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

31 Medium

Understanding +4 | -1

Which among the following is an acidic oxide?

View answer and explanation

Correct option: D

Ge{O_2}

Option D is correct because

GeO_2

is an acidic oxide due to Germanium being a non-metal.

Solution approach

Correct Answer + Core Reason: $Ge{O_2}$ . $GeO_2$ is an acidic oxide because Germanium is a non-metal.

Step-by-Step Solution:

Step 1: Define acidic oxide: An oxide that reacts with water to form an acid or with a base to form a salt.

Step 2: Non-metal oxides are generally acidic.

Step 3: Metal oxides are generally basic.

Step 4: Amphoteric oxides can act as both acids and bases.

Step 5: $SnO_2$ and $PbO_2$ are amphoteric oxides.

Step 6: $PbO$ is a basic oxide.

Step 7: $GeO_2$ is an acidic oxide.

Why other options are wrong:
- Option A: $Sn{O_2}$ is amphoteric, not acidic.
- Option B: $Pb{O_2}$ is amphoteric, not acidic.
- Option C: $PbO$ is basic, not acidic.

Key Concepts:
- Acidic oxides
- Basic oxides
- Amphoteric oxides
- Periodic trends

Memory Aids:
- Remember that non-metal oxides are generally acidic, and metal oxides are generally basic.

Quick Tips:
- Consider the position of the element in the periodic table to determine the nature of its oxide.

32 Medium

32 of 175Question permalink

Understanding +4 | -1

The correct decreasing order of first ionisation enthalpies of five elements of the second period is

View answer and explanation

Correct option: C

F\mathrm{ } > \mathrm{ }N\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }Be\mathrm{ } > \mathrm{ }B

Option C (

F\mathrm{ } > \mathrm{ }N\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }Be\mathrm{ } > \mathrm{ }B

) is correct because it considers the general trend of increasing ionization energy across a period and the exceptions due to stable electronic configurations (half-filled and filled orbitals).

Solution approach

Correct Answer + Core Reason: $F\mathrm{ } > \mathrm{ }N\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }Be\mathrm{ } > \mathrm{ }B$ . This order considers the general trend and exceptions due to stable configurations.

Step-by-Step Solution:
Step 1: Recall the general trend of ionization energy across a period. Ionization energy generally increases as you move from left to right across a period.
Step 2: Identify the positions of Be, B, C, N, and F in the periodic table. They are all in the second period.
Step 3: Write the electronic configurations of the elements:
- Be: $1s^2 2s^2$
- B: $1s^2 2s^2 2p^1$
- C: $1s^2 2s^2 2p^2$
- N: $1s^2 2s^2 2p^3$
- F: $1s^2 2s^2 2p^5$
Step 4: Consider the exceptions to this trend due to electronic configurations.
- Beryllium (Be) has a filled 2s orbital, making it more stable than Boron (B).
- Nitrogen (N) has a half-filled 2p orbital, making it more stable than Oxygen (which is not in the list but would be next).
- Fluorine (F) is the most electronegative element and has a high effective nuclear charge.
Step 5: Arrange the elements in decreasing order of ionization energy, considering the exceptions. F > N > C > Be > B.

Why other options are wrong:
- Option A: $Be\mathrm{ } > \mathrm{ }B\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }N\mathrm{ } > \mathrm{ }F$ : Incorrect, this order does not account for the general trend or the exceptions.
- Option B: $N\mathrm{ } > \mathrm{ }F\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }B\mathrm{ } > \mathrm{ }Be$ : Incorrect, this order does not correctly place F and Be.
- Option D: $N\mathrm{ } > \mathrm{ }F\mathrm{ } > \mathrm{ }B\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }Be$ : Incorrect, this order does not correctly place F, C, and Be.

Key Concepts:
- Ionization energy
- Periodic trends
- Electronic configuration
- Stability of half-filled and filled orbitals

Memory Aids:
- Remember the exceptions to the ionization energy trend due to electronic configurations.

Quick Tips:
- Always consider the electronic configurations when comparing ionization energies of elements in the same period.

33 Medium

33 of 175Question permalink

Remembering +4 | -1

Alkaline earth metals form ions of the formula:

View answer and explanation

Correct option: A

{M^{2 + }}

Option A (

{M^{2 + }}

) is correct because alkaline earth metals have two valence electrons which they lose to form +2 ions.

Solution approach

Correct Answer + Core Reason: ${M^{2 + }}$ because alkaline earth metals lose two electrons to achieve a stable noble gas configuration.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of alkaline earth metals.

Step 2: Alkaline earth metals belong to Group 2 of the periodic table and have two valence electrons in their outermost shell.

Step 3: To achieve a stable noble gas configuration, they lose these two electrons.

Step 4: When an atom loses two electrons, it forms a +2 ion.

Step 5: Therefore, alkaline earth metals form ions of the formula ${M^{2 + }}$ .

Why other options are wrong:
- Option A: ${M^{2 + }}$ is the correct answer as alkaline earth metals lose two electrons.
- Option B: ${M^ + }$ is incorrect because alkaline earth metals lose two electrons, not one.
- Option C: $M$ is incorrect because alkaline earth metals do not exist as neutral atoms in ionic compounds.
- Option D: ${M^{2 - }}$ is incorrect because alkaline earth metals lose electrons, they do not gain electrons.

Key Concepts:
- Alkaline Earth Metals
- Electronic Configuration
- Ion Formation

Memory Aids:
- Group 2 elements form +2 ions.

Quick Tips:
- Remember the group number and the corresponding charge on the ion formed.

34 Medium

34 of 175Question permalink

Remembering +4 | -1

The pair of amphoteric oxides among the following is

View answer and explanation

Correct option: C

{ZnO}, {Al}_2 {O}_3

Option C (

{ZnO}, {Al}_2 {O}_3

) is correct because both Zinc oxide and Aluminum oxide exhibit amphoteric behavior, reacting with both acids and bases.

Solution approach

Correct Answer + Core Reason: ${ZnO}, {Al}_2 {O}_3$ . Amphoteric oxides react with both acids and bases.

