The ionic radii of ${{\rm{N}}^{3 - }},{{\rm{O}}^{2 - }}{\rm{and}}\,{{\rm{F}}^ - }$ are respectively given by:

$\begin{array}{*{20}{l}} {1.71,{\rm{ }}1.40,{\rm{ }}1.36} \end{array}$

$\begin{array}{*{20}{l}} {1.36,{\rm{ }}1.40,{\rm{ }}1.71} \end{array}$

$\begin{array}{*{20}{l}} {1.36,{\rm{ }}1.71,{\rm{ }}1.40} \end{array}$

$\begin{array}{*{20}{l}} {1.71,{\rm{ }}1.36,{\rm{ }}1.40} \end{array}$

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Easy, Page 6

151 Easy

151 of 171Question permalink

Application +4 | -1

The compound of vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula:

View answer and explanation

Correct option: C

VC{l_4}

The correct answer is C (

VC{l_4}

) because a magnetic moment of 1.73 BM corresponds to one unpaired electron, indicating V has a +4 oxidation state in

VC{l_4}

.

Solution approach

Correct Answer + Core Reason: $VC{l_4}$ . A magnetic moment of 1.73 BM indicates one unpaired electron, suggesting V is in the +4 oxidation state.

Step-by-Step Solution:

Step 1: Calculate the number of unpaired electrons using the spin-only formula: $\mu = \sqrt{n(n+2)}$ BM, where $\mu$ is the magnetic moment and $n$ is the number of unpaired electrons.

Step 2: Given $\mu = 1.73$ BM, solve for $n$ : $1.73 = \sqrt{n(n+2)}$ . Squaring both sides, $3 \approx n(n+2)$ . This gives $n=1$ .

Step 3: Determine the oxidation state of V in each compound:
- $VC{l_2}$ : V has +2 oxidation state, electronic configuration is [Ar] 3d³, 3 unpaired electrons.
- $VC{l_3}$ : V has +3 oxidation state, electronic configuration is [Ar] 3d², 2 unpaired electrons.
- $VC{l_4}$ : V has +4 oxidation state, electronic configuration is [Ar] 3d¹, 1 unpaired electron.
- $VC{l_5}$ : V has +5 oxidation state, electronic configuration is [Ar] 3d⁰, 0 unpaired electrons.

Step 4: Since the magnetic moment corresponds to 1 unpaired electron, the correct formula is $VC{l_4}$ .

Why other options are wrong:
- Option A: $VC{l_2}$ has 3 unpaired electrons, which does not match the given magnetic moment.
- Option B: $VC{l_3}$ has 2 unpaired electrons, which does not match the given magnetic moment.
- Option D: $VC{l_5}$ has 0 unpaired electrons, which does not match the given magnetic moment.

Key Concepts:
- Magnetic Moment
- Spin-Only Formula
- Oxidation States
- Electronic Configuration

Memory Aids:
- Remember the spin-only formula: $\mu = \sqrt{n(n+2)}$ BM.

Quick Tips:
- Use the magnetic moment to determine the number of unpaired electrons and then the oxidation state.

152 Easy

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Analyzing +4 | -1

Assertion(A)

{Cu}^{2+}

is diamagnetic.

Reason(R) All the orbitals present in

{Cu}^{2+}

ion are doubly occupied.

View answer and explanation

Correct option: D

If both (A) and (R) are incorrect.

Option D is correct because both the assertion and the reason are incorrect.

{Cu}^{2+}

is paramagnetic due to unpaired electrons, and not all orbitals are doubly occupied.

Solution approach

Correct Answer + Core Reason: If both (A) and (R) are incorrect. ${Cu}^{2+}$ is paramagnetic, and its orbitals are not all doubly occupied.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of Cu.\nThe electronic configuration of Cu is [Ar] 3d¹⁰4s¹.

Step 2: Determine the electronic configuration of Cu²⁺.\nWhen Cu loses two electrons to form Cu²⁺, it loses the 4s electron and one 3d electron. The electronic configuration of Cu²⁺ is [Ar] 3d⁹.

Step 3: Determine if Cu²⁺ is diamagnetic or paramagnetic.\nSince Cu²⁺ has 9 d electrons, there is one unpaired electron. Therefore, Cu²⁺ is paramagnetic.

Step 4: Evaluate the assertion.\nThe assertion states that Cu²⁺ is diamagnetic, which is incorrect.

Step 5: Evaluate the reason.\nThe reason states that all the orbitals present in Cu²⁺ are doubly occupied, which is incorrect because there is one unpaired electron in the 3d orbitals.

Step 6: Conclusion.\nBoth the assertion and the reason are incorrect.

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

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

Memory Aids:
- Paramagnetic substances are attracted to a magnetic field, while diamagnetic substances are repelled.
- Unpaired electrons lead to paramagnetism.

Quick Tips:
- Remember that Cu and Cr are exceptions to the Aufbau principle.

