CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Easy, Page 5

121 Easy

Remembering +4 | -1

The elements which occupy peaks in the atomic volume curves are

View answer and explanation

Correct option: B

Na, K, Rb

Option B (Na, K, Rb) is correct because alkali metals (Na, K, Rb) have the largest atomic volumes in their respective periods, thus occupying the peaks in the atomic volume curves.

Solution approach

Correct Answer + Core Reason: Na, K, Rb. Alkali metals occupy the peaks in the atomic volume curves.

Step-by-Step Solution:

Step 1: Recall the trend of atomic volume in the periodic table.\nAtomic volume generally increases down a group and decreases across a period.

Step 2: Consider the given options.\nNa, K, and Rb are alkali metals, which have the largest atomic volumes in their respective periods.\nCu, Ag, and Au are transition metals.\nCl, Br, and I are halogens.\nFe, Co, and Ni are transition metals.

Step 3: Determine which set occupies the peaks in the atomic volume curves.\nAlkali metals occupy the peaks in the atomic volume curves because they have the largest atomic volumes in their respective periods.

Why other options are wrong:
- Option A: Cu, Ag, and Au are transition metals and do not occupy the peaks in the atomic volume curves.
- Option C: Cl, Br, and I are halogens and do not occupy the peaks in the atomic volume curves.
- Option D: Fe, Co, and Ni are transition metals and do not occupy the peaks in the atomic volume curves.

Key Concepts:
- Atomic volume
- Periodic trends
- Alkali metals

Memory Aids:
- Remember that alkali metals have the largest atomic volumes in their periods.

Quick Tips:
- Know the trends of atomic properties in the periodic table.

122 Easy

122 of 171Question permalink

Remembering +4 | -1

Which of the following shows diagonal relationship

View answer and explanation

Correct option: A

boron and silicon

Option A is correct because boron and silicon exhibit a diagonal relationship due to their similar electronegativities and polarizing power.

Solution approach

Correct Answer + Core Reason: boron and silicon. Boron and silicon show similar properties due to diagonal relationship.

Step-by-Step Solution:

Step 1: Define diagonal relationship.\nDiagonal relationship refers to the similarities in properties between elements located diagonally adjacent to each other in the periodic table. This is primarily observed in the second and third periods.

Step 2: Identify the diagonal relationships in the periodic table.\nThe most well-known diagonal relationships are between:
- Lithium (Li) and Magnesium (Mg)
- Beryllium (Be) and Aluminum (Al)
- Boron (B) and Silicon (Si)

Step 3: Conclusion.\nBoron and silicon show a diagonal relationship.

Why other options are wrong:
- Option A: Correct because boron and silicon show a diagonal relationship.
- Option B: Incorrect because boron and carbon are in the same period and do not show a diagonal relationship.
- Option C: Incorrect because boron and gallium are in the same group and do not show a diagonal relationship.
- Option D: Incorrect because boron and aluminum are in the same group and do not show a diagonal relationship.

Key Concepts:
- Diagonal relationship
- Periodic table

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

Quick Tips:
- Understand the reasons for diagonal relationships, such as similar electronegativities and polarizing power.

123 Easy

123 of 171Question permalink

Understanding +4 | -1

Among

O,\mathrm{ }C,\mathrm{ }F,\mathrm{ }Cl,\mathrm{ }Br

the correct order of increasing atomic radii is:

View answer and explanation

Correct option: A

\mathrm{F} < O < C < Cl < Br

Option A (

\mathrm{F} < O < C < Cl < Br

) is correct because atomic radii increase down a group and generally decrease across a period.

Solution approach

Correct Answer + Core Reason: $\mathrm{F} < O < C < Cl < Br$ . Atomic size increases down the group and decreases along the period.

Step-by-Step Solution:

Step 1: Identify the positions of the elements in the periodic table.

Step 2: Recall the trend of atomic radii: increases down a group, decreases across a period.

Step 3: Arrange the elements according to their positions and trends:
- F and O are in the same period (2nd), with F to the right of O, so F < O.
- C is also in the 2nd period, to the left of O, so F < O < C.
- Cl and Br are in the same group (17), with Cl above Br, so Cl < Br.
- C is in period 2 and Cl is in period 3, so C < Cl.

Step 4: Combine the trends to get the final order: $\mathrm{F} < O < C < Cl < Br$ .

Why other options are wrong:
- Option B: $\mathrm{F} < C < O < Br < Cl$ : Incorrect order of O and C within the period and Cl and Br within the group.
- Option C: $\mathrm{F} < Cl < Br < O < C$ : Incorrect order of elements within periods and groups.
- Option D: $\mathrm{C} < O < F < Cl < Br$ : Incorrect order of F, O, and C within the period.

Key Concepts:
- Periodic trends of atomic radii.
- Position of elements in the periodic table.

Memory Aids:
- Atomic size increases as you go down a group (more electron shells).
- Atomic size decreases as you go across a period (increased nuclear charge).

Quick Tips:
- Always consider both group and period trends when comparing atomic radii.
- Visualize the periodic table to quickly determine relative positions of elements.

124 Easy

124 of 171Question permalink

Understanding +4 | -1

The electron with z = 117 and z = 120 belong to ______ ______ family respectively.

View answer and explanation

Correct option: C

Halogen family and alkali earth metals

Option C (Halogen family and alkali earth metals) is correct because element with Z=117 belongs to the halogen family and Z=120 belongs to alkaline earth metals.

Solution approach

Correct Answer + Core Reason: Halogen family and alkali earth metals. Elements with Z=117 and Z=120 belong to the halogen family and alkali earth metals, respectively.

Step-by-Step Solution:

Step 1: Determine the group of the element with Z = 117.
- Element 117 is Tennessine (Ts), which belongs to Group 17, the halogen family.

Step 2: Determine the group of the element with Z = 120.
- Element 120 is an alkaline earth metal, belonging to Group 2.

