Consider the ions ${{\rm{K}}^ + },{\rm{\;}}{{\rm{S}}^{2 - }},{\rm{C}}{{\rm{l}}^ - }{\rm{\;and\;C}}{{\rm{a}}^{2 + }}$. The radii of these ionic species follow the order

${{\rm{S}}^{2 - }} > C{{\rm{l}}^ - } > {{\rm{K}}^ + } > {\rm{C}}{{\rm{a}}^{2 + }}{\rm{\;}}$

${\rm{C}}{{\rm{a}}^{2 + }} > {{\rm{K}}^ + } > C{{\rm{l}}^ - } > {{\rm{S}}^{2 - }}$

${\rm{C}}{{\rm{l}}^ - } > {{\rm{S}}^{2 - }} > {{\rm{K}}^ + } > {\rm{C}}{{\rm{a}}^{2 + }}$

${\rm{C}}{{\rm{a}}^{2 + }} > C{{\rm{l}}^ - } > K > {{\rm{S}}^{2 - }}$

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

1 Medium

1 of 389Question permalink

Remembering +4 | -1

The elements with zero electron affinity are

View answer and explanation

Correct option: B

Beryllium and Helium

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

Solution approach

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

Step-by-Step Solution:

Step 1: Understand electron affinity.

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

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

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

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

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

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

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

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

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

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

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

2 Medium

2 of 389Question permalink

Analyzing +4 | -1

S-I:

\mathrm{Cu}^{+2}

paramagnetic.

S-II: All the orbitals present in

\mathrm{Cu}^{+2}

ion are doubly occupied.

View answer and explanation

Correct option: C

Only I is true

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

3 Easy

3 of 389Question permalink

Understanding +4 | -1

Which of the following is the most electropositive element?

View answer and explanation

Correct option: C

Mg

Option C (Mg) is correct because electropositivity increases down and to the left of the periodic table, and Mg is more to the left and down compared to P, S, and Al.

Solution approach

Correct Answer + Core Reason: Mg. Electropositivity increases down and to the left of the periodic table.

Step-by-Step Solution:
Step 1: Recall the trend of electropositivity in the periodic table. Electropositivity (metallic character) increases as you move down a group and to the left across a period.
Step 2: Locate the elements P, S, Mg, and Al in the periodic table. P and S are in the same period (Period 3) to the right, Al is also in Period 3 but to the left, and Mg is in Period 3, Group 2, which is further to the left than Al.
Step 3: Apply the trend. Mg is the most electropositive element among the given options because it is located furthest to the left in the periodic table compared to the other elements.

Why other options are wrong:
- Option A: P: Incorrect, P is a nonmetal and is not electropositive.
- Option B: S: Incorrect, S is a nonmetal and is not electropositive.
- Option D: Al: Incorrect, Al is less electropositive than Mg because Mg is further to the left in the periodic table.

Key Concepts:
- Electropositivity (metallic character)
- Periodic trends

Memory Aids:
- Electropositivity increases down and to the left.

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

4 Medium

4 of 389Question permalink

Remembering +4 | -1

The law of Octaves is not applicable to:

View answer and explanation

Correct option: D

Se,Te,As

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

Solution approach

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

Step-by-Step Solution:

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

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

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

Key Concepts:
- Law of Octaves
- Periodic table

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

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

5 Medium

5 of 389Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: B

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

6 Medium

6 of 389Question permalink

Analyzing +4 | -1

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

View answer and explanation

Correct option: C

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

7 Easy

7 of 389Question permalink

Analyzing +4 | -1

Identify the least stable ion among the following

View answer and explanation

Correct option: B

B{e^ - }

The correct answer is B (

B{e^ - }

) because Beryllium (

Be

) has a stable electronic configuration (2s²) and adding an electron to form

Be^-

is energetically unfavorable.

Solution approach

Correct Answer + Core Reason: $B{e^ - }$ . Beryllium has a relatively stable electronic configuration (2s²) and adding an electron is energetically unfavorable.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the neutral atoms:
- Li: 1s² 2s¹
- Be: 1s² 2s²
- B: 1s² 2s² 2p¹
- C: 1s² 2s² 2p²

Step 2: Write the electronic configurations of the anions:
- $L{i^ - }$ : 1s² 2s² (stable)
- $B{e^ - }$ : 1s² 2s² 2p¹ (less stable than Be)
- ${B^ - }$ : 1s² 2s² 2p² (more stable than Be-)
- ${C^ - }$ : 1s² 2s² 2p³ (more stable than Be-)

Step 3: Beryllium has a filled 2s subshell. Adding an electron requires adding it to the higher energy 2p subshell, making it less stable compared to the other options.

Why other options are wrong:
- Option A: $L{i^ - }$ achieves a noble gas configuration (1s² 2s²), making it relatively stable.
- Option C: ${B^ - }$ has a partially filled 2p subshell, which is more stable than adding an electron to Be.
- Option D: ${C^ - }$ has a more filled 2p subshell than B-, making it more stable.

Key Concepts:
- Electronic Configuration
- Stability of Ions
- Octet Rule

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

Quick Tips:
- Consider the electronic configurations of the ions when determining their stability.

8 Easy

8 of 389Question permalink

Application +4 | -1

An element has E.C.

1{S^2}\,\,2{S^2}\,\,2{P^6}\,\,3{S^2}\,\,3{P^3}

predict their period, group and block

View answer and explanation

Correct option: A

period = 3, block = p, group = 15

Option A (period = 3, block = p, group = 15) is correct because the element's electronic configuration ends in 3p3, indicating the 3rd period, p-block, and group 15.

Solution approach

Correct Answer + Core Reason: period = 3, block = p, group = 15. The electronic configuration determines the period, block, and group of the element.

Step-by-Step Solution:

Step 1: Identify the period. The period is determined by the highest principal quantum number (n) in the electronic configuration. In this case, the highest n is 3, so the period is 3.

Step 2: Identify the block. The block is determined by the subshell where the last electron enters. Here, the last electron enters the p-subshell, so the block is p.

Step 3: Identify the group. For p-block elements, the group number is calculated as 10 + the number of electrons in the outermost p-subshell. In this case, there are 3 electrons in the 3p subshell, so the group number is 10 + 5 = 15. (Note: The number of electrons in the outermost p-subshell is 3, plus the 2 electrons in the s-subshell of the same shell, which gives 5. So, the group number is 10 + 5 = 15.)

Step 4: Final Answer: Period = 3, Block = p, Group = 15

Why other options are wrong:
- Option B: This is incorrect because the period is determined by the highest principal quantum number, which is 3, not 5. Also, the block is p, not s.
- Option C: This is incorrect because while the period and block are correctly identified, the group number is miscalculated. The group number for p-block elements is 10 + number of valence electrons in p-orbital which is 10+5=15, not 10.
- Option D: This is incorrect because the period is 3, not 4, and the block is p, not d.

Key Concepts:
- Electronic configuration
- Period, block, and group in the periodic table
- Principal quantum number
- Valence electrons

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:
- The period number is the highest principal quantum number (n) in the electronic configuration.
- The block is determined by the last subshell filled (s, p, d, or f).
- For s-block elements, the group number is the number of valence electrons.
- For p-block elements, the group number is 10 + the number of valence electrons in p-orbital.

9 Hard

9 of 389Question permalink

Understanding +4 | -1

In

{K^ + }\,\,\,{F^ - }

ionic radius of

{F^ - }

is more while atomic radius of

{K^ + }

is

View answer and explanation

Correct option: B

More than

{F^ - }

Option B (More than

{F^ - }

) is correct because the potassium ion has fewer electrons than the fluoride ion, but more protons, resulting in a smaller ionic radius, but the atomic radius of K is larger than F-

Solution approach

Correct Answer + Core Reason: More than ${F^ - }$ . K has more protons and fewer electrons than F-, so its atomic radius is larger.

Step-by-Step Solution:

Step 1: Consider the electronic configurations of K⁺ and F^-.

Step 2: K⁺ has the electronic configuration of Argon (Ar), and F^- also has the electronic configuration of Neon (Ne). However, the question asks about atomic radius of K, not K+.

Step 3: The atomic radius of K is larger than the ionic radius of K+ because K has an additional electron shell. The atomic radius of F is smaller than the ionic radius of F- because F- has more electrons than F.

Step 4: K (atomic number 19) has more protons than F (atomic number 9). Therefore, the atomic radius of K is larger than the ionic radius of F^-.

Why other options are wrong:
- Option A: Less than F^- is incorrect because K has more protons than F, so its atomic radius is larger.
- Option C: Equal to F^- is incorrect because K has more protons than F, so its atomic radius is larger.
- Option D: None of these is incorrect because the atomic radius of K is larger than the ionic radius of F^-.

Key Concepts:
- Atomic radius
- Ionic radius
- Electronic configuration

Memory Aids:
- Cations are smaller than their neutral atoms.
- Anions are larger than their neutral atoms.

Quick Tips:
- Pay attention to whether the question is asking about atomic radius or ionic radius.

10 Hard

10 of 389Question permalink

Understanding +4 | -1

Consider the following statements:
I. Metals comprise more than

78 \%

of all known elements and appear on the left side of the periodic table.
II. Most non-metallic solids are brittle and are neither malleable nor ductile.
III. The elements become more metallic as we go down a group.
IV. The non-metallic character decreases as one goes from left to right across the periodic table.
Choose the option with all correct statements:

View answer and explanation

Correct option: A

I, II and III

Option A (I, II and III) is correct because metals are abundant on the left side of the periodic table, non-metallic solids are brittle, and metallic character increases down a group.

Solution approach

Correct Answer + Core Reason: I, II and III. Metals are abundant, non-metallic solids are brittle, and metallic character increases down a group.

Step-by-Step Solution:

Step 1: Analyze each statement.\nI. Metals comprise more than $78 \%$ of all known elements and appear on the left side of the periodic table. (Correct)\nII. Most non-metallic solids are brittle and are neither malleable nor ductile. (Correct)\nIII. The elements become more metallic as we go down a group. (Correct)\nIV. The non-metallic character decreases as one goes from left to right across the periodic table. (Incorrect)

Step 2: Identify the correct statements.\nStatements I, II, and III are correct.

Why other options are wrong:
- Option B: II, III and IV - Statement IV is incorrect.
- Option C: I and II - Statement III is also correct.
- Option D: III and IV - Statement IV is incorrect, and Statement I and II are correct.

Key Concepts:
- Metals and Non-metals
- Periodic Trends
- Properties of Metals and Non-metals

Memory Aids:
- Metals are on the left side of the periodic table.
- Non-metals are on the right side of the periodic table.

Quick Tips:
- Remember the general properties of metals and non-metals and their positions in the periodic table.

11 Medium

11 of 389Question permalink

Understanding +4 | -1

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

Ti

(II)

Ca

(III)

Sc

View answer and explanation

Correct option: D

(I), (III), (II)

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

12 Easy

12 of 389Question permalink

Understanding +4 | -1

The correct order according to size is

View answer and explanation

Correct option: C

{O^{2 - }} > {O^ - } > O

Option C (

{O^{2 - }} > {O^ - } > O

) is correct because as the negative charge increases, the electron-electron repulsion also increases, leading to an increase in size.

Solution approach

Correct Answer + Core Reason: ${O^{2 - }} > {O^ - } > O$ . Increasing negative charge increases ionic size.

Step-by-Step Solution:

Step 1: Consider the effect of adding electrons to an atom.

Step 2: Adding electrons increases electron-electron repulsion, which causes the electron cloud to expand.

Step 3: Therefore, the more negative the ion, the larger its size.

Step 4: Comparing O, O⁻, and O²⁻, the order of increasing size is O < O⁻ < O²⁻.

Why other options are wrong:
- Option A: This order is incorrect because it suggests that the neutral atom is larger than its negative ions.
- Option B: This order is incorrect because it suggests that O⁻ is larger than O²⁻, which is not true.
- Option D: This order is incorrect because it suggests that the neutral atom is larger than its negative ions, and that O²⁻ is smaller than O⁻.

Key Concepts:
- Ionic size
- Electron-electron repulsion
- Effect of charge on size

Memory Aids:
- Negative charge = bigger size.

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

13 Easy

13 of 389Question permalink

Application +4 | -1

Element with electronic configuration

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

3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^3

belongs to the following group of the periodic table

View answer and explanation

Correct option: B

$5 th$

Option B ($5 th$) is correct because the electronic configuration ends with

5p^3

, which indicates that the element belongs to the 5th group (Group 15) of the p-block elements.

Solution approach

Correct Answer + Core Reason: $5 th$. The electronic configuration ends with $5p^3$ , indicating group 15 (5th group of p-block).

Step-by-Step Solution:

Step 1: Analyze the electronic configuration: $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^3$ .

Step 2: Identify the valence shell: The valence shell is the 5th shell ( $5s^2 5p^3$ ).

Step 3: Determine the number of valence electrons: The number of valence electrons is 2 + 3 = 5.

Step 4: Determine the group number: Since the last electron enters the $p$ -orbital, it's a $p$ -block element. The group number is 10 + number of electrons in the $p$ -orbital = 10 + 5 = 15. The question asks for the group of p-block, so it is the 5th group of p-block elements.

Why other options are wrong:
- Option A: $2 {nd}$ : Incorrect group number.
- Option C: $3 {rd}$ : Incorrect group number.
- Option D: $7 {th}$ : Incorrect group number.

Key Concepts:
- Electronic configuration and its relation to the periodic table.
- Group number determination.

Memory Aids:
- For $p$ -block elements, group number = 10 + number of valence electrons in $p$ -orbital.

Quick Tips:
- Identify the valence shell and the number of valence electrons to determine the group number.

14 Easy

14 of 389Question permalink

Analyzing +4 | -1

On which of the following Dobereiner's Triad law is not applicable?

View answer and explanation

Correct option: C

{F}, {Cl}, {Br}

The correct answer is C (

{F}, {Cl}, {Br}

) because the atomic mass of Chlorine is not approximately the average of Fluorine and Bromine.

Solution approach

Correct Answer + Core Reason: ${F}, {Cl}, {Br}$ . The atomic mass of Cl is not approximately the average of F and Br.

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: Check each option:
- ${Cl}, {Br}, {I}$ : (35.5 + 127)/2 = 81.25 (close to 79.9 for Br)
- ${Ca}, {Sr}, {B}$ : (40 + 137)/2 = 88.5 (not close to 87.6 for Sr - this is a bad triad)
- ${F}, {Cl}, {Br}$ : (19 + 79.9)/2 = 49.45 (not close to 35.5 for Cl)
- ${Li}, {Na}, {K}$ : (7 + 39)/2 = 23 (close to 23 for Na)

Step 3: The triad law is not applicable for ${F}, {Cl}, {Br}$ because the atomic mass of Cl (35.5) is not approximately the average of F (19) and Br (79.9).

Why other options are wrong:
- Option A: ${Cl}, {Br}, {I}$ approximately follows the triad law.
- Option B: ${Ca}, {Sr}, {B}$ approximately follows the triad law.
- Option D: ${Li}, {Na}, {K}$ approximately follows the triad law.

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.

15 Easy

15 of 389Question permalink

Application +4 | -1

The electronegativity of A and B are 0.7 and 4.0 respectively. The bond formed in AB and

{B_2}

molecule will be

View answer and explanation

Correct option: A

Ionic and covalent resp

Option A (Ionic and covalent resp) is correct because the large electronegativity difference between A and B leads to ionic bonding in AB, while identical atoms in B2 form a covalent bond.

Solution approach

Correct Answer + Core Reason: Ionic and covalent resp. Large electronegativity difference leads to ionic bond, similar electronegativity leads to covalent bond.

Step-by-Step Solution:

Step 1: Identify the electronegativity difference between A and B. The electronegativity difference is 4.0 - 0.7 = 3.3.

Step 2: Determine the type of bond formed between A and B. Since the electronegativity difference is greater than 1.7, the bond is considered ionic.

Step 3: Determine the type of bond formed between B and B in the B2 molecule. Since the electronegativity difference is 0, the bond is purely covalent.

Step 4: Conclude the types of bonds formed in AB and B2.

Why other options are wrong:
- Option B: Covalent & ionic resp: This is incorrect because the bond in AB is ionic, not covalent.
- Option C: Ionic and dative resp: Dative bonds are coordinate covalent bonds, not typically formed between elements with such a large electronegativity difference.
- Option D: Metallic & covalent resp: Metallic bonds are formed between metal atoms, not between A and B.

Key Concepts:
- Electronegativity
- Ionic bond
- Covalent bond
- Electronegativity difference and bond type

Memory Aids:
- Large electronegativity difference -> Ionic bond
- Small electronegativity difference -> Covalent bond

Quick Tips:
- Remember the electronegativity difference threshold for ionic vs. covalent bonds (approximately 1.7).

16 Easy

16 of 389Question permalink

Understanding +4 | -1

Which one of the following combinations represents a metallic element?

View answer and explanation

Correct option: A

2, 8, 2

Option A is correct because the electronic configuration 2, 8, 2 represents an element with 2 valence electrons, which is characteristic of metallic elements in Group 2 (alkaline earth metals).

Solution approach

Correct Answer + Core Reason: A. 2, 8, 2. This electronic configuration represents a metallic element because it has 2 valence electrons.

Step-by-Step Solution:

Step 1: Analyze the electronic configuration 2, 8, 2. This means the element has 2 electrons in the first shell, 8 in the second, and 2 in the outermost shell.

Step 2: Elements with 1, 2, or 3 valence electrons tend to be metals.

Step 3: Elements with 5, 6, or 7 valence electrons tend to be non-metals.

Step 4: Elements with 4 valence electrons tend to be metalloids.

Step 5: Elements with 8 valence electrons (except Helium with 2) are noble gases.

Why other options are wrong:
- Option B: 2, 8, 4 could be a metalloid or non-metal.
- Option C: 2, 8, 7 is a halogen, which is a non-metal.
- Option D: 2, 8, 8 is a noble gas, which is a non-metal.

Key Concepts:
- Electronic configuration
- Valence electrons
- Metals, non-metals, and metalloids

Memory Aids:
- Remember the general trend: few valence electrons = metal, many valence electrons = non-metal.

Quick Tips:
- Determine the number of valence electrons to predict the metallic character.

17 Medium

17 of 389Question permalink

Remembering +4 | -1

Which among the following is/are amphoteric oxides

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because

BeO

,

SnO

, and

ZnO

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

Key Concepts:
- Amphoteric Oxides

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

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

18 Medium

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Analyzing +4 | -1

Which one of the following is smallest in size

View answer and explanation

Correct option: C

Na

Option C (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

19 Medium

19 of 389Question permalink

Understanding +4 | -1

Chemical properties of

Li

and

Mg

are similar because:

View answer and explanation

Correct option: B

Shows diagonal relationship

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

20 Easy

20 of 389Question permalink

Remembering +4 | -1

IUPAC name of an element with atomic number 110 is

View answer and explanation

Correct option: A

Ununnillium

Option A (Ununnillium) is correct because it follows the IUPAC nomenclature rules for naming elements with atomic number 110.

Solution approach

Correct Answer + Core Reason: Ununnillium. This name is derived directly from the digits of the atomic number 110 using IUPAC nomenclature.

Step-by-Step Solution:

Step 1: Understand the question: The question asks for the IUPAC name of an element with atomic number 110.

Step 2: Recall IUPAC nomenclature rules: The IUPAC nomenclature for elements with atomic numbers greater than 100 uses the following roots for the digits:
- 0: nil
- 1: un
- 2: bi
- 3: tri
- 4: quad
- 5: pent
- 6: hex
- 7: sept
- 8: oct
- 9: enn

Step 3: Apply the rules to atomic number 110:
- 1: un
- 1: un
- 0: nil
- Add the suffix -ium
- Therefore, the name is Ununnillium.

Why other options are wrong:
- Option B: Unununium is the name for element 111.
- Option C: Unnilennium is not a valid IUPAC name.
- Option D: Unnilunium is not a valid IUPAC name.

Key Concepts:
- IUPAC nomenclature for elements with atomic numbers greater than 100.

Memory Aids:
- Remember the roots for each digit (0-9).

Quick Tips:
- Break down the atomic number into its individual digits and apply the IUPAC rules.

21 Medium

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Analyzing +4 | -1

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

View answer and explanation

Correct option: B

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

Option B (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

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

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

22 Medium

22 of 389Question permalink

Application +4 | -1

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

X

'. If half of the atoms are converted into

X_{(g)}^{+}

by energy

\Delta {H}

. The IE of

{X}

is

View answer and explanation

Correct option: A

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

Option A (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

Key Concepts:
- Ionization Energy
- Avogadro's Number

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

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

23 Medium

23 of 389Question permalink

Application +4 | -1

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

View answer and explanation

Correct option: D

15

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

24 Hard

24 of 389Question permalink

Understanding +4 | -1

The pair of elements having approximately equal ionisation potential is

View answer and explanation

Correct option: A

Al, Ga

Option A (Al, Ga) is correct because of the lanthanide contraction and shielding effects, which cause Ga to have a slightly higher effective nuclear charge, resulting in a similar ionization potential to Al.

Solution approach

Correct Answer + Core Reason: Al, Ga. Due to the lanthanide contraction and shielding effects, Ga has a similar ionization potential to Al.

Step-by-Step Solution:

Step 1: Identify the positions of the elements in the periodic table.\nAl is in Group 13, Period 3.\nGa is in Group 13, Period 4.\nSi is in Group 14, Period 3.\nMg is in Group 2, Period 3.\nB is in Group 13, Period 2.

Step 2: Recall the general trend that ionization energy decreases down a group.

Step 3: However, there are exceptions to this trend. In the case of Al and Ga, the filling of the d-orbitals before Ga leads to poor shielding and an increase in effective nuclear charge. This causes Ga to have a slightly higher ionization energy than expected, making it similar to Al.

Why other options are wrong:
- Option B: Al, Si - Si is in the next group and has a significantly higher ionization energy.
- Option C: Al, Mg - Mg is in Group 2 and has a significantly lower ionization energy than Al.
- Option D: Al, B - B is above Al in the same group and has a higher ionization energy.

Key Concepts:
- Ionization Energy
- Periodic Trends
- Lanthanide Contraction
- Shielding Effect

Memory Aids:
- Remember the exceptions to the general trends due to shielding and effective nuclear charge.

Quick Tips:
- Be aware of the exceptions to the general trends in ionization energy.

25 Medium

25 of 389Question permalink

Application +4 | -1

An element having electronic configuration

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

forms

View answer and explanation

Correct option: B

Basic oxide

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

Solution approach

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

Step-by-Step Solution:

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

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

Step 3: Recall that alkali metals form basic oxides.

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

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

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

Memory Aids:
- Alkali metals form basic oxides

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

26 Easy

26 of 389Question permalink

Understanding +4 | -1

Which of the following element can exhibit more than one oxidation state

View answer and explanation

Correct option: A

Cr

The correct answer is A (Cr) because Chromium (Cr) can exhibit multiple oxidation states due to its electronic configuration.

Solution approach

Correct Answer + Core Reason: Cr. Chromium exhibits variable oxidation states due to its electronic configuration and ability to lose different numbers of electrons.

Step-by-Step Solution:

Step 1: Recall the electronic configurations of the given elements.

Step 2: Cr has an electronic configuration of [Ar] 3d⁵ 4s¹. It can lose electrons from both the 3d and 4s orbitals, leading to multiple oxidation states (e.g., +2, +3, +6).

Step 3: Cs, Sr, and Rb are alkali and alkaline earth metals, which typically exhibit only one oxidation state (+1 and +2, respectively).

Why other options are wrong:
- Option B: Cs (Cesium) is an alkali metal and typically exhibits only a +1 oxidation state.
- Option C: Sr (Strontium) is an alkaline earth metal and typically exhibits only a +2 oxidation state.
- Option D: Rb (Rubidium) is an alkali metal and typically exhibits only a +1 oxidation state.

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

Memory Aids:
- Remember that transition metals often exhibit multiple oxidation states.

Quick Tips:
- Know the common oxidation states of alkali metals, alkaline earth metals, and transition metals.

27 Medium

27 of 389Question permalink

Understanding +4 | -1

The ionic radii of

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

and

B{e^{ + 3}}

follow the order

View answer and explanation

Correct option: D

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

Option D (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

28 Easy

28 of 389Question permalink

Understanding +4 | -1

Consider the ions

{\mathrm{K}^ + },\mathrm{\;}{\mathrm{S}^{2 - }},\mathrm{C}{\mathrm{l}^ - }\mathrm{\;and\;C}{\mathrm{a}^{2 + }}

. The radii of these ionic species follow the order

View answer and explanation

Correct option: D

{\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > \mathrm{C}{\mathrm{a}^{2 + }}\mathrm{\;}

Option D (

{\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > {\mathrm{Ca}^{2 + }}\mathrm{\;}

) is correct because ionic radii decrease with increasing nuclear charge and increase with increasing negative charge.

Solution approach

Correct Answer + Core Reason: ${\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > {\mathrm{Ca}^{2 + }}\mathrm{\;}$ . Ionic radii are affected by nuclear charge and number of electrons.

Step-by-Step Solution:

Step 1: Identify the number of electrons for each ion:
- K+ (19 - 1 = 18 electrons)
- S2- (16 + 2 = 18 electrons)
- Cl- (17 + 1 = 18 electrons)
- Ca2+ (20 - 2 = 18 electrons)

Step 2: All ions are isoelectronic (same number of electrons). Therefore, the ionic radius is determined by the effective nuclear charge (Zeff).

Step 3: Higher the effective nuclear charge, smaller the ionic radius. The nuclear charges are:
- K+ (Z = 19)
- S2- (Z = 16)
- Cl- (Z = 17)
- Ca2+ (Z = 20)

Step 4: Arrange the ions in increasing order of nuclear charge: S2- < Cl- < K+ < Ca2+

Step 5: Since ionic radius decreases with increasing nuclear charge, the order of ionic radii is: S2- > Cl- > K+ > Ca2+

Why other options are wrong:
- Option A: $\mathrm{Ca}^{2 + } > \mathrm{K}^ + } > C\mathrm{l}^ - } > \mathrm{S}^{2 - }}$ : This is incorrect because it places the ion with the highest nuclear charge (Ca2+) as having the largest radius, which is the opposite of the correct trend.
- Option B: $\mathrm{Cl}^ - } > \mathrm{S}^{2 - } > \mathrm{K}^ + } > \mathrm{Ca}^{2 + }}$ : This is incorrect because it incorrectly orders S2- and Cl- relative to each other. S2- has a lower nuclear charge than Cl-, so it should have a larger radius.
- Option C: $\mathrm{Ca}^{2 + } > C\mathrm{l}^ - } > K > \mathrm{S}^{2 - }}$ : This is incorrect because it includes neutral K, which is not isoelectronic with the other ions, and also incorrectly orders the ionic radii based on nuclear charge.

Key Concepts:
- Isoelectronic species
- Effective nuclear charge (Zeff)
- Ionic radius trends

Memory Aids:
- For isoelectronic species, higher positive charge = smaller radius.
- Anions are larger than their neutral atoms; cations are smaller.

Quick Tips:
- When comparing isoelectronic species, focus on the nuclear charge. The ion with the greater nuclear charge will have a smaller radius.

29 Medium

29 of 389Question permalink

Understanding +4 | -1

The correct order of ionic radii is represented in

View answer and explanation

Correct option: B

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

Option B (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

Memory Aids:
- Positive charge = smaller size.

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

30 Easy

30 of 389Question permalink

Remembering +4 | -1

The outer electronic configuration of transition element is

View answer and explanation

Correct option: A

\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}

Option A (

\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}

) is correct because it accurately represents the general outer electronic configuration of transition elements, allowing for variations in the d and s orbital filling.

Solution approach

Correct Answer + Core Reason: $\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}$ . This configuration allows for the filling of the (n-1)d orbitals with 1 to 10 electrons and the ns orbital with 0 to 2 electrons, which is characteristic of transition elements.

Step-by-Step Solution:

Step 1: Recall the definition of transition elements. Transition elements are those elements that have partially filled d orbitals in their elemental or most stable oxidation state.

Step 2: Consider the general electronic configuration of transition elements. The general electronic configuration is $(n-1)d^{1-10} ns^{0-2}$ , where n is the principal quantum number of the outermost shell.

Step 3: Understand the variations in the electronic configuration. The $ns$ orbital can have 0, 1, or 2 electrons due to the small energy difference between the $(n-1)d$ and $ns$ orbitals. This leads to exceptions like Cr and Cu.

Why other options are wrong:
- **Option B ( $n{s^2}\,\,n{p^{1 - 6}}$ ): This is the general electronic configuration of p-block elements, not transition elements.
- Option C ( $\left( {n - 2} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}$ ): The d-orbital being filled is (n-1)d, not (n-2)d. (n-2)f orbitals are filled in inner transition elements (lanthanides and actinides).
- Option D ( $\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^2}$ ): This configuration is not entirely correct as it doesn't account for the possibility of having 0 or 1 electrons in the ns orbital, as seen in elements like Chromium (Cr).

Key Concepts:
- Electronic configuration of transition elements
- Definition of transition elements
- Filling of d-orbitals

Memory Aids:
- Remember that transition elements involve the filling of the (n-1)d orbitals.

Quick Tips:**
- Pay attention to the (n-1) factor for d-orbitals in transition elements.

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