Step-by-Step Solution:

Step 1: Define amphoteric oxides. Amphoteric oxides are oxides that can react with both acids and bases.

Step 2: Analyze each option:
- A: ${H}_2 {O}$ is neutral or weakly amphoteric, and ${SO}_3$ is acidic.
- B: ${Na}_2 {O}$ is basic, and ${ZnO}$ is amphoteric.
- C: ${ZnO}$ and ${Al}_2 {O}_3$ are both amphoteric.
- D: ${Al}_2 {O}_3$ is amphoteric, and ${SO}_3$ is acidic.

Step 3: Conclusion: Only option C contains a pair of amphoteric oxides.

Why other options are wrong:
- Option A: ${H}_2 {O}$ is not strongly amphoteric, and ${SO}_3$ is an acidic oxide.
- Option B: ${Na}_2 {O}$ is a basic oxide, not amphoteric.
- Option D: ${SO}_3$ is an acidic oxide, not amphoteric.

Key Concepts:
- Amphoteric oxides
- Acidic oxides
- Basic oxides

Memory Aids:
- Amphoteric = "ambi" (both) + "phos" (acid/base)

Quick Tips:
- Remember common amphoteric oxides like ZnO, Al2O3, SnO, PbO.

35 Medium

35 of 175Question permalink

Understanding +4 | -1

The characteristic of elements

X,\,Y

and

\,Z

with atomic numbers, respectively, 33, 53 and 83 are

View answer and explanation

Correct option: B

X

is a metalloid,

Y

is a non – metal and

Z

is a metal

Option B is correct because elements with atomic numbers 33, 53, and 83 correspond to a metalloid (As), a non-metal (I), and a metal (Bi), respectively.

Solution approach

Correct Answer + Core Reason: $X$ is a metalloid, $Y$ is a non – metal and $Z$ is a metal because their atomic numbers correspond to Arsenic, Iodine, and Bismuth, respectively.

Step-by-Step Solution:

Step 1: Identify the elements based on their atomic numbers.
$X$ with atomic number 33 is Arsenic (As).\nY with atomic number 53 is Iodine (I).\nZ with atomic number 83 is Bismuth (Bi).

Step 2: Determine the classification of each element.\nArsenic (As) is a metalloid.\nIodine (I) is a non-metal.\nBismuth (Bi) is a metal.

Step 3: Summarize the findings.
$X$ (Arsenic) is a metalloid, $Y$ (Iodine) is a non-metal, and $Z$ (Bismuth) is a metal.

Why other options are wrong:
- Option A: Incorrect because Iodine is a non-metal, not a metalloid.
- Option C: Incorrect because Arsenic and Iodine are not metals.
- Option D: Incorrect because Arsenic is a metalloid, and Bismuth is a metal.

Key Concepts:
- Periodic table and element classification (metals, non-metals, metalloids)
- Atomic number and element identification

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

Quick Tips:
- Memorize the common metalloids: Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), Polonium (Po).

36 Medium

36 of 175Question permalink

Application +4 | -1

An element X which occurs in the first short period has an outer electronic structure

{s^2}{p^1}

. What are the formula and acid-base character of its oxides?

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: ${X_2}{O_3},\; amphoteric. Element X belongs to group 13 and its oxide is amphoteric.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Determine the group of the element X based on its electronic configuration (s²p¹). This configuration indicates that X belongs to group 13.\n\n**Step 2:** Recall that elements in group 13 form oxides with the formula$ {X_2}{O_3} $.\n\n**Step 3:** Recall the nature of oxides of group 13 elements. Generally, they are amphoteric.\n\n**Step 4:** Conclude that the formula and nature of the oxide is$ {X_2}{O_3} $Amphoteric.\n\n**Why other options are wrong:**\n- **Option A:**$ X{O_3}, $basic: Incorrect formula and nature.\n- **Option B:**$ {X_2}{O_3} $, basic: Correct formula, but incorrect nature.\n- **Option D:**$ X{O_2},\; acidic: Incorrect formula and 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.

37 Medium

37 of 175Question permalink

Understanding +4 | -1

Which among the following elements shows diagonal relationship with beryllium?

View answer and explanation

Correct option: C

\mathrm{A}{l}

The correct answer is C (Al) because Aluminum (Al) shows a diagonal relationship with Beryllium (Be) in the periodic table.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{l}$ . Aluminum exhibits a diagonal relationship with beryllium in the periodic table.

Step-by-Step Solution:

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

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

Step 3: Be and Al have similar polarizing power due to similar charge/size ratios, leading to similar properties.

Why other options are wrong:
- Option A: $\mathrm{Na}$ (Sodium) does not have a diagonal relationship with Be.
- Option B: $\mathrm{Li}$ (Lithium) has a diagonal relationship with Mg, not Be.
- Option D: $\mathrm{Si}$ (Silicon) has a diagonal relationship with B, not Be.

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

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

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

38 Medium

38 of 175Question permalink

Analyzing +4 | -1

Which one pair of atoms or ions will have same configuration?

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{l}^ - }\,\,and\,\,\mathrm{Ar}

Option B (

\mathrm{C}{\mathrm{l}^ - }\,\,and\,\,\mathrm{Ar}

) is correct because both the chloride ion and argon have the same number of electrons and thus the same electronic configuration.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}^ - }\,\,and\,\,\mathrm{Ar}$ . Both have the same number of electrons and thus the same electronic configuration.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each atom/ion.

Step 2: Cl has 17 electrons. $\mathrm{C}{\mathrm{l}^ - }$ gains one electron, so it has 18 electrons.

Step 3: Ar has 18 electrons.

Step 4: Therefore, $\mathrm{C}{\mathrm{l}^ - }$ and Ar have the same number of electrons and the same electronic configuration.

Why other options are wrong:
- Option A: $\mathrm{L}{\mathrm{i}^ + }$ has 2 electrons (3-1=2). $\mathrm{H}{\mathrm{e}^ - }$ has 3 electrons (2+1=3). They do not have the same configuration.
- Option C: Na has 11 electrons, and K has 19 electrons. They do not have the same configuration.
- Option D: ${\mathrm{F}^ + }$ has 8 electrons (9-1=8). Ne has 10 electrons. They do not have the same configuration.