Step 3: Match the groups to the given options.

Why other options are wrong:
- Option A: Halogen family and alkali metals: Alkali metals belong to Group 1, not Group 2.
- Option B: Nitrogen family and alkali metals: Nitrogen family belongs to Group 15, and alkali metals belong to Group 1.
- Option D: Chalcogens family and alkali metals: Chalcogens belong to Group 16, and alkali metals belong to Group 1.

Key Concepts:
- Periodic table and group numbers.
- Halogens, alkali metals, alkaline earth metals, chalcogens, and nitrogen family.

Memory Aids:
- Group 1: Alkali metals.
- Group 2: Alkaline earth metals.
- Group 16: Chalcogens.
- Group 17: Halogens.
- Group 15: Pnictogens (Nitrogen family).

Quick Tips:
- Knowing the group numbers and their corresponding families is essential for predicting element properties.

125 Easy

125 of 171Question permalink

Understanding +4 | -1

An element whose IUPAC name is ununtrium (Uut), belongs to

View answer and explanation

Correct option: B

P – block

Option B (P – block) is correct because Ununtrium (Uut), now known as Moscovium (Mc), belongs to group 15, which is in the p-block.

Solution approach

Correct Answer + Core Reason: P – block. Ununtrium (Uut), now known as Moscovium (Mc), belongs to the p-block.

Step-by-Step Solution:

Step 1: Identify the element Ununtrium (Uut). It is now known as Moscovium (Mc) with atomic number 115.

Step 2: Determine the group to which Moscovium belongs. Moscovium belongs to group 15.

Step 3: Relate the group number to the block. Group 15 elements belong to the p-block.

Step 4: Conclude that Ununtrium belongs to the p-block.

Why other options are wrong:
- Option A: S – block elements belong to group 1 and 2. Moscovium does not belong to these groups.
- Option C: d – block elements are transition metals. Moscovium is not a transition metal.
- Option D: Transition elements are d-block elements. Moscovium is not a transition metal.

Key Concepts:
- Periodic table blocks (s, p, d, f)
- Group numbers and their relation to blocks

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

Quick Tips:
- Knowing the group number helps determine the block to which an element belongs.

126 Easy

126 of 171Question permalink

Understanding +4 | -1

The order of stability of metal oxides is

View answer and explanation

Correct option: C

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} < C{\mathrm{r}_2}{\mathrm{O}_3} < A{\mathrm{l}_2}{\mathrm{O}_3} < MgO

Option C (

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} < C{\mathrm{r}_2}{\mathrm{O}_3} < A{\mathrm{l}_2}{\mathrm{O}_3} < MgO

) is correct because the stability of metal oxides generally increases with the electropositivity of the metal.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} < C{\mathrm{r}_2}{\mathrm{O}_3} < A{\mathrm{l}_2}{\mathrm{O}_3} < MgO$ . The stability of oxides increases with the electropositivity of the metal.

Step-by-Step Solution:

Step 1: Understand the question. The question asks for the order of stability of metal oxides.

Step 2: Recall the factors affecting the stability of metal oxides. The stability of metal oxides generally increases with the electropositivity of the metal. More electropositive metals form more stable oxides.

Step 3: Determine the relative electropositivity of the metals. The order of electropositivity is: Fe < Cr < Al < Mg.

Step 4: Arrange the oxides in order of increasing stability based on the electropositivity of the metals: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} < C{\mathrm{r}_2}{\mathrm{O}_3} < A{\mathrm{l}_2}{\mathrm{O}_3} < MgO$

Why other options are wrong:
- Option A: Incorrect order of stability.
- Option B: Incorrect order of stability.
- Option D: Incorrect order of stability.

Key Concepts:
- Stability of Metal Oxides
- Electropositivity of Metals

Memory Aids:
- More electropositive metals form more stable oxides.

Quick Tips:
- Consider the position of the metal in the electrochemical series to determine its electropositivity.

127 Easy

127 of 171Question permalink

Application +4 | -1

In which block does 106th element belong?

View answer and explanation

Correct option: C

d-block

Option C (d-block) is correct because the 106th element, Seaborgium (Sg), is a transition metal and belongs to the d-block.

Solution approach

Correct Answer + Core Reason: d-block. The 106th element is Seaborgium (Sg), which is a d-block element.

Step-by-Step Solution:

Step 1: Identify the 106th element in the periodic table. The 106th element is Seaborgium (Sg).

Step 2: Determine the block to which Seaborgium belongs. Seaborgium is a transition metal and belongs to the d-block.

Why other options are wrong:
- Option A: s-block elements are in groups 1 and 2. Seaborgium is not in these groups.
- Option B: p-block elements are in groups 13-18 (excluding Helium). Seaborgium is not in these groups.
- Option D: f-block elements are lanthanides and actinides. Seaborgium is not a lanthanide or actinide.

Key Concepts:
- Periodic table blocks (s, p, d, f)
- Transition metals

Memory Aids:
- d-block elements are transition metals.

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

128 Easy

128 of 171Question permalink

Analyzing +4 | -1

Statement I:

F

is more electronegative than

Cl.

Statement II:

F

has high electron affinity than

Cl.

View answer and explanation

Correct option: C

If statement 1 is true but statement 2 is false

Option C is correct. Fluorine is more electronegative than chlorine, but chlorine has a higher electron affinity than fluorine due to fluorine's small size and strong interelectronic repulsions.

Solution approach

Correct Answer + Core Reason: If statement 1 is true but statement 2 is false. Fluorine is indeed more electronegative than chlorine, but chlorine has a higher electron affinity.

Step-by-Step Solution:

Step 1: Analyze Statement I: Electronegativity trends generally increase across a period and decrease down a group. Fluorine is above chlorine in the same group (Group 17), so it is more electronegative.