153 Easy

153 of 171Question permalink

Remembering +4 | -1

Which of the following is amphoteric oxide

View answer and explanation

Correct option: B

A{l_2}{O_3}

Option B (

A{l_2}{O_3}

) is correct because amphoteric oxides can react with both acids and bases.

Solution approach

Correct Answer + Core Reason: $A{l_2}{O_3}$ is amphoteric because it reacts with both acids and bases.

Step-by-Step Solution:

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

Step 2: Consider the given options:
- $N{a_2}O$ is a basic oxide.
- $A{l_2}{O_3}$ is an amphoteric oxide.
- $S{O_2}$ is an acidic oxide.
- ${P_2}{O_5}$ is an acidic oxide.

Step 3: $A{l_2}{O_3}$ reacts with acids: $A{l_2}{O_3} + 6HCl \rightarrow 2AlCl_3 + 3H_2O$
$A{l_2}{O_3}$ reacts with bases: $A{l_2}{O_3} + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$

Step 4: Conclusion: $A{l_2}{O_3}$ is the amphoteric oxide.

Why other options are wrong:
- Option A: $N{a_2}O$ is a basic oxide and reacts only with acids.
- Option B: $A{l_2}{O_3}$ is the correct answer as it is amphoteric.
- Option C: $S{O_2}$ is an acidic oxide and reacts only with bases.
- Option D: ${P_2}{O_5}$ is an acidic oxide and reacts only with bases.

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

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

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

154 Easy

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Remembering +4 | -1

The charges on the colloidal

CdS

sol and

Ti{O_2}

sol are, respectively

View answer and explanation

Correct option: D

negative and positive

Option D (negative and positive) is correct because metal sulfides like CdS typically form negatively charged sols, while metal oxides like

Ti{O_2}

usually form positively charged sols.

Solution approach

Correct Answer + Core Reason: negative and positive. $CdS$ sol is negatively charged and $Ti{O_2}$ sol is positively charged.

Step-by-Step Solution:

Step 1: Recall the general properties of colloidal solutions.

Step 2: Consider the nature of the dispersed phase in each sol.

Step 3: $CdS$ is a metal sulfide. Metal sulfides generally form negatively charged colloidal sols due to preferential adsorption of $S^{2-}$ ions.

Step 4: $Ti{O_2}$ is a metal oxide. Metal oxides generally form positively charged colloidal sols due to preferential adsorption of $H^+$ ions.

Step 5: Conclusion: $CdS$ sol is negatively charged, and $Ti{O_2}$ sol is positively charged.

Why other options are wrong:
- Option A: positive and positive - Incorrect because $CdS$ sol is negatively charged.
- Option B: positive and negative - Incorrect because $Ti{O_2}$ sol is positively charged and $CdS$ is negatively charged.
- Option C: negative and negative - Incorrect because $Ti{O_2}$ sol is positively charged.
- Option D: negative and positive - Correct answer.

Key Concepts:
- Colloidal Solutions
- Charge on Colloidal Particles
- Preferential Adsorption

Memory Aids:
- Metal sulfides are generally negative, metal oxides are generally positive.

Quick Tips:
- Remember the common examples of positively and negatively charged sols.

155 Easy

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Remembering +4 | -1

Element with atomic number 35 belongs to

View answer and explanation

Correct option: C

17th group

Option C (17th group) is correct because an element with atomic number 35 has 7 valence electrons, placing it in the 17th group (halogens).

Solution approach

Correct Answer + Core Reason: 17th group. Element with atomic number 35 is Bromine, which belongs to the halogen group (17th group).

Step-by-Step Solution:

Step 1: Identify the element with atomic number 35. The element with atomic number 35 is Bromine (Br).

Step 2: Determine the electronic configuration of Bromine. The electronic configuration of Bromine is [Ar] 3d10 4s2 4p5.

Step 3: Determine the number of valence electrons. Bromine has 7 valence electrons (4s2 4p5).

Step 4: Relate the number of valence electrons to the group number. Elements with 7 valence electrons belong to the 17th group (halogens).

Why other options are wrong:
- Option A (3rd group): Elements in the 3rd group have 3 valence electrons.
- Option B (18th group): Elements in the 18th group (noble gases) have 8 valence electrons (except Helium, which has 2).
- Option D (5th group): Elements in the 5th group have 5 valence electrons.

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

Memory Aids:
- Remember that the group number often corresponds to the number of valence electrons (for main group elements).

Quick Tips:
- Know the electronic configurations of common elements to determine their group number.

156 Easy

156 of 171Question permalink

Remembering +4 | -1

Atomic radii of fluorine and neon in angstrom unit are respectively given by:

View answer and explanation

Correct option: A

0.72,\mathrm{ }1.60

Option A (

0.72,\, 1.60

) is correct because fluorine is a smaller atom with a covalent radius, while neon is larger with a van der Waals radius.

