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 — Medium, Page 6

151 Medium

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Understanding +4 | -1

The correct order of ionisation energy for comparing carbon, nitrogen and oxygen atom is

View answer and explanation

Correct option: C

C < N > O

Option C (C < N > O) is correct because nitrogen has a higher ionization energy than both carbon and oxygen due to its half-filled p-orbital configuration, which provides extra stability.

Solution approach

Correct Answer + Core Reason: C < N > O. Nitrogen has a higher ionization energy than both carbon and oxygen due to its half-filled p-orbital configuration.

Step-by-Step Solution:

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

Step 2: Consider the trends in ionization energy across a period. Ionization energy generally increases from left to right.

Step 3: Note the exception: Nitrogen has a half-filled $2p$ subshell, which is more stable than the partially filled $2p$ subshells of carbon and oxygen. This makes it more difficult to remove an electron from nitrogen.

Step 4: Therefore, the ionization energy of N is higher than both C and O. The ionization energy of C is lower than O because it is to the left of O in the period.

Why other options are wrong:
- Option A: C > N > O is incorrect because N has a higher ionization energy than both C and O.
- Option B: C > N < O is incorrect because C has a lower ionization energy than O.
- Option D: C < N < O is incorrect because N has a higher ionization energy than O.

Key Concepts:
- Ionization Energy
- Electronic Configuration
- Hund's Rule
- Stability of Half-Filled Orbitals

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

Quick Tips:
- Remember the exceptions to the general trends in ionization energy due to stable electronic configurations.

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Given below are two statements:
Statement I

Ce{O_2}

can be used for oxidation of aldehydes and ketones.
Statement II Aqueous solution of

EuS{O_4}

is a strong reducing agent.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: A

Both statement I and statement II are true.

Option A (Both statement I and statement II are true) is correct because

CeO_2

is used as an oxidizing agent and

EuSO_4

is a strong reducing agent.

Solution approach

Correct Answer + Core Reason: Both statement I and statement II are true. $CeO_2$ is an oxidizing agent, and $EuSO_4$ is a strong reducing agent.

Step-by-Step Solution:

Step 1: Evaluate Statement I: $CeO_2$ can be used for oxidation of aldehydes and ketones.
- Cerium(IV) oxide ( $CeO_2$ ) is a known oxidizing agent and can be used in organic reactions to oxidize aldehydes and ketones.
- Therefore, Statement I is true.

Step 2: Evaluate Statement II: Aqueous solution of $EuSO_4$ is a strong reducing agent.
- Europium(II) sulfate ( $EuSO_4$ ) is a strong reducing agent because $Eu^{2+}$ readily oxidizes to $Eu^{3+}$ .
- Therefore, Statement II is true.

Step 3: Determine the correct answer based on the evaluation of both statements.

Why other options are wrong:
- Option B: Both statement I and statement II are false: Both statements are true.
- Option C: Statement I is true but statement II is false: Statement II is also true.
- Option D: Statement I is false but statement II is true: Statement I is also true.

Key Concepts:
- Oxidizing and reducing agents.
- Properties of cerium and europium compounds.

Memory Aids:
- $Ce^{4+}$ is a good oxidizing agent.
- $Eu^{2+}$ is a good reducing agent.

Quick Tips:
- Remember common oxidizing and reducing agents and their properties.

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Remembering +4 | -1

A plot of

\sqrt{v}

(frequency) against atomic number

(z)

gave a :

View answer and explanation

Correct option: A

Straight line

The correct answer is A (Straight line) because Moseley's law states that the square root of the frequency of emitted X-rays is proportional to the atomic number.

Solution approach

Correct Answer + Core Reason: Straight line. Moseley's law establishes a linear relationship between the square root of frequency and atomic number.

Step-by-Step Solution:

Step 1: Recall Moseley's Law: $\sqrt{v} = a(Z - b)$ , where $v$ is the frequency, $Z$ is the atomic number, and $a$ and $b$ are constants.

Step 2: Recognize that this equation represents a linear relationship between $\sqrt{v}$ and $Z$ .

Step 3: A linear relationship, when plotted on a graph, results in a straight line.

Why other options are wrong:
- Option B: A curve would indicate a non-linear relationship, which is not the case according to Moseley's law.
- Option C: A parabola would indicate a quadratic relationship, which is not supported by Moseley's law.
- Option D: A hyperbola would indicate an inverse relationship, which is not consistent with Moseley's law.

Key Concepts:
- Moseley's Law
- Linear Relationships

Memory Aids:
- Remember Moseley's Law: $\sqrt{v} \propto Z$

Quick Tips:
- Understand the mathematical representation of different types of graphs (linear, quadratic, inverse).

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Understanding +4 | -1

The tenth elements in the Periodic Table resembles with the

View answer and explanation

Correct option: B

Second period

Option B (Second period) is correct because the tenth element (Neon) is in the second period and exhibits similar properties to other elements in the same period.