Step 2: Analyze Statement II: Electron affinity is the energy change when an electron is added to a neutral atom. Chlorine has a higher electron affinity than fluorine. This is due to the small size of fluorine, which leads to strong interelectronic repulsions, making it less favorable to add an electron.

Step 3: Combine the analyses: Statement I is true, and Statement II is false.

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

Key Concepts:
- Electronegativity trends
- Electron affinity trends
- Interelectronic repulsions

Memory Aids:
- Electronegativity: FONClBrISCH (most to least electronegative)
- Electron Affinity: Chlorine > Fluorine (an exception to the general trend)

Quick Tips:
- Remember the exceptions to general trends, such as the electron affinity of chlorine and fluorine.

129 Easy

129 of 171Question permalink

Analyzing +4 | -1

Anomalous pair among the following is

View answer and explanation

Correct option: D

Cobalt – Nickel

Option D (Cobalt – Nickel) is correct because Cobalt and Nickel have nearly the same atomic masses, which is an anomaly in the periodic trends.

Solution approach

Correct Answer + Core Reason: Cobalt – Nickel. This pair violates the increasing atomic mass order in Mendeleev's periodic table.

Step-by-Step Solution:

Step 1: Understand the concept of anomalous pairs. Anomalous pairs are elements that do not follow the expected trend of increasing atomic mass in the periodic table.

Step 2: Analyze the given pairs:
- Boron – silicon: These elements belong to different groups and periods and do not form an anomalous pair.
- Beryllium – Indium: These elements belong to different groups and periods and do not form an anomalous pair.
- Aluminium – Galium: These elements belong to the same group and follow the expected trend.
- Cobalt – Nickel: Cobalt (atomic number 27) has a slightly higher atomic mass than Nickel (atomic number 28), which is an anomaly.

Step 3: Identify the anomalous pair. Cobalt and Nickel form an anomalous pair because their atomic masses are not in the expected order.

Why other options are wrong:
- Option A (Boron – silicon): These elements are in different groups and periods and do not exhibit anomalous behavior.
- Option B (Beryllium – Indium): These elements are in different groups and periods and do not exhibit anomalous behavior.
- Option C (Aluminium – Galium): These elements are in the same group and follow the expected trend of increasing atomic mass.
- Option D (Cobalt – Nickel): This is the correct anomalous pair.

Key Concepts:
- Mendeleev's periodic table
- Atomic mass trends
- Anomalous pairs

Memory Aids:
- Remember that Cobalt and Nickel have atomic masses that are close, leading to the anomaly.

Quick Tips:
- Know the general trends in the periodic table and be aware of exceptions like the Cobalt-Nickel pair.

130 Easy

130 of 171Question permalink

Analyzing +4 | -1

Assertion (A): In a triad, the three elements present have different gaps of atomic masses.

Reason (R): Elements in a triad have different properties.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C is correct because the assertion about triads having different gaps in atomic masses is true, but the reason that elements in a triad have different properties is not the correct explanation for the assertion.

Solution approach

Correct Answer + Core Reason: (A) is true but (R) is false. The atomic mass gaps in triads are not uniform, and the reason provided is a general statement, not a specific explanation.

Step-by-Step Solution:

Step 1: Analyze the Assertion (A): In a triad, the three elements present have different gaps of atomic masses. This is generally true; the differences in atomic masses between elements in a triad are not constant.

Step 2: Analyze the Reason (R): Elements in a triad have different properties. This is also true; elements have different properties.

Step 3: Determine if (R) is the correct explanation of (A). The reason (R) does not explain why the atomic mass gaps are different. The different properties of elements in a triad do not directly cause the varying atomic mass gaps.

Step 4: Conclusion: Assertion (A) is true, but Reason (R) is false, or at least not the correct explanation.

Why other options are wrong:
- Option A: Both (A) & (R) are true and the (R) is the correct explanation of the (A): This is incorrect because while both statements are true, the reason does not explain the assertion.
- Option B: Both (A) & (R) are true but the (R) is not the correct explanation of the (A): This is incorrect because (R) is not true.
- Option D: Both (A) and (R) are false: This is incorrect because (A) is true.

Key Concepts:
- Dobereiner's Triads
- Atomic Mass
- Periodic Properties

Memory Aids:
- Remember that triads were an early attempt to classify elements based on atomic mass relationships.

Quick Tips:
- When analyzing assertion-reason questions, first determine if each statement is true or false independently. Then, consider if the reason logically explains the assertion.

131 Easy

131 of 171Question permalink

Remembering +4 | -1

Eka aluminum and Eka silicon are known as

View answer and explanation

Correct option: A

Ga and Ge

Option A is correct. Eka-aluminum was predicted by Mendeleev to be Gallium (Ga), and Eka-silicon was predicted to be Germanium (Ge).

Solution approach

Correct Answer + Core Reason: Ga and Ge. Eka-aluminum refers to Gallium (Ga), and Eka-silicon refers to Germanium (Ge), as predicted by Mendeleev.

Step-by-Step Solution:

Step 1: Recall Mendeleev's predictions: Mendeleev predicted the existence and properties of several elements that were not yet discovered at the time. He named them using the prefix "eka-" (Sanskrit for "one") before the name of the element above them in the periodic table.

Step 2: Identify Eka-aluminum: Eka-aluminum was the element predicted to be below aluminum in the periodic table, which is Gallium (Ga).

Step 3: Identify Eka-silicon: Eka-silicon was the element predicted to be below silicon in the periodic table, which is Germanium (Ge).

Why other options are wrong:
- Option B: Al and Si are the elements used as the basis for the "eka-" naming convention, not the predicted elements themselves.
- Option C: Fe and S are not related to Mendeleev's eka-aluminum or eka-silicon predictions.
- Option D: N and Mg are not related to Mendeleev's eka-aluminum or eka-silicon predictions.