Solution approach

Correct Answer + Core Reason: $0.72,\, 1.60$ . Fluorine has a smaller covalent radius, while Neon has a larger van der Waals radius.

Step-by-Step Solution:

Step 1: Understand atomic radii trends.

Step 2: Atomic radii decrease across a period due to increasing nuclear charge.

Step 3: Noble gases are an exception; their atomic radii are measured using van der Waals radii, which are larger than covalent radii.

Step 4: Fluorine is a halogen and its atomic radius is measured as a covalent radius.

Step 5: Neon is a noble gas and its atomic radius is measured as a van der Waals radius.

Step 6: Therefore, fluorine will have a smaller atomic radius than neon.

Step 7: The atomic radius of fluorine is approximately 0.72 Å.

Step 8: The atomic radius of neon is approximately 1.60 Å.

Why other options are wrong:
- Option B: ${1.60,\, 1.60}$ : Both values are the same, which is incorrect because Neon is larger than Fluorine.
- Option C: ${0.72,\, 0.72}$ : Both values are the same, which is incorrect because Neon is larger than Fluorine.
- Option D: ${1.60,\, 0.72}$ : The order is reversed. Neon should be larger than Fluorine.

Key Concepts:
- Atomic radius
- Covalent radius
- Van der Waals radius
- Periodic trends

Memory Aids:
- Van der Waals radii are larger than covalent radii.

Quick Tips:
- Remember that noble gas atomic radii are measured using van der Waals radii, which are larger than covalent radii.

157 Easy

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Understanding +4 | -1

An element whose IUPAC name is ununtrium (Uut), belongs to

View answer and explanation

Correct option: B

P – block

Option B (P – block) is correct because Ununtrium (Uut), now known as Moscovium (Mc), belongs to group 15, which is in the p-block.

Solution approach

Correct Answer + Core Reason: P – block. Ununtrium (Uut), now known as Moscovium (Mc), belongs to the p-block.

Step-by-Step Solution:

Step 1: Identify the element Ununtrium (Uut). It is now known as Moscovium (Mc) with atomic number 115.

Step 2: Determine the group to which Moscovium belongs. Moscovium belongs to group 15.

Step 3: Relate the group number to the block. Group 15 elements belong to the p-block.

Step 4: Conclude that Ununtrium belongs to the p-block.

Why other options are wrong:
- Option A: S – block elements belong to group 1 and 2. Moscovium does not belong to these groups.
- Option C: d – block elements are transition metals. Moscovium is not a transition metal.
- Option D: f – block elements are lanthanides and actinides. Moscovium is not a lanthanide or actinide.

Key Concepts:
- Periodic table blocks (s, p, d, f)
- Group numbers and their relation to blocks

Memory Aids:
- s-block: Groups 1 and 2
- p-block: Groups 13 to 18
- d-block: Groups 3 to 12
- f-block: Lanthanides and Actinides

Quick Tips:
- Knowing the group number helps determine the block to which an element belongs.

158 Easy

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Understanding +4 | -1

An element has the configuration

1{s^2},2{s^2},2{p^6},3{s^2},3{p^2}.\,

To which block in the long form of the periodic table, does this belong?

View answer and explanation

Correct option: B

p-block

Option B (p-block) is correct because the last electron enters the p-orbital in the given electronic configuration.

Solution approach

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

Step-by-Step Solution:

Step 1: The given electronic configuration is $1{s^2},2{s^2},2{p^6},3{s^2},3{p^2}$ .

Step 2: The last electron enters the 3p orbital.

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

Why other options are wrong:
- Option A (s-block): s-block elements have their last electron in the s-orbital.
- Option C (d-block): d-block elements have their last electron in the d-orbital.
- Option D (f-block): f-block elements have their last electron in the f-orbital.

Key Concepts:
- Electronic Configuration
- Periodic Table Blocks

Memory Aids:
- The block is determined by the type of orbital the last electron occupies.

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

159 Easy

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Application +4 | -1

If each orbital can take maximum of three electrons, the number of elements in the third period of the periodic table will be :

View answer and explanation

Correct option: A

12

Option A (12) is correct because if each orbital can take a maximum of three electrons, the third period will have 12 elements.

Solution approach

Correct Answer + Core Reason: 12. The third period has s and p orbitals, which can hold a total of 12 electrons if each orbital holds 3 electrons.

Step-by-Step Solution:

Step 1: Identify the orbitals present in the third period. The third period consists of the 3s and 3p orbitals.

Step 2: Determine the number of orbitals. There is one 3s orbital and three 3p orbitals, totaling 4 orbitals.

Step 3: Calculate the maximum number of electrons. If each orbital can hold 3 electrons, then 4 orbitals can hold 4 * 3 = 12 electrons.

Step 4: Relate the number of electrons to the number of elements. The number of elements in the period is equal to the maximum number of electrons that can be accommodated in the orbitals of that period. Therefore, the number of elements is 12.