Solution approach

Correct Answer + Core Reason: Second period. The tenth element is Neon, which belongs to the second period.

Step-by-Step Solution:

Step 1: Identify the tenth element in the periodic table. The tenth element is Neon (Ne).

Step 2: Determine the period to which Neon belongs. Neon belongs to the second period.

Step 3: Conclude that the tenth element resembles elements in the second period.

Why other options are wrong:
- Option A: First period contains only Hydrogen and Helium.
- Option C: Fourth period contains elements with different properties than Neon.
- Option D: Ninth period does not exist in the current periodic table.

Key Concepts:
- Periodic table periods
- Element properties within a period

Memory Aids:
- Remember the first few elements and their periods: H (1), He (1), Li (2), Be (2), B (2), C (2), N (2), O (2), F (2), Ne (2).

Quick Tips:
- Knowing the element's position in the periodic table helps determine its properties and the period it belongs to.

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Analyzing +4 | -1

Which of the following configuration has strongest metallic bonding?

View answer and explanation

Correct option: B

[\mathrm{Ar}] 3 d^5 4 s^1

Option B is correct. Strongest metallic bonding occurs when there are more unpaired electrons available for metallic bonding. [Ar] 3d⁵ 4s¹ has 6 unpaired electrons.

Solution approach

Correct Answer + Core Reason: $[\mathrm{Ar}] 3 d^5 4 s^1$ . Strongest metallic bonding is generally observed when there are more unpaired electrons available for metallic bonding.

Step-by-Step Solution:

Step 1: Determine the number of unpaired electrons in each configuration:\nA. $[\mathrm{Ar}] 3 d^7 4 s^2$ : 3 unpaired electrons in 3d (7 electrons fill as 2 paired and 3 unpaired)\nB. $[\mathrm{Ar}] 3 d^5 4 s^1$ : 6 unpaired electrons (5 in 3d and 1 in 4s)\nC. $[\mathrm{Ar}] 3 d^6 4 s^2$ : 4 unpaired electrons in 3d (6 electrons fill as 2 paired and 4 unpaired)\nD. $[\mathrm{Ar}] 3 d^3 4 s^2$ : 3 unpaired electrons in 3d

Step 2: Compare the number of unpaired electrons: Configuration B has the most unpaired electrons.

Step 3: Conclusion: The configuration with the most unpaired electrons will exhibit the strongest metallic bonding.

Why other options are wrong:
- Option A: Has fewer unpaired electrons than option B.
- Option C: Has fewer unpaired electrons than option B.
- Option D: Has fewer unpaired electrons than option B.

Key Concepts:
- Metallic Bonding
- Unpaired Electrons
- Electronic Configuration

Memory Aids:
- More unpaired electrons = stronger metallic bond.

Quick Tips:
- Determine the number of unpaired electrons by drawing orbital diagrams.

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Remembering +4 | -1

The energy released when a neutral gaseous atom takes up an electron is called:

View answer and explanation

Correct option: D

Electron affinity

Option D (Electron affinity) is correct because electron affinity is defined as the energy released when a neutral gaseous atom accepts an electron.

Solution approach

Correct Answer + Core Reason: Electron affinity is the energy released when a neutral gaseous atom takes up an electron.

Step-by-Step Solution:

Step 1: Define the terms:
- Ionization energy: The energy required to remove an electron from a neutral gaseous atom.
- Solvation energy: The energy released when ions are solvated by solvent molecules.
- Electronegativity: The tendency of an atom to attract a shared pair of electrons in a chemical bond.
- Electron affinity: The energy change when an electron is added to a neutral gaseous atom.

Step 2: The question asks for the energy released when a neutral gaseous atom takes up an electron.

Step 3: This definition matches the definition of electron affinity.

Step 4: Conclusion: Electron affinity is the correct answer.

Why other options are wrong:
- Option A: Ionization energy is the energy required to remove an electron, not the energy released when an electron is added.
- Option B: Solvation energy is related to the interaction of ions with solvent molecules, not the addition of an electron to an atom.
- Option C: Electronegativity is a measure of the attraction of an atom for electrons in a chemical bond, not the energy change when an electron is added.
- Option D: Electron affinity is the correct answer.

Key Concepts:
- Ionization Energy
- Solvation Energy
- Electronegativity
- Electron Affinity

Memory Aids:
- Electron Affinity: "Affinity" implies attraction, so it's about adding an electron.

Quick Tips:
- Know the definitions of these fundamental terms in chemistry.

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Understanding +4 | -1

A

p

-block element in which last electron enters into

s

-orbitals of valence shell instead of

p

-orbital is:

View answer and explanation

Correct option: D

\mathrm{He}

The correct answer is D (He) because Helium is a p-block element where the last electron enters the s-orbital instead of the p-orbital.

Solution approach

Correct Answer + Core Reason: $\mathrm{He}$ . Helium is a p-block element with electronic configuration $1s^2$ where the last electron enters into s-orbital.

Step-by-Step Solution:

Step 1: Identify the p-block elements and their electronic configurations.

Step 2: Analyze the electronic configurations of the given options.

Step 3: As has the electronic configuration [Ar] 3d10 4s2 4p3. The last electron enters into the p-orbital.

Step 4: Ga has the electronic configuration [Ar] 3d10 4s2 4p1. The last electron enters into the p-orbital.

Step 5: He has the electronic configuration 1s2. It is placed in the p-block but the last electron enters into the s-orbital.

Step 6: Conclusion: He is the correct answer.

Why other options are wrong:
- Option A: As is a p-block element, but its last electron enters the p-orbital, not the s-orbital.
- Option B: Ga is a p-block element, but its last electron enters the p-orbital, not the s-orbital.
- Option C: This statement is incorrect because Helium exists and fits the description.

Key Concepts:
- Electronic configuration of elements
- p-block elements
- s-orbitals and p-orbitals

Memory Aids:
- Remember that Helium is an exception in the p-block.

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

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Understanding +4 | -1

The correct order of ionic radii is

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 on the aluminum ion increases, the ionic radius decreases due to an increase in 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.

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Consider the following statements:

(i) The discovery of inert gases later on did not disturb Mendeleev's arrangement.

(ii) In the present periodic table, periodicity in the properties of elements is related to the periodicity in their electronic configurations.

Which of these statement(s) is/are correct?

View answer and explanation

Correct option: C

Both (i) and (ii)

Option C (Both (i) and (ii)) is correct because the discovery of inert gases fit into Mendeleev's table without disrupting it, and the modern periodic table's properties are linked to electronic configurations.

Solution approach

Correct Answer + Core Reason: Both (i) and (ii). Both statements about the periodic table are correct.

Step-by-Step Solution:

Step 1: Evaluate statement (i): The discovery of inert gases did not disturb Mendeleev's arrangement because they could be added as a new group.

Step 2: Evaluate statement (ii): Periodicity in the properties of elements is indeed related to the periodicity in their electronic configurations.

Step 3: Conclude that both statements are correct.

Why other options are wrong:
- Option A ((i) only): Statement (ii) is also correct.
- Option B ((ii) only): Statement (i) is also correct.
- Option D (Neither (i) nor (ii)): Both statements are correct.

Key Concepts:
- Mendeleev's Periodic Table
- Inert Gases
- Electronic Configuration
- Periodic Properties

Memory Aids:
- Remember that inert gases were a later addition to the periodic table.

Quick Tips:
- Understand the historical development of the periodic table and its connection to electronic structure.

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Application +4 | -1

Aqueous solutions of two compounds

M-{O}-{H}

and

M^{\prime}-{O}-{H}

have been prepared in two different beakers. If the electronegativity of

M=3.5, M^{\prime}=1.72

,

{O}=3.0

and

{H}=2.1

, then the solutions respectively are

View answer and explanation

Correct option: B

acidic, basic

Option B (acidic, basic) is correct because the electronegativity difference between M and O is higher, leading to acidic properties, while a lower electronegativity difference leads to basic properties.

Solution approach

Correct Answer + Core Reason: acidic, basic. Higher electronegativity of M leads to acidic character, lower electronegativity leads to basic character.

Step-by-Step Solution:

Step 1: Analyze the compound M-O-H.\nThe electronegativity of M is 3.5, O is 3.0, and H is 2.1.\nThe electronegativity difference between M and O is |3.5 - 3.0| = 0.5.\nThe electronegativity difference between O and H is |3.0 - 2.1| = 0.9.\nSince the electronegativity difference between O and H is greater than that between M and O, the M-O bond is more likely to break, releasing $H^+$ ions, making the solution acidic.

Step 2: Analyze the compound M'-O-H.\nThe electronegativity of M' is 1.72, O is 3.0, and H is 2.1.\nThe electronegativity difference between M' and O is |1.72 - 3.0| = 1.28.\nThe electronegativity difference between O and H is |3.0 - 2.1| = 0.9.\nSince the electronegativity difference between M' and O is less than that between O and H, the O-H bond is more likely to break, releasing $OH^-$ ions, making the solution basic.

Why other options are wrong:
- Option A: acidic, acidic: This is incorrect because M'-O-H is basic, not acidic.
- Option C: basic, basic: This is incorrect because M-O-H is acidic, not basic.
- Option D: basic, acidic: This is incorrect because the order is reversed.

Key Concepts:
- Electronegativity
- Acid-base character

Memory Aids:
- Higher electronegativity of M leads to acidic character.
- Lower electronegativity of M leads to basic character.

Quick Tips:
- Consider the electronegativity differences to determine the polarity of bonds and predict the dissociation of the compound.

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Application +4 | -1

The amount of energy released when

10^6

atoms of iodine in vapour state is converted into

{I}^{-}

ion is

4.8 \times 10^{-13} \mathrm{~J}

. What is the electron affinity of iodine in

\mathrm{kJ} / \mathrm{mole}?

View answer and explanation

Correct option: B

289 \mathrm{~kJ}

Option B (289 kJ/mole) is correct because it accurately converts the energy released per atom to energy released per mole, which represents the electron affinity.

Solution approach

Correct Answer + Core Reason: $289 \mathrm{~kJ}$ . This is the electron affinity of iodine in kJ/mole.

Step-by-Step Solution:

Step 1: Calculate the energy released per atom.\nEnergy released per $10^6$ atoms = $4.8 \times 10^{-13} \text{ J}$ \nEnergy released per atom = $\frac{4.8 \times 10^{-13} \text{ J}}{10^6 \text{ atoms}} = 4.8 \times 10^{-19} \text{ J/atom}$

Step 2: Convert energy per atom to energy per mole.\nAvogadro's number, $N_A = 6.022 \times 10^{23} \text{ atoms/mol}$ \nEnergy released per mole = $4.8 \times 10^{-19} \text{ J/atom} \times 6.022 \times 10^{23} \text{ atoms/mol} = 289056 \text{ J/mol}$

Step 3: Convert J/mol to kJ/mol.\nEnergy released per mole = $\frac{289056 \text{ J/mol}}{1000 \text{ J/kJ}} = 289.056 \text{ kJ/mol}$

Step 4: Round to the nearest whole number.\nElectron affinity = 289 kJ/mol

Why other options are wrong:
- Option A: $489 \mathrm{~kJ}$ : This is likely a calculation error, possibly multiplying by a factor of 1000 incorrectly.
- Option C: $259 \mathrm{~kJ}$ : This is likely a calculation error involving incorrect multiplication or division.
- Option D: $389 \mathrm{~kJ}$ : This is likely a calculation error involving incorrect multiplication or division.

Key Concepts:
- Electron affinity
- Avogadro's number
- Unit conversion

Memory Aids:
- Remember Avogadro's number and the conversion factor between J and kJ.

Quick Tips:
- Always convert the energy to per mole to find the electron affinity in kJ/mol.
- Pay attention to units and make sure they are consistent.

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Understanding +4 | -1

The electronic configuration of the element with maximum electron affinity is

View answer and explanation

Correct option: C

1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}

Option C (

1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}

) is correct because elements with electronic configurations close to noble gases (like halogens) have high electron affinity.

Solution approach

Correct Answer + Core Reason: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}$ . This configuration represents a halogen, which has a high electron affinity.

Step-by-Step Solution:
Step 1: Understand the concept of electron affinity. Electron affinity is the energy change that occurs when an electron is added to a gaseous atom.
Step 2: Recall the trend of electron affinity in the periodic table. Electron affinity generally increases as you move from left to right across a period and decreases as you move down a group. Halogens have the highest electron affinities.
Step 3: Analyze the given electronic configurations:
- A: $1{s^2},2{s^2},2{p^3}$ (Nitrogen group)
- B: $1{s^2},2{s^2},2{p^5}$ (Fluorine)
- C: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}$ (Chlorine)
- D: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^3}$ (Phosphorus)
Step 4: Determine which configuration represents the element with the maximum electron affinity. Halogens have the highest electron affinities because adding an electron to a halogen results in a stable noble gas configuration. Among the given options, Chlorine ( $1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}$ ) is the halogen with the given electronic configuration.

Why other options are wrong:
- Option A: $1{s^2},2{s^2},2{p^3}$ : Incorrect, this is a nitrogen group element and has a lower electron affinity than halogens.
- Option B: $1{s^2},2{s^2},2{p^5}$ : Incorrect, this is Fluorine, but Chlorine has a higher electron affinity due to its larger size and less electron-electron repulsion.
- Option D: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^3}$ : Incorrect, this is a phosphorus group element and has a lower electron affinity than halogens.

Key Concepts:
- Electron affinity
- Periodic trends
- Electronic configuration
- Halogens

Memory Aids:
- Halogens have the highest electron affinities.

Quick Tips:
- Look for halogens when determining which element has the maximum electron affinity.

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Application +4 | -1

The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is

View answer and explanation

Correct option: D

0.736

Option D (0.736) is correct because the percentage of ionic character can be estimated using the Hannay-Smith equation or a similar empirical relationship based on the electronegativity difference.

Solution approach

Correct Answer + Core Reason: 0.736. The percentage of ionic character is calculated using the electronegativity difference.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference between A and B: 4.0 - 1.20 = 2.8.

Step 2: Use the Hannay-Smith equation (or a similar empirical relationship) to estimate the percentage of ionic character. A common approximation is:\nPercentage ionic character = $16 \times |\Delta EN| + 3.5 \times |\Delta EN|^2$ , where $\Delta EN$ is the electronegativity difference.

Step 3: Substitute the electronegativity difference into the equation:\nPercentage ionic character = $16 \times 2.8 + 3.5 \times (2.8)^2 = 44.8 + 3.5 \times 7.84 = 44.8 + 27.44 = 72.24$ \%

Step 4: Divide the percentage by 100 to get the fraction: 72.24/100 = 0.7224. This is close to 0.736, the given answer.\nAnother approximation is the Pauling equation: $1 - e^{-0.25(X_A - X_B)^2} = 1 - e^{-0.25(2.8)^2} = 1 - e^{-1.96} = 1 - 0.141 = 0.859$ . This result is not among the options.\nUsing the approximation: % ionic character = (0.16 * (4.0-1.2) + 0.035 * (4.0-1.2)²) * 100 = (0.16 * 2.8 + 0.035 * 7.84) * 100 = (0.448 + 0.2744) * 100 = 0.7224 * 100 = 72.24% = 0.7224

Step 5: The closest answer to the calculated value is 0.736.

Why other options are wrong:
- Option A: 0.583: This value is too low and does not accurately reflect the significant electronegativity difference.
- Option B: 0.48: This value is also too low and likely results from an incorrect application of the electronegativity difference formula.
- Option C: 0.796: This value is higher than the calculated value but still not the closest to the correct approximation.

Key Concepts:
- Electronegativity
- Electronegativity difference
- Percentage ionic character
- Hannay-Smith equation (or similar empirical relationship)

Memory Aids:
- Remember the approximate formula for calculating the percentage of ionic character based on electronegativity difference.

Quick Tips:
- Be familiar with the common formulas used to estimate the percentage of ionic character.

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Remembering +4 | -1

In Mendeleev's periodic table, iodine with lower atomic weight than that of tellurium was placed alongwith

View answer and explanation

Correct option: B

Fluorine, chlorine and bromine

Option B (Fluorine, chlorine and bromine) is correct because Iodine was placed with the halogen family (F, Cl, Br) due to its similar chemical properties, despite its atomic weight being less than Tellurium.

Solution approach

Correct Answer + Core Reason: Fluorine, chlorine and bromine. Iodine was placed with the halogens due to similar properties.

Step-by-Step Solution:

Step 1: Recall Mendeleev's periodic table and the placement of Iodine.

Step 2: Remember that Iodine is a halogen and shares similar chemical properties with Fluorine, Chlorine, and Bromine.

Step 3: Conclude that Iodine was placed alongside F, Cl, and Br.

Why other options are wrong:
- Option A (Iron, cobalt and nickel): These are transition metals and do not share the same properties as Iodine.
- Option C (Oxygen, sulphur and selenium): These are chalcogens and have different properties than Iodine.
- Option D (Carbon, silicon and germanium): These are group 14 elements and have different properties than Iodine.

Key Concepts:
- Mendeleev's Periodic Table
- Halogens
- Chemical Properties

Memory Aids:
- Remember that halogens are in Group 17 and share similar reactivity.

Quick Tips:
- Know the common groups in the periodic table and their properties.

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Understanding +4 | -1

Zinc is not considered as a transition metal because

View answer and explanation

Correct option: C

It has no unpaired electrons

The correct answer is C (It has no unpaired electrons) because Zinc is not considered a transition metal since it has a completely filled d-orbital (

d^{10}

) and thus no unpaired electrons in its common oxidation states.

Solution approach

Correct Answer + Core Reason: It has no unpaired electrons. Zinc has a completely filled d-orbital ( $d^{10}$ ) and thus no unpaired electrons in its common oxidation states.

Step-by-Step Solution:

Step 1: Define transition metals.

Step 2: Transition metals are elements with partially filled d-orbitals in their elemental or common oxidation states.

Step 3: Zinc has the electronic configuration [Ar] $3d^{10}4s^2$ . In its common oxidation state ( $Zn^{2+}$ ), it has the configuration [Ar] $3d^{10}$ .

Step 4: Since $Zn^{2+}$ has a completely filled d-orbital, it does not meet the definition of a transition metal.

Why other options are wrong:
- Option A: While Zinc is diamagnetic due to having all paired electrons, this is a consequence of having a filled d-orbital, not the primary reason it's not considered a transition metal.
- Option B: The ability to form alloys is not a defining characteristic of transition metals. Many metals can form alloys.
- Option D: The color of Zinc compounds is not the reason it is not considered a transition metal. The electronic configuration is the determining factor.

Key Concepts:
- Transition metals
- Electronic configuration
- Partially filled d-orbitals

Memory Aids:
- Remember that Zinc, Cadmium, and Mercury are not considered transition metals.

Quick Tips:
- Check the electronic configuration of the ion in its common oxidation state.

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Understanding +4 | -1

Chlorine atom tends to acquire the structure of:

View answer and explanation

Correct option: C

\mathrm{Ar}

Option C (

\mathrm{Ar}

) is correct because chlorine has 17 electrons and needs to gain one electron to achieve the stable electronic configuration of Argon (18 electrons).

Solution approach

Correct Answer + Core Reason: $\mathrm{Ar}$ . Chlorine gains one electron to achieve the stable electronic configuration of Argon.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of chlorine (Cl): Cl has 17 electrons, so its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$ .

Step 2: Determine the electronic configurations of the noble gases:
- He: $1s^2$ (2 electrons)
- Ne: $1s^2 2s^2 2p^6$ (10 electrons)
- Ar: $1s^2 2s^2 2p^6 3s^2 3p^6$ (18 electrons)
- Kr: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$ (36 electrons)

Step 3: Determine which noble gas configuration chlorine is closest to achieving: Chlorine needs to gain one electron to achieve the configuration of Argon.

Why other options are wrong:
- Option A: $\mathrm{He}$ : Chlorine would need to lose many electrons to achieve the He configuration.
- Option B: $\mathrm{Ne}$ : Chlorine would need to lose many electrons to achieve the Ne configuration.
- Option D: $\mathrm{Kr}$ : Chlorine would need to gain many electrons to achieve the Kr configuration.

Key Concepts:
- Electronic configuration of elements.
- Noble gas configurations.
- Octet rule.

Memory Aids:
- Noble gases have complete valence shells, making them stable.

Quick Tips:
- Elements tend to gain or lose electrons to achieve the nearest noble gas configuration.

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Analyzing +4 | -1

Which of the following relation is correct?

View answer and explanation

Correct option: A

{\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} > {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}

Option A (

{\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} > {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}

) is correct because carbon has a higher effective nuclear charge and a more stable electronic configuration compared to boron, resulting in a higher first ionization energy.

Solution approach

Correct Answer + Core Reason: ${\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} > {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}$ . Carbon has a higher first ionization energy than Boron due to its higher effective nuclear charge and electronic configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations of B and C.\nB: $1s^22s^22p^1$ \nC: $1s^22s^22p^2$

Step 2: Consider the first ionization energy.\nFirst ionization energy is the energy required to remove the first electron from a neutral atom.

Step 3: Compare the effective nuclear charge and electronic configuration.\nCarbon has a higher effective nuclear charge than Boron. Also, removing an electron from Boron results in a fully filled s-orbital, which is relatively stable. However, the higher nuclear charge of Carbon outweighs this effect.

Step 4: Therefore, the first ionization energy of Carbon is greater than that of Boron.

Why other options are wrong:
- Option B: ${\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} < {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}$ is incorrect because Carbon has a higher first ionization energy than Boron.
- Option C: $\mathrm{I}{\mathrm{I}^{\mathrm{nd}}}\mathrm{\;IE\;of\;C} > \mathrm{I}{\mathrm{I}^{\mathrm{nd}}}\mathrm{\;IE\;of\;B}$ is incorrect. After removing one electron, B $^+$ has a configuration of $1s^22s^2$ , while C $^+$ has a configuration of $1s^22s^22p^1$ . Removing another electron from B $^+$ requires more energy because it is removing from a stable, fully filled s-orbital.
- Option D: Both (b) and (c ) is incorrect because both options B and C are incorrect.

Key Concepts:
- Ionization Energy
- Effective Nuclear Charge
- Electronic Configuration

Memory Aids:
- Higher effective nuclear charge leads to higher ionization energy.

Quick Tips:
- Consider the effective nuclear charge and electronic configuration when comparing ionization energies.

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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.

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Remembering +4 | -1

Which of the following elements are found in pitch blende?

View answer and explanation

Correct option: D

Both (1) and (2)

Option D is correct. Pitchblende is a uranium-rich ore that contains Actinium and Protoactinium, as well as Neptunium and Plutonium in trace amounts due to the radioactive decay of Uranium.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). Pitchblende contains Actinium and Protoactinium, Neptunium and Plutonium.

Step-by-Step Solution:

Step 1: Understand Pitchblende: Pitchblende is a uranium-rich ore.

Step 2: Identify the elements found in Pitchblende:
- Uranium is the primary element.
- Actinium and Protoactinium are present as decay products of Uranium.
- Neptunium and Plutonium can also be present in trace amounts due to nuclear reactions.

Step 3: Conclusion: Pitchblende contains Actinium, Protoactinium, Neptunium, and Plutonium.

Why other options are wrong:
- Option A: Only mentions Actinium and Protoactinium, but not Neptunium and Plutonium.
- Option B: Only mentions Neptunium and Plutonium, but not Actinium and Protoactinium.
- Option C: Only mentions Actinium.

Key Concepts:
- Pitchblende Composition
- Radioactive Decay
- Uranium Ore

Memory Aids:
- Remember that Pitchblende is a uranium ore and contains its decay products.

Quick Tips:
- Know the composition of common ores.

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Analyzing +4 | -1

The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine 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, removing an electron from N+ gives it a stable half-filled configuration, making it 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.

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Application +4 | -1

If

E A_1

and

E A_2

for oxygen atom

-142 \mathrm{~kJ ~mol}^{-1}

and

+844 \mathrm{~kJ ~mol}^{-1}

. The energy released from

2 O+2 e^{-} \rightarrow 2 O^{-}

will be

View answer and explanation

Correct option: C

-284 \mathrm{~kJ ~mol}^{-1}

Option C (-284 kJ/mol) is correct because it correctly sums the electron affinities for the two oxygen atoms.

Solution approach

Correct Answer + Core Reason: $-284 \mathrm{~kJ ~mol}^{-1}$ . This is the energy released from $2 O+2 e^{-} \rightarrow 2 O^{-}$ .

Step-by-Step Solution:

Step 1: Write the reaction for one oxygen atom.
$O + e^- \rightarrow O^- \quad EA_1 = -142 \text{ kJ/mol}$

Step 2: Multiply the energy by 2 since there are two oxygen atoms.\nEnergy released for 2 oxygen atoms = $2 \times EA_1 = 2 \times (-142 \text{ kJ/mol}) = -284 \text{ kJ/mol}$

Step 3: Note that the second electron affinity is not relevant since the product is $O^-$ , not $O^{2-}$ .

Why other options are wrong:
- Option A: $-986 \mathrm{~kJ ~mol}^{-1}$ : This incorrectly uses both $EA_1$ and $EA_2$ and performs an incorrect calculation.
- Option B: $-702 \mathrm{~kJ ~mol}^{-1}$ : This is likely a calculation error involving incorrect multiplication or division.
- Option D: $-1688 \mathrm{~kJ ~mol}^{-1}$ : This incorrectly uses both $EA_1$ and $EA_2$ and performs an incorrect calculation.

Key Concepts:
- Electron affinity

Memory Aids:
- Remember that electron affinity is the energy change when an electron is added to a gaseous atom.

Quick Tips:
- Carefully read the question to determine which electron affinities are relevant.
- Pay attention to the stoichiometry of the reaction.

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Analyzing +4 | -1

The electronic configuration of four elements are given below. Which elements does not belong to the same family as others?

View answer and explanation

Correct option: C

[\mathrm{Ne}] 3 s^2 3 p^5

Option C ([Ne] 3s² 3p⁵) is correct because it represents a halogen, while the others represent elements in the same group (group 12).

Solution approach

Correct Answer + Core Reason: [Ne] 3s² 3p⁵. This configuration represents a halogen (Group 17), while the others represent elements of Group 12.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations:
- A: $[\mathrm{Xe}] 4 f^{14} 5 d^{10} 1 s^2$ - This is likely a typo, should be $6s^2$ . It has a filled d-orbital and 2 s-electrons, suggesting Group 12.
- B: $[\mathrm{Kr}] 4 d^{10} 5 s^2$ - This also has a filled d-orbital and 2 s-electrons, suggesting Group 12.
- C: $[\mathrm{Ne}] 3 s^2 3 p^5$ - This has 7 valence electrons (2 in s and 5 in p), suggesting Group 17 (a halogen).
- D: $[\mathrm{Ar}] 3 d^{10} 4 s^2$ - This also has a filled d-orbital and 2 s-electrons, suggesting Group 12.

Step 2: Identify the outlier: Configurations A, B, and D represent elements in the same group (Group 12), while configuration C represents an element in Group 17.

Why other options are wrong:
- Option A: $[\mathrm{Xe}] 4 f^{14} 5 d^{10} 1 s^2$ : This configuration (corrected to $6s^2$ ) belongs to the same family (Group 12) as B and D.
- Option B: $[\mathrm{Kr}] 4 d^{10} 5 s^2$ : This configuration belongs to the same family (Group 12) as A and D.
- Option D: $[\mathrm{Ar}] 3 d^{10} 4 s^2$ : This configuration belongs to the same family (Group 12) as A and B.

Key Concepts:
- Electronic configuration
- Group identification in the periodic table
- Valence electrons

Memory Aids:
- Group 12 elements often have a $d^{10}s^2$ configuration.
- Halogens (Group 17) have 7 valence electrons.

Quick Tips:
- Focus on the outermost electronic configuration to determine the group.
- Recognize common group configurations (e.g., halogens, alkali metals).

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Remembering +4 | -1

Which element exist as a solid at

{25^0}C

and 1 atm pressure among the following?

View answer and explanation

Correct option: D

P

Option D (P) is correct because Phosphorus (P) exists as a solid at 25°C and 1 atm pressure.

Solution approach

Correct Answer + Core Reason: P. Phosphorus exists as a solid at standard conditions.

Step-by-Step Solution:

Step 1: Consider the state of each element at 25°C and 1 atm:
- Br (Bromine): Liquid.
- Cl (Chlorine): Gas.
- Hg (Mercury): Liquid.
- P (Phosphorus): Solid.

Step 2: Identify the element that is a solid: Phosphorus.

Why other options are wrong:
- Option A (Br): Bromine is a liquid at room temperature.
- Option B (CI): Chlorine is a gas at room temperature.
- Option C (Hg): Mercury is a liquid at room temperature.

Key Concepts:
- States of Matter
- Periodic Table

Memory Aids:
- Remember the common states of elements at room temperature.

Quick Tips:
- Know the physical states of common elements.

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Understanding +4 | -1

Which main group elements have a different number of outermost electrons than their group number?

View answer and explanation

Correct option: B

Noble gases

Option B is correct because noble gases have 8 outermost electrons (except for Helium which has 2), while their group number is 18 (or 8A), thus having a different number of outermost electrons than their group number.

Solution approach

Correct Answer + Core Reason: Noble gases. Noble gases have a different number of outermost electrons than their group number (except Helium).

Step-by-Step Solution:

Step 1: Define group number: The vertical column in the periodic table.

Step 2: Define outermost electrons: The electrons in the valence shell.

Step 3: Alkali metals (Group 1) have 1 outermost electron.

Step 4: Halogens (Group 17) have 7 outermost electrons.

Step 5: Noble gases (Group 18) have 8 outermost electrons (except Helium, which has 2).

Step 6: Therefore, noble gases have a different number of outermost electrons than their group number (18) when considering the number of valence electrons (8).

Why other options are wrong:
- Option A: Alkali metals have the same number of outermost electrons as their group number (1).
- Option C: Halogens have the same number of outermost electrons as their group number (17, considering only the last digit 7).
- Option D: Noble gases are the correct answer.

Key Concepts:
- Group number
- Outermost electrons
- Noble gases

Memory Aids:
- Remember that noble gases have a full octet (8 valence electrons), except for Helium (2 valence electrons).

Quick Tips:
- Consider the electronic configuration of each group to determine the number of outermost electrons.

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Application +4 | -1

The first, second and third ionisation energies of

{Al}

are 578,1817 and

2745 \mathrm{~kJ} \mathrm{~mol}^{-1}

respectively. Calculate the energy required to convert all the atoms of

{Al}

to

{Al}^{+3}

present in

270{mg}

of

{Al}

vapours

View answer and explanation

Correct option: B

51.40 \mathrm{~kJ}

Option B (

51.40 \mathrm{~kJ}

) is correct because it involves calculating the total ionization energy for converting all Al atoms to Al+3 in the given mass.

Solution approach

Correct Answer + Core Reason: $51.40 \mathrm{~kJ}$ . The total energy required is the sum of the first three ionization energies multiplied by the number of moles of Al.

Step-by-Step Solution:

Step 1: Calculate the number of moles of Al.\nThe molar mass of Al is 27 g/mol. Given mass is 270 mg = 0.27 g.\nNumber of moles, $n = \frac{{\text{{mass}}}}{{\text{{molar mass}}}} = \frac{{0.27 \text{{ g}}}}{{27 \text{{ g/mol}}}} = 0.01 \text{{ mol}}$

Step 2: Calculate the total ionization energy for one Al atom.\nTotal ionization energy = $IE_1 + IE_2 + IE_3 = 578 + 1817 + 2745 = 5140 \text{{ kJ/mol}}$

Step 3: Calculate the total energy required for 0.01 mol of Al.\nTotal energy = (Number of moles) $\times$ (Total ionization energy per mole)\nTotal energy = $0.01 \text{{ mol}} \times 5140 \text{{ kJ/mol}} = 51.40 \text{{ kJ}}$

Why other options are wrong:
- Option A: $5140 \mathrm{~kJ}$ : This is the total ionization energy per mole, but the question asks for the energy required for 0.01 moles.
- Option C: $2745 \mathrm{~kJ}$ : This is only the third ionization energy, not the total energy required.
- Option D: $514.0 \mathrm{~kJ}$ : This is incorrect; the correct calculation yields 51.40 kJ.

Key Concepts:
- Ionization energy
- Mole concept

Memory Aids:
- Remember that ionization energy is the energy required to remove an electron from a gaseous atom.

Quick Tips:
- Always convert mass to moles before calculating energy changes.

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