Key Concepts:
- Mendeleev's Periodic Table
- Predictions of New Elements
- Eka-aluminum (Gallium)
- Eka-silicon (Germanium)

Memory Aids:
- Remember that "eka-" means "one after" in Sanskrit.

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

132 Easy

132 of 171Question permalink

Understanding +4 | -1

Periodic classification of elements is used to examine the

View answer and explanation

Correct option: C

Both (1) and (2)

The correct answer is C (Both (1) and (2)) because the periodic classification is used to examine both physical and chemical properties of elements.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). Periodic classification helps in understanding trends in both physical and chemical properties of elements.

Step-by-Step Solution:

Step 1: Recall the purpose of the periodic table.

Step 2: The periodic table organizes elements based on their atomic number and recurring chemical properties.

Step 3: The periodic table also reveals trends in physical properties such as atomic size, ionization energy, and electronegativity.

Why other options are wrong:
- Option A: Periodic trends in physical properties of elements is partially correct but incomplete.
- Option B: Periodic trends in chemical properties of elements is partially correct but incomplete.
- Option D: None of the above is incorrect because both (1) and (2) are correct.

Key Concepts:
- Periodic Trends
- Physical Properties
- Chemical Properties

Memory Aids:
- The periodic table is a tool for understanding both physical and chemical behavior.

Quick Tips:
- Remember that the periodic table is organized based on recurring properties.

133 Easy

133 of 171Question permalink

Remembering +4 | -1

The general electronic configuration of the transition elements is

View answer and explanation

Correct option: D

\left( {n - 1} \right){d^{1 - 10}},\;n{s^{1 - 2}}

Option D ((n - 1)d^(1-10), ns^(1-2)) is correct because it accurately represents the filling of d orbitals and s orbitals in transition elements.

Solution approach

Correct Answer + Core Reason: $\left( {n - 1} \right){d^{1 - 10}},\;n{s^{1 - 2}}$ . This configuration correctly describes the filling of d and s orbitals in transition elements.

Step-by-Step Solution:

Step 1: Define transition elements: Transition elements are elements that have partially filled d orbitals in their elemental state or in one or more of their common oxidation states.

Step 2: Recall the general electronic configuration of transition elements: The general electronic configuration is (n-1)d^(1-10) ns^(1-2), where n is the principal quantum number of the outermost shell.

Step 3: Analyze the options:\nA. $\left( {n - 1} \right){d^{10}},\mathrm{\;}\left( {n + 1} \right){s^2}\mathrm{\;}$ : Incorrect. The s orbital should have the principal quantum number n, not (n+1).\nB. $\left( {n - 1} \right){d^{1 - 10}},\left( {n + 1} \right){s^{1 - 2}}$ : Incorrect. The s orbital should have the principal quantum number n, not (n+1).\nC. $\left( {n - 1} \right){d^{1 - 10}},n{p^6},\;n{s^2}$ : Incorrect. This configuration includes a filled p orbital, which is not generally part of the transition metal configuration. Also, the s orbital can have 1 or 2 electrons.\nD. $\left( {n - 1} \right){d^{1 - 10}},\;n{s^{1 - 2}}$ : Correct. This configuration accurately represents the filling of d orbitals and s orbitals in transition elements.

Why other options are wrong:
- Option A: $\left( {n - 1} \right){d^{10}},\mathrm{\;}\left( {n + 1} \right){s^2}\mathrm{\;}$ : The (n+1)s² part is incorrect. The s orbital belongs to the nth shell.
- Option B: $\left( {n - 1} \right){d^{1 - 10}},\left( {n + 1} \right){s^{1 - 2}}$ : The (n+1)s^(1-2) part is incorrect. The s orbital belongs to the nth shell.
- Option C: $\left( {n - 1} \right){d^{1 - 10}},n{p^6},\;n{s^2}$ : The inclusion of np⁶ is incorrect for the general transition metal configuration.

Key Concepts:
- Transition Elements
- Electronic Configuration
- d-block Elements

Memory Aids:
- Remember that transition elements have partially filled d orbitals.

Quick Tips:
- The general electronic configuration of transition elements is (n-1)d^(1-10) ns^(1-2).

134 Easy

134 of 171Question permalink

Remembering +4 | -1

Which one of the following is not a transition metal

View answer and explanation

Correct option: D

Cd

Option D is correct. Cadmium (Cd) is not considered a typical transition metal because it has a completely filled d-orbital in its common oxidation states.

Solution approach

Correct Answer + Core Reason: Cd. Cadmium is not considered a typical transition metal because it has a completely filled d-orbital in its common oxidation states.

Step-by-Step Solution:

Step 1: Define transition metals: Transition metals are elements that have partially filled d orbitals in at least one of their common oxidation states.

Step 2: Analyze the electronic configurations of the given elements:\nA. Mn: $[\mathrm{Ar}] 3 d^5 4 s^2$ - Transition metal\nB. Cr: $[\mathrm{Ar}] 3 d^5 4 s^1$ - Transition metal\nC. Cu: $[\mathrm{Ar}] 3 d^{10} 4 s^1$ - Transition metal (Cu2+ has a d9 configuration)\nD. Cd: $[\mathrm{Kr}] 4 d^{10} 5 s^2$ - Not a transition metal (Cd2+ has a d10 configuration)

Step 3: Conclusion: Cd has a completely filled d-orbital (d10) in its common oxidation state (+2), so it is not considered a typical transition metal.

Why other options are wrong:
- Option A: Mn has partially filled d orbitals.
- Option B: Cr has partially filled d orbitals.
- Option C: Cu has partially filled d orbitals in some oxidation states.

Key Concepts:
- Transition Metals
- Electronic Configuration
- Oxidation States

Memory Aids:
- Transition metals have partially filled d orbitals.

Quick Tips:
- Check the electronic configuration of the element in its common oxidation states.