Why other options are wrong:
- Option B (6): This would be the number of elements if each orbital held only 1.5 electrons, which is not the case.
- Option C (8): This is the number of elements in the third period under normal conditions (2 electrons per orbital), not when each orbital holds 3 electrons.
- Option D (24): This would be the number of elements if there were 8 orbitals (incorrect) and each held 3 electrons.

Key Concepts:
- Electronic configuration
- Orbitals and electron capacity
- Periodic table structure

Memory Aids:
- s orbital holds 1 orbital, p orbital holds 3 orbitals, d orbital holds 5 orbitals, f orbital holds 7 orbitals.

Quick Tips:
- Remember the number of orbitals in each subshell (s, p, d, f) to determine the electron capacity.

160 Easy

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Understanding +4 | -1

In the Periodic Table, going down in the fluorine group

View answer and explanation

Correct option: B

Ionic radii will increases

Option B (Ionic radii will increases) is correct because as you go down a group, the number of electron shells increases, leading to larger ionic radii.

Solution approach

Correct Answer + Core Reason: Ionic radii will increases. As you go down a group, the number of electron shells increases, leading to larger ionic radii.

Step-by-Step Solution:

Step 1: Recall the trend of ionic radii down a group in the periodic table. Ionic radii generally increase as you move down a group.

Step 2: Explain the reason for this trend. As you go down a group, the number of electron shells increases. This means that the outermost electrons are farther from the nucleus, and the ionic radius increases.

Why other options are wrong:
- Option A: Stability of hydrides will increases is incorrect because the stability of hydrides generally decreases down the group due to decreasing bond strength.
- Option C: Electronegativity will increases is incorrect because electronegativity generally decreases down the group due to increasing atomic size and decreasing effective nuclear charge.
- Option D: IE will increases is incorrect because ionization energy generally decreases down the group due to increasing atomic size and easier removal of outer electrons.

Key Concepts:
- Ionic radius trend down a group
- Electronegativity trend down a group
- Ionization energy trend down a group
- Stability of hydrides

Memory Aids:
- Down a group: atomic/ionic radius increases, electronegativity decreases, ionization energy decreases.

Quick Tips:
- Remember the general trends in the periodic table for ionic radius, electronegativity, and ionization energy.

161 Easy

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Understanding +4 | -1

The smallest among the following ions is

View answer and explanation

Correct option: D

\mathrm{A}{\mathrm{l}^{3 + }}

Option D (

\mathrm{A}{\mathrm{l}^{3 + }}

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

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{l}^{3 + }}$ . For isoelectronic species, the ion with the highest positive charge has the smallest size.

Step-by-Step Solution:

Step 1: Identify isoelectronic species. Isoelectronic species have the same number of electrons.

Step 2: Determine the number of electrons for each ion:
- $\mathrm{N}{\mathrm{a}^ + }$ : 10 electrons
- $\mathrm{M}{\mathrm{g}^{2 + }}$ : 10 electrons
- $\mathrm{B}{\mathrm{a}^{2 + }}$ : 54 electrons
- $\mathrm{A}{\mathrm{l}^{3 + }}$ : 10 electrons

Step 3: Note that $\mathrm{B}{\mathrm{a}^{2 + }}$ is not isoelectronic with the others. Compare $\mathrm{N}{\mathrm{a}^ + }$ , $\mathrm{M}{\mathrm{g}^{2 + }}$ , and $\mathrm{A}{\mathrm{l}^{3 + }}$ .

Step 4: For isoelectronic species, the ionic radius decreases with increasing nuclear charge (atomic number - number of electrons).
- Na (Z=11), Mg (Z=12), Al (Z=13)

Step 5: Therefore, the order of size is: $\mathrm{N}{\mathrm{a}^ + } > \mathrm{M}{\mathrm{g}^{2 + }} > \mathrm{A}{\mathrm{l}^{3 + }}$ .

Step 6: Conclusion: $\mathrm{A}{\mathrm{l}^{3 + }}$ is the smallest.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{a}^ + }$ has a smaller nuclear charge than $\mathrm{M}{\mathrm{g}^{2 + }}$ and $\mathrm{A}{\mathrm{l}^{3 + }}$ , so it's larger.
- Option B: $\mathrm{M}{\mathrm{g}^{2 + }}$ has a smaller nuclear charge than $\mathrm{A}{\mathrm{l}^{3 + }}$ , so it's larger.
- Option C: $\mathrm{B}{\mathrm{a}^{2 + }}$ is not isoelectronic with the other ions and is much larger due to having more electron shells.

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

Memory Aids:
- More positive charge, smaller size (for isoelectronic species).

Quick Tips:
- Identify isoelectronic species first. Then, compare nuclear charges.

162 Easy

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Analyzing +4 | -1

The least stable ion among the following is

View answer and explanation

Correct option: B

B{e^ - }

Option B (

Be^−

) is correct because adding an electron to Be, which already has a stable, filled s-orbital configuration (

2s^2

), requires significant energy.

Solution approach

Correct Answer + Core Reason: $Be^-$ . Beryllium has a stable electronic configuration of $1s^22s^2$ . Adding an electron disrupts this stability, making $Be^-$ the least stable ion.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the neutral atoms.\nLi: $1s^22s^1$ \nBe: $1s^22s^2$ \nB: $1s^22s^22p^1$ \nC: $1s^22s^22p^2$

Step 2: Write the electronic configurations of the negative ions.
$Li^-$ : $1s^22s^2$
$Be^-$ : $1s^22s^22p^1$
$B^-$ : $1s^22s^22p^2$
$C^-$ : $1s^22s^22p^3$

Step 3: Analyze the stability of the negative ions.\nAdding an electron to Be requires adding it to the 2p orbital, which is higher in energy and less stable than filling the 2s orbital. $Li^-$ achieves a noble gas configuration. $B^-$ and $C^-$ also gain stability by filling the p orbitals.

Step 4: Final Answer: $Be^-$ is the least stable ion.

Why other options are wrong:
- Option A: $Li^-$ : Lithium gains a stable $1s^22s^2$ configuration, similar to Helium.
- Option C: $B^-$ : Boron becomes more stable by filling its p orbitals.
- Option D: $C^-$ : Carbon becomes more stable by filling its p orbitals.

Key Concepts:
- Electronic configuration
- Stability of ions
- Octet rule

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

Quick Tips:
- Consider the electronic configuration of the neutral atom before adding an electron.

163 Easy

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Application +4 | -1

The formula of the binary compound formed by the element with atomic number 120 and sulphur is:

View answer and explanation

Correct option: A

XS

Option A (

XS

) is correct because element 120 belongs to the alkaline earth metals (Group 2), forming a +2 ion, while sulfur forms a -2 ion, resulting in a 1:1 ratio.

Solution approach

Correct Answer + Core Reason: $XS$ . Element 120 belongs to group 2 and forms a +2 cation, while sulfur forms a -2 anion, resulting in a 1:1 ratio.

Step-by-Step Solution:

Step 1: Determine the group of element with atomic number 120.\nElement 120 would be in the alkaline earth metals (Group 2) since it follows the pattern of filling electron shells.

Step 2: Determine the charge of the ion formed by element 120.\nGroup 2 elements form +2 ions.

Step 3: Determine the charge of the sulfide ion.\nSulfur (S) belongs to Group 16 and forms a -2 ion ( $S^{2-}$ ).

Step 4: Determine the formula of the binary compound.\nSince the charges are +2 and -2, the formula will be $XS$ .

Why other options are wrong:
- **Option B ( ${X}_2 {S}$ ): This would be the formula if X had a +1 charge and S had a -2 charge, which is not the case.
- Option C ( ${XS}_2$ ): This would be the formula if X had a +4 charge and S had a -2 charge, which is not the case.
- Option D ( $X_3 S$ ): This formula is not possible with the valencies of X (+2) and S (-2).

Key Concepts:
- Periodic Trends
- Group valency
- Formation of ionic compounds

Memory Aids:
- Group 1 elements form +1 ions, Group 2 elements form +2 ions, Group 16 elements form -2 ions, Group 17 elements form -1 ions.

Quick Tips:**
- Remember the charges of common ions to predict the formulas of ionic compounds.

164 Easy

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Understanding +4 | -1

The ionic radii of

{\mathrm{N}^{3 - }},{\mathrm{O}^{2 - }}\mathrm{and}\,{\mathrm{F}^ - }

are respectively given by:

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {1.71,\mathrm{ }1.40,\mathrm{ }1.36} \end{array}

Option C (

\begin{array}{llllllllllllllllllll}{1.71,\mathrm{ }1.40,\mathrm{ }1.36}\end{array}

) is correct because as the negative charge increases, the ionic radii increase due to greater electron repulsion, and all are isoelectronic.

Solution approach

Correct Answer + Core Reason: $\begin{array}{llllllllllllllllllll}{1.71,\mathrm{ }1.40,\mathrm{ }1.36}\end{array}$ . For isoelectronic species, ionic radii decrease with increasing nuclear charge.

Step-by-Step Solution:

Step 1: Define isoelectronic species. Isoelectronic species are atoms or ions that have the same number of electrons.

Step 2: Identify the number of electrons in each ion.
- ${\mathrm{N}^{3 - }}$ has 7 + 3 = 10 electrons.
- ${\mathrm{O}^{2 - }}$ has 8 + 2 = 10 electrons.
- ${\mathrm{F}^ - }$ has 9 + 1 = 10 electrons.

Step 3: Determine the nuclear charge of each ion.
- ${\mathrm{N}^{3 - }}$ has a nuclear charge of +7.
- ${\mathrm{O}^{2 - }}$ has a nuclear charge of +8.
- ${\mathrm{F}^ - }$ has a nuclear charge of +9.

Step 4: Relate ionic radii to nuclear charge for isoelectronic species. For isoelectronic species, the ionic radius decreases as the nuclear charge increases. This is because the increased nuclear charge pulls the electrons closer to the nucleus, reducing the ionic radius.

Step 5: Arrange the ions in order of decreasing ionic radii. Since ${\mathrm{N}^{3 - }}$ has the smallest nuclear charge, it will have the largest ionic radius. ${\mathrm{F}^ - }$ has the largest nuclear charge, so it will have the smallest ionic radius. ${\mathrm{O}^{2 - }}$ will have an intermediate ionic radius.

Step 6: Therefore, the order of ionic radii is ${\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - }$ .

Step 7: Match the order with the given options. The correct order is 1.71, 1.40, 1.36.

Why other options are wrong:
- Option A: This order is incorrect because it places ${\mathrm{F}^ - }$ as having a larger ionic radius than ${\mathrm{O}^{2 - }}$ , which is not true for isoelectronic species.
- Option B: This order is incorrect because it places ${\mathrm{O}^{2 - }}$ as having a larger ionic radius than ${\mathrm{N}^{3 - }}$ , which is not true for isoelectronic species.
- Option D: This order is incorrect because it places ${\mathrm{O}^{2 - }}$ as having a smaller ionic radius than ${\mathrm{F}^ - }$ , which is not true for isoelectronic species.

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

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

Quick Tips:
- Remember that for isoelectronic species, the ion with the most protons (highest nuclear charge) will have the smallest radius.

165 Easy

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Understanding +4 | -1

Metallic nature increases on moving down the group because

View answer and explanation

Correct option: C

Both 1 and 2

Option C is correct because as you move down a group, the nuclear charge increases, but the shielding effect also increases, leading to a weaker effective nuclear charge and easier loss of electrons, hence increasing metallic nature.

Solution approach

Correct Answer + Core Reason: Both 1 and 2. Metallic nature increases down the group due to increased shielding and nuclear charge.

Step-by-Step Solution:

Step 1: Define metallic nature: The tendency of an element to lose electrons and form positive ions.

Step 2: Nuclear charge increases down the group because the number of protons increases.

Step 3: Shielding increases down the group because the number of inner electron shells increases.

Step 4: Increased shielding reduces the effective nuclear charge experienced by the valence electrons.

Step 5: A weaker effective nuclear charge makes it easier for the valence electrons to be lost, increasing metallic nature.

Why other options are wrong:
- Option A: Nuclear charge increase alone would make it harder to lose electrons, decreasing metallic nature.
- Option B: Shielding increases alone would make it easier to lose electrons, increasing metallic nature, but the nuclear charge also plays a role.
- Option D: Both factors contribute to the increase in metallic nature.

Key Concepts:
- Metallic nature
- Nuclear charge
- Shielding effect
- Effective nuclear charge

Memory Aids:
- Remember that increased shielding weakens the pull of the nucleus on valence electrons.

Quick Tips:
- Consider both nuclear charge and shielding when evaluating trends in metallic nature.

166 Easy

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Understanding +4 | -1

The electronic configuration of most electronegative elements is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because this electronic configuration corresponds to a halogen, which are the most electronegative elements.

Solution approach

Correct Answer + Core Reason: $1{s^2},2{s^2},2{p^5}$ . This configuration corresponds to a halogen.

Step-by-Step Solution:

Step 1: Recall the electronic configurations of highly electronegative elements.\nHighly electronegative elements are halogens, which have 7 valence electrons.

Step 2: Analyze the given electronic configurations.
- $1{s^2},2{s^2},2{p^5}$ has 7 valence electrons (2+5).
- $1{s^2},2{s^2},2{p^4},3{s^1}$ has 7 valence electrons (4+1+2), but it's not a typical halogen configuration.
- $1{s^2},2{s^2},2{p^6},3{s^1},3{p^1}$ has 2 valence electrons (1+1).
- $1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}$ has 7 valence electrons (2+5), but it's in the third period.

Step 3: Conclusion.\nTherefore, the electronic configuration of most electronegative elements is $1{s^2},2{s^2},2{p^5}$ .

Why other options are wrong:
- Option B: $1{s^2},2{s^2},2{p^4},3{s^1}$ is not a typical halogen configuration.
- Option C: $1{s^2},2{s^2},2{p^6},3{s^1},3{p^1}$ has only 2 valence electrons.
- Option D: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^5}$ is a halogen but not the most electronegative.

Key Concepts:
- Electronegativity
- Electronic configuration
- Halogens

Memory Aids:
- Halogens have 7 valence electrons and are highly electronegative.

Quick Tips:
- Remember the electronic configurations of halogens.

167 Easy

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Analyzing +4 | -1

The ionization energy will be maximum for the process:

View answer and explanation

Correct option: B

\mathrm{Be} \to \mathrm{B}{\mathrm{e}^{2 + }}

Option B (

\mathrm{Be} \to \mathrm{B}{\mathrm{e}^{2 + }}

) is correct because removing two electrons from Be requires significantly more energy due to its small size and relatively high nuclear charge compared to the other options.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be} \to \mathrm{B}{\mathrm{e}^{2 + }}$ . Beryllium has a smaller size and higher effective nuclear charge compared to Ba, Cs, and Li, making it require maximum ionization energy to remove two electrons.

Step-by-Step Solution:

Step 1: Analyze the ionization processes.\nBa $\to$ Ba $^{2+}$ : Removing two electrons from Barium.\nBe $\to$ Be $^{2+}$ : Removing two electrons from Beryllium.\nCs $\to$ Cs $^+$ : Removing one electron from Cesium.\nLi $\to$ Li $^+$ : Removing one electron from Lithium.

Step 2: Consider the size and nuclear charge of the elements.\nBeryllium is smaller and has a higher nuclear charge compared to Barium, Cesium, and Lithium.

Step 3: Removing two electrons from a smaller atom with a higher nuclear charge requires more energy than removing one electron from larger atoms with lower nuclear charges.

Why other options are wrong:
- Option A: $\mathrm{Ba} \to \mathrm{B}{\mathrm{a}^{2 + }}$ is incorrect because Barium is a larger atom with a lower nuclear charge compared to Beryllium. Therefore, it requires less energy to remove two electrons.
- Option C: $\mathrm{Cs} \to \mathrm{C}{\mathrm{s}^ + }$ is incorrect because Cesium is a large atom with a low nuclear charge. It requires the least amount of energy to remove one electron.
- Option D: $\mathrm{Li} \to \mathrm{L}{\mathrm{i}^ + }$ is incorrect because Lithium is larger and has a lower nuclear charge compared to Beryllium. Therefore, it requires less energy to remove one electron.

Key Concepts:
- Ionization Energy
- Atomic Size
- Nuclear Charge

Memory Aids:
- Smaller atoms with higher nuclear charges have higher ionization energies.

Quick Tips:
- Compare the size and nuclear charge of the elements when determining ionization energy.

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Analyzing +4 | -1

Which of the following species has highest ionisation energy

View answer and explanation

Correct option: D

M{g^{2 + }}

Option A is incorrect because ionic radii decrease with increasing nuclear charge for isoelectronic species. The correct order should be Sc3+ > Mn3+ > Fe3+ > Cr3+.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}{\mathrm{c}^{3 + }} > C{\mathrm{r}^{3 + }} > F{\mathrm{e}^{3 + }} > M{\mathrm{n}^{3 + }} -$ ionic radii. The order of ionic radii is incorrect because it doesn't consider the effect of increasing nuclear charge on isoelectronic species.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations and charges of the ions:
- Sc³⁺: [Ar]
- Cr³⁺: [Ar] 3d³
- Fe³⁺: [Ar] 3d⁵
- Mn³⁺: [Ar] 3d⁴

Step 2: Consider the isoelectronic nature: Sc³⁺ has a noble gas configuration. The others are not isoelectronic.

Step 3: Analyze the trends in ionic radii: For isoelectronic species, ionic radii decrease with increasing nuclear charge. However, since Sc³⁺ is not isoelectronic with the others, we must consider the effect of d-electrons.

Step 4: Evaluate Option A: $\mathrm{S}{\mathrm{c}^{3 + }} > C{\mathrm{r}^{3 + }} > F{\mathrm{e}^{3 + }} > M{\mathrm{n}^{3 + }} -$ ionic radii. This order is incorrect. The correct order should be Sc³⁺ > Mn³⁺ > Fe³⁺ > Cr³⁺. Higher the nuclear charge, smaller the ionic radii for isoelectronic species. Since these are not all isoelectronic, the trend is more complex, but the given order is still incorrect.

Step 5: Evaluate Option B: $\mathrm{Sc} < Ti < Cr < Mn -$ density. This order is generally correct as density increases with increasing atomic mass and decreasing atomic radius.

Step 6: Evaluate Option C: $\mathrm{M}{\mathrm{n}^{2 + }} > N{\mathrm{i}^{2 + }} > C{\mathrm{o}^{2 + }} < F{\mathrm{e}^{2 + }} -$ ionic radii. This order is generally correct.

Step 7: Evaluate Option D: $\mathrm{FeO} < CaO < MnO < CuO -$ basic nature. This order is generally correct as basic nature increases with decreasing electronegativity of the metal.

Step 8: Identify the incorrect order: The incorrect order is Option A.

Why other options are wrong:
- Option B: $\mathrm{Sc} < Ti < Cr < Mn -$ density: This order is generally correct.
- Option C: $\mathrm{M}{\mathrm{n}^{2 + }} > N{\mathrm{i}^{2 + }} > C{\mathrm{o}^{2 + }} < F{\mathrm{e}^{2 + }} -$ ionic radii: This order is generally correct.
- Option D: $\mathrm{FeO} < CaO < MnO < CuO -$ basic nature: This order is generally correct.

Key Concepts:
- Ionic Radii
- Density
- Basic Nature of Oxides
- Periodic Trends

Memory Aids:
- For isoelectronic species, ionic radii decrease with increasing nuclear charge.

Quick Tips:
- Be careful when comparing ionic radii of non-isoelectronic species. Consider the electronic configurations and effective nuclear charge.

169 Easy

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Understanding +4 | -1

Which has the largest first ionisation energy?

View answer and explanation

Correct option: A

Li

Option A (Li) is correct because ionization energy decreases down a group, and Li is the first element in the group.

Solution approach

Correct Answer + Core Reason: Li. Ionization energy decreases down a group.

Step-by-Step Solution:
Step 1: Recall the trend of ionization energy within a group. Ionization energy generally decreases as you move down a group in the periodic table.
Step 2: Identify the positions of the given elements in the periodic table. Li, Na, K, and Rb are all alkali metals (Group 1), with Li at the top and Rb at the bottom.
Step 3: Apply the trend. Since ionization energy decreases down the group, Li will have the highest ionization energy among the given elements.

Why other options are wrong:
- Option B: Na: Incorrect, Na is below Li in Group 1, so it has a lower ionization energy.
- Option C: K: Incorrect, K is below Na in Group 1, so it has a lower ionization energy.
- Option D: Rb: Incorrect, Rb is the lowest in Group 1, so it has the lowest ionization energy.

Key Concepts:
- Ionization energy
- Periodic trends
- Group 1 elements (alkali metals)

Memory Aids:
- Ionization energy decreases down a group.

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

170 Easy

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Application +4 | -1

The electronegativity difference between two atoms A and B is 2, the percentage of covalent

View answer and explanation

Correct option: A

54\%

Option A (

54\%

) is correct. The percentage of ionic character can be estimated using the Hannay-Smith equation: % ionic character = 16 * (electronegativity difference) + 3.5 * (electronegativity difference)².

Solution approach

Correct Answer + Core Reason: $54\%$ . The percentage of ionic character can be estimated using the Hannay-Smith equation.

Step-by-Step Solution:

Step 1: Recall the Hannay-Smith equation for estimating the percentage of ionic character:
$\text{{Percentage ionic character}} = 16 \times |{{\chi _A} - {\chi _B}}| + 3.5 \times |{{\chi _A} - {\chi _B}}{|^2}$ \nWhere ${\chi _A}$ and ${\chi _B}$ are the electronegativities of atoms A and B.

Step 2: In this problem, the electronegativity difference between A and B is 2, so $|{{\chi _A} - {\chi _B}}| = 2$ .

Step 3: Substitute the given value into the Hannay-Smith equation:
$\text{{Percentage ionic character}} = 16 \times 2 + 3.5 \times {2^2}$

Step 4: Calculate the percentage ionic character:
$\text{{Percentage ionic character}} = 32 + 3.5 \times 4 = 32 + 14 = 46\%$

Step 5: The question asks for the percentage of *covalent* character. Since the total percentage is 100%, the percentage of covalent character is:
$100\% - 46\% = 54\%$

Why other options are wrong:
- Option B: $46\%$ - This is the percentage of ionic character, not covalent character.
- Option C: $23\%$ - Incorrect calculation.
- Option D: $72\%$ - Incorrect calculation.

Key Concepts:
- Electronegativity
- Ionic Character
- Covalent Character
- Hannay-Smith Equation

Memory Aids:
- Remember the Hannay-Smith equation for estimating the percentage of ionic character.

Quick Tips:
- Be familiar with the Hannay-Smith equation and its application.

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Remembering +4 | -1

The elements with atomic numbers 58 to 71 are called

View answer and explanation

Correct option: C

Lanthanides

Option C (Lanthanides) is correct because elements with atomic numbers 58 to 71 are defined as Lanthanides.

Solution approach

Correct Answer + Core Reason: Lanthanides. Elements with atomic numbers 58 to 71 are defined as lanthanides.

Step-by-Step Solution:

Step 1: Recall the definition of Lanthanides: These are elements with atomic numbers 58 (Cerium) to 71 (Lutetium).

Step 2: Consider the other options:
- Normal elements: A vague term, not a specific group.
- Transition elements: Elements in the d-block of the periodic table.
- Actinides: Elements with atomic numbers 90 to 103.

Step 3: Conclusion: The elements with atomic numbers 58 to 71 are Lanthanides.

Why other options are wrong:
- Option A (Normal elements): This is not a specific classification of elements.
- Option B (Transition elements): Transition elements are in the d-block, not the f-block where lanthanides are found.
- Option D (Actinides): Actinides have atomic numbers 90 to 103.

Key Concepts:
- Lanthanides
- Actinides
- Periodic Table

Memory Aids:
- Lanthanides follow Lanthanum (La) in the periodic table.

Quick Tips:
- Know the positions and definitions of lanthanides and actinides in the periodic table.

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