Key Concepts:
- Electronic configuration
- Number of electrons in ions and atoms

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

Quick Tips:
- Remember to account for the charge when determining the number of electrons in an ion.

39 Medium

39 of 175Question permalink

Remembering +4 | -1

Which element has highest electronegativity?

View answer and explanation

Correct option: A

F

Option A (F) is correct because Fluorine is the most electronegative element in the periodic table.

Solution approach

Correct Answer + Core Reason: F. Fluorine is the most electronegative element.

Step-by-Step Solution:

Step 1: Recall the electronegativity trend in the periodic table.\nElectronegativity generally increases across a period and decreases down a group.

Step 2: Consider the positions of the elements in the periodic table.
- F is in group 17 (halogens) and period 2.
- He and Ne are noble gases and generally do not form bonds, so electronegativity is not typically considered.
- Na is in group 1 (alkali metals) and period 3.

Step 3: Conclusion.\nTherefore, Fluorine has the highest electronegativity among the given elements.

Why other options are wrong:
- Option B: Helium is a noble gas and has very low electronegativity.
- Option C: Neon is a noble gas and has very low electronegativity.
- Option D: Sodium is an alkali metal and has low electronegativity.

Key Concepts:
- Electronegativity
- Periodic trends

Memory Aids:
- Fluorine is the most electronegative element.

Quick Tips:
- Remember that electronegativity increases towards the top right of the periodic table (excluding noble gases).

40 Medium

40 of 175Question permalink

Application +4 | -1

Assign the position of the element having outer electronic configuration

n s^2 n p^4

for

n=3

View answer and explanation

Correct option: A

Third period, 16th group

The correct answer is A (Third period, 16th group) because with

n=3

, the electronic configuration is

3s^2 3p^4

, indicating it's in the third period and group 16 (chalcogens).

Solution approach

Correct Answer + Core Reason: Third period, 16th group. The electronic configuration $3s^2 3p^4$ indicates the element's position in the periodic table.

Step-by-Step Solution:

Step 1: Determine the period number from the value of 'n'.

Step 2: Given $n=3$ , the element is in the third period.

Step 3: Determine the group number from the electronic configuration $ns^2np^4$ .

Step 4: The number of valence electrons is 2 + 4 = 6.

Step 5: For p-block elements, the group number is 10 + number of valence electrons = 10 + 6 = 16.

Step 6: Therefore, the element is in the third period and 16th group.

Why other options are wrong:
- Option B: Second period, 15th group is incorrect because the element is in the third period, not the second, and has 6 valence electrons, not 5.
- Option C: Third period, 14th group is incorrect because the element has 6 valence electrons, placing it in group 16, not 14.
- Option D: Fourth period, 15th group is incorrect because the element is in the third period, not the fourth, and has 6 valence electrons, not 5.

Key Concepts:
- Electronic configuration
- Periodic table groups and periods
- Valence electrons

Memory Aids:
- Remember that the period number corresponds to the highest principal quantum number (n).

Quick Tips:
- Relate the number of valence electrons to the group number.

41 Medium

41 of 175Question permalink

Analyzing +4 | -1

Which statement is wrong?

View answer and explanation

Correct option: C

Maximum electron affinity exists for

\mathrm{F}

Option C is wrong because maximum electron affinity exists for

\mathrm{Cl}

not

\mathrm{F}

due to its smaller size and strong interelectronic repulsions.

Solution approach

Correct Answer + Core Reason: Maximum electron affinity exists for $\mathrm{F}$ . Electron affinity is maximum for chlorine (Cl) due to smaller size of fluorine and strong interelectronic repulsions.

Step-by-Step Solution:

Step 1: Analyze the given statements.

Step 2: Evaluate each statement:
- A: 2nd ionization energy shows a jump in alkali metals because after losing one electron, they attain a stable noble gas configuration.
- B: 2nd electron affinity for halogens is zero because after gaining one electron, they attain a stable noble gas configuration, and adding another electron is highly unfavorable.
- C: Maximum electron affinity exists for F. This is incorrect. Electron affinity generally increases across a period, but due to its small size and strong interelectronic repulsions, fluorine has a lower electron affinity than chlorine.
- D: Maximum ionization energy exists for He because it has a stable $1s^2$ configuration and a high nuclear charge.

Step 3: Identify the incorrect statement. Statement C is incorrect.

Why other options are wrong:
- Option A: 2nd ionization energy shows jump in alkali metals. This is correct because removing a second electron from an alkali metal requires breaking into a stable noble gas configuration.
- Option B: 2nd electron affinity for halogens is zero. This is correct because halogens readily accept one electron to achieve a stable noble gas configuration, but adding a second electron is highly unfavorable.
- Option D: Maximum ionization energy exists for $\mathrm{He}$ . This is correct because helium has a very small size and a stable electronic configuration ( $1s^2$ ), making it difficult to remove an electron.

Key Concepts:
- Ionization Energy
- Electron Affinity
- Electronic Configuration
- Periodic Trends

Memory Aids:
- Noble gases have high ionization energies.
- Halogens have high electron affinities.

Quick Tips:
- Remember the exceptions to the general trends in ionization energy and electron affinity due to electronic configuration and interelectronic repulsions.

42 Medium

42 of 175Question permalink

Analyzing +4 | -1

Which of the following does not represents the correct order of the property indicated?

View answer and explanation

Correct option: A

\mathrm{S}{\mathrm{c}^{3 + }} > C{\mathrm{r}^{3 + }} > F{\mathrm{e}^{3 + }} > M{\mathrm{n}^{3 + }} -

ionic radii

Option C is incorrect because the first ionization energy of Nitrogen is higher than Oxygen due to the stable half-filled p-orbital configuration of Nitrogen.

Solution approach

Correct Answer + Core Reason: $\mathrm{B} < C < N < O$ increasing first ionisation enthalphy. The ionization energy of N is higher than O due to the half-filled p-orbital.

Step-by-Step Solution:

Step 1: Analyze the trends in metallic radius, electron gain enthalpy, ionization enthalpy, and ionic size.