135 Easy

135 of 171Question permalink

Remembering +4 | -1

Atomic radii of fluorine and neon in angstrom unit are respectively given by:

View answer and explanation

Correct option: A

0.72,1.60

Option A (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. Fluorine is a halogen and Neon is a noble gas. 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 B: 1.60,1.60 is incorrect because Fluorine has a smaller atomic radius than Neon.
- Option C: 0.72,0.72 is incorrect because Neon has a larger atomic radius than Fluorine.
- Option D: 1.60,0.72 is incorrect because Fluorine has a smaller atomic radius than Neon.

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.

136 Easy

136 of 171Question permalink

Remembering +4 | -1

Which of the scientists given below discovered that periodic table should be based on the atomic number?

View answer and explanation

Correct option: C

Moseley

Option C (Moseley) is correct because Moseley discovered that the periodic table should be based on atomic number.

Solution approach

Correct Answer + Core Reason: Moseley. Moseley discovered the relationship between atomic number and the properties of elements.

Step-by-Step Solution:

Step 1: Recall the contributions of each scientist:
- Mendeleev: Arranged elements by atomic weight.
- Newlands: Law of Octaves.
- Moseley: Established atomic number as the basis for the periodic table.
- Lothar Meyer: Developed a periodic table based on physical properties.

Step 2: Identify the scientist who based the periodic table on atomic number: Moseley.

Why other options are wrong:
- Option A (Mendeleev): Mendeleev used atomic weight.
- Option B (Newlands): Newlands proposed the Law of Octaves.
- Option D (Lothar Meyer): Lothar Meyer focused on physical properties and atomic volume.

Key Concepts:
- Atomic Number
- Periodic Table
- History of Chemistry

Memory Aids:
- Moseley = Atomic Number.

Quick Tips:
- Remember the key contributions of each scientist to the development of the periodic table.

137 Easy

137 of 171Question permalink

Analyzing +4 | -1

Identify the least stable ion amongst the following:

View answer and explanation

Correct option: B

\mathrm{B}{\mathrm{e}^ - }

Option B (

\mathrm{B}{\mathrm{e}^ - }

) is the least stable ion because adding an electron to Beryllium (Be) disrupts its stable, completely filled s-orbital configuration.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{e}^ - }$ . Adding an electron to Be disrupts its stable configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the neutral atoms.\nLi: $1s^22s^1$ \nBe: $1s^22s^2$ \nB: $1s^22s^22p^1$ \nC: $1s^22s^22p^2$

Step 2: Write the electronic configurations of the negative ions.
$Li^-$ : $1s^22s^2$
$Be^-$ : $1s^22s^22p^1$
$B^-$ : $1s^22s^22p^2$
$C^-$ : $1s^22s^22p^3$

Step 3: Analyze the stability of the negative ions.
$Li^-$ has a stable filled $2s$ orbital.
$Be^-$ has an electron in the $2p$ orbital, which is less stable than a filled $s$ orbital.
$B^-$ has two electrons in the $2p$ orbital.
$C^-$ has three electrons in the $2p$ orbital.\nAdding an electron to Be disrupts its stable, completely filled s-orbital configuration. Therefore, $Be^-$ is the least stable.

Why other options are wrong:
- Option A: $\mathrm{L}{\mathrm{i}^ - }$ : This is incorrect because $Li^-$ has a stable filled $2s$ orbital.
- Option C: ${\mathrm{B}^ - }$ : This is incorrect because $B^-$ has two electrons in the $2p$ orbital.
- Option D: ${\mathrm{C}^ - }$ : This is incorrect because $C^-$ has three electrons in the $2p$ orbital.

Key Concepts:
- Electronic configuration
- Stability of ions

Memory Aids:
- Filled and half-filled orbitals are more stable.

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

138 Easy

138 of 171Question permalink

Remembering +4 | -1

Differentiating electron in inner transition elements enters the………….. orbital.

View answer and explanation

Correct option: D

f

Option D is correct because the differentiating electron in inner transition elements (lanthanides and actinides) enters the f-orbital.

Solution approach

Correct Answer + Core Reason: f. Inner transition elements are characterized by the filling of f-orbitals.

Step-by-Step Solution:

Step 1: Define inner transition elements.\nInner transition elements are those in which the last electron enters the (n-2)f orbitals. These elements comprise the lanthanides and actinides.

Step 2: Identify the orbital being filled.\nThe differentiating electron (the last electron added) in inner transition elements enters the f-orbital.

Step 3: Conclusion.\nThe differentiating electron in inner transition elements enters the f-orbital.

Why other options are wrong:
- Option A: Incorrect because the s-orbital is filled in alkali and alkaline earth metals.
- Option B: Incorrect because the p-orbital is filled in p-block elements.
- Option C: Incorrect because the d-orbital is filled in transition elements.
- Option D: Correct because the f-orbital is filled in inner transition elements.

Key Concepts:
- Inner transition elements
- f-orbitals

Memory Aids:
- Remember that f-block elements are also known as inner transition elements.

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

139 Easy

139 of 171Question permalink

Understanding +4 | -1

All elements in the third period have

View answer and explanation

Correct option: B

Three complete subshells

The correct answer is B (Three complete subshells) because elements in the third period have electrons filling the 3s, 3p, and 3d subshells (though 3d is not filled for all elements).

Solution approach

Correct Answer + Core Reason: Three complete subshells. Elements in the third period have electrons in the n=3 shell, which has 3s, 3p, and 3d subshells.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of elements in the third period (Na to Ar).

Step 2: The third period elements have electrons in the 3s and 3p subshells. Although the 3d subshell exists, it is not filled in the third period (it starts filling in the fourth period with Sc).

Step 3: The 3s, 3p, and 3d are the three subshells associated with the third energy level, so they have three complete subshells.

Why other options are wrong:
- Option A: Three complete shells is incorrect. They have three shells, but only the first two are complete.
- Option C: Three valence electrons is incorrect. The number of valence electrons varies across the period (from 1 to 8).
- Option D: Three electrons less than octet is incorrect. This only applies to Group 5A elements.

Key Concepts:
- Electronic Configuration
- Shells and Subshells
- Periodic Trends

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

Quick Tips:
- Understand the relationship between the period number and the principal quantum number (n).

140 Easy

140 of 171Question permalink

Remembering +4 | -1

Diagonal relationship is for

View answer and explanation

Correct option: D

B-Si

Option D (B-Si) is correct because Boron and Silicon exhibit similar properties due to their diagonal relationship.

Solution approach

Correct Answer + Core Reason: B-Si. Diagonal relationship exists between elements diagonally adjacent to each other in the periodic table.

Step-by-Step Solution:

Step 1: Recall the concept of diagonal relationship in the periodic table. Elements in the second and third periods exhibit a diagonal relationship due to the similarity in their charge/size ratio.

Step 2: Examine the given options:
- Li-Na: These are in the same group (alkali metals), not diagonally related.
- Be-Mg: These are in the same group (alkaline earth metals), not diagonally related.
- Si-C: These are in the same group, not diagonally related.
- B-Si: Boron (B) and Silicon (Si) are diagonally adjacent to each other.

Step 3: Therefore, the diagonal relationship is for B-Si.

Why other options are wrong:
- Option A: Li-Na is incorrect because Li and Na are in the same group (alkali metals), not diagonally related.
- Option B: Be-Mg is incorrect because Be and Mg are in the same group (alkaline earth metals), not diagonally related.
- Option C: Si-C is incorrect because Si and C are in the same group, not diagonally related.

Key Concepts:
- Diagonal relationship
- Periodic table trends

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

Quick Tips:
- Diagonal relationships occur between elements in the second and third periods.

141 Easy

141 of 171Question permalink

Remembering +4 | -1

The starting element of fourth period is

View answer and explanation

Correct option: A

K

Option A (K) is correct because Potassium (K) is the first element in the fourth period.

Solution approach

Correct Answer + Core Reason: K. Potassium is the first element of the fourth period.

Step-by-Step Solution:

Step 1: The fourth period starts after Argon (Ar), which completes the third period.

Step 2: The element following Argon is Potassium (K), with atomic number 19.

Step 3: Potassium's electronic configuration is [Ar] 4s1, placing it as the first element in the fourth period.

Why other options are wrong:
- Option B (Rb): Rubidium (Rb) is the first element of the fifth period.
- Option C (Kr): Krypton (Kr) is the last element of the fourth period (a noble gas).
- Option D (Xe): Xenon (Xe) is the last element of the fifth period (a noble gas).

Key Concepts:
- Periodic Table
- Periods
- Electronic Configuration

Memory Aids:
- Remember the order of elements in the periodic table to identify the starting element of each period.

Quick Tips:
- Knowing the atomic numbers and positions of common elements helps in quickly identifying the period they belong to.

142 Easy

142 of 171Question permalink

Remembering +4 | -1

Which one these basic

View answer and explanation

Correct option: D

N{a_2}O

Option D (

N{a_2}O

) is correct because oxides of metals are generally basic.

Solution approach

Correct Answer + Core Reason: $N{a_2}O$ is a basic oxide because it is an oxide of a highly electropositive metal (Sodium).

Step-by-Step Solution:

Step 1: Identify the nature of the elements in each oxide.

Step 2: Recall that oxides of metals are generally basic, while oxides of non-metals are acidic or neutral.

Step 3: $Si{O_2}$ , $S{O_2}$ , and $C{O_2}$ are oxides of non-metals (Silicon, Sulfur, and Carbon, respectively), and are therefore acidic or neutral.

Step 4: $N{a_2}O$ is an oxide of a metal (Sodium), and is therefore basic.

Step 5: Conclusion: $N{a_2}O$ is the basic oxide.

Why other options are wrong:
- Option A: $Si{O_2}$ is an oxide of a non-metal (Silicon) and is acidic.
- Option B: $S{O_2}$ is an oxide of a non-metal (Sulfur) and is acidic.
- Option C: $C{O_2}$ is an oxide of a non-metal (Carbon) and is acidic.
- Option D: $N{a_2}O$ is the correct answer as it is a basic oxide.

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

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

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

143 Easy

143 of 171Question permalink

Understanding +4 | -1

If the differentiating electron enters (n-1)d – subshell, the element is

View answer and explanation

Correct option: D

A transition element

Option D is correct because transition elements are defined by having their differentiating electron enter the (n-1)d subshell.

Solution approach

Correct Answer + Core Reason: D. A transition element. Transition elements are characterized by the filling of the (n-1)d subshell.

Step-by-Step Solution:

Step 1: Define transition elements: Elements in which the (n-1)d orbitals are being filled.

Step 2: The differentiating electron is the last electron added to the electronic configuration.

Step 3: If the differentiating electron enters the (n-1)d subshell, the element is a transition element.

Why other options are wrong:
- Option A: Representative elements have their differentiating electron entering the s or p subshells.
- Option B: Noble gases have completely filled s and p subshells.
- Option C: Alkali metals have their differentiating electron entering the ns subshell.

Key Concepts:
- Transition elements
- Differentiating electron
- Electronic configuration

Memory Aids:
- Remember that transition metals are in the d-block, and the d-orbitals are being filled.

Quick Tips:
- Consider the subshell being filled by the differentiating electron to classify the element.

144 Easy

144 of 171Question permalink

Analyzing +4 | -1

Which of the following is a false statement?

View answer and explanation

Correct option: C

Lithium is amphoteric

Option C (Lithium is amphoteric) is the false statement because Lithium is an alkali metal and forms basic oxides, not amphoteric ones.

Solution approach

Correct Answer + Core Reason: Lithium is amphoteric. Lithium is an alkali metal and forms basic oxides.

Step-by-Step Solution:

Step 1: Evaluate each statement.\nStatement A: Fluorine is more electronegative than chlorine. This is true because electronegativity decreases down the group.\nStatement B: Nitrogen has greater $\mathrm{I}{\mathrm{E}_1}$ than oxygen. This is true because nitrogen has a stable half-filled electronic configuration.\nStatement C: Lithium is amphoteric. This is false because lithium is an alkali metal and forms basic oxides. Only Be among the group 2 elements is amphoteric.\nStatement D: Chlorine is an oxidising agent. This is true because chlorine has a high electron affinity and readily accepts electrons.

Why other options are wrong:
- Option A: Fluorine is more electronegative than chlorine: This statement is true.
- Option B: Nitrogen has greater $\mathrm{I}{\mathrm{E}_1}$ than oxygen: This statement is true.
- Option D: Chlorine is an oxidising agent: This statement is true.

Key Concepts:
- Electronegativity
- Ionization energy
- Amphoteric oxides
- Oxidising agents

Memory Aids:
- Remember the trends in electronegativity and ionization energy in the periodic table.

Quick Tips:
- Know the properties of common elements and their oxides.

145 Easy

145 of 171Question permalink

Remembering +4 | -1

To which block of the periodic table does element with atomic number 55 belong

View answer and explanation

Correct option: A

s – block

Option A (s – block) is correct because element with atomic number 55 is Cesium, which is an alkali metal and belongs to the s-block.

Solution approach

Correct Answer + Core Reason: s – block. Element with atomic number 55 is Cesium (Cs), an alkali metal, which belongs to the s-block.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 55. The element with atomic number 55 is Cesium (Cs).

Step 2: Determine the electronic configuration of Cesium. The electronic configuration of Cesium is [Xe] 6s1.

Step 3: Identify the block to which Cesium belongs. Since the last electron enters the s-orbital, Cesium belongs to the s-block.

Why other options are wrong:
- Option B (p – block): Elements in the p-block have their last electron entering a p-orbital.
- Option C (d – block): Elements in the d-block have their last electron entering a d-orbital.
- Option D (f – block): Elements in the f-block have their last electron entering an f-orbital.

Key Concepts:
- Electronic configuration
- s, p, d, and f blocks in the periodic table

Memory Aids:
- Remember that alkali metals and alkaline earth metals belong to the s-block.

Quick Tips:
- Know the general electronic configurations of elements in each block to quickly identify the block to which an element belongs.

146 Easy

146 of 171Question permalink

Application +4 | -1

Set of isoelectronic species is

View answer and explanation

Correct option: C

{N}_2, {CO}, {CN}^{-}, {O}_2^{+2}

Option C (

{N}_2, {CO}, {CN}^{-}, {O}_2^{+2})

is correct because all species have the same number of electrons (14).

Solution approach

Correct Answer + Core Reason: ${N}_2, {CO}, {CN}^{-}, {O}_2^{+2}$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each species for option C:
- ${N}_2$ : 2 * 7 = 14 electrons
- ${CO}$ : 6 + 8 = 14 electrons
- ${CN}^{-}$ : 6 + 7 + 1 = 14 electrons
- ${O}_2^{+2}$ : (2 * 8) - 2 = 14 electrons

Step 2: Check the other options:
- Option A: ${N}_2$ (14), ${CO}_2$ (22), ${CN}^{-}$ (14), ${O}_2$ (16) - Not isoelectronic
- Option B: ${N}_2$ (14), ${H}_2 {~S}$ (18), ${CO}$ (14) - Not isoelectronic
- Option D: ${Ca}$ (20), ${Mg}$ (12), ${Cl}$ (17) - Not isoelectronic

Step 3: Only option C contains species with the same number of electrons.

Why other options are wrong:
- Option A: ${CO}_2$ and ${O}_2$ have different numbers of electrons than ${N}_2$ and ${CN}^{-}$ .
- Option B: ${H}_2 {~S}$ has a different number of electrons than ${N}_2$ and ${CO}$ .
- Option D: ${Ca}$ , ${Mg}$ , and ${Cl}$ all have different numbers of electrons.

Key Concepts:
- Isoelectronic species
- Counting electrons in molecules and ions

Memory Aids:
- Remember to add electrons for negative charges and subtract for positive charges.

Quick Tips:
- Carefully count the number of electrons, considering the charge of the species.

147 Easy

147 of 171Question permalink

Understanding +4 | -1

Variable oxidation state and degenerated orbital shows

View answer and explanation

Correct option: C

d-block elements

Option C is correct because d-block elements are known for exhibiting variable oxidation states and having degenerate d-orbitals.

Solution approach

Correct Answer + Core Reason: C. d-block elements. d-block elements exhibit variable oxidation states and have degenerate d-orbitals.

Step-by-Step Solution:

Step 1: Define variable oxidation state: The ability of an element to exhibit multiple oxidation states in its compounds.

Step 2: Define degenerate orbitals: Orbitals having the same energy level.

Step 3: d-block elements have partially filled d-orbitals, allowing them to lose different numbers of electrons, resulting in variable oxidation states.

Step 4: The five d-orbitals are degenerate in the absence of external fields.

Why other options are wrong:
- Option A: s-block elements typically have fixed oxidation states (e.g., +1 for alkali metals, +2 for alkaline earth metals) and have only s-orbitals, which are non-degenerate.
- Option B: p-block elements can have variable oxidation states, but not as readily as d-block elements. Also, the p-orbitals are degenerate, but the variable oxidation states are more characteristic of d-block.
- Option D: While some p-block elements can show variable oxidation states, it is not as characteristic as for d-block elements. s-block elements do not show variable oxidation states.

Key Concepts:
- d-block elements
- Variable oxidation states
- Degenerate orbitals

Memory Aids:
- Remember that d-block elements are in the middle of the periodic table and often form colorful compounds due to their variable oxidation states.

Quick Tips:
- Consider the electronic configuration of the elements when determining their properties.

148 Easy

148 of 171Question permalink

Remembering +4 | -1

Which of the following is a transition element?

View answer and explanation

Correct option: C

Ni

Option C (

Ni

) is correct because Nickel is a d-block element and exhibits properties characteristic of transition elements.

Solution approach

Correct Answer + Core Reason: $Ni$ . Nickel is a d-block element and exhibits properties of transition elements.

Step-by-Step Solution:

Step 1: Define transition elements. Transition elements are elements that have partially filled d-orbitals in their elemental or ionic forms.

Step 2: Analyze the given options:
- $Al$ : Aluminum is a p-block element and is not a transition element.
- $As$ : Arsenic is a p-block element and is not a transition element.
- $Ni$ : Nickel is a d-block element and is a transition element.
- $Rb$ : Rubidium is an s-block element and is not a transition element.

Step 3: Identify the transition element. Nickel (Ni) is a transition element because it has partially filled d-orbitals.

Why other options are wrong:
- **Option A ( $Al$ ): Aluminum is a p-block element and not a transition element.
- Option B ( $As$ ): Arsenic is a p-block element and not a transition element.
- Option D ( $Rb$ ): Rubidium is an s-block element and not a transition element.

Key Concepts:
- Transition elements
- d-block elements

Memory Aids:
- Remember that transition elements are typically found in the d-block of the periodic table.

Quick Tips:**
- Know the location of the d-block in the periodic table to quickly identify transition elements.

149 Easy

149 of 171Question permalink

Analyzing +4 | -1

The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is:

View answer and explanation

Correct option: C

O>F>N>C

Option C (O>F>N>C) is correct because after the first ionization, oxygen attains a more stable half-filled configuration, requiring more energy for the second ionization.

Solution approach

Correct Answer + Core Reason: O>F>N>C. After the first ionization, oxygen achieves a more stable electronic configuration than fluorine, nitrogen, or carbon, leading to a higher second ionization potential.

Step-by-Step Solution:

Step 1: Write the electronic configurations of C, N, O, and F.\nC: $1s^22s^22p^2$ \nN: $1s^22s^22p^3$ \nO: $1s^22s^22p^4$ \nF: $1s^22s^22p^5$

Step 2: Write the electronic configurations of the ions after the first ionization.\nC $^+$ : $1s^22s^22p^1$ \nN $^+$ : $1s^22s^22p^2$ \nO $^+$ : $1s^22s^22p^3$ (half-filled, relatively stable)\nF $^+$ : $1s^22s^22p^4$

Step 3: Consider the second ionization potential. Removing an electron from O $^+$ (half-filled) requires more energy than removing an electron from F $^+$ , N $^+$ , or C $^+$ .

Step 4: F $^+$ is next because it has a higher effective nuclear charge than N $^+$ and C $^+$ .

Step 5: N $^+$ is next because it has a higher effective nuclear charge than C $^+$ .

Why other options are wrong:
- Option A: C>N>O>F is incorrect because it doesn't account for the stability of the half-filled configuration of O $^+$ after the first ionization.
- Option B: O>N>F>C is incorrect because F $^+$ has a higher effective nuclear charge than N $^+$ .
- Option D: F>O>N>C is incorrect because O $^+$ has a stable half-filled configuration, making its second ionization potential higher than that of F $^+$ .

Key Concepts:
- Ionization Potential
- Electronic Configuration
- Stability of Half-Filled Orbitals
- Effective Nuclear Charge

Memory Aids:
- Half-filled and fully-filled orbitals are more stable.

Quick Tips:
- Consider the electronic configuration of the ion after the first ionization when determining the second ionization potential.

150 Easy

150 of 171Question permalink

Analyzing +4 | -1

Be and Al exhibit diagonal relationship. Which of the following statement about them is/are not true?
I. Both react with HCl to liberate

{H_2}

II.They are made passive by

HN{O_3}

III.Their carbides given acetylene on treatment with water
IV.Their oxides are amphoteric

View answer and explanation

Correct option: D

(iii) only

Option D ((iii) only) is correct because Be and Al carbides produce methane, not acetylene, upon hydrolysis.

Solution approach

Correct Answer + Core Reason: (iii) only. Beryllium and Aluminium carbides produce methane ( $CH_4$ ) upon hydrolysis, not acetylene.

Step-by-Step Solution:

Step 1: Analyze each statement.\nI. Both react with HCl to liberate $H_2$ : This is true due to their electropositive nature.\nII. They are made passive by $HNO_3$ : This is also true because of the formation of a protective oxide layer.\nIII. Their carbides give acetylene on treatment with water: This is false. $Be_2C$ and $Al_4C_3$ give methane ( $CH_4$ ) upon hydrolysis.\nIV. Their oxides are amphoteric: This is true; $BeO$ and $Al_2O_3$ react with both acids and bases.

Step 2: Identify the incorrect statement.\nStatement III is incorrect.

Step 3: Final Answer: The correct answer is (iii) only.

Why other options are wrong:
- Option A: (iii) and (iv): Statement (iv) is true, so this option is incorrect.
- Option B: (i) and (iii): Statement (i) is true, so this option is incorrect.
- Option C: (i) only: Statement (i) is true, so this option is incorrect.

Key Concepts:
- Diagonal relationship between Be and Al
- Properties of Be and Al compounds
- Hydrolysis of carbides
- Amphoteric oxides

Memory Aids:
- Be and Al show similar properties due to their diagonal relationship.

Quick Tips:
- Remember that $Be_2C$ and $Al_4C_3$ produce methane upon hydrolysis.

Your Progress

Classification of elements and periodicity in properties practice · Easy

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice