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

31 Hard

Remembering +4 | -1

Which of the following is an amphoteric oxide?

View answer and explanation

Correct option: C

\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}

Option C (

\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}

) is correct because aluminum oxide is a well-known amphoteric oxide, reacting with both acids and bases.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ is amphoteric because it reacts with both acids and bases.

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:
- $\mathrm{S}{\mathrm{O}_3}$ is an acidic oxide.
- MgO is a basic oxide.
- $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ is an amphoteric oxide.
- ${\mathrm{P}_4}{\mathrm{O}_{10}}$ is an acidic oxide.

Step 3: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ reacts with acids: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3} + 6HCl \rightarrow 2AlCl_3 + 3H_2O$
$\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ reacts with bases: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3} + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$

Step 4: Conclusion: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ is the amphoteric oxide.

Why other options are wrong:
- Option A: $\mathrm{S}{\mathrm{O}_3}$ is an acidic oxide and reacts only with bases.
- Option B: MgO is a basic oxide and reacts only with acids.
- Option C: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ is the correct answer as it is amphoteric.
- Option D: ${\mathrm{P}_4}{\mathrm{O}_{10}}$ is an acidic oxide and reacts only with bases.

Key Concepts:
- Amphoteric Oxides
- Acidic Oxides
- Basic Oxides

Memory Aids:
- Aluminum oxide is a common example of an amphoteric oxide.

Quick Tips:
- Remember common amphoteric oxides like $A{l_2}{O_3}$ and $ZnO$ .

32 Hard

32 of 43Question permalink

Understanding +4 | -1

The correct order of ionic radii for the ions,

{P^{3 - }},\,{S^{2 - }},\,C{a^{2 + }},\,{K^ + },\,C{l^ - }

is

View answer and explanation

Correct option: A

{P^{3 - }} > {S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}

Option A (

{P^{3 - }} > {S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}

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

Solution approach

Correct Answer + Core Reason: ${P^{3 - }} > {S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}$ . For isoelectronic species, ionic radii decrease with increasing nuclear charge.

Step-by-Step Solution:

Step 1: Identify that all the given ions are isoelectronic, meaning they have the same number of electrons.

Step 2: Count the number of electrons in each ion: ${P^{3 - }}$ (18), ${S^{2 - }}$ (18), $C{l^ - }$ (18), ${K^ + }$ (18), $C{a^{2 + }}$ (18).

Step 3: For isoelectronic species, ionic radius decreases with increasing nuclear charge (atomic number).

Step 4: Arrange the ions in order of increasing atomic number: P (15), S (16), Cl (17), K (19), Ca (20).

Step 5: Therefore, the order of ionic radii is ${P^{3 - }} > {S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}$ .

Why other options are wrong:
- Option B: $C{l^ - } > {S^{2 - }} > {P^{3 - }} > C{a^{2 + }} > {K^ + }$ is incorrect because it reverses the order based on nuclear charge.
- Option C: ${P^{3 - }} > {S^{2 - }} > C{l^ - } > C{a^{2 + }} > {K^ + }$ is incorrect because it incorrectly places Calcium before Potassium.
- Option D: ${K^ + } > C{a^{2 + }} > {P^{3 - }} > {S^{2 - }} > C{l^ - }$ is incorrect because it reverses the order based on nuclear charge.

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

Memory Aids:
- Remember that for isoelectronic species, ionic radius decreases with increasing nuclear charge.

Quick Tips:
- Always check if the given ions are isoelectronic before comparing their ionic radii.

33 Hard

33 of 43Question permalink

Application +4 | -1

The 71st electron of an element

X

with an atomic number of 71 enters into the orbital

View answer and explanation

Correct option: C

5d

The correct answer is C (

5d

) because after filling the

4f

orbitals, the 71st electron of element X (Lutetium) enters the

5d

orbital.

Solution approach

Correct Answer + Core Reason: $5d$ . The 71st electron enters the $5d$ orbital after filling the $4f$ orbitals.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 71.

Step 2: The element with atomic number 71 is Lutetium (Lu).

Step 3: Write the electronic configuration of Lutetium.

Step 4: The electronic configuration of Lutetium is $[Xe] 4f^{14} 5d^1 6s^2$ .

Step 5: Determine the orbital in which the 71st electron enters.

Step 6: The 71st electron enters the $5d$ orbital.

Why other options are wrong:
- Option A: $4f$ orbitals are filled before the $5d$ orbital, so the 71st electron does not enter the $4f$ orbital.
- Option B: $4p$ orbitals are filled much earlier in the electronic configuration, so the 71st electron does not enter the $4p$ orbital.
- Option D: $6s$ orbitals are filled before the $4f$ and $5d$ orbitals, so the 71st electron does not enter the $6s$ orbital.

Key Concepts:
- Electronic configuration
- Aufbau principle
- Filling of orbitals

Memory Aids:
- Remember the order of filling orbitals: $ns$ , $(n-2)f$ , $(n-1)d$ , $np$ .

Quick Tips:
- Know the order of filling orbitals according to the Aufbau principle.

34 Hard

34 of 43Question permalink

Application +4 | -1

General electronic configuration of actionoids is

(n-2) f^{1-14}(n-1) d^{0-2} n s^2

. Which of the following actinoids have one electron in

6d

orbital?

View answer and explanation

Correct option: C

{Cm}

(Atomic no. 96)

Option C is correct because Curium (Cm) with atomic number 96 has an electronic configuration that includes one electron in the 6d orbital.

Solution approach

Correct Answer + Core Reason: ${Cm}$ (Atomic no. 96). Curium has one electron in the 6d orbital.

Step-by-Step Solution:

Step 1: The general electronic configuration of actinoids is $(n-2) f^{1-14}(n-1) d^{0-2} n s^2$ .

Step 2: For actinoids, $n = 7$ , so the configuration is $(5f)^{1-14}(6d)^{0-2}(7s)^2$ .

Step 3: Consider Curium (Cm) with atomic number 96. Its electronic configuration is $[Rn] 5f^7 6d^1 7s^2$ .

Step 4: Consider Plutonium (Pu) with atomic number 94. Its electronic configuration is $[Rn] 5f^6 7s^2$ .

Step 5: Consider Americium (Am) with atomic number 95. Its electronic configuration is $[Rn] 5f^7 7s^2$ .

Step 6: Consider Berkelium (Bk) with atomic number 97. Its electronic configuration is $[Rn] 5f^9 7s^2$ .

Step 7: Only Curium has one electron in the 6d orbital.

Why other options are wrong:
- Option A: $Pu$ (Atomic no. 94) does not have one electron in the 6d orbital.
- Option B: $Am$ (Atomic no. 95) does not have one electron in the 6d orbital.
- Option D: $Bk$ (Atomic no. 97) does not have one electron in the 6d orbital.

Key Concepts:
- Electronic configuration
- Actinoids
- f-block elements

Memory Aids:
- Remember the general electronic configuration of actinoids.

Quick Tips:
- Determine the electronic configuration of each element to identify the number of electrons in the 6d orbital.

35 Hard

35 of 43Question permalink

Understanding +4 | -1

The set representing the correct order of ionic radius is

View answer and explanation

Correct option: B

N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}

Option B (

N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}

) is correct because ionic radius decreases with increasing nuclear charge for isoelectronic species and increases down a group.

Solution approach

Correct Answer + Core Reason: $N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}$ . Ionic radius decreases with increasing nuclear charge for isoelectronic species and increases down a group.

Step-by-Step Solution:

Step 1: Identify the electronic configurations of the ions.
- $Li^+$ : 2 electrons
- $Be^{2+}$ : 2 electrons
- $Na^+$ : 10 electrons
- $Mg^{2+}$ : 10 electrons

Step 2: Compare isoelectronic species. For ions with the same number of electrons (isoelectronic), the ionic radius decreases with increasing nuclear charge (more protons).
- $Li^+$ and $Be^{2+}$ are isoelectronic (2 electrons). $Li^+$ has 3 protons, and $Be^{2+}$ has 4 protons. Therefore, $Li^+ > Be^{2+}$ .
- $Na^+$ and $Mg^{2+}$ are isoelectronic (10 electrons). $Na^+$ has 11 protons, and $Mg^{2+}$ has 12 protons. Therefore, $Na^+ > Mg^{2+}$ .

Step 3: Compare ions in the same group. Ionic radius increases down a group.
- $Li^+$ and $Na^+$ are in the same group (Group 1). $Na^+$ is below $Li^+$ , so $Na^+ > Li^+$ .
- $Be^{2+}$ and $Mg^{2+}$ are in the same group (Group 2). $Mg^{2+}$ is below $Be^{2+}$ , so $Mg^{2+} > Be^{2+}$ .

Step 4: Combine the information to determine the overall order. $Na^+ > Li^+$ and $Mg^{2+} > Be^{2+}$ . Also, $Na^+$ and $Mg^{2+}$ have more electrons than $Li^+$ and $Be^{2+}$ , so they will be larger. The correct order is $Na^+ > Li^+ > Mg^{2+} > Be^{2+}$ .

Why other options are wrong:
- Option A: $L{i^ + } > B{e^{2 + }} > N{a^ + } > M{g^{2 + }}$ : Incorrect. $Na^+$ should be larger than $Li^+$ .
- Option C: $L{i^{2 + }} > N{a^ + } > M{g^{2 + }} > B{e^{2 + }}$ : Incorrect. $Li^{2+}$ is not a common ion, and the order is incorrect.
- Option D: $M{g^{2 + }} > B{e^{2 + }} > L{i^ + } > N{a^ + }$ : Incorrect. $Na^+$ should be the largest, and $Li^+$ is larger than $Mg^{2+}$ and $Be^{2+}$ .

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

Memory Aids:
- For isoelectronic species, more protons = smaller radius.
- Ionic radius increases down a group.

Quick Tips:
- Consider the number of electrons and protons when comparing ionic radii.
- Remember the trends in the periodic table.

36 Hard

36 of 43Question permalink

Analyzing +4 | -1

The successive ionisation energy values for an element ‘X’ are given below XV. Ist ionisation energy =410 kJ

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

XVI. 2nd ionisation energy =820 kJ

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

XVII. 3rd ionisation energy =1100 kJ

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

XVIII. 4th ionisation energy =1500 kJ

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

XIX. 5th ionisation energy =3200 kJ

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

Find out the number of valence electron for the atom ‘X ’

View answer and explanation

Correct option: A

4

Option A (4) is correct because the large jump in ionization energy between the 4th and 5th ionization energies indicates that the element has 4 valence electrons.

Solution approach

Correct Answer + Core Reason: 4. The large jump in ionization energy between the 4th and 5th ionization energies indicates 4 valence electrons.

Step-by-Step Solution:

Step 1: Analyze the successive ionization energies.

Step 2: 1st IE = 410 kJ/mol

Step 3: 2nd IE = 820 kJ/mol

Step 4: 3rd IE = 1100 kJ/mol

Step 5: 4th IE = 1500 kJ/mol

Step 6: 5th IE = 3200 kJ/mol

Step 7: Observe the significant jump between the 4th and 5th ionization energies.

Step 8: This jump indicates that after removing 4 electrons, it becomes much more difficult to remove the 5th electron.

Step 9: Therefore, the element has 4 valence electrons.

Why other options are wrong:
- Option B: 3: If there were 3 valence electrons, the large jump would be between the 3rd and 4th ionization energies.
- Option C: 5: If there were 5 valence electrons, the large jump would be between the 5th and 6th ionization energies.
- Option D: 2: If there were 2 valence electrons, the large jump would be between the 2nd and 3rd ionization energies.

Key Concepts:
- Ionization energy
- Valence electrons

Memory Aids:
- A large jump in ionization energy indicates the number of valence electrons.

Quick Tips:
- Look for significant jumps in ionization energy values to determine the number of valence electrons.

37 Hard

37 of 43Question permalink

Application +4 | -1

The electronic configuration of an element is

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

. The atomic number and the group number of the element '

\mathrm{X}

' which is just below the above element in the periodic table are respectively

View answer and explanation

Correct option: C

33 and 15

Option C (33 and 15) is correct. Adding one more shell to the given configuration results in an element with atomic number 33, which is in group 15.

Solution approach

Correct Answer + Core Reason: 33 and 15. The element below will have one more shell and belong to the same group.

Step-by-Step Solution:

Step 1: Identify the electronic configuration of element X: $1 s^2 2 s^2 2 p^6 3 s^2 3 p^3$ .

Step 2: Calculate the atomic number of element X: 2 + 2 + 6 + 2 + 3 = 15.

Step 3: Determine the group number of element X: Since the last electron enters the p-orbital, it's a p-block element. The group number is 10 + number of valence electrons = 10 + 5 = 15.

Step 4: The element just below X in the periodic table will have one more energy level (shell). The next element in the same group will have the electronic configuration: $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^{10} 4 p^3$ .

Step 5: Calculate the atomic number of the element below X: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33.

Step 6: The element below X is in group 15 because it has 5 valence electrons (3 from 4p and 2 from 4s).

Why other options are wrong:
- Option A (23 and 5): The atomic number and group number are incorrect.
- Option B (23 and 15): The atomic number is incorrect, but the group number is correct.
- Option D (33 and 5): The atomic number is correct, but the group number is incorrect.

Key Concepts:
- Electronic configuration
- Periodic table trends
- Group and period relationships

Memory Aids:
- Elements in the same group have similar valence electronic configurations.

Quick Tips:
- Adding a new shell to an electronic configuration means adding electrons to the next energy level.

38 Hard

38 of 43Question permalink

Analyzing +4 | -1

The third ionisation enthalpy is minimum for:

View answer and explanation

Correct option: D

Fe

Option D (Fe) has the minimum third ionization enthalpy because after losing two electrons, it attains a relatively stable d6 configuration, making it easier to remove the third electron compared to the other options.

Solution approach

Correct Answer + Core Reason: $Fe$ . The third ionization enthalpy is minimum for Fe because it achieves a relatively stable electronic configuration after losing two electrons.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the elements:\nMn: [Ar] $3d^54s^2$ \nNi: [Ar] $3d^84s^2$ \nCo: [Ar] $3d^74s^2$ \nFe: [Ar] $3d^64s^2$

Step 2: Write the electronic configurations after losing two electrons:
$Mn^{2+}$ : [Ar] $3d^5$
$Ni^{2+}$ : [Ar] $3d^8$
$Co^{2+}$ : [Ar] $3d^7$
$Fe^{2+}$ : [Ar] $3d^6$

Step 3: Consider the stability of the resulting configurations after the third ionization:
$Mn^{3+}$ : [Ar] $3d^4$
$Ni^{3+}$ : [Ar] $3d^7$
$Co^{3+}$ : [Ar] $3d^6$
$Fe^{3+}$ : [Ar] $3d^5$ (half-filled, relatively stable)

Step 4: Removing an electron from $Fe^{2+}$ to form $Fe^{3+}$ results in a half-filled $d^5$ configuration, which is relatively stable. Therefore, the third ionization enthalpy is lower for Fe.

Why other options are wrong:
- Option A: $Mn^{3+}$ does not achieve a particularly stable configuration.
- Option B: $Ni^{3+}$ does not achieve a particularly stable configuration.
- Option C: $Co^{3+}$ does not achieve a particularly stable configuration.

Key Concepts:
- Ionization enthalpy
- Electronic configuration
- Stability of half-filled and fully filled orbitals

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

Quick Tips:
- Consider the electronic configurations of the ions when comparing ionization energies.

39 Hard

39 of 43Question permalink

Understanding +4 | -1

The correct electronegativity order is:

View answer and explanation

Correct option: C

\mathrm{Si},\mathrm{P},\mathrm{C},\mathrm{N}

Option C (Si, P, C, N) is the correct electronegativity order. Electronegativity increases across a period and decreases down a group. N is the most electronegative among the given elements.

Solution approach

Correct Answer + Core Reason: $\mathrm{Si},\mathrm{P},\mathrm{C},\mathrm{N}$ . Electronegativity 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.\nC and N are in Period 2.\nSi and P are in Period 3.

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

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

Step 4: Therefore, N is more electronegative than C, and C is more electronegative than Si and P. Also, P is less electronegative than Si.

Step 5: The correct order is Si < P < C < N.

Why other options are wrong:
- Option A: $\mathrm{C},\mathrm{N},\mathrm{Si},\mathrm{P}$ - Incorrect order.
- Option B: $\mathrm{N},\mathrm{Si},\mathrm{C},\mathrm{P}$ - Incorrect order.
- Option D: $\mathrm{P},\mathrm{Si},\mathrm{N},\mathrm{C}$ - Incorrect order.

Key Concepts:
- Electronegativity
- Periodic Trends

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

Quick Tips:
- Remember the general trends in electronegativity across and down the periodic table.

40 Hard

40 of 43Question permalink

Analyzing +4 | -1

Out of

Ne,\,\,N{a^ + },\,\,O,\,\,Na

and

N,

the species having maximum and minimum ionization energies respectively are

View answer and explanation

Correct option: C

N{a^ + },\,\,Na

Option C (

N{a^ + },\,Na

) is correct because

N{a^ + }

has a noble gas configuration and a high nuclear charge, making it difficult to remove an electron, while Na has a single valence electron that is easily removed.

Solution approach

Correct Answer + Core Reason: $N{a^ + },\,Na$ . $N{a^ + }$ has a stable noble gas configuration and high effective nuclear charge, while Na has a single valence electron.

Step-by-Step Solution:

Step 1: Consider the electronic configurations of each species.

Step 2: Ne: $1s^2 2s^2 2p^6$ (Noble gas, high ionization energy)

Step 3: $N{a^ + }$ : $1s^2 2s^2 2p^6$ (Isoelectronic with Ne, but higher nuclear charge, even higher ionization energy)

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

Step 5: Na: $1s^2 2s^2 2p^6 3s^1$ (Alkali metal, low ionization energy)

Step 6: N: $1s^2 2s^2 2p^3$

Step 7: $N{a^ + }$ has the highest ionization energy because it has a noble gas configuration and a higher nuclear charge than Ne.

Step 8: Na has the lowest ionization energy because it only has one valence electron that is easily removed.

Why other options are wrong:
- Option A: $O,\,Na$ : Oxygen has a higher ionization energy than Na.
- Option B: $Ne,\,N{a^ + }$ : $N{a^ + }$ has higher ionization energy than Ne.
- Option D: $N,\,Na$ : Nitrogen has a higher ionization energy than Na.

Key Concepts:
- Ionization energy
- Electronic configuration
- Effective nuclear charge

Memory Aids:
- Noble gases have high ionization energies.
- Alkali metals have low ionization energies.

Quick Tips:
- Consider the electronic configuration and nuclear charge of each species.

41 Hard

41 of 43Question permalink

Understanding +4 | -1

The basic character of

\mathrm{MgO}, \mathrm{BaO}, \mathrm{Na}_2 \mathrm{O}

and

\mathrm{FeO}

follow the order

View answer and explanation

Correct option: C

\mathrm{FeO} < \mathrm{MgO} < \mathrm{BaO} < \mathrm{Na}_2 \mathrm{O}

Option C (FeO < MgO < BaO < Na2O) is correct because basic character increases with decreasing electronegativity of the metal and increasing electropositivity.

Solution approach

Correct Answer + Core Reason: $\mathrm{FeO} < \mathrm{MgO} < \mathrm{BaO} < \mathrm{Na}_2 \mathrm{O}$ . This is the correct order of basic character.

Step-by-Step Solution:

Step 1: Consider the metallic character of the elements.\nNa is the most metallic, followed by Ba, Mg, and Fe.

Step 2: Relate metallic character to basicity.\nThe more metallic the element, the more basic its oxide.

Step 3: Explain the trend within a group.\nFor elements within the same group (e.g., Mg and Ba), basicity increases down the group because metallic character increases.\nTherefore, BaO is more basic than MgO.

Step 4: Explain the trend across a period.\nFor elements across a period, basicity decreases as you move from left to right.\nNa is more metallic than Mg, so $Na_2O$ is more basic than MgO.\nFe is a transition metal and its oxide is generally less basic than oxides of Group 1 and Group 2 elements.

Step 5: Combine the trends to determine the overall order.
$FeO < MgO < BaO < Na_2O$

Why other options are wrong:
- Option A: $\mathrm{MgO} < \mathrm{FeO} < \mathrm{BaO} < \mathrm{Na}_2 \mathrm{O}$ : Incorrect because FeO is less basic than MgO.
- Option B: $\mathrm{FeO} < \mathrm{MgO} < \mathrm{Na}_2 \mathrm{O} < \mathrm{BaO}$ : Incorrect because BaO is more basic than $Na_2O$ .
- Option D: $\mathrm{Na}_2 \mathrm{O} < \mathrm{MgO} < \mathrm{FeO} < \mathrm{BaO}$ : Completely incorrect.

Key Concepts:
- Basic character of oxides
- Metallic character
- Periodic trends

Memory Aids:
- Remember that basicity increases with increasing metallic character.

Quick Tips:
- Consider the position of the element in the periodic table to determine its metallic character and the basicity of its oxide.

42 Hard

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Understanding +4 | -1

The correct order of electron affinity is

View answer and explanation

Correct option: C

Al < N < O < Cl

Option C (

Al < N < O < Cl

) is correct because electron affinity generally increases across a period and decreases down a group, with exceptions due to electronic configurations.

Solution approach

Correct Answer + Core Reason: $Al < N < O < Cl$ . Electron affinity increases across a period and decreases down a group, with exceptions.

Step-by-Step Solution:

Step 1: Understand the general trends of electron affinity in the periodic table. Electron affinity generally increases across a period (from left to right) and decreases down a group (from top to bottom).

Step 2: Consider the positions of the given elements (N, O, Cl, Al) in the periodic table. N and O are in the second period, Cl is in the third period, and Al is in the third period.

Step 3: Account for exceptions. Nitrogen has a relatively low electron affinity due to its stable half-filled $p$ orbital configuration ( $2p^3$ ). Oxygen has a higher electron affinity than nitrogen.

Step 4: Chlorine has a high electron affinity because it needs only one more electron to achieve a stable noble gas configuration.

Step 5: Aluminum has a lower electron affinity than chlorine because it is further to the left in the periodic table and has a less favorable effective nuclear charge.

Step 6: Combine the trends and exceptions to determine the correct order: $Al < N < O < Cl$ .

Why other options are wrong:
- Option A: $N < O < Cl < Al$ - Incorrect because Al has the lowest electron affinity.
- Option B: $O < N < Al < Cl$ - Incorrect because N has a lower electron affinity than O, and Al has a lower electron affinity than both N and O.
- Option D: $Cl < N < O < Al$ - Incorrect because Cl has the highest electron affinity among the given elements.

Key Concepts:
- Electron Affinity Trends
- Periodic Table
- Electronic Configuration

Memory Aids:
- Electron affinity increases across a period (with exceptions).
- Electron affinity decreases down a group.

Quick Tips:
- Remember the exceptions to the general trends, especially for elements with stable electronic configurations (half-filled or fully-filled orbitals).

43 Hard

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Understanding +4 | -1

The ionisation potential of '

{X^ + }

' ion is equal to

View answer and explanation

Correct option: D

The EA of

{X^{2 + }}

ion

The correct answer is the EA of

{X^{2 + }}

ion because the ionization potential of

{X^ + }

is the energy required to remove an electron from

{X^ + }

, which is the same as the electron affinity of

{X^{2 + }}

.

Solution approach

Correct Answer + Core Reason: The EA of ${X^{2 + }}$ ion. The ionization potential of ${X^ + }$ is the energy required to remove an electron from ${X^ + }$ , which is the same as the electron affinity of ${X^{2 + }}$ .

Step-by-Step Solution:

Step 1: Define Ionization Potential (IP). The ionization potential is the energy required to remove an electron from a gaseous atom or ion.

Step 2: Define Electron Affinity (EA). The electron affinity is the energy change when an electron is added to a gaseous atom or ion.

Step 3: Consider the process of ionization of ${X^ + }$ . ${X^ + }(g) \rightarrow {X^{2 + }}(g) + e^-$ . The energy required for this process is the ionization potential of ${X^ + }$ .

Step 4: Consider the process of electron addition to ${X^{2 + }}$ . ${X^{2 + }}(g) + e^- \rightarrow {X^ + }(g)$ . The energy change for this process is the electron affinity of ${X^{2 + }}$ .

Step 5: Recognize that the ionization potential of ${X^ + }$ is the reverse process of the electron affinity of ${X^{2 + }}$ , but with opposite signs. Therefore, the magnitude of the ionization potential of ${X^ + }$ is equal to the magnitude of the electron affinity of ${X^{2 + }}$ .

Why other options are wrong:
- Option A: The EA of ‘X’ atom - Incorrect because the ionization potential of ${X^ + }$ involves removing an electron from ${X^ + }$ , not adding one to X.
- Option B: The EN of ‘X’ atom - Incorrect because electronegativity is a measure of the ability of an atom to attract electrons in a chemical bond, not the energy change associated with adding or removing electrons.
- Option C: The IE of ‘X’ atom - Incorrect because the ionization energy of X is the energy required to remove an electron from X, not ${X^ + }$ .

Key Concepts:
- Ionization Potential
- Electron Affinity

Memory Aids:
- Ionization Potential: Removing an electron.
- Electron Affinity: Adding an electron.

Quick Tips:
- Understand the definitions of ionization potential and electron affinity clearly.
- Remember that ionization is an endothermic process (energy is required), while electron affinity can be exothermic or endothermic.

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