Step 2: Evaluate Option A: $\mathrm{Li} < Na < K < Rb$ increasing metallic radius. This order is correct as metallic radius increases down a group.

Step 3: Evaluate Option B: $\mathrm{I} < Br < F < Cl$ increasing electron gain enthaly (with negative sign). This order is correct. Chlorine has a higher electron affinity than fluorine due to its smaller size and greater electron density.

Step 4: Evaluate Option C: $\mathrm{B} < C < N < O$ increasing first ionisation enthalphy. This order is incorrect. The correct order should be $\mathrm{B} < C < O < N$ because Nitrogen has a higher ionization energy than Oxygen due to the half-filled p-orbital configuration of Nitrogen.

Step 5: Evaluate Option D: $\mathrm{A}{\mathrm{l}^{3 + }} < \mathrm{M}{\mathrm{g}^{2 + }} < \mathrm{N}{\mathrm{a}^ + } < {\mathrm{F}^ - }$ increasing ionic size. This order is correct for isoelectronic species; ionic size increases with decreasing nuclear charge.

Step 6: Identify the incorrect order: The incorrect order is Option C.

Why other options are wrong:
- Option A: $\mathrm{Li} < Na < K < Rb$ increasing metallic radius: This order is correct; metallic radius increases down the group.
- Option B: $\mathrm{I} < Br < F < Cl$ increasing electron gain enthaly (with negative sign): This order is correct, considering the anomaly between F and Cl.
- Option D: $\mathrm{A}{\mathrm{l}^{3 + }} < \mathrm{M}{\mathrm{g}^{2 + }} < \mathrm{N}{\mathrm{a}^ + } < {\mathrm{F}^ - }$ increasing ionic size: This order is correct for isoelectronic species.

Key Concepts:
- Ionization Energy
- Electron Affinity
- Metallic Radius
- Ionic Size
- Periodic Trends

Memory Aids:
- Remember the exception for N and O in ionization energy trends due to half-filled orbitals.

Quick Tips:
- Be aware of exceptions to periodic trends, such as the higher ionization energy of nitrogen compared to oxygen.

43 Medium

43 of 175Question permalink

Remembering +4 | -1

The atomic radius of Ag is closest to

View answer and explanation

Correct option: B

Au

Option B is correct because Au (Gold) has an atomic radius closest to that of Ag (Silver) due to the lanthanide contraction affecting the size of gold.

Solution approach

Correct Answer + Core Reason: Au. Due to lanthanide contraction, Au has an atomic radius similar to Ag.

Step-by-Step Solution:

Step 1: Understand the concept of atomic radius.\nAtomic radius is the distance from the nucleus to the outermost electron of an atom.

Step 2: Consider the position of Ag and Au in the periodic table.\nAg (Silver) is in the 5th period, and Au (Gold) is in the 6th period. Generally, atomic radius increases down a group.

Step 3: Account for the lanthanide contraction.\nThe lanthanide contraction affects the size of the elements in the 6th period, particularly after the lanthanides. The 4f electrons in the lanthanides have a poor shielding effect, leading to an increased effective nuclear charge. This causes the atomic radius of Au to be smaller than expected, making it closer to the atomic radius of Ag.

Step 4: Conclusion.\nThe atomic radius of Ag is closest to Au.

Why other options are wrong:
- Option A: Incorrect because Hg (Mercury) has a significantly larger atomic radius than Ag.
- Option B: Correct because Au (Gold) has an atomic radius closest to that of Ag due to the lanthanide contraction.
- Option C: Incorrect because Ni (Nickel) has a smaller atomic radius than Ag.
- Option D: Incorrect because Cu (Copper) has a smaller atomic radius than Ag.

Key Concepts:
- Atomic radius
- Lanthanide contraction
- Periodic trends

Memory Aids:
- Remember that lanthanide contraction affects the size of elements in the 6th period.

Quick Tips:
- Consider the effects of lanthanide contraction when comparing the atomic radii of elements in the 6th period.

44 Medium

44 of 175Question permalink

Understanding +4 | -1

Which of the following ion is the smallest ion?

View answer and explanation

Correct option: B

O_2^ +

Option B (

O_2^+

) is correct because removing an electron from a molecule reduces electron-electron repulsion and increases the effective nuclear charge, leading to a smaller size.

Solution approach

Correct Answer + Core Reason: $O_2^+$ . Removing electrons decreases electron-electron repulsion and increases effective nuclear charge, thus decreasing size.

Step-by-Step Solution:

Step 1: Consider the number of electrons in each species.

Step 2: $O_2$ has a certain number of electrons. $O_2^+$ has one less electron, $O_2^-$ has one more electron, and $O_2^{2-}$ has two more electrons.

Step 3: Removing electrons reduces electron-electron repulsion and increases the effective nuclear charge experienced by the remaining electrons.

Step 4: This leads to a contraction of the electron cloud and a smaller size.

Step 5: Adding electrons increases electron-electron repulsion and decreases the effective nuclear charge, leading to an expansion of the electron cloud and a larger size.

Step 6: Therefore, $O_2^+$ is the smallest ion among the given options.

Why other options are wrong:
- Option A: $O_2$ is larger than $O_2^+$ because it has more electrons.
- Option C: $O_2^-$ is larger than $O_2$ and $O_2^+$ because it has more electrons.
- Option D: $O_2^{2-}$ is the largest ion because it has the most electrons.

Key Concepts:
- Ionic size
- Electron-electron repulsion
- Effective nuclear charge

Memory Aids:
- Positive ions are smaller than their neutral atoms/molecules; negative ions are larger.

Quick Tips:
- When comparing the sizes of ions/molecules, consider the number of electrons and the nuclear charge.

45 Medium

45 of 175Question permalink

Understanding +4 | -1

The following ion is colourless in aqueous solution

View answer and explanation

Correct option: D

All the above

The correct answer is D (All the above) because

Ca^{2+}

,

Sc^{3+}

, and

Zn^{2+}

are all colorless in aqueous solution due to having no d-d transitions.

Solution approach

Correct Answer + Core Reason: All the above. $C{a^{ + 2}}$ , $S{c^{3 + }}$ and $Z{n^{ + 2}}$ are colorless because they do not have unpaired d-electrons for d-d transitions.

Step-by-Step Solution:

Step 1: Recall the reason for color in transition metal ions.

Step 2: Color arises from d-d transitions, which require partially filled d-orbitals.

Step 3: $C{a^{ + 2}}$ has the electronic configuration [Ar], with no d-electrons. Therefore, it is colorless.

Step 4: $S{c^{3 + }}$ has the electronic configuration [Ar], with no d-electrons. Therefore, it is colorless.

Step 5: $Z{n^{ + 2}}$ has the electronic configuration [Ar] $3d^{10}$ , with completely filled d-orbitals. Therefore, it is colorless.

Step 6: Since all the given ions are colorless, the correct answer is All the above.

Why other options are wrong:
- Option A: $C{a^{ + 2}}$ is colorless, but this option doesn't include the other colorless ions.
- Option B: $S{c^{3 + }}$ is colorless, but this option doesn't include the other colorless ions.
- Option C: $Z{n^{ + 2}}$ is colorless, but this option doesn't include the other colorless ions.

Key Concepts:
- Color of transition metal ions
- d-d transitions
- Electronic configuration

Memory Aids:
- Remember that ions with no d-electrons or completely filled d-orbitals are generally colorless.

Quick Tips:
- Check the electronic configuration to determine the presence of unpaired d-electrons.

46 Medium

46 of 175Question permalink

Understanding +4 | -1

The tendency towards formation of coloured ions is the maximum in:

View answer and explanation

Correct option: B

d-block elements

The correct answer is B (d-block elements) because d-block elements have unpaired d-electrons which allow for d-d transitions, leading to the formation of colored ions.

Solution approach

Correct Answer + Core Reason: d-block elements. d-block elements have unpaired d-electrons that allow for d-d transitions, leading to colored ions.

Step-by-Step Solution:

Step 1: Recall the origin of color in ions.

Step 2: Color in ions is due to electronic transitions, specifically d-d transitions in transition metal ions.

Step 3: d-block elements (transition metals) have partially filled d-orbitals, allowing for d-d transitions when electrons absorb energy and move between d-orbitals.

Step 4: s-block and p-block elements generally do not have partially filled d-orbitals and therefore do not exhibit d-d transitions.

Step 5: f-block elements also exhibit color due to f-f transitions, but the tendency is less pronounced than in d-block elements.

Why other options are wrong:
- Option A: s-block elements typically form colorless ions because they do not have d-electrons.
- Option C: p-block elements generally do not have partially filled d-orbitals and therefore do not exhibit d-d transitions.
- Option D: f-block elements do form colored ions, but the tendency is less pronounced compared to d-block elements.

Key Concepts:
- Color of ions
- d-d transitions
- Electronic configuration

Memory Aids:
- Remember that transition metal ions are often colored.

Quick Tips:
- Relate the color of ions to the presence of partially filled d-orbitals.

47 Medium

47 of 175Question permalink

Remembering +4 | -1

The electron affinity of for the inert gases is

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because inert gases have a stable electronic configuration and do not tend to accept electrons, resulting in zero electron affinity.

Solution approach

Correct Answer + Core Reason: Zero. Inert gases have a stable electronic configuration and do not accept electrons.

Step-by-Step Solution:

Step 1: Understand electron affinity.

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

Step 3: Inert gases have a stable, completely filled valence shell.

Step 4: Adding an electron to an inert gas requires a significant amount of energy to overcome the stable configuration.

Step 5: Therefore, the electron affinity of inert gases is essentially zero.

Why other options are wrong:
- Option B: High: Inert gases do not have a high electron affinity.
- Option C: Negative: Inert gases do not readily accept electrons, so their electron affinity is not negative (exothermic).
- Option D: Positive: A positive electron affinity would mean energy is required to add an electron, but the value is essentially zero for inert gases.

Key Concepts:
- Electron affinity
- Inert gases
- Stable electronic configuration

Memory Aids:
- Inert gases have zero electron affinity.

Quick Tips:
- Remember that inert gases have stable electronic configurations and do not readily accept electrons.

48 Medium

48 of 175Question permalink

Remembering +4 | -1

The electronegativity order of

O

,

F

,

Cl

and

Br

is:

View answer and explanation

Correct option: A

\mathrm{F} > O > Cl > Br

Option A (

\mathrm{F} > O > Cl > Br

) is correct because it represents the correct order of electronegativity for the given elements.

Solution approach

Correct Answer + Core Reason: $\mathrm{F} > O > Cl > Br$ . This is the correct electronegativity order.

Step-by-Step Solution:

Step 1: Recall the electronegativity trend in the periodic table.\nElectronegativity generally increases across a period and decreases down a group.

Step 2: Consider the positions of the elements in the periodic table.
- F and O are in the second period.
- Cl and Br are in the third and fourth periods, respectively.
- F > O because electronegativity increases across a period.
- F > Cl > Br because electronegativity decreases down a group.
- O > Cl > Br.

Step 3: Conclusion.\nTherefore, the electronegativity order is $\mathrm{F} > O > Cl > Br$ .

Why other options are wrong:
- Option B: $\mathrm{F} > Cl\left\langle {Br} \right\rangle O$ is incorrect.
- Option C: $\mathrm{Br} > Cl > F > O$ is incorrect because Fluorine is the most electronegative.
- Option D: $\mathrm{F} < Cl < Br < O$ is incorrect because Fluorine is the most electronegative.

Key Concepts:
- Electronegativity
- Periodic trends

Memory Aids:
- Remember the electronegativity trends.

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

49 Medium

49 of 175Question permalink

Remembering +4 | -1

Electron affinity is measured in

View answer and explanation

Correct option: D

Both 2 and 3

Option D (Both 2 and 3) is correct because electron affinity is an energy change, and energy is commonly measured in both

K\,cal\,mo{l^{ - 1}}

and

kJ\,mo{l^{ - 1}}

.

Solution approach

Correct Answer + Core Reason: Both 2 and 3. Electron affinity is an energy change, measured in both $K\,cal\,mo{l^{ - 1}}$ and $kJ\,mo{l^{ - 1}}$ .

Step-by-Step Solution:

Step 1: Understand electron affinity.

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

Step 3: Energy changes are measured in units of energy per mole.

Step 4: Common units for energy are kilocalories (kcal) and kilojoules (kJ).

Step 5: Therefore, electron affinity is measured in both $K\,cal\,mo{l^{ - 1}}$ and $kJ\,mo{l^{ - 1}}$ .

Why other options are wrong:
- Option A: No units: Electron affinity is an energy change and has units.
- Option B: $K\,cal\,mo{l^{ - 1}}$ : This is a valid unit, but not the only one.
- Option C: $kJ\,mo{l^{ - 1}}$ : This is a valid unit, but not the only one.

Key Concepts:
- Electron affinity
- Units of energy

Memory Aids:
- Electron affinity is an energy change.

Quick Tips:
- Remember that electron affinity is an energy change and has units of energy per mole.

50 Medium

50 of 175Question permalink

Remembering +4 | -1

The elements with atomic numbers 9, 17, 35, 53, 85 are all

View answer and explanation

Correct option: A

Halogens

Option A (Halogens) is correct because elements with atomic numbers 9 (Fluorine), 17 (Chlorine), 35 (Bromine), 53 (Iodine), and 85 (Astatine) are all halogens.

Solution approach

Correct Answer + Core Reason: Halogens. Elements with atomic numbers 9, 17, 35, and 53 are all halogens.

Step-by-Step Solution:

Step 1: Identify the elements: The elements with the given atomic numbers are Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), and Astatine (At).

Step 2: Determine their group: These elements all belong to Group 17 of the periodic table.

Step 3: Classify the group: Group 17 elements are known as halogens.

Step 4: Conclusion: The correct answer is Halogens.

Why other options are wrong:
- Option B: Noble gases: Noble gases are in Group 18 of the periodic table.
- Option C: Heavy metals: This is a general term and not specific to this group of elements.
- Option D: Light metals: This is a general term and not specific to this group of elements.

Key Concepts:
- Periodic Table groups
- Halogens

Memory Aids:
- Remember that halogens are in Group 17 and include F, Cl, Br, I, and At.

Quick Tips:
- Know the common group names in the periodic table.

51 Medium

51 of 175Question permalink

Remembering +4 | -1

Atomic number of element Ununnillium is

View answer and explanation

Correct option: B

110

Option B (110) is correct because Ununnillium is the IUPAC name for the element with atomic number 110.

Solution approach

Correct Answer + Core Reason: 110. Ununnillium is the IUPAC name for the element with atomic number 110.

Step-by-Step Solution:

Step 1: Recall the IUPAC name for element 110: The IUPAC name for element 110 is Ununnillium.

Step 2: Conclusion: The correct answer is 110.

Why other options are wrong:
- Option A: 101: This is Mendelevium.
- Option C: 111: This is Unununium.
- Option D: 100: This is Fermium.

Key Concepts:
- IUPAC nomenclature
- Element atomic numbers

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

Quick Tips:
- Memorize the IUPAC names for elements with atomic numbers greater than 100.

52 Medium

52 of 175Question permalink

Understanding +4 | -1

Which of the atomic number pairs represents elements of s-block?

View answer and explanation

Correct option: D

3,12

Option D (3,12) is correct because element with atomic number 3 (Lithium) has electronic configuration

1s^22s^1

and element with atomic number 12 (Magnesium) has electronic configuration

1s^22s^22p^63s^2

, both having their last electron in s-orbital.

Solution approach

Correct Answer + Core Reason: 3,12. Elements with atomic numbers 3 and 12 have their last electron in the s-orbital.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the element with atomic number 3. The electronic configuration is $1s^22s^1$ .

Step 2: Determine the electronic configuration of the element with atomic number 12. The electronic configuration is $1s^22s^22p^63s^2$ .

Step 3: Identify the last electron entering the orbital for both elements. For both elements, the last electron enters the s-orbital.

Step 4: Conclude that both elements belong to the s-block.

Why other options are wrong:
- Option A: 7 (Nitrogen) and 15 (Phosphorus) are p-block elements.
- Option B: 5 (Boron) is a p-block element, and 12 (Magnesium) is an s-block element.
- Option C: 9 (Fluorine) and 17 (Chlorine) are p-block elements.

Key Concepts:
- Electronic configuration
- Periodic table blocks (s, p, d, f)

Memory Aids:
- s-block elements have their last electron entering the s-orbital.

Quick Tips:
- Knowing the electronic configuration helps determine the block to which an element belongs.

53 Medium

53 of 175Question permalink

Analyzing +4 | -1

Given below are two statements:
Statement I: The pentavalent oxide of group-15 element,

{E_2}{O_5}

, is less acidic than trivalent oxide,

{E_2}{O_3}

, of the same element.
Statement II: The acidic character of trivalent oxide of group 15 elements,

{E_2}{O_3}

, decreases down the group.
In light of the above statements, choose most appropriate answer from the options given below:

View answer and explanation

Correct option: D

Statement I false, but Statement II is true

Option D (Statement I false, but Statement II is true) is correct because the pentavalent oxide is more acidic than the trivalent oxide, and the acidic character of trivalent oxides decreases down the group.

Solution approach

Correct Answer + Core Reason: Statement I false, but Statement II is true. Pentavalent oxides are more acidic than trivalent oxides, and the acidic character of trivalent oxides decreases down the group.

Step-by-Step Solution:

Step 1: Analyze Statement I.\nStatement I: The pentavalent oxide of group-15 element, ${E_2}{O_5}$ , is less acidic than trivalent oxide, ${E_2}{O_3}$ , of the same element. This is false. Higher oxidation states generally lead to more acidic oxides.

Step 2: Analyze Statement II.\nStatement II: The acidic character of trivalent oxide of group 15 elements, ${E_2}{O_3}$ , decreases down the group. This is true. As you move down the group, the elements become more metallic, and their oxides become less acidic (more basic).

Step 3: Determine the correct answer.\nStatement I is false, and Statement II is true.

Step 4: Final Answer: Statement I false, but Statement II is true.

Why other options are wrong:
- Option A: Both Statement I and Statement II are true: Statement I is false.
- Option B: Both Statement I and Statement II are false: Statement II is true.
- Option C: Statement I true, but Statement II is false: Statement I is false and Statement II is true.

Key Concepts:
- Acidic character of oxides
- Oxidation states
- Periodic trends

Memory Aids:
- Higher oxidation state leads to more acidic oxides.
- Acidic character decreases down a group.

Quick Tips:
- Remember the general trends in acidity and basicity of oxides.

54 Medium

54 of 175Question permalink

Understanding +4 | -1

Elements of group 1 and group VI in the periodic table have one thing common. That is with the increasing atomic number, the:

View answer and explanation

Correct option: C

Atomic radius increases

Option C (Atomic radius increases) is correct because, within a group, atomic radius generally increases with increasing atomic number due to the addition of electron shells.

Solution approach

Correct Answer + Core Reason: Atomic radius increases. As atomic number increases down a group, atomic radius increases due to the addition of electron shells.

Step-by-Step Solution:

Step 1: Recall the trends in the periodic table for group 1 and group 16 (VI).

Step 2: Consider the trend for atomic radius. Atomic radius increases down a group due to the addition of electron shells.

Step 3: Consider the trend for maximum valency. Maximum valency remains constant or decreases down a group.

Step 4: Consider the trend for reactivity. Reactivity varies depending on the group. For group 1, reactivity increases down the group. For group 16, reactivity generally decreases down the group.

Step 5: Consider the trend for oxidizing power. Oxidizing power generally decreases down a group.

Step 6: Conclude that atomic radius increases with increasing atomic number for both group 1 and group 16.

Why other options are wrong:
- Option A: Maximum valency does not necessarily increase with increasing atomic number in both groups.
- Option B: Reactivity does not necessarily increase with increasing atomic number in both groups. It increases for group 1 but generally decreases for group 16.
- Option D: Oxidizing power generally decreases with increasing atomic number in a group.

Key Concepts:
- Periodic trends: atomic radius, valency, reactivity, oxidizing power

Memory Aids:
- Atomic radius increases down a group.

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

55 Medium

55 of 175Question permalink

Analyzing +4 | -1

Which of the following halides is most acidic?

View answer and explanation

Correct option: B

\mathrm{PC}{\mathrm{l}_3}

Option B (PCl3) is the most acidic because the central atom (P) has the highest oxidation state and electronegativity among the given options, leading to a greater polarization of the P-Cl bond and easier release of H+ ions upon hydrolysis.

Solution approach

Correct Answer + Core Reason: $\mathrm{PC}{\mathrm{l}_3}$ . The acidity of halides increases with the electronegativity and oxidation state of the central atom.

Step-by-Step Solution:

Step 1: Consider the hydrolysis of each halide:
$\mathrm{CC}{\mathrm{l}_4} + 2{\mathrm{H}_2}\mathrm{O} \to {\mathrm{CO}_2} + 4\mathrm{HCl}$
$\mathrm{PC}{\mathrm{l}_3} + 3{\mathrm{H}_2}\mathrm{O} \to {\mathrm{H}_3}\mathrm{PO}_3 + 3\mathrm{HCl}$
$\mathrm{BiC}{\mathrm{l}_3} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{BiOCl} + 2\mathrm{HCl}$
$\mathrm{SbC}{\mathrm{l}_3} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{SbOCl} + 2\mathrm{HCl}$

Step 2: Compare the acidity of the resulting solutions. The acidity depends on the ability of the central atom to polarize the bond with chlorine. Higher oxidation state and electronegativity of the central atom lead to greater polarization and thus higher acidity.

Step 3: Oxidation states: C(+4), P(+3), Bi(+3), Sb(+3). Electronegativity: P > Bi ≈ Sb.

Step 4: $\mathrm{PC}{\mathrm{l}_3}$ produces ${\mathrm{H}_3}\mathrm{PO}_3$ which is more acidic than the other products.

Why other options are wrong:
- Option A: $\mathrm{CC}{\mathrm{l}_4}$ forms ${\mathrm{CO}_2}$ which is weakly acidic.
- Option C: $\mathrm{BiC}{\mathrm{l}_3}$ and $\mathrm{SbC}{\mathrm{l}_3}$ form oxychlorides which are less acidic than the phosphorous acid.
- Option D: $\mathrm{SbC}{\mathrm{l}_3}$ forms oxychlorides which are less acidic than the phosphorous acid.

Key Concepts:
- Hydrolysis of halides
- Acidity of oxides and halides
- Electronegativity and oxidation state

Memory Aids:
- Higher oxidation state = higher acidity

Quick Tips:
- Consider the products of hydrolysis when comparing acidity.

56 Medium

56 of 175Question permalink

Understanding +4 | -1

Which of the following oxides is most basic?

View answer and explanation

Correct option: A

\mathrm{N}{\mathrm{a}_2}\mathrm{O}

Option A (Na₂O) is the most basic oxide because oxides of Group 1 elements are strongly basic.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}_2}\mathrm{O}$ . Oxides of Group 1 elements are strongly basic.

Step-by-Step Solution:

Step 1: Identify the elements forming the oxides.\nNa, Si, S

Step 2: Recall the trend in basicity of oxides across the periodic table. Basicity decreases from left to right.

Step 3: Group 1 oxides are strongly basic. Group 14 and 16 oxides are acidic or amphoteric.

Step 4: $\mathrm{N}{\mathrm{a}_2}\mathrm{O}$ is a Group 1 oxide and is therefore the most basic.
$\mathrm{Si}{\mathrm{O}_2}$ is a Group 14 oxide and is acidic.
$\mathrm{S}{\mathrm{O}_2}$ is a Group 16 oxide and is acidic.

Why other options are wrong:
- Option B: $\mathrm{Si}{\mathrm{O}_2}$ is acidic, not basic.
- Option C: $\mathrm{S}{\mathrm{O}_2}$ is acidic, not basic.
- Option D: All are not equally basic. $\mathrm{N}{\mathrm{a}_2}\mathrm{O}$ is significantly more basic than the others.

Key Concepts:
- Basicity of Oxides
- Periodic Trends

Memory Aids:
- Metal oxides are generally basic, non-metal oxides are generally acidic.

Quick Tips:
- Remember the general trends in basicity and acidity of oxides across the periodic table.

57 Medium

57 of 175Question permalink

Understanding +4 | -1

Which of the following obeys triad law :

View answer and explanation

Correct option: D

Both (1) and (3)

The correct answer is D (Both (1) and (3)) because both Li, Na, K and Be, Mg, Ca follow Dobereiner's Triad law.

Solution approach

Correct Answer + Core Reason: Both (1) and (3). Both Li, Na, K and Be, Mg, Ca follow Dobereiner's Triad law.

Step-by-Step Solution:

Step 1: Recall Dobereiner's Triad law, which states that the atomic weight of the middle element is approximately the average of the atomic weights of the other two elements.

Step 2: For Li, Na, K: Atomic weight of Na (23) is approximately the average of Li (7) and K (39): (7+39)/2 = 23.

Step 3: For Be, Mg, Ca: Atomic weight of Mg (24) is approximately the average of Be (9) and Ca (40): (9+40)/2 = 24.5 (close enough).

Step 4: For H, Li, K: Atomic weight of Li (7) is not the average of H (1) and K (39): (1+39)/2 = 20.

Why other options are wrong:
- Option A: $Li, Na, K$ is partially correct, but incomplete.
- Option B: ${H}, {Li}, {K}$ is incorrect because Li is not the average of H and K.
- Option C: ${Be}, {Mg}, {Ca}$ is partially correct, but incomplete.

Key Concepts:
- Dobereiner's Triad Law
- Atomic Weights

Memory Aids:
- Remember the definition of Dobereiner's Triad Law.

Quick Tips:
- Check if the atomic weight of the middle element is approximately the average of the other two.

58 Medium

58 of 175Question permalink

Application +4 | -1

If each orbital can hold a maximum of 3 electrons,
number of elements in fourth period of periodic table
is :

View answer and explanation

Correct option: C

27

Option C (27) is correct because with each orbital holding 3 electrons, the filling order changes, leading to a different number of elements in the fourth period.

Solution approach

Correct Answer + Core Reason: 27. With each orbital holding 3 electrons, the fourth period would have a different number of elements.

Step-by-Step Solution:

Step 1: Determine the orbitals in the fourth period: 4s, 3d, and 4p.

Step 2: Determine the number of orbitals in each subshell: 1 (s), 5 (d), and 3 (p).

Step 3: Calculate the total number of orbitals: $1 + 5 + 3 = 9$ orbitals.

Step 4: Calculate the total number of electrons that can be accommodated: $9 \text{ orbitals} \times 3 \text{ electrons/orbital} = 27$ electrons.

Step 5: Since each element corresponds to adding one electron, the number of elements in the fourth period is 27.

Why other options are wrong:
- Option A (48): This is incorrect. It likely arises from an incorrect calculation of the number of electrons.
- Option B (54): This is incorrect. It likely arises from an incorrect calculation of the number of electrons.
- Option D (36): This is incorrect. It likely arises from an incorrect calculation of the number of electrons.

Key Concepts:
- Electronic Configuration
- Orbitals and Subshells
- Periodic Table

Memory Aids:
- Remember the number of orbitals in each subshell (s, p, d, f).

Quick Tips:
- Know the filling order of electrons in orbitals.

59 Medium

59 of 175Question permalink

Remembering +4 | -1

Which oxide is amphoteric in nature?

View answer and explanation

Correct option: A

ZnO

Option A (ZnO) is correct because Zinc oxide can act as both an acid and a base.

Solution approach

Correct Answer + Core Reason: ZnO. Zinc oxide is amphoteric in nature.

Step-by-Step Solution:

Step 1: Define amphoteric nature.\nAn amphoteric oxide can react with both acids and bases.

Step 2: Consider the nature of the given oxides.
- ZnO is amphoteric.
- CaO is basic.
- $Na_2O$ is basic.
- BaO is basic.

Step 3: Conclusion.\nTherefore, ZnO is amphoteric in nature.

Why other options are wrong:
- Option B: CaO is a basic oxide.
- Option C: $Na_2O$ is a basic oxide.
- Option D: BaO is a basic oxide.

Key Concepts:
- Amphoteric oxides
- Acidic oxides
- Basic oxides

Memory Aids:
- ZnO is amphoteric.

Quick Tips:
- Remember the common amphoteric oxides.

60 Medium

60 of 175Question permalink

Remembering +4 | -1

The compounds of the

s

-block elements, with the exception of lithium and beryllium are predominantly

View answer and explanation

Correct option: A

Ionic

Option A (Ionic) is correct because compounds of s-block elements (except Li and Be) are predominantly ionic due to their low ionization energies.

Solution approach

Correct Answer + Core Reason: Ionic. Compounds of s-block elements (except Li and Be) are predominantly ionic.

Step-by-Step Solution:

Step 1: s-block elements (Groups 1 and 2) have low ionization energies.

Step 2: When they react with electronegative elements, they readily lose electrons to form positive ions (cations).

Step 3: The resulting compounds are held together by strong electrostatic forces between the cations and anions, forming ionic bonds.

Step 4: Lithium and Beryllium are exceptions due to their small size and high polarizing power, leading to more covalent character in their compounds.

Why other options are wrong:
- Option B (Covalent): Covalent compounds involve sharing of electrons, which is less common for s-block elements (except Li and Be) due to their low ionization energies.
- Option C (Complex): Complex compounds involve coordinate covalent bonds, which are not the primary bonding type for most s-block elements.
- Option D (Both (1) and (2)): While some s-block elements can form covalent compounds (Li and Be), the majority form ionic compounds.

Key Concepts:
- Ionic Bonding
- Covalent Bonding
- s-block elements
- Ionization Energy

Memory Aids:
- Remember that s-block elements (except Li and Be) tend to form ionic compounds.

Quick Tips:
- Consider the electronegativity difference between elements when predicting bond type.

Your Progress

Classification of elements and periodicity in properties practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice