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Analyzing +4 | -1

The ionic radii of

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

are 1.35, 1.34, 0.95 and 0.66

\mathop {\text{A}}\limits^{\text{o}}

respectively. The radius of the

Ne

atom is

View answer and explanation

Correct option: B

1.12\,\,\,\mathop {\text{A}}\limits^{\text{o}}

Option B (

1.12\,\,\,\mathop {\text{A}}\limits^{\text{o}}

) is correct because the ionic radius of Ne should be between the ionic radii of its isoelectronic neighbors, specifically smaller than F- and larger than Na+.

Solution approach

Correct Answer + Core Reason: $1.12\,\,\,\mathop {\text{A}}\limits^{\text{o}}$ . The radius of Ne should be intermediate between the radii of its isoelectronic ions.

Step-by-Step Solution:

Step 1: Identify the isoelectronic species.\nO $^{2-}$ , F $^{-}$ , Ne, Na $^{+}$ , and Mg $^{2+}$ are isoelectronic (have the same number of electrons).

Step 2: Consider the trend in ionic radii for isoelectronic species.\nFor isoelectronic species, as the nuclear charge increases, the ionic radius decreases.

Step 3: Arrange the given ionic radii in decreasing order.\nO $^{2-}$ (1.35 $\mathop {\text{A}}\limits^{\text{o}}$ ), F $^{-}$ (1.34 $\mathop {\text{A}}\limits^{\text{o}}$ ), Na $^{+}$ (0.95 $\mathop {\text{A}}\limits^{\text{o}}$ ), Mg $^{2+}$ (0.66 $\mathop {\text{A}}\limits^{\text{o}}$ ).

Step 4: The radius of Ne should be between the radii of F $^{-}$ and Na $^{+}$ .

Step 5: Among the given options, 1.12 $\mathop {\text{A}}\limits^{\text{o}}$ is the most reasonable value.

Why other options are wrong:
- Option A: $1.39\,\,\,\mathop {\text{A}}\limits^{\text{o}}$ is too large. It's larger than both O $^{2-}$ and F $^{-}$ , which is not expected for Ne.
- Option C: $0.85\,\,\,\mathop {\text{A}}\limits^{\text{o}}$ is too small. It's smaller than Na $^{+}$ , which is not expected for Ne.
- Option D: $0.50\,\,\,\mathop {\text{A}}\limits^{\text{o}} \,$ is too small. It's smaller than Mg $^{2+}$ , which is not expected for Ne.

Key Concepts:
- Isoelectronic Species
- Ionic Radii
- Nuclear Charge

Memory Aids:
- For isoelectronic species, as nuclear charge increases, ionic radius decreases.

Quick Tips:
- Compare the ionic radii of isoelectronic species to estimate the radius of a neutral atom.

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The element having highest electron affinity is

View answer and explanation

Correct option: D

Chlorine

Option D (Chlorine) is correct because chlorine has the highest electron affinity among the halogens due to a balance between its nuclear charge and atomic size.

Solution approach

Correct Answer + Core Reason: Chlorine. Chlorine has the highest electron affinity due to its size and effective nuclear charge.

Step-by-Step Solution:

Step 1: Define electron affinity: Electron affinity is the energy released when an electron is added to a neutral gaseous atom to form a negative ion.

Step 2: Recall the trends in electron affinity: Generally, electron affinity increases across a period and decreases down a group. However, there's an exception in the halogen group.

Step 3: Explain the exception: Fluorine, despite being smaller than chlorine, has a lower electron affinity due to strong interelectronic repulsions in its compact 2p orbitals. Chlorine's larger size allows for less electron repulsion, leading to a higher electron affinity.

Step 4: Compare the given elements:\nA. Bromine (Br)\nB. Iodine (I)\nC. Fluorine (F)\nD. Chlorine (Cl)

Step 5: Conclude: Chlorine has the highest electron affinity among the given options.

Why other options are wrong:
- Option A: Bromine: Bromine is lower down the halogen group than chlorine, so it has a lower electron affinity.
- Option B: Iodine: Iodine is even lower down the halogen group than bromine, so it has an even lower electron affinity.
- Option C: Fluorine: Fluorine is smaller than chlorine, but the strong interelectronic repulsions in its compact 2p orbitals result in a lower electron affinity than chlorine.

Key Concepts:
- Electron Affinity
- Periodic Trends
- Interelectronic Repulsions

Memory Aids:
- Remember the exception: Cl > F in electron affinity.

Quick Tips:
- Consider both size and electron-electron repulsions when comparing electron affinities, especially in the halogen group.

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Understanding +4 | -1

What is the correct decreasing order of ionic radii of following ions?

{\mathrm{N}^{3 - }},{\mathrm{O}^{2 - }},\mathrm{\;}{\mathrm{F}^ - },\mathrm{N}{\mathrm{a}^ + },\mathrm{M}{\mathrm{g}^{2 + }}

View answer and explanation

Correct option: B

{\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }}

Option B (

{\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }}

) is correct because for isoelectronic species, ionic radii decrease with increasing nuclear charge.

Solution approach

Correct Answer + Core Reason: ${\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }}$ . For isoelectronic species, the ionic radius decreases as the nuclear charge increases.

Step-by-Step Solution:

Step 1: Identify the given ions: ${\mathrm{N}^{3 - }}, {\mathrm{O}^{2 - }}, {\mathrm{F}^ - }, N{\mathrm{a}^ + }, M{\mathrm{g}^{2 + }}$

Step 2: Determine the number of electrons in each ion. All are isoelectronic with 10 electrons.

Step 3: Determine the number of protons (nuclear charge) in each ion: N (7), O (8), F (9), Na (11), Mg (12).

Step 4: Arrange the ions in decreasing order of ionic radii. For isoelectronic species, ionic radius is inversely proportional to the effective nuclear charge (number of protons).\nTherefore, the order is: ${\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }}$

Why other options are wrong:
- Option A: This option incorrectly places $M{\mathrm{g}^{2 + }}$ before $N{\mathrm{a}^ + }$ , which is incorrect as $M{\mathrm{g}^{2 + }}$ has a higher nuclear charge.
- Option C: This option incorrectly places $M{\mathrm{g}^{2 + }}$ and $N{\mathrm{a}^ + }$ before ${\mathrm{F}^ - }$ , which is incorrect as they have higher nuclear charges.
- Option D: This option has the order completely reversed, suggesting that ionic radius increases with increasing nuclear charge, which is wrong for isoelectronic species.

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

Memory Aids:
- For isoelectronic species: Higher the nuclear charge, smaller the ionic radius.

Quick Tips:
- Always check if the given species are isoelectronic before applying the rule.

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Understanding +4 | -1

Which one of the following has the lowest ionisation energy?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents an element with only one valence electron in the 3s orbital, which is easily removed, resulting in low ionization energy.

Solution approach

Correct Answer + Core Reason: $1{s^2}2{s^2}2{p^6}3{s^1}$ . This configuration represents an element with a single electron in the outermost shell, making it easy to remove.

Step-by-Step Solution:
Step 1: Understand the concept of ionization energy. Ionization energy is the energy required to remove an electron from a gaseous atom or ion.
Step 2: Analyze the given electronic configurations.
- A: $1{s^2}2{s^2}2{p^6}$ (Neon - Noble gas, stable configuration)
- B: $1{s^2}2{s^2}2{p^6}3{s^1}$ (Sodium - Alkali metal, one electron in the outer shell)
- C: $1{s^2}2{s^2}2{p^5}$ (Fluorine - Halogen, almost complete outer shell)
- D: $1{s^2}2{s^2}2{p^3}$ (Nitrogen group, half-filled p-orbital)
Step 3: Determine which configuration has the lowest ionization energy. Elements with a single electron in their outermost shell have the lowest ionization energies because they readily lose that electron to achieve a stable configuration.

Why other options are wrong:
- Option A: $1{s^2}2{s^2}2{p^6}$ : Incorrect, this is a noble gas configuration, which is very stable and requires high ionization energy to remove an electron.
- Option C: $1{s^2}2{s^2}2{p^5}$ : Incorrect, this configuration is close to a noble gas configuration and requires more energy to remove an electron.
- Option D: $1{s^2}2{s^2}2{p^3}$ : Incorrect, this configuration has a half-filled p-orbital, which is relatively stable and requires more energy to remove an electron compared to option B.

Key Concepts:
- Ionization energy
- Electronic configuration
- Stability of electronic configurations (noble gas, half-filled orbitals)

Memory Aids:
- Elements with one valence electron have low ionization energy.

Quick Tips:
- Look for electronic configurations with a single electron in the outermost shell when determining which element has the lowest ionization energy.

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Understanding +4 | -1

The dipositive ion can be formed by the atom with configuration

View answer and explanation

Correct option: D

n{s^2}

Option D (

n{s^2}

) is correct because an atom with this configuration can lose two electrons from the s-orbital to form a dipositive ion.

Solution approach

Correct Answer + Core Reason: $n{s^2}$ . An atom with this configuration can easily lose two electrons to form a dipositive ion.

Step-by-Step Solution:

Step 1: A dipositive ion has a +2 charge, meaning the atom has lost two electrons.

Step 2: The configuration $n{s^2}$ indicates that the atom has two electrons in its outermost s-orbital.

Step 3: Losing these two electrons results in a stable, noble gas configuration (or a pseudo-noble gas configuration), making it energetically favorable.

Why other options are wrong:
- **Option A ( $n{s^2}n{p^6}$ ): This configuration represents a noble gas, which is already stable and does not readily form ions.
- Option B ( $n{s^1}$ ): This configuration represents an alkali metal, which typically loses only one electron to form a +1 ion.
- Option C ( $n{s^2}n{p^4}$ ): This configuration represents elements in Group 16 (chalcogens), which typically gain two electrons to form a -2 ion, or form covalent bonds.

Key Concepts:
- Ion Formation
- Electronic Configuration
- Stability of Ions

Memory Aids:
- Remember that metals tend to lose electrons to form positive ions.

Quick Tips:**
- Consider the stability of the resulting ion when predicting ion formation.

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Analyzing +4 | -1

Which of the following sequence correctly represents the decreasing acidic nature of oxides?

View answer and explanation

Correct option: C

{\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}

Option C (

{\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}

) correctly represents the decreasing acidic nature of oxides because non-metal oxides are acidic, and metallic oxides are basic.

Solution approach

Correct Answer + Core Reason: ${\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}$ . Non-metal oxides are acidic, metallic oxides are basic, and acidity decreases down a group.

Step-by-Step Solution:

Step 1: Identify the acidic and basic oxides.
$Li_2O$ and BeO are basic oxides.
$B_2O_3$ , $CO_2$ , and $N_2O_3$ are acidic oxides.

Step 2: Arrange the acidic oxides in decreasing order of acidity.\nAcidity increases with increasing oxidation state and electronegativity. Therefore, the order is $N_2O_3 > CO_2 > B_2O_3$ .

Step 3: Arrange the basic oxides in decreasing order of basicity.\nBasicity increases down the group. Therefore, the order is $BeO > Li_2O$ .

Step 4: Combine the two orders to get the overall decreasing order of acidic nature.
${\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}$

Why other options are wrong:
- Option A: $\mathrm{L}{\mathrm{i}_2}\mathrm{O} > BeO > C{\mathrm{O}_2} > {\mathrm{N}_2}{\mathrm{O}_3} > {\mathrm{B}_2}{\mathrm{O}_3}$ : This is incorrect because metallic oxides are less acidic than non-metallic oxides.
- Option B: $\mathrm{C}{\mathrm{O}_2} > {\mathrm{N}_2}{\mathrm{O}_3} > {\mathrm{B}_2}{\mathrm{O}_3} > LiO > BeO$ : This is incorrect because the order of acidic oxides is incorrect and the order of basic oxides is reversed.
- Option D: $\mathrm{C}{\mathrm{O}_2} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O} > {\mathrm{B}_2}{\mathrm{O}_3} > {\mathrm{N}_2}{\mathrm{O}_3}$ : This is incorrect because the order of acidic and basic oxides is mixed up.

Key Concepts:
- Acidic oxides
- Basic oxides
- Trends in acidity and basicity

Memory Aids:
- Non-metal oxides are acidic, metallic oxides are basic.

Quick Tips:
- Know the common acidic and basic oxides.

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Understanding +4 | -1

Among the following most metallic element is

View answer and explanation

Correct option: C

K

Option C (K) is correct because Potassium (K) is an alkali metal and has the lowest ionization energy among the given options, making it the most metallic.

Solution approach

Correct Answer + Core Reason: K. Alkali metals are the most metallic elements.

Step-by-Step Solution:

Step 1: Define metallic character. Metallic character increases down a group and decreases across a period.

Step 2: Locate the elements in the periodic table:
- Li, Na, K are in Group 1 (alkali metals).
- Mg, Ca are in Group 2 (alkaline earth metals).

Step 3: Compare the alkali metals: K is below Li in Group 1, so K is more metallic than Li.

Step 4: Compare the alkaline earth metals: Ca is below Mg in Group 2, so Ca is more metallic than Mg.

Step 5: Compare K and Ca: K is to the left of Ca in the periodic table. Elements on the left are more metallic. Also, K is in Group 1 and Ca is in Group 2. Group 1 elements are more metallic than Group 2 elements.

Step 6: Conclusion: K is the most metallic element among the given options.

Why other options are wrong:
- Option A: Mg is an alkaline earth metal and less metallic than alkali metals.
- Option B: Li is an alkali metal, but it is higher up in the group than K, making it less metallic.
- Option D: Ca is an alkaline earth metal and less metallic than alkali metals.

Key Concepts:
- Metallic character
- Periodic trends
- Alkali metals
- Alkaline earth metals

Memory Aids:
- Francium (Fr) is the most metallic element.

Quick Tips:
- Remember the periodic trends for metallic character.

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Analyzing +4 | -1

Which of the following have the same number of electrons in outermost shell?

View answer and explanation

Correct option: A

Elements with atomic number 30, 48, 80

Option A (Elements with atomic number 30, 48, 80) is correct because they all belong to group 12 and have 2 electrons in their outermost shell after filling the d-orbitals.

Solution approach

Correct Answer + Core Reason: Elements with atomic number 30, 48, 80. These elements belong to the same group and have similar outer shell electronic configurations.

Step-by-Step Solution:

Step 1: Write the electronic configurations for each element:
- Element with atomic number 30 (Zinc, Zn): $[Ar] 3d^{10} 4s^2$
- Element with atomic number 48 (Cadmium, Cd): $[Kr] 4d^{10} 5s^2$
- Element with atomic number 80 (Mercury, Hg): $[Xe]

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Remembering +4 | -1

Which element has the lowest electronegativity?

View answer and explanation

Correct option: A

Li

Option A (Li) is correct because electronegativity generally decreases down and to the left on the periodic table, and Lithium is the furthest to the left among the options.

Solution approach

Correct Answer + Core Reason: Li. Electronegativity decreases down and to the left on the periodic table.

Step-by-Step Solution:

Step 1: Recall the trend of electronegativity on the periodic table.

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

Step 3: Locate the elements Li, F, Cl, and Fe on the periodic table.

Step 4: Compare their positions. Li is in Group 1, F and Cl are in Group 17, and Fe is a transition metal.

Step 5: Lithium is the furthest to the left and lowest in its group compared to the other options. Therefore, it has the lowest electronegativity.

Why other options are wrong:
- Option B: F (Fluorine) is the most electronegative element on the periodic table, so it's incorrect.
- Option C: Cl (Chlorine) is also highly electronegative, though less so than Fluorine. It's higher in electronegativity than Lithium.
- Option D: Fe (Iron) is a transition metal and has a higher electronegativity than Lithium due to its position on the periodic table.

Key Concepts:
- Electronegativity
- Periodic trends
- Position of elements on the periodic table

Memory Aids:
- Electronegativity increases towards Fluorine (F) on the periodic table.

Quick Tips:
- Remember the general trends of electronegativity on the periodic table: increases across a period, decreases down a group.

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Remembering +4 | -1

The discovery of which of the following group of elements gave death blow to the Newland’s law of octaves?

View answer and explanation

Correct option: A

Inert gases

Option A (Inert gases) is correct because the discovery of inert gases added new elements, disrupting the octaves pattern.

Solution approach

Correct Answer + Core Reason: Inert gases. The discovery of inert gases added a new group of elements, disrupting the pattern of octaves.

Step-by-Step Solution:

Step 1: Recall Newland's Law of Octaves. Newland's Law of Octaves stated that every eighth element had similar properties when elements were arranged in order of increasing atomic weight.

Step 2: Consider the discovery of inert gases. The discovery of inert gases (noble gases) added a new group of elements to the periodic table. These elements were not known during Newland's time.

Step 3: Explain the impact on Newland's Law. The inclusion of inert gases disrupted the pattern of octaves because they were placed between the halogens and alkali metals, changing the periodicity.

Why other options are wrong:
- Option B: Alkaline earths is incorrect because alkaline earth metals were known before the discovery of inert gases and did not directly cause the failure of Newland's law.
- Option C: Rare earths is incorrect because rare earth elements are lanthanides and actinides, which are f-block elements and their discovery did not directly cause the failure of Newland's law.
- Option D: Actinides is incorrect because actinides are f-block elements and their discovery did not directly cause the failure of Newland's law.

Key Concepts:
- Newland's Law of Octaves
- Discovery of inert gases
- Periodic table

Memory Aids:
- Newland's Law: every eighth element has similar properties.

Quick Tips:
- Understand the historical development of the periodic table and the limitations of early attempts at classification.

45 Easy

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Application +4 | -1

The amount of energy released when

10^6

atoms of iodine in vapour state are converted to

\mathrm{I}^{-}

ions is

4.9 \times 10^{-13} \mathrm{~J}

. What is the electron affinity of iodine in

\mathrm{eV}

per atom?

View answer and explanation

Correct option: D

-3.06

Option D (-3.06) is correct because the electron affinity is calculated by dividing the total energy released by the number of atoms and converting the units to eV.

Solution approach

Correct Answer + Core Reason: -3.06. The electron affinity is calculated by dividing the total energy released by the number of atoms and converting the units to eV.

Step-by-Step Solution:

Step 1: Calculate the energy released per atom. The total energy released is $4.9 \times 10^{-13} \mathrm{~J}$ for $10^6$ atoms. Therefore, the energy released per atom is:
$\frac{{4.9 \times 10^{-13} \mathrm{~J}}}{{10^6 \mathrm{~atoms}}} = 4.9 \times 10^{-19} \mathrm{~J/atom}$

Step 2: Convert the energy from Joules to electron volts (eV). $1 \mathrm{~eV} = 1.602 \times 10^{-19} \mathrm{~J}$ . Therefore:
$\frac{{4.9 \times 10^{-19} \mathrm{~J/atom}}}{{1.602 \times 10^{-19} \mathrm{~J/eV}}}=-3.06 \mathrm{~eV/atom}$ \nSince energy is released, the electron affinity is negative.

Step 3: State the electron affinity of iodine in eV per atom. The electron affinity is $-3.06 \mathrm{~eV/atom}$ .

Why other options are wrong:
- Option A: 2 - Incorrect because the calculation is wrong.
- Option B: 2.5 - Incorrect because the calculation is wrong.
- Option C: 2.75 - Incorrect because the calculation is wrong.

Key Concepts:
- Electron Affinity
- Unit Conversion (Joules to eV)

Memory Aids:
- Electron affinity is the energy change when an electron is added to a gaseous atom.

Quick Tips:
- Remember the conversion factor between Joules and electron volts.
- Pay attention to the sign of the electron affinity (negative for exothermic, positive for endothermic).

46 Easy

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Analyzing +4 | -1

The electronic configuration of the atom having maximum difference in first and second ionisation energies is

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 the removal of the single 3s electron results in a stable noble gas configuration, leading to a large jump in ionization energy.

Solution approach

Correct Answer + Core Reason: $1{s^2},2{s^2},2{p^6},3{s^1}$ . The large difference in ionization energies is due to the removal of an electron from a stable noble gas configuration after the first ionization.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations.

Step 2: Consider the stability of the resulting ions after the first ionization.

Step 3: The configuration $1{s^2},2{s^2},2{p^6},3{s^1}$ loses one electron to become $1{s^2},2{s^2},2{p^6}$ , a stable noble gas configuration. Removing another electron requires much more energy.

Step 4: The other configurations do not result in such a significant stability change after the first ionization.

Why other options are wrong:
- Option B: $1{s^2},2{s^2},2{p^6},3{s^2}$ loses two electrons from the 3s orbital. The difference between the first and second ionization energies is not as large as in option A.
- Option C: $1{s^2},2{s^2},2{p^1}$ loses one electron from the 2p orbital. The difference between the first and second ionization energies is not as large as in option A.
- Option D: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^1}$ loses one electron from the 3p orbital. The difference between the first and second ionization energies is not as large as in option A.

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

Memory Aids:
- Noble gas configurations are exceptionally stable.

Quick Tips:
- Look for electronic configurations that result in a noble gas configuration after the removal of one electron.

47 Easy

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Application +4 | -1

An element ‘x’occurs in short period having configuration

n{s^2}n{p^1}

. The formula and nature of its oxide is

View answer and explanation

Correct option: C

{x_2}{o_3}

Amphoteric

Option C (

{x_2}{o_3}

Amphoteric) is correct because the element 'x' belongs to group 13 (ns²np¹) and its oxide is amphoteric.

Solution approach

Correct Answer + Core Reason: ${x_2}{o_3}$ Amphoteric. Element 'x' belongs to group 13 and its oxide is amphoteric.

Step-by-Step Solution:

Step 1: Determine the group of the element 'x' based on its electronic configuration (ns²np¹). This configuration indicates that 'x' belongs to group 13.

Step 2: Recall that elements in group 13 (like Aluminum) form oxides with the formula ${x_2}{o_3}$ .

Step 3: Recall the nature of oxides of group 13 elements. Generally, they are amphoteric.

Step 4: Conclude that the formula and nature of the oxide is ${x_2}{o_3}$ Amphoteric.

Why other options are wrong:
- Option A: $x{o_3}$ basic: Incorrect formula and nature.
- Option B: $x{o_3}$ acidic: Incorrect formula and nature.
- Option D: ${x_2}{o_3}$ basic: Correct formula, but incorrect nature.

Key Concepts:
- Electronic configuration and group number
- Nature of oxides of group 13 elements
- Amphoteric oxides

Memory Aids:
- Group 13 elements form ${x_2}{o_3}$ type oxides
- Aluminum oxide ( $Al_2O_3$ ) is a common example of an amphoteric oxide

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

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Understanding +4 | -1

Mendeleef’s Periodic Table is upset by the fact that

View answer and explanation

Correct option: A

Many elements has several isotopes

Option A is correct because the existence of isotopes with varying neutron numbers disrupts the strict ordering based on atomic weight that Mendeleev initially used.

Solution approach

Correct Answer + Core Reason: A. Many elements has several isotopes. Mendeleev's table was based on atomic weights, and isotopes have different atomic weights.

Step-by-Step Solution:

Step 1: Mendeleev's periodic table was primarily organized by atomic weight.

Step 2: Isotopes of an element have the same atomic number but different atomic masses due to varying numbers of neutrons.

Step 3: The existence of isotopes means that elements don't have a single, fixed atomic weight, which can lead to inconsistencies in Mendeleev's original ordering.

Why other options are wrong:
- Option B: Noble gases not forming compounds doesn't directly upset the ordering of elements in the table, but rather their placement and understanding of their properties.
- Option C: Subgroups A and B were later additions to the periodic table to accommodate different properties within a group, but this doesn't fundamentally upset the original table's organization.
- Option D: Atomic weights not being whole numbers is a consequence of isotopes and doesn't directly upset the table as much as the existence of multiple isotopes for the same element.

Key Concepts:
- Mendeleev's Periodic Table
- Isotopes
- Atomic Weight

Memory Aids:
- Remember that Mendeleev organized by atomic weight, and isotopes affect atomic weight.

Quick Tips:
- Consider the basis of Mendeleev's table (atomic weight) when evaluating factors that might disrupt it.

49 Easy

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Remembering +4 | -1

Atomic radii of

F

and

Ne

in

{A^0}

, are given by

View answer and explanation

Correct option: B

0.72, 1.60

Option B (0.72, 1.60) is correct because atomic radii decrease across a period, and noble gases have larger van der Waals radii.

Solution approach

Correct Answer + Core Reason: 0.72, 1.60. Atomic radii decrease across a period, and noble gases have larger van der Waals radii.

Step-by-Step Solution:

Step 1: Recall the trend of atomic radii across a period. Atomic radii generally decrease from left to right across a period due to increasing effective nuclear charge.

Step 2: Consider the position of Fluorine (F) and Neon (Ne) in the periodic table. Fluorine is a halogen, and Neon is a noble gas. Neon is to the right of Fluorine in the same period.

Step 3: Noble gases have larger atomic radii than the preceding halogens because their radii are van der Waals radii, while the radii of other elements are covalent radii. Van der Waals radii are larger than covalent radii.

Step 4: Therefore, the atomic radius of Fluorine is smaller than that of Neon. The given values are approximately 0.72 Å for Fluorine and 1.60 Å for Neon.

Why other options are wrong:
- Option A: 0.72, 0.71 is incorrect because Neon has a larger atomic radius than Fluorine.
- Option C: 1.6, 1.58 is incorrect because Fluorine has a smaller atomic radius than Neon.
- Option D: 0.71, 0.72 is incorrect because Neon has a larger atomic radius than Fluorine.

Key Concepts:
- Atomic radius trend across a period
- Van der Waals radius
- Covalent radius

Memory Aids:
- Atomic radius decreases across a period.
- Noble gases have larger van der Waals radii.

Quick Tips:
- Remember that noble gas radii are van der Waals radii, which are significantly larger than covalent radii.

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Application +4 | -1

The electro negativity of

\mathrm{H}, \mathrm{X}

and

\mathrm{O}

are

2.1,0.8

and

3.5

respectively on Pauling scale. The nature of compound

\mathrm{HOX}

is:

View answer and explanation

Correct option: B

Basic

Option B (Basic) is correct because the electronegativity difference between M' and O is smaller than between H and O, indicating a higher tendency for M'-O bond to break, releasing OH- ions.

Solution approach

Correct Answer + Core Reason: Basic. The electronegativity difference suggests HOX will dissociate to produce OH-.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference between H and O: 3.5 - 2.1 = 1.4.

Step 2: Calculate the electronegativity difference between X and O: 3.5 - 0.8 = 2.7.

Step 3: Analyze the bond polarity. Since the electronegativity difference between X and O is larger than that between H and O, the O-X bond is more polar than the O-H bond.

Step 4: Predict the dissociation of HOX. The compound HOX will tend to dissociate as $H^+ + OX^-$ , which is not basic. However, since X is very electropositive, the compound is better written as XOH. The electronegativity difference between X and O is 2.7, which is large, so the X-O bond is more likely to be ionic. Thus, the compound will dissociate as $X^+ + OH^-$ , making it basic.

Why other options are wrong:
- Option A: Acidic: If HOX were acidic, it would dissociate as $H^+ + OX^-$ . This is less likely given the electronegativity differences.
- Option C: Neutral: HOX is unlikely to be neutral due to the significant electronegativity differences.
- Option D: Amphoteric: Amphoteric compounds can act as both acids and bases. While possible, the strong electropositivity of X makes it more likely to be basic.

Key Concepts:
- Electronegativity
- Bond polarity
- Acid-base character

Memory Aids:
- Higher electronegativity difference between O and X favors basic character.

Quick Tips:
- Consider the electronegativity differences to determine the polarity of bonds and predict the dissociation of the compound.

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Remembering +4 | -1

The element which occupies peak of Lother Meyer curve is

View answer and explanation

Correct option: C

K

Option C (

K

) is correct because alkali metals like Potassium (K) occupy the peaks in the Lothar Meyer curve.

Solution approach

Correct Answer + Core Reason: $K$ . Alkali metals with large atomic volumes occupy the peaks of the Lothar Meyer curve.

Step-by-Step Solution:

Step 1: Understand the Lothar Meyer Curve: It plots atomic volume against atomic weight.

Step 2: Identify the elements at the peaks: The peaks are occupied by alkali metals (Group 1 elements).

Step 3: Analyze the options:
- V (Vanadium) is a transition metal.
- Se (Selenium) is a non-metal.
- K (Potassium) is an alkali metal.
- La (Lanthanum) is a lanthanide.

Step 4: Conclusion: Potassium (K) is the alkali metal among the options and thus occupies the peak.

Why other options are wrong:
- **Option A ( $V$ ): Vanadium is a transition metal and does not occupy the peaks of the Lothar Meyer curve.
- Option B ( $Se$ ): Selenium is a non-metal and does not occupy the peaks of the Lothar Meyer curve.
- Option D ( $La$ ): Lanthanum is a lanthanide and does not occupy the peaks of the Lothar Meyer curve.

Key Concepts:
- Lothar Meyer Curve
- Periodic Trends
- Alkali Metals

Memory Aids:
- Alkali metals have the largest atomic volumes in their respective periods.

Quick Tips:**
- Remember the position of alkali metals in the periodic table and their characteristic properties.

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Understanding +4 | -1

Which pair has both members from the same period of Periodic Table?

View answer and explanation

Correct option: D

Na, Cl

Option D (Na, Cl) is correct because both Sodium (Na) and Chlorine (Cl) are in the same period (Period 3) of the Periodic Table.

Solution approach

Correct Answer + Core Reason: Na, Cl. Both Sodium and Chlorine are in the same period.

Step-by-Step Solution:

Step 1: Identify the period of Sodium (Na): Sodium is in Period 3.

Step 2: Identify the period of Chlorine (Cl): Chlorine is in Period 3.

Step 3: Identify the period of Calcium (Ca): Calcium is in Period 4.

Step 4: Identify the period of Bromine (Br): Bromine is in Period 4.

Step 5: Conclusion: The correct answer is Na, Cl because they are both in Period 3.

Why other options are wrong:
- Option A: Cl, Br: Chlorine is in Period 3, and Bromine is in Period 4.
- Option B: Ca, Cl: Calcium is in Period 4, and Chlorine is in Period 3.
- Option C: Na, Ca: Sodium is in Period 3, and Calcium is in Period 4.
- Option D: Na, Cl: Sodium and Chlorine are both in Period 3.

Key Concepts:
- Periodic Table structure
- Periodicity

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

Quick Tips:
- Use a periodic table to quickly determine the period of an element.

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Analyzing +4 | -1

Which one of the following statement is false?

View answer and explanation

Correct option: A

The electron affinity of chlorine is less than that of fluorine.

Option A (The electron affinity of chlorine is less than that of fluorine) is correct because chlorine has a higher electron affinity than fluorine due to smaller size and higher interelectronic repulsion in fluorine.

Solution approach

Correct Answer + Core Reason: The electron affinity of chlorine is less than that of fluorine. Chlorine has a higher electron affinity than fluorine due to fluorine's small size and strong interelectronic repulsions.

Step-by-Step Solution:

Step 1: Define electron affinity.\nElectron affinity is the energy change when an electron is added to a neutral gaseous atom.

Step 2: Analyze the trends in electron affinity.\nGenerally, electron affinity increases across a period and decreases down a group. However, chlorine has a higher electron affinity than fluorine due to the small size of fluorine and the resulting interelectronic repulsions.

Step 3: Evaluate the given statements.\nA. The electron affinity of chlorine is less than that of fluorine: This is false.\nB. The electronegativity of fluorine is more than that of chlorine: This is true.\nC. The electron affinity of bromine is less than that of chlorine: This is true.\nD. The electronegativity of chlorine is more than that of bromine: This is true.

Step 4: Final Answer: The false statement is A.

Why other options are wrong:
- Option B: The electronegativity of fluorine is more than that of chlorine: This is a true statement.
- Option C: The electron affinity of bromine is less than that of chlorine: This is a true statement.
- Option D: The electronegativity of chlorine is more than that of bromine: This is a true statement.

Key Concepts:
- Electron affinity
- Electronegativity
- Periodic trends

Memory Aids:
- Remember that chlorine has a higher electron affinity than fluorine.

Quick Tips:
- Be aware of exceptions to general periodic trends.

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Remembering +4 | -1

Which of the following element has less electro negativity

View answer and explanation

Correct option: B

S

Option B (S) is correct because electronegativity decreases down a group in the periodic table, and sulfur (S) is below oxygen (O) in Group 16.

Solution approach

Correct Answer + Core Reason: S. Electronegativity decreases down the group.

Step-by-Step Solution:

Step 1: Understand electronegativity trends in the periodic table.

Step 2: Electronegativity increases across a period (from left to right).

Step 3: Electronegativity decreases down a group (from top to bottom).

Step 4: Locate the elements in the periodic table.

Step 5: O, S are in Group 16 (Chalcogens).

Step 6: F is in Group 17 (Halogens).

Step 7: N is in Group 15 (Pnictogens).

Step 8: Since electronegativity decreases down a group, S is less electronegative than O.

Step 9: F is the most electronegative element.

Step 10: N is more electronegative than S.

Why other options are wrong:
- Option A: O: Oxygen is more electronegative than Sulfur.
- Option C: F: Fluorine is the most electronegative element.
- Option D: N: Nitrogen is more electronegative than Sulfur.

Key Concepts:
- Electronegativity
- Periodic trends

Memory Aids:
- Electronegativity increases across a period, decreases down a group.

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

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Remembering +4 | -1

Neutral oxide among the following is

View answer and explanation

Correct option: C

CO

Option C (

CO

) is correct because carbon monoxide (CO) does not react with either acids or bases, making it a neutral oxide.

Solution approach

Correct Answer + Core Reason: $CO$ . Carbon monoxide is a neutral oxide as it does not react with acids or bases.

Step-by-Step Solution:

Step 1: Understand the definition of neutral oxides.

Step 2: Neutral oxides are oxides that do not react with either acids or bases to form salts.

Step 3: $C{O_2}$ is an acidic oxide as it reacts with bases to form carbonates.

Step 4: $A{l_2}{O_3}$ is an amphoteric oxide as it reacts with both acids and bases.

Step 5: $CO$ is a neutral oxide as it does not react with either acids or bases under normal conditions.

Step 6: $N{a_2}O$ is a basic oxide as it reacts with acids to form salts.

Why other options are wrong:
- Option A: $C{O_2}$ : Carbon dioxide is an acidic oxide.
- Option B: $A{l_2}{O_3}$ : Aluminum oxide is an amphoteric oxide.
- Option D: $N{a_2}O$ : Sodium oxide is a basic oxide.

Key Concepts:
- Neutral oxide
- Acidic oxide
- Basic oxide
- Amphoteric oxide

Memory Aids:
- CO is a neutral oxide.

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

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Understanding +4 | -1

Born Haber cycle is used to determine:

View answer and explanation

Correct option: D

Either of them

Option D (Either of them) is correct because the Born-Haber cycle relates lattice energy, electron affinity, ionization energy, and other thermodynamic quantities, allowing for the determination of any one of them if the others are known.

Solution approach

Correct Answer + Core Reason: Either of them. The Born-Haber cycle can be used to determine lattice energy, electron affinity, or ionization energy.

Step-by-Step Solution:

Step 1: Define the Born-Haber cycle: The Born-Haber cycle is a thermodynamic cycle that relates the lattice energy of an ionic compound to its enthalpy of formation and other thermodynamic quantities such as ionization energy, electron affinity, sublimation energy, and bond dissociation energy.

Step 2: The cycle is based on Hess's law, which states that the enthalpy change for a reaction is independent of the path taken.

Step 3: By applying Hess's law to the formation of an ionic compound, we can relate the lattice energy to other thermodynamic quantities.

Step 4: Therefore, if we know all the other quantities, we can determine the lattice energy. Conversely, if we know the lattice energy and all other quantities except one, we can determine that unknown quantity (which could be electron affinity or ionization energy).

Why other options are wrong:
- Option A: Lattice energy is a common application, but not the only one.
- Option B: Electron affinity can be determined using the cycle, but it's not the only application.
- Option C: Ionization energy can be determined using the cycle, but it's not the only application.
- Option D: Either of them is the correct answer as the cycle can be used to determine any of the thermodynamic quantities involved.

Key Concepts:
- Born-Haber Cycle
- Lattice Energy
- Electron Affinity
- Ionization Energy
- Hess's Law

Memory Aids:
- The Born-Haber cycle is a thermodynamic cycle that relates various energy terms.

Quick Tips:
- Understand the relationship between the different energy terms in the Born-Haber cycle.

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Remembering +4 | -1

Which of the following set of elements follows dobereiner law of triads

View answer and explanation

Correct option: B

Li, Na, K

Option B is correct. The Dobereiner's law of triads states that the atomic mass of the middle element is approximately the arithmetic mean of the atomic masses of the other two elements. Li, Na, and K follow this law.

Solution approach

Correct Answer + Core Reason: Li, Na, K. Dobereiner's law of triads states that the atomic mass of the middle element is approximately the arithmetic mean of the atomic masses of the other two elements.

Step-by-Step Solution:

Step 1: Recall Dobereiner's Law of Triads: In a triad, the atomic weight of the middle element is approximately the average of the atomic weights of the first and third elements.

Step 2: Check the given options:\nA. Fe, Co, Ni: Atomic masses are approximately 56, 59, 59. (56+59)/2 != 59\nB. Li, Na, K: Atomic masses are approximately 7, 23, 39. (7+39)/2 = 23\nC. Ru, Rh, Pd: Atomic masses are approximately 101, 103, 106. (101+106)/2 != 103\nD. Os, Ir, Pt: Atomic masses are approximately 190, 192, 195. (190+195)/2 != 192

Step 3: Conclusion: Only Li, Na, and K follow Dobereiner's law of triads.

Why other options are wrong:
- Option A: Fe, Co, Ni do not follow the law of triads.
- Option C: Ru, Rh, Pd do not follow the law of triads.
- Option D: Os, Ir, Pt do not follow the law of triads.

Key Concepts:
- Dobereiner's Law of Triads
- Atomic Mass

Memory Aids:
- Remember the triad of Li, Na, K as a classic example.

Quick Tips:
- Calculate the average atomic mass of the first and third elements and compare it to the atomic mass of the middle element.

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Analyzing +4 | -1

In which of the following arrangements the order is not correct according to property indicated against it?

View answer and explanation

Correct option: B

Increasing

I{E_1}\mathrm{\;}:\mathrm{B} < C < N < O

Option B (Increasing

I{E_1}\mathrm{\;}:\mathrm{B} < C < N < O

) is correct because the first ionization energy of Oxygen is less than Nitrogen due to the extra stability of half-filled p-orbitals in Nitrogen.

Solution approach

Correct Answer + Core Reason: Increasing $I{E_1}\mathrm{\;}:\mathrm{B} < C < N < O$ . The ionization energy of N is greater than O due to the half-filled p-orbital configuration of N.

