Identify the correct order of solubility of ${\rm{N}}{{\rm{a}}_2}{\rm{S}},{\rm{CuS}}$ and ${\rm{ZnS}}$ in aqueous medium:

${\rm{N}}{{\rm{a}}_2}{\rm{S}} > ZnS > CuS$

${\rm{CuS}} > ZnS > N{{\rm{a}}_2}{\rm{S}}$

${\rm{ZnS}} > N{{\rm{a}}_2}{\rm{S}} > CuS$

${\rm{N}}{{\rm{a}}_2}{\rm{S}} > CuS > ZnS$

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

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

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

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

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

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

1 Hard

1 of 43Question 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.

2 Hard

2 of 43Question 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.

3 Hard

3 of 43Question 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.

4 Hard

4 of 43Question permalink

Remembering +4 | -1

The electronegativity follows the order

View answer and explanation

Correct option: A

{F} > {O} > {Cl} > {Br}

Option A (

{F} > {O} > {Cl} > {Br}

) is correct because it represents the correct order of electronegativity.

Solution approach

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

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 ${F} > {O} > {Cl} > {Br}$ .

Why other options are wrong:
- Option B: ${F} > {Cl} > {Br} > {O}$ is incorrect because Oxygen is more electronegative than Chlorine and Bromine.
- Option C: ${O} > {F} > {Cl} > {Br}$ is incorrect because Fluorine is the most electronegative.
- Option D: ${Cl} > {F} > {O} > {Br}$ 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.

5 Hard

5 of 43Question permalink

Understanding +4 | -1

The IUPAC nomenclature of an element with electronic configuration

[R n] 5 f^{14} 6 d^1 7 s^2

is

View answer and explanation

Correct option: D

Unniltrium

Option D (Unniltrium) is correct because it follows the IUPAC nomenclature rules for elements with atomic number 103.

Solution approach

Correct Answer + Core Reason: Unniltrium. The IUPAC name is derived from the digits of the atomic number (103).

Step-by-Step Solution:

Step 1: Determine the atomic number of the element. The electronic configuration $[Rn] 5f^{14} 6d^1 7s^2$ indicates that the element has 103 protons.

Step 2: Apply the IUPAC nomenclature rules for elements with atomic numbers greater than 100. The digits are named as follows: 0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn. The name is formed by concatenating the names of the digits in the atomic number, followed by "ium".

Step 3: For atomic number 103, the digits are 1, 0, and 3. The corresponding names are un, nil, and tri. Therefore, the IUPAC name is Unniltrium.

Why other options are wrong:
- Option A (Unnilbium): This would correspond to element 102.
- Option B (Unnilunium): This would correspond to element 101.
- Option C (Unnilquadium): This would correspond to element 104.

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

Memory Aids:
- Remember the digit-to-name mapping: 0=nil, 1=un, 2=bi, 3=tri, 4=quad, 5=pent, 6=hex, 7=sept, 8=oct, 9=enn.

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

6 Hard

6 of 43Question permalink

Remembering +4 | -1

The pair of amphoteric hydroxides is

View answer and explanation

Correct option: D

\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}

Option D (

\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}

) is correct because both beryllium hydroxide and zinc hydroxide are amphoteric, reacting with both acids and bases.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}$ are amphoteric hydroxides.

Step-by-Step Solution:

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

Step 2: Consider the given options:
- $\mathrm{LiOH}$ is a strong base.
- $\mathrm{Al}{\left( {\mathrm{OH}} \right)_3}$ is amphoteric.
- $\mathrm{Be}{\left( {\mathrm{OH}} \right)_2}$ is amphoteric.
- $\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2}$ is a weak base.
- $\mathrm{B}{\left( {\mathrm{OH}} \right)_3}$ is a weak acid.
- $\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}$ is amphoteric.

Step 3: $\mathrm{Be}{\left( {\mathrm{OH}} \right)_2}$ and $\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}$ react with both acids and bases.

Step 4: Conclusion: $\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}$ is the correct pair of amphoteric hydroxides.

Why other options are wrong:
- Option A: $\mathrm{LiOH}$ is a strong base and not amphoteric.
- Option B: $\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2}$ is a weak base and not amphoteric.
- Option C: $\mathrm{B}{\left( {\mathrm{OH}} \right)_3}$ is a weak acid and not amphoteric.
- Option D: $\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}$ is the correct answer as both are amphoteric.

Key Concepts:
- Amphoteric Hydroxides
- Acidic Hydroxides
- Basic Hydroxides

Memory Aids:
- Remember common amphoteric hydroxides: $Be(OH)_2$ , $Zn(OH)_2$ , $Al(OH)_3$ .

Quick Tips:
- Learn the common examples of amphoteric, acidic, and basic hydroxides.

7 Hard

7 of 43Question permalink

Understanding +4 | -1

The radii of

F,{F^ - },O

and

{O^{ - 2}}

are in the order of

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because anions are larger than their neutral atoms, and size decreases with increasing nuclear charge for isoelectronic species.

Solution approach

Correct Answer + Core Reason: ${O^{ - 2}} > {F^ - } > O > F$ . Anions are larger than neutral atoms, and for isoelectronic species, size decreases with increasing nuclear charge.

Step-by-Step Solution:

Step 1: Identify the species: $F, {F^-}, O, {O^{2-}}$ .

Step 2: Understand the trend: Anions are larger than their parent atoms due to increased electron repulsion and effective nuclear charge decrease. For isoelectronic species (same number of electrons), the species with a smaller nuclear charge will have a larger radius.

Step 3: Compare O and O^2-: ${O^{2-}}$ is larger than O because it's an anion.

Step 4: Compare F and F^-: ${F^-}$ is larger than F because it's an anion.

Step 5: Compare ${O^{2-}}$ and ${F^-}$ : Both are isoelectronic (10 electrons). Oxygen has a nuclear charge of +8, and Fluorine has a nuclear charge of +9. Therefore, ${O^{2-}}$ is larger than ${F^-}$ .

Step 6: Compare O and F: Oxygen has a nuclear charge of +8 and Fluorine has a nuclear charge of +9. Since they are neutral atoms in the same period, Fluorine is smaller than Oxygen due to greater effective nuclear charge.

Step 7: Combine the trends: ${O^{2-}} > {F^-} > O > F$ .

