${\rm{I}}{{\rm{P}}_1}{\rm{and I}}{{\rm{P}}_2}$ of ${\rm{Mg}}$ are 178 and 348 kcal ${\rm{mo}}{{\rm{l}}^{ - 1}}$. The energy required for the reaction, ${\rm{Mg}} \to {\rm{M}}{{\rm{g}}^{2 + }} + 2{{\rm{e}}^ - }$ is:

$\begin{array}{*{20}{l}} { + 526{\rm{ }}kcal} \end{array}$

$\begin{array}{*{20}{l}} { + 170{\rm{ }}kcal} \end{array}$

$\begin{array}{*{20}{l}} { - 170{\rm{ }}kcal} \end{array}$

$\begin{array}{*{20}{l}} { - 526{\rm{ }}kcal} \end{array}$

The metallic radius of sodium is $1.86\,\,\mathop {\rm{A}}\limits^{\rm{o}} ,$ where as its cation radius from its vapour state is

$1.54\,\,\mathop {\rm{A}}\limits^{\rm{o}} $

$1.86\,\,\mathop {\rm{A}}\limits^{\rm{o}} $

$1.94\,\,\mathop {\rm{A}}\limits^{\rm{o}} $

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Medium, Page 4

91 Medium

91 of 175Question permalink

Understanding +4 | -1

The elements having the electronic configuration

\left[ {\mathrm{Kr}} \right]\;4{d^{10}}{f^{14}},\;5{s^2}{p^6}{d^2},\;6{s^2}

belongs to

View answer and explanation

Correct option: C

d-block

Option C (d-block) is correct because the last electron enters the d-orbital.

Solution approach

Correct Answer + Core Reason: d-block. The last electron enters the d-orbital.

Step-by-Step Solution:

Step 1: Identify the last electron added in the given electronic configuration: $Kr 4d^{10} 5s^2 5p^6 5d^2 6s^2$ .

Step 2: The last electrons are added to the 5d orbital.

Step 3: Elements in which the last electron enters the d-orbital belong to the d-block.

Step 4: Therefore, the element belongs to the d-block.

Why other options are wrong:
- Option A: s-block elements have their last electron in the s-orbital.
- Option B: p-block elements have their last electron in the p-orbital.
- Option D: f-block elements have their last electron in the f-orbital.

Key Concepts:
- Electronic configuration
- Block elements (s, p, d, f)

Memory Aids:
- s-block: Groups 1 and 2
- p-block: Groups 13-18
- d-block: Transition metals
- f-block: Lanthanides and Actinides

Quick Tips:
- Determine the block of an element by looking at the orbital where the last electron is added.

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Remembering +4 | -1

The first ionisation potential in electron volts of nitrogen and oxygen atoms are respectively given by

View answer and explanation

Correct option: A

14.6,13.6

Option A (

14.6,13.6

) is correct because nitrogen has a higher ionization potential than oxygen due to its half-filled p-orbital stability.

Solution approach

Correct Answer + Core Reason: $14.6,13.6$ . Nitrogen has a higher ionization potential than oxygen due to its half-filled p-orbital stability.

Step-by-Step Solution:

Step 1: Write the electronic configurations of Nitrogen (N) and Oxygen (O):
- N: $1s^22s^22p^3$
- O: $1s^22s^22p^4$

Step 2: Explain the stability of half-filled orbitals. Nitrogen has a half-filled $2p$ orbital ( $2p^3$ ), which is more stable than the partially filled $2p$ orbital of Oxygen ( $2p^4$ ). This extra stability makes it more difficult to remove an electron from Nitrogen.

Step 3: Compare the ionization potentials. Due to the greater stability of the half-filled $2p$ orbital in Nitrogen, it has a higher ionization potential than Oxygen.

Step 4: The approximate first ionization potentials are: Nitrogen (14.6 eV) and Oxygen (13.6 eV).

Why other options are wrong:
- Option B: $13.6,14.6$ is incorrect because Nitrogen has a higher ionization potential than Oxygen.
- Option C: $13.6,13.6$ is incorrect because Nitrogen has a higher ionization potential than Oxygen.
- Option D: $14.6,14.6$ is incorrect because Oxygen has a lower ionization potential than Nitrogen.

Key Concepts:
- Ionization potential
- Electronic configuration
- Stability of half-filled orbitals

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

Quick Tips:
- Remember that exceptions to ionization potential trends occur due to the stability of half-filled and fully-filled orbitals.

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Analyzing +4 | -1

Which one of the following elements has the highest ionisation energy?

View answer and explanation

Correct option: B

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

Option B ([Ne]3s²3p³) is correct because half-filled p orbitals have extra stability, leading to a higher ionization energy.

Solution approach

Correct Answer + Core Reason: [Ne]3s²3p³. Half-filled electronic configurations have extra stability and thus higher ionization energy.

Step-by-Step Solution:

Step 1: Identify the electronic configurations given.

Step 2: Recall the trends in ionization energy: Ionization energy generally increases across a period and decreases down a group. Half-filled and fully filled orbitals have extra stability, leading to higher ionization energies.

Step 3: Analyze the configurations:\nA. [Ne]3s²3p¹: Not a stable configuration.\nB. [Ne]3s²3p³: Half-filled p orbitals (more stable).\nC. [Ne]3s²3p²: Not a stable configuration.\nD. [Ar] 3d¹⁰ 4s²4p²: This element is in the 4th period, while A, B, and C are in the 3rd period. Elements in higher periods generally have lower ionization energies due to increased atomic size and shielding.

Step 4: Compare the configurations. The half-filled p³ configuration is expected to have a higher ionization energy than the other configurations in the same period.

Step 5: Conclusion: [Ne]3s²3p³ has the highest ionization energy.

Why other options are wrong:
- Option A: [Ne]3s²3p¹: This configuration is not particularly stable and has a lower ionization energy than a half-filled p orbital.
- Option C: [Ne]3s²3p²: This configuration is also not particularly stable and has a lower ionization energy than a half-filled p orbital.
- Option D: [Ar] 3d¹⁰ 4s²4p²: This element is in the 4th period and generally has a lower ionization energy than elements in the 3rd period.

Key Concepts:
- Ionization Energy Trends
- Electronic Configuration
- Stability of Half-filled and Fully Filled Orbitals

Memory Aids:
- HAF (Half-filled, Always, Favored)

Quick Tips:
- Remember that half-filled and fully filled orbitals are more stable and require more energy to remove an electron.

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Remembering +4 | -1

Which elements in the periodic table should be in liquid state at

35^{\circ} {C}

?

View answer and explanation

Correct option: D

{Hg}, {Br_2}, {Cs}, {Ga}

Option D is correct. At 35°C, Mercury (Hg) and Bromine (Br₂) are liquids. Cesium (Cs) and Gallium (Ga) also have melting points below 35°C, so they are also liquids at this temperature.

Solution approach

Correct Answer + Core Reason: ${Hg}, {Br_2}, {Cs}, {Ga}$ . These elements are in liquid state at $35^{\circ} {C}$ because their melting points are below this temperature.

Step-by-Step Solution:

Step 1: Recall the elements that are liquid at or near room temperature:
- Mercury (Hg): Melting point = -38.83 °C
- Bromine (Br₂): Melting point = -7.2 °C
- Cesium (Cs): Melting point = 28.4 °C
- Gallium (Ga): Melting point = 29.8 °C

Step 2: Check if the melting points are below $35^{\circ} {C}$ :
- Hg, Br₂, Cs, and Ga all have melting points below $35^{\circ} {C}$ .

Step 3: Conclusion: All four elements are in the liquid state at $35^{\circ} {C}$ .

Why other options are wrong:
- Option A: Misses Cesium and Gallium.
- Option B: Misses Bromine and Gallium.
- Option C: Misses Gallium.

Key Concepts:
- Physical States of Elements
- Melting Point

Memory Aids:
- Remember that Hg and Br₂ are liquids at room temperature.

Quick Tips:
- Know the melting points of common elements.

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Analyzing +4 | -1

In which of the following arrangements the order is not according to the property indicated against it?

View answer and explanation

Correct option: C

\mathrm{B} < C < N < O

increasing first ionisation enthalphy

Option C is correct because the first ionization enthalpy of Nitrogen is higher than Oxygen due to the stable half-filled electronic configuration of Nitrogen.

Solution approach

Correct Answer + Core Reason: $\mathrm{B} < C < N < O$ increasing first ionisation enthalphy. The IE of N is higher than O due to half-filled stable configuration.

Step-by-Step Solution:

Step 1: Analyze Option A: $\mathrm{Li} < Na < K < Rb$ increasing metallic radius. As we go down the group, the metallic radius increases. This order is correct.

Step 2: Analyze Option B: $\mathrm{I} < Br < F < Cl$ increasing electron gain enthalpy (with negative sign). Electron gain enthalpy generally increases across a period and decreases down a group. However, due to small size of F, it has lower electron gain enthalpy than Cl. So, the correct order is $\mathrm{I} < Br < Cl < F$ . This order is incorrect.

Step 3: Analyze Option C: $\mathrm{B} < C < N < O$ increasing first ionisation enthalphy. Generally, ionization enthalpy increases across a period. However, Nitrogen has a half-filled $p$ orbital configuration ( $p^3$ ), which is more stable than Oxygen's ( $p^4$ ). Therefore, Nitrogen has a higher ionization enthalpy than Oxygen. The correct order should be $\mathrm{B} < C < O < N$ . This order is incorrect.

Step 4: Analyze Option D: $\mathrm{A}{\mathrm{l}^{3 + }} < \mathrm{M}{\mathrm{g}^{2 + }} < \mathrm{N}{\mathrm{a}^ + } < {\mathrm{F}^ - }$ increasing ionic size. These are isoelectronic species. For isoelectronic species, the ionic size decreases with increasing nuclear charge. The order of nuclear charge is $\mathrm{F} < \mathrm{Na} < \mathrm{Mg} < \mathrm{Al}$ . Therefore, the order of ionic size is $\mathrm{A}{\mathrm{l}^{3 + }} < \mathrm{M}{\mathrm{g}^{2 + }} < \mathrm{N}{\mathrm{a}^ + } < {\mathrm{F}^ - }$ . This order is correct.

Step 5: Identify the incorrect order. Options B and C are incorrect. The question asks for the order that is *not* according to the property indicated. Option C is chosen because the question asks for the order that is NOT according to the property indicated.

Why other options are wrong:
- Option A: The metallic radius increases down the group, so this order is correct.
- Option B: The electron gain enthalpy order is incorrect due to the small size of Fluorine. However, the question asks for first ionisation enthalpy, so this is not the primary reason.
- Option D: The ionic size order is correct for isoelectronic species.

Key Concepts:
- Metallic radius
- Electron gain enthalpy
- Ionization enthalpy
- Isoelectronic species
- Hund's rule

Memory Aids:
- Down the group, metallic radius increases.
- Across the period, ionization enthalpy generally increases.
- Half-filled and fully-filled configurations are more stable.

Quick Tips:
- Remember the general trends of periodic properties.
- Consider exceptions to the trends due to electronic configurations.
- For isoelectronic species, ionic size decreases with increasing nuclear charge.

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Remembering +4 | -1

14 elements of 6th period and 14 elements of 7th period in the periodic table are termed as respectively

View answer and explanation

Correct option: A

Lanthanoids, actinoids

Option A is correct. The 14 elements of the 6th period are called Lanthanoids, and the 14 elements of the 7th period are called Actinoids.

Solution approach

Correct Answer + Core Reason: Lanthanoids, actinoids. The 14 elements of the 6th period are Lanthanoids, and the 14 elements of the 7th period are Actinoids.

Step-by-Step Solution:

Step 1: Define Lanthanoids: Lanthanoids are the 14 elements following Lanthanum (La) in the 6th period.

Step 2: Define Actinoids: Actinoids are the 14 elements following Actinium (Ac) in the 7th period.

Step 3: Conclusion: The correct order is Lanthanoids, then Actinoids.

Why other options are wrong:
- Option B: The order is reversed.
- Option C: Chalcogens and halogens are groups, not series of elements.
- Option D: Actinoids and halogens are not the correct series.

Key Concepts:
- Lanthanoids
- Actinoids
- Periodic Table

Memory Aids:
- Remember the location of Lanthanoids and Actinoids in the periodic table.

Quick Tips:
- Know the periods and groups in the periodic table.

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Remembering +4 | -1

Pauling’s equation for determining the electronegativity of an element, is

{X_A},\;{X_B} =

electronegativity values of elements and

\Delta =

represents polarity of A-B bond

View answer and explanation

Correct option: A

{X_A} - {X_B} = 0.208\sqrt \Delta

Option A (

{X_A} - {X_B} = 0.208\sqrt \Delta

) is correct because it represents Pauling's equation for determining electronegativity difference.

Solution approach

Correct Answer + Core Reason: ${X_A} - {X_B} = 0.208\sqrt \Delta$ . This is Pauling's equation for electronegativity difference.

Step-by-Step Solution:

Step 1: Recall Pauling's equation for electronegativity.\nPauling's equation relates the difference in electronegativity between two atoms to the polarity of the bond between them.

Step 2: Identify the correct equation.\nThe correct equation is ${X_A} - {X_B} = 0.208\sqrt \Delta$ , where ${X_A}$ and ${X_B}$ are the electronegativity values of elements A and B, and $\Delta$ represents the polarity of the A-B bond.

Step 3: Conclusion.\nTherefore, Pauling's equation is ${X_A} - {X_B} = 0.208\sqrt \Delta$ .

Why other options are wrong:
- Option B: ${X_A} + {X_B} = 0.208\sqrt \Delta$ is incorrect because it uses addition instead of subtraction.
- Option C: ${X_A} - {X_B} = 0.208{\Delta ^2}$ is incorrect because it uses $\Delta ^2$ instead of $\sqrt \Delta$ .
- Option D: ${X_A} - {X_B} = \sqrt \Delta$ is incorrect because it omits the 0.208 factor.

Key Concepts:
- Electronegativity
- Pauling scale

Memory Aids:
- Remember Pauling's equation for electronegativity: ${X_A} - {X_B} = 0.208\sqrt \Delta$ .

Quick Tips:
- Familiarize yourself with Pauling's equation and its components.

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Remembering +4 | -1

Element having atomic no. of 56 belongs to which of the following block of periodic table?

View answer and explanation

Correct option: D

s

-block

Option D (

s

-block) is correct because elements with atomic number 56 (Barium) belongs to the s-block.

Solution approach

Correct Answer + Core Reason: $s$ -block. Element with atomic number 56 is Barium (Ba), which belongs to the s-block.

Step-by-Step Solution:

Step 1: The element with atomic number 56 is Barium (Ba).

Step 2: Barium is in Group 2 (alkaline earth metals) of the periodic table.

Step 3: Group 2 elements have their valence electrons in the s-orbital, hence they belong to the s-block.

Why other options are wrong:
- **Option A ( $p$ -block): The p-block elements are in groups 13-18.
- Option B ( $d$ -block): The d-block elements are the transition metals, typically with partially filled d-orbitals.
- Option C ( $f$ -block): The f-block elements are the lanthanides and actinides, with partially filled f-orbitals.

Key Concepts:
- Periodic Table
- s-block elements
- Electronic Configuration

Memory Aids:
- Remember that Group 1 and Group 2 elements belong to the s-block.

Quick Tips:**
- Knowing the group numbers and their corresponding blocks helps in quickly identifying the block of an element.

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Application +4 | -1

The first and second ionization energies of

{Mn}

are in the ratio

0.475: 1

. If the sum of the two energy values is

2226 \mathrm{~kJ} / \mathrm{mol}

, calculate the second ionization neergy (in

\mathrm{kJ} / \mathrm{mol}

).

View answer and explanation

Correct option: D

1509

Option D (1509) is correct because it correctly calculates the second ionization energy based on the given ratio and the sum of the first and second ionization energies.

Solution approach

Correct Answer + Core Reason: 1509. The second ionization energy is calculated using the given ratio and the sum of the two energies.

Step-by-Step Solution:

Step 1: Define the variables: Let $IE_1$ be the first ionization energy and $IE_2$ be the second ionization energy.

Step 2: Write the given ratio: $\frac{IE_1}{IE_2} = 0.475$ .

Step 3: Write the given sum: $IE_1 + IE_2 = 2226 \text{ kJ/mol}$ .

Step 4: Express $IE_1$ in terms of $IE_2$ : $IE_1 = 0.475 \times IE_2$ .

Step 5: Substitute this expression into the sum equation: $0.475 \times IE_2 + IE_2 = 2226$ .

Step 6: Combine terms: $1.475 \times IE_2 = 2226$ .

Step 7: Solve for $IE_2$ : $IE_2 = \frac{2226}{1.475} = 1509.15 \approx 1509 \text{ kJ/mol}$ .

Why other options are wrong:
- Option A (1095): This is incorrect. It likely arises from an error in the calculation or a misunderstanding of how to use the ratio.
- Option B (1960): This is incorrect. It likely arises from an error in the calculation or a misunderstanding of how to use the ratio.
- Option C (1069): This is incorrect. It likely arises from an error in the calculation or a misunderstanding of how to use the ratio.

Key Concepts:
- Ionization Energy
- Ratios and Proportions
- Solving Equations

Memory Aids:
- Carefully set up the equations based on the given information.

Quick Tips:
- Double-check your calculations to avoid errors.

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Understanding +4 | -1

The correct order of variation in sizes of atoms is

View answer and explanation

Correct option: C

Be > C > F < Ne

Option C is correct because atomic size decreases across a period due to increasing nuclear charge, but Neon is larger than Fluorine due to increased electron-electron repulsion in the fully filled electron configuration.

Solution approach

Correct Answer + Core Reason: $Be > C > F < Ne$ . Atomic size decreases across a period, but noble gases have larger sizes due to electron-electron repulsion.

Step-by-Step Solution:

Step 1: Atomic size generally decreases across a period from left to right due to increasing nuclear charge.

Step 2: Beryllium (Be) is in Group 2, Carbon (C) is in Group 14, Fluorine (F) is in Group 17, and Neon (Ne) is in Group 18.

Step 3: Based on the trend, the size order should be $Be > C > F$ .

Step 4: However, noble gases like Neon have larger sizes than their preceding halogens due to increased electron-electron repulsion in their fully filled electron configurations.

Step 5: Therefore, the correct order is $Be > C > F < Ne$ .

Why other options are wrong:
- Option A: $Be > C > F > Ne$ is incorrect because Neon is larger than Fluorine.
- Option B: $Be < C < F < Ne$ is incorrect because size decreases from Be to F.
- Option D: $F > Ne > Be > C$ is incorrect because size decreases from Be to F and Neon is larger than Fluorine.

Key Concepts:
- Atomic size
- Periodic trends
- Nuclear charge
- Electron-electron repulsion

Memory Aids:
- Remember that atomic size decreases across a period, but noble gases are an exception.

Quick Tips:
- Consider the position of the elements in the periodic table and the effect of electron-electron repulsion.

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Remembering +4 | -1

Among the following elements, the most electronegative is:

View answer and explanation

Correct option: D

Fluorine

Option D (Fluorine) is correct because Fluorine is the most electronegative element.

Solution approach

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

Step-by-Step Solution:

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

Step 2: Consider the positions of the elements in the periodic table.
- Oxygen, Chlorine, Nitrogen, and Fluorine are all non-metals.
- Fluorine is the most electronegative element.

Step 3: Conclusion.\nTherefore, Fluorine is the most electronegative among the given elements.

Why other options are wrong:
- Option A: Oxygen is less electronegative than Fluorine.
- Option B: Chlorine is less electronegative than Fluorine.
- Option C: Nitrogen is less electronegative than Fluorine.

Key Concepts:
- Electronegativity
- Periodic trends

Memory Aids:
- Fluorine is the most electronegative element.

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

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Remembering +4 | -1

The lightest liquid metal is

View answer and explanation

Correct option: C

{Cs}

Option C (

{Cs}

) is correct because Cesium is the lightest liquid metal at or near room temperature.

Solution approach

Correct Answer + Core Reason: ${Cs}$ . Cesium has the lowest melting point among the given options and is a liquid near room temperature.

Step-by-Step Solution:

Step 1: Identify the melting points of the given metals:
- Mercury (Hg): -38.83 °C
- Gallium (Ga): 29.76 °C
- Cesium (Cs): 28.44 °C
- Francium (Fr): Estimated to be around 27 °C (but highly radioactive and not well-studied)

Step 2: Compare the melting points. Cesium has the lowest melting point among the well-characterized metals.

Step 3: Cesium is a liquid near room temperature and is lighter than Mercury.

Why other options are wrong:
- **Option A ( ${Hg}$ ): Mercury is a liquid at room temperature but is much denser and heavier than Cesium.
- Option B ( ${Ga}$ ): Gallium is a liquid at temperatures slightly above room temperature, but its melting point is higher than Cesium.
- Option D ( $Fr$ ): Francium is highly radioactive and not well-studied. While its melting point is estimated to be close to Cesium, it is not commonly considered due to its instability.

Key Concepts:
- Melting Point
- Physical Properties of Metals
- Periodic Trends

Memory Aids:
- Remember that Cesium has a very low melting point and is a liquid near room temperature.

Quick Tips:**
- Knowing the melting points of common metals helps in answering such questions.

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Understanding +4 | -1

The increasing order of first ionisation enthalpies of the elements B, P, S and F (lowest first) is

View answer and explanation

Correct option: D

B < S < P < F

Option D (

B < S < P < F

) is correct because ionization energy generally increases across a period, with exceptions due to half-filled orbitals (P).

Solution approach

Correct Answer + Core Reason: $B < S < P < F$ . Ionization energy generally increases across a period, with exceptions for half-filled orbitals.

Step-by-Step Solution:
Step 1: Recall the general trend of ionization energy across a period. Ionization energy generally increases as you move from left to right across a period.
Step 2: Identify the positions of B, P, S, and F in the periodic table. B is in Group 13 (Period 2), P is in Group 15 (Period 3), S is in Group 16 (Period 3), and F is in Group 17 (Period 2).
Step 3: Consider the exceptions to this trend due to electronic configurations.
- P has a half-filled p-orbital ( $3p^3$ ), which is more stable than S's ( $3p^4$ ). Therefore, P has a higher ionization energy than S.
Step 4: Arrange the elements in increasing order of ionization energy, considering the exception. B < S < P < F.

Why other options are wrong:
- Option A: $F < S < P < B$ : Incorrect, this is the reverse order and does not account for the exception.
- Option B: $P < S < B < F$ : Incorrect, this does not account for the exception and incorrectly places B and P.
- Option C: $B < P < S < F$ : Incorrect, this does not account for the exception between P and S.

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

Memory Aids:
- Remember the exception between P and S: P > S

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

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Remembering +4 | -1

In the Mendeleev periodic table, which of the following element instead of having lower atomic weight was placed after the element of higher atomic weight thereby ignoring the order of increasing atomic weights.

View answer and explanation

Correct option: A

Iodine

Option A (Iodine) is correct because Iodine, despite having a lower atomic weight than Tellurium, was placed after Tellurium in Mendeleev's periodic table due to its chemical properties.

Solution approach

Correct Answer + Core Reason: Iodine. Iodine was placed after Tellurium despite having a lower atomic weight.

Step-by-Step Solution:

Step 1: Recall Mendeleev's periodic table and its arrangement based on atomic weight.

Step 2: Remember that Iodine (I) was placed after Tellurium (Te) due to similar chemical properties with other halogens, even though Te has a higher atomic weight.

Step 3: Conclude that Iodine is the correct answer.

Why other options are wrong:
- Option B (Antimony): Antimony's placement in the periodic table is generally consistent with increasing atomic weight relative to its neighbors.
- Option C (Bromine): Bromine's placement follows the increasing atomic weight order.
- Option D (Molybdenum): Molybdenum's placement follows the increasing atomic weight order.

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

Memory Aids:
- Remember the Te-I pair as a key example of Mendeleev prioritizing properties over atomic weight.

Quick Tips:
- Focus on the exceptions to the atomic weight arrangement in Mendeleev's table.

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Understanding +4 | -1

The elements with atomic numbers 101 and 104 belong to, respectively,

View answer and explanation

Correct option: A

actinoids and group 4

Option A (actinoids and group 4) is correct because element 101 is an actinoid (Mendelevium) and element 104 is in group 4 (Rutherfordium).

Solution approach

Correct Answer + Core Reason: actinoids and group 4. Element 101 is an actinoid, and element 104 is in group 4.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 101: Mendelevium (Md). It belongs to the actinoid series.

Step 2: Identify the element with atomic number 104: Rutherfordium (Rf). It belongs to Group 4.

Step 3: Match the elements to the given options.

Why other options are wrong:
- Option B: actinoids and group 6: Element 104 does not belong to group 6.
- Option C: group 6 and actinoids: Element 101 is an actinoid, and element 104 does not belong to group 6.
- Option D: group 11 and group 4: Element 101 is not in group 11.

Key Concepts:
- Periodic table and group numbers.
- Actinoid series.

Memory Aids:
- Know the positions of the actinoids and the group numbers of common elements.

Quick Tips:
- Refer to a periodic table to quickly identify the group and series of elements.

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Application +4 | -1

An atom has electronic configuration

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

, you will place it in which group of periodic table?

View answer and explanation

Correct option: A

Fifth

Option A (Fifth) is correct. The electronic configuration ends in

3d^3 4s^2

, indicating that it's a d-block element in the 5th group (3+2 = 5).

Solution approach

Correct Answer + Core Reason: Fifth. The element belongs to the d-block, and the group number is determined by adding the number of d-electrons and s-electrons in the outermost shell.

Step-by-Step Solution:

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

Step 2: Determine the block: The last electron enters the d-orbital, so it's a d-block element.

Step 3: Determine the period: The highest principal quantum number is 4, so it's in the 4th period.

Step 4: Determine the group number: For d-block elements, the group number is the sum of the number of electrons in the (n-1)d orbitals and the ns orbitals. In this case, it's 3 (from $3d^3$ ) + 2 (from $4s^2$ ) = 5.

Step 5: Therefore, the element belongs to the 5th group.

Why other options are wrong:
- Option B (Second): This is incorrect as the element is a d-block element and not an s-block element.
- Option C (Fifteenth): This would be the group number if it were a p-block element with 5 valence electrons, but it's a d-block element.
- Option D (Third): This is incorrect; the group number is 5, not 3.

Key Concepts:
- Electronic configuration
- d-block elements
- Periodic table trends
- Group number determination

Memory Aids:
- Remember the rules for determining group numbers based on electronic configuration.

Quick Tips:
- For d-block elements, add the number of d and s electrons in the outermost shell to find the group number.

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For alkali metals, which one of the following trends is incorrect?

View answer and explanation

Correct option: C

Density : Li < Na < K < Rb

Option C (Density : Li < Na < K < Rb) is correct because the density of K is actually less than Na due to its larger size and weaker metallic bonding.

Solution approach

Correct Answer + Core Reason: Density : Li < Na < K < Rb. The density of potassium is less than that of sodium.

Step-by-Step Solution:
Step 1: Recall the properties of alkali metals.
Step 2: Consider the trend of hydration energy. Hydration energy decreases down the group due to increasing ionic size. Li+ has the highest hydration energy.
Step 3: Consider the trend of ionization energy. Ionization energy decreases down the group due to increasing atomic size and shielding effect. Li has the highest ionization energy.
Step 4: Consider the trend of atomic size. Atomic size increases down the group due to the addition of new electron shells.
Step 5: Consider the trend of density. Density generally increases down the group, but potassium (K) is an exception. Potassium is less dense than sodium (Na) due to its larger atomic size and weaker metallic bonding.

Why other options are wrong:
- Option A: Hydration energy : Li > Na > K > Rb: Incorrect, this trend is correct.
- Option B: Ionisation energy : Li > Na > K > Rb: Incorrect, this trend is correct.
- Option D: Atomic size : Li < Na < K < Rb: Incorrect, this trend is correct.

Key Concepts:
- Properties of alkali metals
- Hydration energy
- Ionization energy
- Atomic size
- Density

Memory Aids:
- Remember that potassium is less dense than sodium.

Quick Tips:
- Be aware of exceptions to general trends in the periodic table.

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\mathrm{I}{\mathrm{P}_1}\mathrm{and I}{\mathrm{P}_2}

of

\mathrm{Mg}

are 178 and 348 kcal

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

. The energy required for the reaction,

\mathrm{Mg} \to \mathrm{M}{\mathrm{g}^{2 + }} + 2{\mathrm{e}^ - }

is:

View answer and explanation

Correct option: B

\begin{array}{llllllllllllllllllll} { + 526\mathrm{ }kcal} \end{array}

Option B (+526 kcal) is correct because it represents the sum of the first and second ionization energies, which is the total energy required to remove two electrons from Mg.

Solution approach

Correct Answer + Core Reason: +526 kcal. The energy required is the sum of the first and second ionization energies.

Step-by-Step Solution:

Step 1: Identify the given information: $IP_1 = 178 \text{ kcal/mol}$ and $IP_2 = 348 \text{ kcal/mol}$ .

Step 2: Determine the required energy. The reaction is $Mg \rightarrow Mg^{2+} + 2e^-$ , which involves removing two electrons.

Step 3: Calculate the total energy required: $E = IP_1 + IP_2$ .

Step 4: Substitute the given values: $E = 178 + 348 = 526 \text{ kcal/mol}$ .

Step 5: The energy required is +526 kcal/mol.

Why other options are wrong:
- Option A (+170 kcal): This is incorrect because it subtracts the ionization energies instead of adding them.
- Option C (-170 kcal): This is incorrect because it subtracts the ionization energies and has the wrong sign. Ionization energies are always positive.
- Option D (-526 kcal): This is incorrect because it has the wrong sign. Ionization energies are always positive.

Key Concepts:
- Ionization Energy
- Successive Ionization Energies

Memory Aids:
- Ionization energy is always positive (energy is required to remove an electron).

Quick Tips:
- Remember to add successive ionization energies to find the total energy required for multiple ionizations.

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In the long form of periodic table, elements are arranged according to

View answer and explanation

Correct option: A

Increasing atomic number

Option A (Increasing atomic number) is correct because elements are arranged in order of increasing atomic number in the long form of the periodic table.

Solution approach

Correct Answer + Core Reason: Increasing atomic number. The long form of the periodic table is arranged by increasing atomic number.

Step-by-Step Solution:

Step 1: Recall the arrangement of elements in the modern periodic table: Elements are arranged in order of increasing atomic number.

Step 2: Consider the options:
- Increasing atomic number: Correct arrangement.
- Decreasing atomic number: Incorrect arrangement.
- Increasing atomic mass: Used in earlier versions but not the modern table.
- Decreasing atomic mass: Incorrect arrangement.

Step 3: Conclusion: Elements are arranged by increasing atomic number.

Why other options are wrong:
- Option B (Decreasing atomic number): The periodic table is not arranged in decreasing order of atomic number.
- Option C (Increasing atomic mass): While Mendeleev's table was based on atomic mass, the modern table uses atomic number.
- Option D (Decreasing atomic mass): The periodic table is not arranged in decreasing order of atomic mass.

Key Concepts:
- Periodic Table
- Atomic Number
- Atomic Mass

Memory Aids:
- The modern periodic table is based on atomic number.

Quick Tips:
- Remember the basis of the modern periodic table.

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

View answer and explanation

Correct option: B

Halogens

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

Solution approach

Correct Answer + Core Reason: Halogens. Elements with atomic numbers 9, 17, 35, 53, and 85 are all halogens.

Step-by-Step Solution:

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

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

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

Step 4: Conclusion: The correct answer is Halogens.

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

Key Concepts:
- Periodic Table groups
- Halogens

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

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

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The behaviour of lithium and magnesium is similar to each other. This suggests that both

\mathrm{Li}

and

\mathrm{Mg}

View answer and explanation

Correct option: A

Have diagonal relationship

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

Solution approach

Correct Answer + Core Reason: Have diagonal relationship. The similar behavior of Li and Mg is due to their diagonal relationship.

Step-by-Step Solution:

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

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

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

Why other options are wrong:
- Option B: Belong to same group is incorrect. Li and Mg belong to different groups (Group 1 and Group 2, respectively).
- Option C: Belong to groups at extreme ends is incorrect. They are not at extreme ends of the same period.
- Option D: Both (1) and (2) is incorrect because only (1) is correct.

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

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

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

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Assertion (A): Helium is placed in ‘p’ block of modern periodic table

Reason (R): Differentiating electron in Helium enters into ‘p’ orbital

View answer and explanation

Correct option: C

A is true but R is false

Option C is correct because Helium is placed in the 's' block, not the 'p' block, and its differentiating electron enters the 's' orbital.

Solution approach

Correct Answer + Core Reason: A is true but R is false. Helium is in the s-block, and its differentiating electron enters the s orbital.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Helium.\nThe electronic configuration of Helium (He) is 1s².

Step 2: Determine the block to which Helium belongs.\nSince the last electron enters the s-orbital, Helium belongs to the s-block.

Step 3: Evaluate the assertion.\nThe assertion states that Helium is placed in the 'p' block of the modern periodic table, which is incorrect.

Step 4: Evaluate the reason.\nThe reason states that the differentiating electron in Helium enters into the 'p' orbital, which is incorrect. The differentiating electron enters the 's' orbital.

Step 5: Conclusion.\nThe assertion is false, and the reason is false.

Why other options are wrong:
- Option A: Incorrect because both the assertion and reason are not true.
- Option B: Incorrect because both the assertion and reason are not true.
- Option C: Correct because the assertion is false, and the reason is false.
- Option D: Incorrect because the assertion is not true.

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

Memory Aids:
- Remember that Helium has only two electrons, both in the 1s orbital.

Quick Tips:
- Review the block elements in the periodic table and their corresponding electronic configurations.

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The magnitude of screening effect depends upon the number of

View answer and explanation

Correct option: A

Inner electrons

Option A (Inner electrons) is correct because inner electrons shield the outer electrons from the full nuclear charge.

Solution approach

Correct Answer + Core Reason: Inner electrons. The magnitude of the screening effect depends on the number of inner electrons.

Step-by-Step Solution:

Step 1: Define the screening effect (shielding effect). The screening effect is the reduction in the effective nuclear charge experienced by outer electrons due to the presence of inner electrons.

Step 2: Explain how inner electrons cause the screening effect. Inner electrons shield the outer electrons from the full positive charge of the nucleus. This shielding reduces the attractive force between the nucleus and the outer electrons.

Step 3: Explain why the number of inner electrons matters. The more inner electrons there are, the greater the shielding effect, and the lower the effective nuclear charge experienced by the outer electrons.

Why other options are wrong:
- Option B: Outer electrons is incorrect because outer electrons do not contribute significantly to the screening effect. They are the ones being screened.
- Option C: Bond order is incorrect because bond order is related to the number of chemical bonds between atoms and does not directly affect the screening effect.
- Option D: Both 1 and 2 is incorrect because only inner electrons contribute significantly to the screening effect.

Key Concepts:
- Screening effect (shielding effect)
- Effective nuclear charge
- Electronic configuration

Memory Aids:
- Inner electrons shield outer electrons.

Quick Tips:
- Remember that the screening effect increases with the number of inner electrons.

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Which among the following elements has lowest value of ionisation energy?

View answer and explanation

Correct option: B

Sn

Option B (Sn) has the lowest ionization energy because ionization energy generally increases up a group, and Sn is below Pb, Si, and C in Group 14.

Solution approach

Correct Answer + Core Reason: Sn. Ionization energy generally increases up a group.

Step-by-Step Solution:

Step 1: Identify the group to which each element belongs.\nC, Si, Ge, Sn, Pb are in Group 14.

Step 2: Recall the trend in ionization energy down a group. Ionization energy decreases down a group.

Step 3: Therefore, the element lowest in the group will have the lowest ionization energy.

Step 4: Sn is below Pb, Si, and C in Group 14, thus it has the lowest ionization energy among the given elements.

Why other options are wrong:
- Option A: Pb has a higher ionization energy than Sn because it is above Sn in the group.
- Option C: Si has a higher ionization energy than Sn because it is above Sn in the group.
- Option D: C has a higher ionization energy than Sn because it is above Sn in the group.

Key Concepts:
- Ionization Energy
- Periodic Trends

Memory Aids:
- Ionization energy decreases down a group.

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

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Which of the following statements is wrong?

View answer and explanation

Correct option: D

Non-metals are more than metals.

Option D (Non-metals are more than metals) is the wrong statement because metals constitute a larger portion of the periodic table than non-metals.

Solution approach

Correct Answer + Core Reason: Non-metals are more than metals. This statement is incorrect because metals are more abundant than non-metals in the periodic table.

Step-by-Step Solution:

Step 1: Analyze the given statements:\nA. Metals are more than non-metals.\nB. There are only few metalloids.\nC. Hydrogen can be placed with alkali metals as well as with halogens in the Periodic Table.\nD. Non-metals are more than metals.

Step 2: Evaluate each statement:\nA. Metals occupy a larger portion of the periodic table compared to non-metals. This statement is correct.\nB. Metalloids (or semi-metals) are elements with properties intermediate between those of metals and non-metals. There are only a few metalloids (e.g., B, Si, Ge, As, Sb, Te). This statement is correct.\nC. Hydrogen exhibits properties similar to both alkali metals (due to its electronic configuration of $1s^1$ ) and halogens (due to its ability to gain one electron to achieve a stable configuration). This statement is correct.\nD. Non-metals are fewer in number compared to metals. This statement is incorrect.

Step 3: Conclusion: The wrong statement is that non-metals are more than metals.

Why other options are wrong:
- Option A: Metals are indeed more numerous than non-metals in the periodic table.
- Option B: The number of metalloids is limited compared to metals and non-metals.
- Option C: Hydrogen's unique properties allow it to be placed with both alkali metals and halogens.

Key Concepts:
- Metals, non-metals, and metalloids
- Periodic table trends
- Properties of hydrogen

Memory Aids:
- Metals are generally located on the left side of the periodic table, while non-metals are on the right.

Quick Tips:
- Remember the general distribution of metals, non-metals, and metalloids in the periodic table.

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The electronic configuration,

1{s^2},\;2{s^2}2{p^6},\;3{s^2}\;3{p^6}3{d^9}

represents a

View answer and explanation

Correct option: D

Metallic cation

Option D is correct because the electronic configuration indicates a loss of electrons, resulting in a positive charge, thus a metallic cation.

Solution approach

Correct Answer + Core Reason: Metallic cation. The electronic configuration suggests a metal atom that has lost electrons to achieve a stable noble gas configuration.

Step-by-Step Solution:

Step 1: Analyze the electronic configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$ .

Step 2: Determine the number of electrons. The total number of electrons is 2 + 2 + 6 + 2 + 6 + 9 = 27.

Step 3: Identify the element. The element with 27 electrons is Cobalt (Co), which is a metal. However, a neutral Cobalt atom has 27 electrons, with the configuration $[Ar] 4s^2 3d^7$ . The given configuration has 27 electrons, but it's missing two electrons from the 4s orbital and has two extra electrons in the 3d orbital. It looks like Copper (Cu) which has 29 electrons, but it has lost two electrons to become Cu $^{2+}$ .

Step 4: Determine the charge. The closest stable configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}$ , which is achieved by losing one more electron. This indicates the formation of a cation.

Step 5: Conclusion: The electronic configuration represents a metallic cation.

Why other options are wrong:
- Option A: Metal atom: This is incorrect because the configuration represents an ion, not a neutral atom.
- Option B: Non-metal atom: This is incorrect because the element is a metal.
- Option C: Non-metallic anion: This is incorrect because the configuration represents a cation, not an anion.

Key Concepts:
- Electronic Configuration
- Cations and Anions
- Metals and Non-metals

Memory Aids:
- Remember that metals tend to lose electrons to form cations, while non-metals tend to gain electrons to form anions.

Quick Tips:
- When analyzing electronic configurations, determine the number of electrons and compare it to the number of protons to determine the charge.

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Which of the following groups in modern periodic table have largest number of elements

View answer and explanation

Correct option: C

III B

Option C (III B) is correct because Group IIIB (also known as Group 3) has the largest number of elements due to the inclusion of Lanthanides and Actinides.

Solution approach

Correct Answer + Core Reason: III B. Group IIIB includes the lanthanides and actinides, making it the group with the largest number of elements.

Step-by-Step Solution:

Step 1: Analyze the groups:
- I A (Alkali metals): Contains a relatively small number of elements.
- III A: Contains a moderate number of elements.
- III B: Contains the transition metals plus the lanthanides and actinides (f-block elements).
- Zero group (Noble gases): Contains a relatively small number of elements.

Step 2: Identify the group with the largest number of elements: Group III B.

Why other options are wrong:
- Option A (I A): Group IA has fewer elements than Group IIIB.
- Option B (III A): Group IIIA has fewer elements than Group IIIB.
- Option D (Zero group): The Zero group (Noble gases) has fewer elements than Group IIIB.

Key Concepts:
- Periodic Table
- Group Properties
- Lanthanides and Actinides

Memory Aids:
- Group IIIB includes the f-block elements.

Quick Tips:
- Remember the position of the f-block elements in the periodic table.

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The correct ionic radii order is:

View answer and explanation

Correct option: A

{\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }} > A{\mathrm{l}^{3 + }}

Option A (

{\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }} > A{\mathrm{l}^{3 + }}

) is correct because it correctly orders the isoelectronic species based on their nuclear charge; higher the nuclear charge, smaller the ionic radii.

Solution approach

Correct Answer + Core Reason: ${\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }} > A{\mathrm{l}^{3 + }}$ . For isoelectronic species, higher nuclear charge means smaller ionic radius.

Step-by-Step Solution:

Step 1: Identify that all the given ions are isoelectronic, each having 10 electrons.

Step 2: List the ions with their atomic numbers (number of protons):
- N (7)
- O (8)
- F (9)
- Na (11)
- Mg (12)
- Al (13)

Step 3: For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases.

Step 4: Arrange the ions in decreasing order of ionic radii (increasing order of atomic number): ${\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }} > A{\mathrm{l}^{3 + }}$

Why other options are wrong:
- Option B: This order is incorrect because it places Na⁺ between N³⁻ and O²⁻, which violates the trend of decreasing ionic radius with increasing nuclear charge.
- Option C: This order is incorrect because it places Na⁺ at the beginning, which violates the trend of decreasing ionic radius with increasing nuclear charge.
- Option D: This order is incorrect because it places O²⁻ at the beginning and N³⁻ in the middle, which violates the trend of decreasing ionic radius with increasing nuclear charge.

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

Memory Aids:
- For isoelectronic species, more protons = smaller size.

Quick Tips:
- Always identify isoelectronic species first and then compare their nuclear charges.

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Which of the following set of oxides are amphoteric

View answer and explanation

Correct option: D

A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}

Option D (

A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}

) is correct because all the oxides in this set can react with both acids and bases.

Solution approach

Correct Answer + Core Reason: $A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}$ are amphoteric oxides.

Step-by-Step Solution:

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

Step 2: Consider the given options:
- $A{l_2}{O_3}$ (Aluminum oxide) is an amphoteric oxide.
- $ZnO$ (Zinc oxide) is an amphoteric oxide.
- $Sn{O_2}$ (Tin(IV) oxide) is an amphoteric oxide.
- $S{b_4}{O_6}$ (Antimony(III) oxide) is an amphoteric oxide.

Step 3: Analyze the other options:
- Option A contains acidic oxides.
- Option B contains basic oxides.
- Option C contains neutral and acidic oxides.

Step 4: Conclusion: $A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}$ is the correct set of amphoteric oxides.

Why other options are wrong:
- Option A: $C{O_2},Si{O_2},{P_4}{O_{10}}$ are acidic oxides, not amphoteric.
- Option B: ${K_2}O,BaO,CrO,FeO$ are basic oxides, not amphoteric.
- Option C: ${N_2}{O_5},CO,NO,{N_2}O$ are acidic and neutral oxides, not amphoteric.
- Option D: $A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}$ is the correct answer as all are amphoteric oxides.

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

Memory Aids:
- Remember common amphoteric oxides: $BeO$ , $ZnO$ , $A{l_2}{O_3}$ , $SnO$ , $PbO$ , $S{b_2}{O_3}$ .

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

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The metallic radius of sodium is

1.86\,\,\mathop \mathrm{A}\limits^\mathrm{o} ,

where as its cation radius from its vapour state is

View answer and explanation

Correct option: C

1.54\,\,\mathop \mathrm{A}\limits^\mathrm{o}

Option C is correct because the cation radius is smaller than the metallic radius due to increased effective nuclear charge and removal of an electron.

Solution approach

Correct Answer + Core Reason: $1.54\mathop \mathrm{A}\limits^\mathrm{o}$ because the cation radius is smaller than the neutral atom's radius due to increased effective nuclear charge.

Step-by-Step Solution:

Step 1: Understand the concept of ionic radius.\nWhen an atom loses an electron to form a cation, the effective nuclear charge increases, pulling the remaining electrons closer to the nucleus. This results in a smaller ionic radius compared to the metallic radius.

Step 2: Consider the given metallic radius of sodium.\nThe metallic radius of sodium (Na) is $1.86\mathop \mathrm{A}\limits^\mathrm{o}$ .

Step 3: Determine the approximate cation radius of sodium.\nSince the cation radius is smaller than the metallic radius, options A and B can be eliminated. The cation radius of $Na^+$ is approximately $1.54\mathop \mathrm{A}\limits^\mathrm{o}$ .

Why other options are wrong:
- Option A: Incorrect because the cation radius is always smaller than the metallic radius.
- Option B: Incorrect because the cation radius is always smaller than the metallic radius.
- Option D: Incorrect because option C is the correct cation radius.

Key Concepts:
- Metallic radius
- Ionic radius
- Effective nuclear charge

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

Quick Tips:
- Remember that cations are smaller and anions are larger than their corresponding neutral atoms.

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