Step-by-Step Solution:

Step 1: Analyze Option A: Increasing size $:\mathrm{A}{\mathrm{l}^{3 + }} < M{\mathrm{g}^{2 + }} < N{\mathrm{a}^ + } < {\mathrm{F}^ - }$ . This is correct. All are isoelectronic species. For isoelectronic species, as the nuclear charge increases, the size decreases. The order of nuclear charge is: Al > Mg > Na > F. Therefore, the order of size is: $\mathrm{A}{\mathrm{l}^{3 + }} < M{\mathrm{g}^{2 + }} < N{\mathrm{a}^ + } < {\mathrm{F}^ - }$ .

Step 2: Analyze Option B: Increasing $I{E_1}\mathrm{\;}:\mathrm{B} < C < N < O$ . This is incorrect. The general trend is that ionization energy increases across a period. However, there is an exception between N and O. The electronic configuration of N is $1s^2 2s^2 2p^3$ (half-filled p-orbitals), which is more stable than the electronic configuration of O ( $1s^2 2s^2 2p^4$ ). Therefore, it requires more energy to remove an electron from N than from O. The correct order is: B < C < O < N.

Step 3: Analyze Option C: Increasing $E{A_1}\mathrm{\;}:\mathrm{I} < Br < F < Cl$ . This is correct. Electron affinity generally increases across a period and decreases down a group. However, chlorine has a higher electron affinity than fluorine due to its smaller size and greater electron density which causes greater repulsion in fluorine.

Step 4: Analyze Option D: Increasing metallic radius : $\mathrm{Li} < Na < K < Rb$ . This is correct. Metallic radius increases down a group.

Why other options are wrong:
- Option A: Increasing size $:\mathrm{A}{\mathrm{l}^{3 + }} < M{\mathrm{g}^{2 + }} < N{\mathrm{a}^ + } < {\mathrm{F}^ - }$ : This order is correct for isoelectronic species.
- Option C: Increasing $E{A_1}\mathrm{\;}:\mathrm{I} < Br < F < Cl$ : This order is correct due to the anomalous behavior of F.
- Option D: Increasing metallic radius : $\mathrm{Li} < Na < K < Rb$ : This order is correct as metallic radius increases down the group.

Key Concepts:
- Ionization energy trends
- Electron affinity trends
- Atomic size trends
- Electronic configuration

Memory Aids:
- Remember the exceptions in ionization energy and electron affinity trends.

Quick Tips:
- Always consider electronic configuration when comparing ionization energies.

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Application +4 | -1

Maximum number of orbitals in

3^{\text {rd }}

excited state of

L i^{+2}=x

. Electronegativity (Pauling scale) of most electronegative element

=y

. Then find value of

\left(\frac{x}{y}\right)

.

View answer and explanation

Correct option: A

4

Option A (4) is correct because the value of x is 16 and the electronegativity of fluorine is 4, so x/y = 16/4 = 4.

Solution approach

Correct Answer + Core Reason: 4. The number of orbitals in the 3rd excited state of $Li^{+2}$ is 16, and the electronegativity of Fluorine is 4. Thus, 16/4 = 4.

Step-by-Step Solution:

Step 1: Determine the principal quantum number (n) for the 3rd excited state. The ground state is n=1, so the 3rd excited state is n=1+3 = 4.

Step 2: Calculate the total number of orbitals in the n=4 energy level. The number of orbitals is given by $n^2$ , so $4^2 = 16$ . Therefore, x = 16.

Step 3: Identify the most electronegative element. The most electronegative element is Fluorine (F).

Step 4: Determine the electronegativity of Fluorine on the Pauling scale. The electronegativity of Fluorine is approximately 4.0, so y = 4.

Step 5: Calculate x/y. x/y = 16/4 = 4.

Why other options are wrong:
- Option B (6): This is incorrect because it doesn't correctly calculate the number of orbitals in the 3rd excited state or the final division.
- Option C (8): This is incorrect because it doesn't correctly calculate the number of orbitals in the 3rd excited state or the final division.
- Option D (10): This is incorrect because it doesn't correctly calculate the number of orbitals in the 3rd excited state or the final division.

Key Concepts:
- Quantum numbers
- Electronic configuration
- Electronegativity
- Pauling scale

Memory Aids:
- The number of orbitals in a given energy level n is $n^2$ .

Quick Tips:
- Remember the electronegativity values of common elements, especially Fluorine.

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Remembering +4 | -1

The only non-metal which is liquid at ordinary temperature is

View answer and explanation

Correct option: B

B{r_2}

Option B (

B{r_2}

) is correct because Bromine (

B{r_2}

) is the only non-metal that exists as a liquid at room temperature.

Solution approach

Correct Answer + Core Reason: $B{r_2}$ . Bromine is the only non-metal that is liquid at room temperature.

Step-by-Step Solution:

Step 1: Recall the physical states of common non-metals at room temperature: Most non-metals are gases (e.g., nitrogen, oxygen, chlorine) or solids (e.g., carbon, sulfur, phosphorus).

Step 2: Identify the non-metal that is liquid: Bromine ( $B{r_2}$ ) is a reddish-brown liquid at room temperature.

Step 3: Conclusion: The correct answer is $B{r_2}$ .

Why other options are wrong:
- Option A: Hg: Mercury (Hg) is a metal that is liquid at room temperature, not a non-metal.
- Option C: $N{H_3}$ : Ammonia ( $N{H_3}$ ) is a gas at room temperature.
- Option D: None of these: This is incorrect because Bromine is the correct answer.

Key Concepts:
- Physical states of elements
- Non-metals

Memory Aids:
- Remember that Bromine is the only non-metal that is liquid at room temperature.

Quick Tips:
- Know the common physical states of elements at room temperature.

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Classification of elements and periodicity in properties practice · Easy

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