Why other options are wrong:
- Option A: ${O^{ - 2}} > {F^ - } > F > O$ - Incorrect because F is smaller than O.
- Option B: ${F^ - } > {O^{2 - }} > O > F$ - Incorrect because ${O^{2-}}$ is larger than ${F^-}$ .
- Option C: ${O^{2 - }} > O > {F^ - } > F$ - Incorrect because ${F^-}$ is larger than O.

Key Concepts:
- Ionic Radius Trends
- Isoelectronic Species
- Effective Nuclear Charge

Memory Aids:
- Anions are larger than neutral atoms.
- For isoelectronic species, higher nuclear charge means smaller size.

Quick Tips:
- Always consider the charge when comparing ionic radii.
- Remember the trends for isoelectronic species.

8 Hard

8 of 43Question permalink

Application +4 | -1

The electronegativity of A and B are 1.20 and 4.0 respectively. Therefore, ionic character in A - B bond will be

View answer and explanation

Correct option: D

72.23\%

Option D (

72.23\%

) is correct because it is the ionic character calculated using the Pauling electronegativity difference formula.

Solution approach

Correct Answer + Core Reason: $72.23\%$ . The ionic character is calculated using the Pauling electronegativity difference formula.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference.
$\Delta \chi = |\chi_A - \chi_B| = |1.20 - 4.0| = 2.8$

Step 2: Calculate the percentage ionic character using Pauling's formula.
$\mathrm{\% Ionic Character} = (1 - e^{-0.25(\Delta \chi)^2}) \times 100$
$\mathrm{\% Ionic Character} = (1 - e^{-0.25(2.8)^2}) \times 100 = (1 - e^{-0.25(7.84)}) \times 100 = (1 - e^{-1.96}) \times 100$
$\mathrm{\% Ionic Character} = (1 - 0.141) \times 100 = 0.859 \times 100 = 85.9 \%$

Step 3: Calculate the percentage ionic character using Hannay-Smyth equation.
$\mathrm{\% Ionic Character} = 16|\chi_A - \chi_B| + 3.5|\chi_A - \chi_B|^2$
$\mathrm{\% Ionic Character} = 16(2.8) + 3.5(2.8)^2 = 44.8 + 3.5(7.84) = 44.8 + 27.44 = 72.24 \%$

Why other options are wrong:
- **Option A ( $50\%$ ): This is incorrect as it likely involves an incorrect calculation or a misunderstanding of the relationship between electronegativity difference and ionic character.
- Option B ( $43\%$ ): This is incorrect as it likely involves an incorrect calculation or a misunderstanding of the relationship between electronegativity difference and ionic character.
- Option C ( $53.3\%$ ): This is incorrect as it likely involves an incorrect calculation or a misunderstanding of the relationship between electronegativity difference and ionic character.

Key Concepts:
- Electronegativity
- Ionic Character
- Pauling scale

Memory Aids:
- Remember the Pauling electronegativity difference formula.

Quick Tips:**
- Use the appropriate formula to calculate the percentage ionic character based on the electronegativity difference.

9 Hard

9 of 43Question permalink

Application +4 | -1

One mole of magnesium in the vapour state absorbed 1200

\mathrm{kJmo}{\mathrm{l}^{ - 1}}

of energy. If the first and second ionisation energies of Mg are 750 and 1450

\mathrm{kJmo}{\mathrm{l}^{ - 1}}

respectively, the final composition of the mixture is

View answer and explanation

Correct option: B

69\mathrm{\% M}{\mathrm{g}^ + } + 31\mathrm{\% M}{\mathrm{g}^{2 + }}

Option B (

69\% \mathrm{Mg}^+ + 31\% \mathrm{Mg}^{2+}

) is correct because it accurately calculates the percentage of Mg+ and Mg2+ ions formed based on the energy absorbed and the ionization energies.

Solution approach

Correct Answer + Core Reason: $69\mathrm{\% M}{\mathrm{g}^ + } + 31\mathrm{\% M}{\mathrm{g}^{2 + }}$ . This is the final composition of the mixture.

Step-by-Step Solution:

Step 1: Calculate the energy required for complete ionization to Mg+.\nEnergy required for Mg to Mg+ = 750 kJ/mol

Step 2: Calculate the remaining energy after forming Mg+.\nRemaining energy = 1200 kJ/mol - 750 kJ/mol = 450 kJ/mol

Step 3: Calculate the fraction of Mg+ that is further ionized to Mg2+.\nEnergy required for Mg+ to Mg2+ = 1450 kJ/mol\nFraction of Mg+ ionized to Mg2+ = $\frac{450}{1450} = 0.3103$

Step 4: Calculate the percentage of Mg2+.\nPercentage of Mg2+ = 0.3103 $\times$ 100% = 31.03% ≈ 31%

Step 5: Calculate the percentage of Mg+.\nPercentage of Mg+ = 100% - 31% = 69%

Why other options are wrong:
- Option A: $31\mathrm{\% M}{\mathrm{g}^ + } + 69\mathrm{\% M}{\mathrm{g}^{2 + }}$ : This reverses the percentages of Mg+ and Mg2+.
- Option C: $86\mathrm{\% M}{\mathrm{g}^ + } + 14\mathrm{\% M}{\mathrm{g}^{2 + }}$ : This is likely a calculation error involving incorrect division or subtraction.
- Option D: $14\mathrm{\% M}{\mathrm{g}^ + } + 86\mathrm{\% M}{\mathrm{g}^{2 + }}$ : This is likely a calculation error involving incorrect division or subtraction.

Key Concepts:
- Ionization energy
- Mole concept

Memory Aids:
- Remember that ionization energies are required to remove electrons.

Quick Tips:
- Calculate the amount of each ion formed based on the energy absorbed and the ionization energies.
- Pay attention to the order of ionization.

10 Hard

10 of 43Question permalink

Analyzing +4 | -1

The correct order of ionic radius is

View answer and explanation

Correct option: D

{\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}

Option D (

{\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}

) is correct because a smaller positive charge results in a larger ionic radius due to reduced effective nuclear charge.

Solution approach

Correct Answer + Core Reason: ${\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}$ . The ionic radius decreases with increasing positive charge due to a greater effective nuclear charge.

Step-by-Step Solution:

Step 1: Understand the concept of ionic radius and its relationship with nuclear charge and number of electrons.

Step 2: Recall that as the positive charge on an ion increases, the effective nuclear charge experienced by the remaining electrons increases. This pulls the electrons closer to the nucleus, resulting in a smaller ionic radius.

Step 3: Compare ${\mathrm{P}^{3 + }}$ and ${\mathrm{P}^{5 + }}$ . ${\mathrm{P}^{5 + }}$ has a higher positive charge than ${\mathrm{P}^{3 + }}$ .

Step 4: Therefore, ${\mathrm{P}^{5 + }}$ will have a smaller ionic radius than ${\mathrm{P}^{3 + }}$ .

Step 5: Conclusion: ${\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}$

Why other options are wrong:
- Option A: $\mathrm{T}{\mathrm{i}^{4 + }} < M{\mathrm{n}^{7 + }}$ : ${\mathrm{Mn}^{7+}}$ has a much higher positive charge than ${\mathrm{Ti}^{4+}}$ , so it will have a smaller ionic radius. The given order is incorrect.
- Option B: ${\mathrm{\;}^{35}}\mathrm{C}{\mathrm{l}^ - }{ > ^{37}}\mathrm{C}{\mathrm{l}^ - }$ : Isotopes of the same element have the same electronic structure and nearly the same ionic radii. The difference in mass number is negligible in determining ionic radius.
- Option C: ${\mathrm{K}^ + } > C{\mathrm{l}^ - }$ : ${\mathrm{K}^ + }$ and ${\mathrm{Cl}^ - }$ are isoelectronic species. For isoelectronic species, the ion with greater nuclear charge will have smaller ionic radius. Cl has a greater nuclear charge, so ${\mathrm{Cl}^ - }$ should be smaller than ${\mathrm{K}^ + }$ . The given order is incorrect.

Key Concepts:
- Ionic Radius
- Effective Nuclear Charge
- Isoelectronic Species

Memory Aids:
- Cations are smaller than their parent atoms, and anions are larger.
- For isoelectronic species, greater nuclear charge means smaller radius.

Quick Tips:
- When comparing ionic radii, first consider the charge. Higher positive charge means smaller radius, and higher negative charge means larger radius.
- If the ions are isoelectronic, compare their nuclear charges.

11 Hard

11 of 43Question permalink

Understanding +4 | -1

Arrange the elements

N,\,\,\,P,\,\,\,O

and

S

in the order of increasing non-metallic nature.

View answer and explanation

Correct option: B

P < S < N < O

Option B (

P < S < N < O

) is correct because non-metallic character increases across a period and decreases down a group.

Solution approach

Correct Answer + Core Reason: $P < S < N < O$ . Non-metallic character increases across a period and decreases down a group.

Step-by-Step Solution:

Step 1: Understand the trend of non-metallic character in the periodic table. Non-metallic character generally increases across a period (from left to right) and decreases down a group (from top to bottom).

Step 2: Locate the elements N, P, O, and S in the periodic table. N and O are in the second period, while P and S are in the third period. N and P are in group 15, and O and S are in group 16.

Step 3: Compare the non-metallic character of N and P. Since N is above P in group 15, N is more non-metallic than P.

Step 4: Compare the non-metallic character of O and S. Since O is above S in group 16, O is more non-metallic than S.

Step 5: Compare the non-metallic character of S and P. Since S is to the right of P in the third period, S is more non-metallic than P.

Step 6: Compare the non-metallic character of N and S. Since N is in the second period and S is in the third period, and both are in different groups, we compare their positions relative to the trends. N is more non-metallic than S.

Step 7: Combine the comparisons to determine the correct order: $P < S < N < O$ .

Why other options are wrong:
- Option A: $P < N < O < S$ - Incorrect because S is less non-metallic than N.
- Option C: $O < N < S < P$ - Incorrect because P is the least non-metallic and O is the most non-metallic.
- Option D: $S > O > N > P$ - Incorrect because O is the most non-metallic and P is the least non-metallic.

Key Concepts:
- Non-metallic Character Trends
- Periodic Table

Memory Aids:
- Non-metallic character increases across a period.
- Non-metallic character decreases down a group.

Quick Tips:
- Remember the general trends of non-metallic character in the periodic table.
- Compare the positions of the elements in the periodic table to determine their relative non-metallic character.

12 Hard

12 of 43Question permalink

Understanding +4 | -1

Which of the following orders of ionic radius is correctly represented?

View answer and explanation

Correct option: A

\mathrm{H}^{-} > \mathrm{H} > \mathrm{H}^{+}

Option A (

\mathrm{H}^{-} > \mathrm{H} > \mathrm{H}^{+}

) is correct because adding electrons increases the ionic radius, while removing electrons decreases it.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}^{-} > \mathrm{H} > \mathrm{H}^{+}$ . Adding electrons increases the radius due to increased electron repulsion, while removing electrons decreases the radius due to increased effective nuclear charge.

Step-by-Step Solution:

Step 1: Consider the hydrogen atom and its ions. H has 1 proton and 1 electron. H- has 1 proton and 2 electrons. H+ has 1 proton and 0 electrons.

Step 2: Adding an electron to form H- increases electron-electron repulsion, causing the electron cloud to expand, thus increasing the ionic radius.

Step 3: Removing an electron to form H+ removes the electron cloud entirely, leaving only the bare nucleus (proton), which has a much smaller radius.

Step 4: Therefore, the order of ionic radii is H- > H > H+

Why other options are wrong:
- Option B: $\mathrm{Na}^{+} > \mathrm{F}^{-} > \mathrm{O}^{2-}$ is incorrect. For isoelectronic species, increasing nuclear charge decreases ionic radius. The correct order would be O2- > F- > Na+
- Option C: $\mathrm{F}^{-} > \mathrm{O}^{2-} > \mathrm{Na}^{+}$ is incorrect. For isoelectronic species, increasing nuclear charge decreases ionic radius. The correct order would be O2- > F- > Na+
- Option D: $\mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{N}^{3-}$ is incorrect. For isoelectronic species, increasing nuclear charge decreases ionic radius. The correct order would be N3- > Mg2+ > Al3+

Key Concepts:
- Ionic radii
- Effect of electron gain/loss on ionic radius
- Isoelectronic species

Memory Aids:
- Anions (negative ions) are larger than their neutral atoms.
- Cations (positive ions) are smaller than their neutral atoms.

Quick Tips:
- When comparing ions of the same element, the more negative the charge, the larger the size, and the more positive the charge, the smaller the size.
- For isoelectronic species, the greater the nuclear charge, the smaller the ionic radius.

13 Hard

13 of 43Question permalink

Application +4 | -1

If the electronegativity difference between two atoms

A

and

B

is 2.0, then the percentage of covalent character in the molecule is

View answer and explanation

Correct option: A

54\%

Option A (

54\%

) is correct because the percentage of covalent character can be estimated using the Pauling equation, which relates electronegativity difference to ionic character, and then subtracting from 100%.

Solution approach

Correct Answer + Core Reason: $54\%$ . The percentage of covalent character is estimated using the relationship between electronegativity difference and ionic character.

Step-by-Step Solution:

Step 1: Determine the percentage of ionic character using the Pauling equation (approximate). While there isn't a direct formula to calculate % covalent character from electronegativity difference, we can approximate using the relationship between electronegativity difference and ionic character. A difference of 1.7 corresponds to 50% ionic character. A difference of 2.0 will have a higher ionic character.

Step 2: Use the Hannay-Smyth equation (more accurate, but not always necessary for multiple choice). The Hannay-Smyth equation provides a more accurate estimation:
% Ionic Character = $16|X_A - X_B| + 3.5|X_A - X_B|^2$ \nWhere $|X_A - X_B|$ is the electronegativity difference.
% Ionic Character = $16(2.0) + 3.5(2.0)^2 = 32 + 3.5(4) = 32 + 14 = 46\%$ (approximately)

Step 3: Calculate the percentage of covalent character.
% Covalent Character = $100\% - \text{{% Ionic Character}} = 100\% - 46\% = 54\%$

Why other options are wrong:
- Option B: $46\%$ : This is the approximate percentage of ionic character, not covalent character.
- Option C: $23\%$ : This value doesn't correspond to any logical calculation based on electronegativity difference.
- Option D: $72\%$ : This value is too high for the covalent character given an electronegativity difference of 2.0.

Key Concepts:
- Electronegativity difference
- Ionic character
- Covalent character
- Hannay-Smyth equation (optional)

Memory Aids:
- Remember that as electronegativity difference increases, ionic character increases and covalent character decreases.

Quick Tips:
- Know the general trend: higher electronegativity difference means higher ionic character.

14 Hard

14 of 43Question permalink

Understanding +4 | -1

The elements X,Y,Z and T have the indicated electronic configuration. Starting with the innermost shell, which is the most metallic element?

View answer and explanation

Correct option: C

Z=2,8,8,1

Option C (Z=2,8,8,1) is correct because metallic character increases with the ease of losing electrons, and this configuration has only one valence electron, making it the most metallic.

Solution approach

Correct Answer + Core Reason: Z=2,8,8,1. Metallic character increases with the ease of losing electrons.

Step-by-Step Solution:
Step 1: Recall the trend of metallic character in the periodic table. Metallic character increases as you move down a group and to the left across a period.
Step 2: Analyze the given electronic configurations:
- X = 2, 8, 4 (Group 14 element)
- Y = 2, 8, 8 (Noble gas)
- Z = 2, 8, 8, 1 (Group 1 element)
- T = 2, 8, 8, 7 (Group 17 element)
Step 3: Determine which configuration represents the most metallic element. Metallic elements tend to lose electrons easily to form positive ions. The element with the fewest valence electrons and the largest atomic size will be the most metallic. Among the given options, Z (2, 8, 8, 1) has only one valence electron, making it the most metallic.

Why other options are wrong:
- Option A: X=2,8,4: Incorrect, this element has 4 valence electrons and is less metallic than Z.
- Option B: Y=2,8,8: Incorrect, this is a noble gas and is not metallic.
- Option D: T=2,8,8,7: Incorrect, this element has 7 valence electrons and is a nonmetal (halogen).

Key Concepts:
- Metallic character
- Periodic trends
- Electronic configuration
- Valence electrons

Memory Aids:
- Metals lose electrons easily.

Quick Tips:
- Look for elements with few valence electrons when determining which element is the most metallic.

15 Hard

15 of 43Question permalink

Understanding +4 | -1

Identify the correct order of solubility of

\mathrm{N}{\mathrm{a}_2}\mathrm{S},\mathrm{CuS}

and

\mathrm{ZnS}

in aqueous medium:

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS

Option D (

\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS

) is correct because alkali metal sulfides are generally soluble, while transition metal sulfides are generally insoluble, with CuS being the least soluble due to its high lattice energy.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS$ . Alkali metal sulfides are soluble, transition metal sulfides are less soluble.

Step-by-Step Solution:

Step 1: Consider the solubility rules for sulfides.

Step 2: Alkali metal sulfides (like Na₂S) are generally soluble in water.

Step 3: Transition metal sulfides (like CuS and ZnS) are generally insoluble in water.

Step 4: Compare the solubility of CuS and ZnS. CuS is less soluble than ZnS due to higher lattice energy and lower $K_{sp}$ value.

Step 5: Therefore, the order of solubility is $\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS$ .

Why other options are wrong:
- Option A: This order is incorrect because it places CuS as the most soluble, which is wrong.
- Option B: This order is incorrect because it places ZnS as more soluble than Na₂S, which is wrong.
- Option C: This order is incorrect because it places CuS as more soluble than ZnS, which is wrong.

Key Concepts:
- Solubility rules
- Lattice energy
- Solubility product ( $K_{sp}$ )

Memory Aids:
- Alkali metals are soluble, many transition metals are not.

Quick Tips:
- Remember the general solubility rules for ionic compounds.

16 Hard

16 of 43Question permalink

Analyzing +4 | -1

The decreasing order of the second ionization energy of

K

,

\mathrm{Ca}

and

\mathrm{Ba}

is:

View answer and explanation

Correct option: A

\mathrm{K} > Ca > Ba

Option A (

\mathrm{K} > Ca > Ba

) is correct because after the first ionization, K attains a noble gas configuration, requiring significantly more energy for the second ionization. Ca and Ba lose electrons from the s-orbital.

Solution approach

Correct Answer + Core Reason: $\mathrm{K} > Ca > Ba$ . The second ionization energy depends on the stability of the resulting ion after the first ionization.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the elements.\nK: $[Ar]4s^1$ \nCa: $[Ar]4s^2$ \nBa: $[Xe]6s^2$

Step 2: Write the electronic configurations after the first ionization.
$K^+$ : $[Ar]$ (Noble gas configuration)
$Ca^+$ : $[Ar]4s^1$
$Ba^+$ : $[Xe]

17 Hard

17 of 43Question permalink

Understanding +4 | -1

The radii of

\mathrm{F},{\mathrm{F}^ - },\mathrm{O}

and

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

are in the order of:

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because anions are larger than their neutral atoms, and for isoelectronic species, the species with lower nuclear charge has a larger radius.

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F$ . Anions are larger than neutral atoms; isoelectronic species with lower nuclear charge are larger.

Step-by-Step Solution:

Step 1: Recognize that anions are larger than their corresponding neutral atoms. Therefore, F^- > F and O^2- > O.

Step 2: Compare F^- and O^2-. They are isoelectronic (both have 10 electrons). O^2- has a smaller nuclear charge (8) than F^- (9). Therefore, O^2- is larger than F^-.

Step 3: Compare O and F. O is to the left of F in the periodic table, so O is larger than F.

Step 4: Combining these relationships, the correct order is O^2- > F^- > O > F.

Why other options are wrong:
- Option A: Incorrect order of neutral atoms.
- Option B: Incorrect order of ions.
- Option C: Incorrect order of neutral atoms and ions.

Key Concepts:
- Ionic radius
- Isoelectronic species
- Periodic trends

Memory Aids:
- Anions are always larger than their neutral atoms.
- For isoelectronic species, lower nuclear charge = larger size.

Quick Tips:
- When comparing sizes, first consider whether the species are ions or neutral atoms. Then, consider isoelectronic relationships and periodic trends.

18 Hard

18 of 43Question permalink

Understanding +4 | -1

Which of the following statements are correct regarding electronegativity?

View answer and explanation

Correct option: C

Decrease in electronegativity down a group in the periodic table confirms decreases in non-metallic properties

Option C (Decrease in electronegativity down a group in the periodic table confirms decreases in non-metallic properties) is correct because electronegativity is directly related to non-metallic character; as electronegativity decreases, non-metallic properties decrease.

Solution approach

Correct Answer + Core Reason: Decrease in electronegativity down a group in the periodic table confirms decreases in non-metallic properties. Electronegativity is directly related to non-metallic character.

Step-by-Step Solution:

Step 1: Recall the definition of electronegativity. It is the ability of an atom to attract electrons in a chemical bond.

Step 2: Recall the trend in electronegativity across a period. Electronegativity generally increases from left to right.

Step 3: Recall the trend in electronegativity down a group. Electronegativity generally decreases down a group.

Step 4: Relate electronegativity to metallic and non-metallic properties. High electronegativity is associated with non-metallic character, and low electronegativity is associated with metallic character.

Step 5: Analyze each statement.\nStatement A: Electronegativity generally increases across a period in the periodic table from left to right. (Incorrect)\nStatement B: Electronegativity value decreases with the increase in atomic radii down a group. (Incorrect)\nStatement C: Decrease in electronegativity down a group in the periodic table confirms decreases in non-metallic properties. (Correct)\nStatement D: In a period of the periodic table electronegativity increases with increase in metallic character. (Incorrect)

Why other options are wrong:
- Option A: Electronegativity generally increases across a period.
- Option B: Electronegativity decreases down a group.
- Option D: Electronegativity increases with increasing non-metallic character.

Key Concepts:
- Electronegativity
- Periodic Trends
- Metallic and Non-metallic Properties

Memory Aids:
- High electronegativity = non-metallic character.
- Low electronegativity = metallic character.

Quick Tips:
- Remember the relationship between electronegativity and metallic/non-metallic character.

19 Hard

19 of 43Question permalink

Understanding +4 | -1

The second electron gain enthalpies (in kJ/mol) of oxygen and Sulphur respectively are

View answer and explanation

Correct option: D

+780, +590

Option D (+780, +590) is correct because the second electron gain enthalpy is always positive due to electron-electron repulsion, and oxygen is smaller than sulfur, leading to greater repulsion.

Solution approach

Correct Answer + Core Reason: +780, +590. The second electron gain enthalpy is always positive due to electron-electron repulsion.

Step-by-Step Solution:
Step 1: Understand the concept of electron gain enthalpy. Electron gain enthalpy is the energy change that occurs when an electron is added to a gaseous atom.
Step 2: Recall that the first electron gain enthalpy can be negative (exothermic) or positive (endothermic), but the second electron gain enthalpy is always positive.
Step 3: Explain why the second electron gain enthalpy is always positive. Adding a second electron to a negatively charged ion requires energy to overcome the electrostatic repulsion between the electron and the ion.
Step 4: Compare the sizes of oxygen and sulfur. Oxygen is smaller than sulfur.
Step 5: Explain why oxygen has a higher second electron gain enthalpy than sulfur. Due to its smaller size, oxygen experiences greater electron-electron repulsion when a second electron is added, requiring more energy. Therefore, oxygen has a higher second electron gain enthalpy than sulfur.

Why other options are wrong:
- Option A: -780, +590: Incorrect, the second electron gain enthalpy cannot be negative.
- Option B: -590, +780: Incorrect, the second electron gain enthalpy cannot be negative, and oxygen has a higher value than sulfur.
- Option C: +590, +780: Incorrect, oxygen has a higher second electron gain enthalpy than sulfur.

Key Concepts:
- Electron gain enthalpy
- Electron-electron repulsion
- Atomic size

Memory Aids:
- The second electron gain enthalpy is always positive.

Quick Tips:
- Remember that the second electron gain enthalpy is always positive due to electron-electron repulsion.

20 Hard

20 of 43Question permalink

Understanding +4 | -1

Which of the following isoelectronic ions has lowest ionisation energy?

View answer and explanation

Correct option: D

{S^{2 - }}

Option D (

{S^{2 - }}

) is correct because for isoelectronic ions, the ion with the least nuclear charge has the lowest ionization energy.

Solution approach

Correct Answer + Core Reason: ${S^{2 - }}$ . For isoelectronic ions, the lowest nuclear charge has the lowest ionization energy.

Step-by-Step Solution:

Step 1: Determine if the ions are isoelectronic. Cl^-, Ca²⁺, K⁺, and S^2- all have the same number of electrons (18).

Step 2: Identify the number of protons (nuclear charge) for each ion: Cl^- (17), Ca²⁺ (20), K⁺ (19), S^2- (16).

Step 3: The ion with the lowest nuclear charge will have the lowest ionization energy because the electrons are held less tightly.

Step 4: S^2- has the lowest nuclear charge (16), so it has the lowest ionization energy.

Why other options are wrong:
- Option A: Cl^- has a higher nuclear charge (17) than S^2-, so it has a higher ionization energy.
- Option B: Ca²⁺ has a higher nuclear charge (20) than S^2-, so it has a higher ionization energy.
- Option C: K⁺ has a higher nuclear charge (19) than S^2-, so it has a higher ionization energy.

Key Concepts:
- Isoelectronic species
- Ionization energy
- Nuclear charge

Memory Aids:
- For isoelectronic species, lower nuclear charge = lower ionization energy.

Quick Tips:
- When comparing ionization energies of isoelectronic species, focus on the nuclear charge.

21 Hard

21 of 43Question permalink

Application +4 | -1

The energy required to convert all atoms present in

1.2 \mathrm{~g}

magnesium to

{Mg}^{2+}

ions if

\mathrm{IE}

, and

\mathrm{IE}_2

of magnesium are

120 \mathrm{~kJ} \mathrm{~mol}^{-1}

and

240 \mathrm{~kJ} \mathrm{~mol}^{-1}

respectively

View answer and explanation

Correct option: A

18 \mathrm{~kJ}

Option A (18 kJ) is correct because it calculates the total energy required to convert 1.2g of magnesium to Mg2+ ions using the given ionization energies and the number of moles of magnesium.

Solution approach

Correct Answer + Core Reason: $18 \mathrm{~kJ}$ . This is the energy required to convert all atoms present in $1.2 \mathrm{~g}$ magnesium to ${Mg}^{2+}$ ions.

Step-by-Step Solution:

Step 1: Calculate the number of moles of magnesium.\nMolar mass of magnesium (Mg) = 24 g/mol\nMoles of magnesium = $\frac{1.2 \text{ g}}{24 \text{ g/mol}} = 0.05 \text{ mol}$

Step 2: Calculate the total ionization energy required.
$IE_1$ = 120 kJ/mol
$IE_2$ = 240 kJ/mol\nTotal ionization energy per mole = $IE_1 + IE_2 = 120 \text{ kJ/mol} + 240 \text{ kJ/mol} = 360 \text{ kJ/mol}$

Step 3: Calculate the total energy required for 0.05 mol of magnesium.\nTotal energy = 0.05 mol $\times$ 360 kJ/mol = 18 kJ

Why other options are wrong:
- Option B: $36 \mathrm{~kJ}$ : This is likely a calculation error, possibly multiplying the total ionization energy per mole by 0.1 instead of 0.05.
- Option C: $360 \mathrm{~kJ}$ : This is the total ionization energy per mole, not the total energy for 1.2g of magnesium.
- Option D: $24 \mathrm{~kJ}$ : This is likely a calculation error involving incorrect multiplication or division.

Key Concepts:
- Ionization energy
- Mole concept

Memory Aids:
- Remember that ionization energies are required to remove electrons.

Quick Tips:
- Always convert mass to moles before calculating energy changes.
- Pay attention to the stoichiometry of the reaction.

22 Hard

22 of 43Question permalink

Understanding +4 | -1

Considering the elements

\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{F}

and

\mathrm{Si}

, the correct order of their non-metallic character is

View answer and explanation

Correct option: C

F

>

N

>

C

>

B

>

Si

Option C (F > N > C > B > Si) is correct because non-metallic character increases across a period and decreases down a group.

Solution approach

Correct Answer + Core Reason: F > N > C > B > Si. Non-metallic character increases across a period and decreases down a group.

Step-by-Step Solution:

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

Step 2: Recall the trend of non-metallic character: increases across a period (left to right) and decreases down a group (top to bottom).

Step 3: Arrange the elements according to their positions and the trend: F (most non-metallic), N, C, B, Si (least non-metallic).

Step 4: Write the order: F > N > C > B > Si

Why other options are wrong:
- Option A: B > C > Si > N > F: Incorrect order, does not follow the periodic trends.
- Option B: Si > C > B > N > F: Incorrect order, does not follow the periodic trends.
- Option D: F > N > C > Si > B: Incorrect order, Si and B are swapped.

Key Concepts:
- Periodic trends: Non-metallic character
- Position of elements in the periodic table

Memory Aids:
- Non-metallic character increases across a period (think of electronegativity)
- Non-metallic character decreases down a group

Quick Tips:
- Always consider the position of the elements in the periodic table when comparing their properties.

23 Hard

23 of 43Question permalink

Understanding +4 | -1

The set representing the correct order of first ionisation energy is

View answer and explanation

Correct option: B

\mathrm{Be} > Mg > Ca

Option B (

\mathrm{Be} > Mg > Ca

) is correct because ionization energy decreases down a group, and Be, Mg, and Ca are in Group 2.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be} > Mg > Ca$ . Ionization energy decreases down a group.

Step-by-Step Solution:
Step 1: Recall the trend of ionization energy within a group. Ionization energy generally decreases as you move down a group in the periodic table.
Step 2: Identify the positions of the given elements in the periodic table.
- Be, Mg, and Ca are all alkaline earth metals (Group 2), with Be at the top, Mg in the middle, and Ca below Mg.
- K, Na, and Li are alkali metals (Group 1), with Li at the top, Na in the middle, and K below Na. The correct order is Li > Na > K.
- B, C, and N are in the same period (Period 2). The general trend is B < C < N, but there's an exception between N and O. The correct order is N > C > B (considering the question asks for first ionization energy).
- Ge, Si, and C are in Group 14, with C at the top, Si in the middle, and Ge below Si. The correct order is C > Si > Ge.
Step 3: Apply the trend to the given options. Since ionization energy decreases down the group, the correct order for Group 2 elements is Be > Mg > Ca.

Why other options are wrong:
- Option A: $\mathrm{K} > Na > Li$ : Incorrect, the correct order is Li > Na > K.
- Option C: $\mathrm{B} > C > N$ : Incorrect, the correct order is N > C > B.
- Option D: $\mathrm{Ge} > Si > C$ : Incorrect, the correct order is C > Si > Ge.

Key Concepts:
- Ionization energy
- Periodic trends
- Group 1 (alkali metals) and Group 2 (alkaline earth metals) elements

Memory Aids:
- Ionization energy decreases down a group.

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

24 Hard

24 of 43Question permalink

Understanding +4 | -1

The ionisation potential values of an element are in the following order

{I_1} < {I_2} < < < < {I_3} < {I_4} < {I_5},

The element is

View answer and explanation

Correct option: D

Alkaline earth metal

Option D (Alkaline earth metal) is correct because the large jump between I3 and I4 indicates that the element has two valence electrons, characteristic of alkaline earth metals.

Solution approach

Correct Answer + Core Reason: Alkaline earth metal. The large jump between I3 and I4 indicates 2 valence electrons.

Step-by-Step Solution:

Step 1: Analyze the ionization potential values. The ionization potential increases as more electrons are removed from an atom.

Step 2: Look for a significant jump in ionization potential values. A large jump indicates that an electron is being removed from a core shell, which requires much more energy.

Step 3: In this case, the large jump is between I3 and I4. This means that after removing 3 electrons, it becomes significantly more difficult to remove the 4th electron.

Step 4: This suggests that the element has 3 valence electrons that are relatively easy to remove, and the 4th electron is a core electron. This is incorrect. The large jump between I3 and I4 indicates that the element has 3 valence electrons. The correct answer is that the large jump between I2 and I3 indicates that the element has 2 valence electrons.

Step 5: An element with 2 valence electrons is an alkaline earth metal.

Why other options are wrong:
- Option A: Alkali metals have 1 valence electron.
- Option B: Chalcogens have 6 valence electrons.
- Option C: Halogens have 7 valence electrons.

Key Concepts:
- Ionization potential
- Valence electrons
- Periodic table groups

Memory Aids:
- Group 1: Alkali metals (1 valence electron)
- Group 2: Alkaline earth metals (2 valence electrons)
- Group 16: Chalcogens (6 valence electrons)
- Group 17: Halogens (7 valence electrons)

Quick Tips:
- Look for large jumps in ionization potential to determine the number of valence electrons.

25 Hard

25 of 43Question permalink

Understanding +4 | -1

The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species,

{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{B}{\mathrm{e}^{2 + }}

?

View answer and explanation

Correct option: C

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

Option C (

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

) is correct because polarizing power increases with increasing charge and decreasing size. Be²⁺ has the highest charge/size ratio and K⁺ has the lowest.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}$ . Polarizing power increases with charge and decreases with size.

Step-by-Step Solution:

Step 1: Define polarizing power. Polarizing power is the ability of a cation to distort the electron cloud of an anion.

Step 2: Polarizing power is directly proportional to the charge of the cation and inversely proportional to its size.

Step 3: Compare the charges and sizes of the given cations:
- K⁺ (charge +1, relatively large size)
- Ca²⁺ (charge +2, relatively large size)
- Mg²⁺ (charge +2, smaller size than Ca²⁺)
- Be²⁺ (charge +2, smallest size)

Step 4: Based on charge/size ratio, the order of increasing polarizing power is: K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺.

Why other options are wrong:
- Option A: This order is incorrect because it places Mg²⁺ before Be²⁺, which is wrong as Be²⁺ is smaller and has a higher charge/size ratio.
- Option B: This order is incorrect because it places Be²⁺ before K⁺ and Ca²⁺, which is wrong as K⁺ and Ca²⁺ have lower charge/size ratios.
- Option D: This order is incorrect because it places Ca²⁺ before Mg²⁺ and Be²⁺, which is wrong as Mg²⁺ and Be²⁺ have higher charge/size ratios.

Key Concepts:
- Polarizing power
- Charge/size ratio
- Fajan's rules

Memory Aids:
- High charge, small size = high polarizing power.

Quick Tips:
- Remember Fajan's rules for polarizing power.

26 Hard

26 of 43Question permalink

Application +4 | -1

If the radius of the

{3^{{\text{rd }}}}

Bohr's orbit of hydrogen atom is

{r_3}

and the radius of

{4^{{\text{th}}}}

Bohr's orbit is

{r_4}.

Then:

View answer and explanation

Correct option: B

{r_4} = \frac{{16}}{9}{r_3}

Option B (

{r_4} = \frac{{16}}{9}{r_3})

is correct because the radius of the nth Bohr orbit is proportional to

n^2

.

Solution approach

Correct Answer + Core Reason: ${r_4} = \frac{{16}}{9}{r_3}$ . The radius of the nth Bohr orbit is proportional to $n^2$ .

Step-by-Step Solution:

Step 1: Recall the formula for the radius of the nth Bohr orbit: $r_n = k \times n^2$ , where k is a constant.

Step 2: Write the expressions for ${r_3}$ and ${r_4}$ :
- ${r_3} = k \times 3^2 = 9k$
- ${r_4} = k \times 4^2 = 16k$

Step 3: Find the ratio of ${r_4}$ to ${r_3}$ : $\frac{{r_4}}{{r_3}} = \frac{{16k}}{{9k}} = \frac{{16}}{9}$

Step 4: Express ${r_4}$ in terms of ${r_3}$ : ${r_4} = \frac{{16}}{9}{r_3}$

Why other options are wrong:
- Option A: ${r_4} = \frac{9}{{16}}{r_3}$ is incorrect. It's the inverse of the correct ratio.
- Option C: ${r_4} = \frac{3}{4}{r_3}$ is incorrect. It uses the ratio of the principal quantum numbers directly instead of their squares.
- Option D: ${r_4} = \frac{4}{3}{r_3}$ is incorrect. It uses the ratio of the principal quantum numbers directly instead of their squares.

Key Concepts:
- Bohr's model of the atom
- Radius of nth Bohr orbit: $r_n \propto n^2$

Memory Aids:
- Remember that the radius is proportional to the square of the principal quantum number.

Quick Tips:
- Always use the correct formula for Bohr radius and remember the proportionality to $n^2$ .

27 Hard

27 of 43Question permalink

Analyzing +4 | -1

Correct order of second ionisation potential of C, N, O, F is

View answer and explanation

Correct option: C

\mathrm{O > F > N > C}

Option C (

\mathrm{O > F > N > C}

) is correct because after the first ionization, nitrogen achieves a stable half-filled configuration, making the second ionization easier than removing from F+.

Solution approach

Correct Answer + Core Reason: $\mathrm{O > F > N > C}$ . After the first ionization, nitrogen achieves a stable half-filled configuration, making the second ionization easier.

Step-by-Step Solution:

Step 1: Write the electronic configurations of C, N, O, and F.

Step 2: C: $1s^2 2s^2 2p^2$

Step 3: N: $1s^2 2s^2 2p^3$

Step 4: O: $1s^2 2s^2 2p^4$

Step 5: F: $1s^2 2s^2 2p^5$

Step 6: After the first ionization:

Step 7: $\mathrm{C}^+$ : $1s^2 2s^2 2p^1$

Step 8: $\mathrm{N}^+$ : $1s^2 2s^2 2p^2$

Step 9: $\mathrm{O}^+$ : $1s^2 2s^2 2p^3$ (half-filled, relatively stable)

Step 10: $\mathrm{F}^+$ : $1s^2 2s^2 2p^4$

Step 11: Removing an electron from $\mathrm{O}^+$ requires more energy because it disrupts the half-filled configuration. Removing an electron from $\mathrm{F}^+$ is easier than from $\mathrm{O}^+$ , but harder than from $\mathrm{N}^+$ . Removing from $\mathrm{N}^+$ is easier than $\mathrm{F}^+$ because it is closer to a stable configuration after losing the second electron.

Step 12: Therefore, the order is $\mathrm{O} > F > N > C$ .

Why other options are wrong:
- Option A: $\mathrm{C > N > O > F}$ : Incorrect order. Does not consider the stability of half-filled configurations.
- Option B: $\mathrm{O > N > F > C}$ : Incorrect order. F should be greater than N.
- Option D: $\mathrm{F > O > N > C}$ : Incorrect order. O should be greater than F due to the stability of the half-filled configuration of O+.

Key Concepts:
- Second ionization potential
- Electronic configuration
- Stability of half-filled and fully-filled orbitals

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

Quick Tips:
- Write the electronic configurations of the ions after the first ionization to determine the relative stability and ease of removing the second electron.

28 Hard

28 of 43Question permalink

Analyzing +4 | -1

The correct order of acidic strength

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > \mathrm{S}{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}

Option A is correct because the acidic strength of oxides increases with the electronegativity and oxidation state of the central atom: Cl2O7 > SO2 > P4O10.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > \mathrm{S}{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}$ . Acidic strength increases with the electronegativity and oxidation state of the central atom.

Step-by-Step Solution:

Step 1: Analyze Option A: The oxidation states are Cl(+7), S(+4), P(+5). Electronegativity: Cl > S > P. Therefore, the acidic strength order is $\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > \mathrm{S}{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}$ . This is correct.

Step 2: Analyze Option B: Basic strength increases down a group. Therefore, the acidic strength order should be $MgO > CaO > {\mathrm{K}_2}\mathrm{O}$ . This option is incorrect.

Step 3: Analyze Option C: The oxidation states are C(+4), N(+5), S(+6). Electronegativity: S > N > C. The acidic strength order should be $\mathrm{S}{\mathrm{O}_3} > {\mathrm{N}_2}{\mathrm{O}_5} > \mathrm{C}{\mathrm{O}_2}$ . This option is incorrect.

Step 4: Analyze Option D: Basic strength increases down a group and decreases across a period. Therefore, the acidic strength order should be $A{\mathrm{l}_2}{\mathrm{O}_3} > MgO > \mathrm{N}{\mathrm{a}_2}\mathrm{O}$ . This option is incorrect.

Why other options are wrong:
- Option B: The basic strength order is incorrect, so the acidic strength order is also incorrect.
- Option C: The acidic strength order is incorrect.
- Option D: The basic strength order is incorrect, so the acidic strength order is also incorrect.

Key Concepts:
- Acidic strength of oxides
- Electronegativity
- Oxidation state
- Basic strength of oxides

Memory Aids:
- Higher oxidation state and electronegativity = higher acidity

Quick Tips:
- Consider both electronegativity and oxidation state when comparing acidic strength.

29 Hard

29 of 43Question permalink

Understanding +4 | -1

In which pair, the first atom or ion is not larger than the second?

View answer and explanation

Correct option: C

\mathrm{O},\mathrm{S}

Option C (O, S) is correct because sulfur is below oxygen in the same group, so it has a larger atomic radius.

Solution approach

Correct Answer + Core Reason: $\mathrm{O},\mathrm{S}$ . Sulfur is below oxygen in the same group, so it is larger.

Step-by-Step Solution:

Step 1: Analyze each pair of atoms or ions.

Step 2: For option A (N, F), F is to the right of N in the same period, so F is smaller than N.

Step 3: For option B (Cl^-, Cl), Cl^- is an anion, so it is larger than Cl.

Step 4: For option C (O, S), S is below O in the same group, so S is larger than O. This means O is NOT larger than S.

Step 5: For option D (Fe²⁺, Fe³⁺), Fe³⁺ has a higher positive charge, so it is smaller than Fe²⁺.

Step 6: The question asks for the pair where the first atom or ion is *not* larger than the second. Only option C fits this condition.

Why other options are wrong:
- Option A: N is larger than F.
- Option B: Cl^- is larger than Cl.
- Option D: Fe²⁺ is larger than Fe³⁺.

Key Concepts:
- Atomic radius
- Ionic radius
- Periodic trends

Memory Aids:
- Anions are larger than their neutral atoms.
- Cations are smaller than their neutral atoms.
- Atomic size increases down a group.
- Atomic size decreases across a period.

Quick Tips:
- When comparing sizes, consider whether the species are ions or neutral atoms. Then, consider their positions in the periodic table.

30 Hard

30 of 43Question permalink

Analyzing +4 | -1

Assertion : Electronegativity of nitrogen is greater than carbon.

Reason : Nitrogen has stable half filled electronic configuration.

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

Option B is correct because both the assertion and reason are true, but the reason (stable half-filled configuration) is not the correct explanation for the assertion (electronegativity). Electronegativity is primarily related to nuclear charge and atomic size.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion. Nitrogen is more electronegative than carbon, and nitrogen has a stable half-filled electronic configuration, but the latter doesn't explain the former.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Electronegativity increases across a period. Nitrogen is to the right of carbon in the periodic table, so nitrogen is more electronegative than carbon. The assertion is true.

Step 2: Analyze the Reason: Nitrogen has a half-filled $2p$ subshell ( $2p^3$ ), which provides some extra stability. The reason is true.

Step 3: Determine if the reason explains the assertion: The electronegativity difference between nitrogen and carbon is primarily due to the difference in their effective nuclear charges and atomic sizes, not the stability of the half-filled configuration. The reason is not the correct explanation.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The reason is not false.
- Option D: The assertion is not false.

Key Concepts:
- Electronegativity trends
- Stability of half-filled electronic configurations

Memory Aids:
- Electronegativity increases across a period.

Quick Tips:
- Distinguish between factors affecting electronegativity and factors affecting ionization energy.

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Classification of elements and periodicity in properties practice · Hard

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