CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Easy, Page 3

61 Easy

Application +4 | -1

Find the element having outer electronic configuration

(n-1) d^2 n s^2

for

n=4

.

View answer and explanation

Correct option: C

Titanium

Option C (Titanium) is correct because Titanium (Ti) has an atomic number of 22, and its electronic configuration matches the given formula for n=4.

Solution approach

Correct Answer + Core Reason: Titanium. The electronic configuration of Titanium matches the given general configuration for n=4.

Step-by-Step Solution:

Step 1: Understand the general electronic configuration: $(n-1)d^2 ns^2$ .

Step 2: Substitute n=4: $(4-1)d^2 4s^2 = 3d^2 4s^2$ .

Step 3: Identify the element with the electronic configuration $3d^2 4s^2$ .

Step 4: Titanium (Ti) has an atomic number of 22. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2$ .

Step 5: Therefore, Titanium matches the given configuration.

Why other options are wrong:
- Option A (Scandium): Scandium (Sc) has an electronic configuration of $3d^1 4s^2$ , which does not match the $3d^2 4s^2$ configuration.
- Option B (Vanadium): Vanadium (V) has an electronic configuration of $3d^3 4s^2$ , which does not match the $3d^2 4s^2$ configuration.
- Option D (Tungsten): Tungsten (W) has a more complex electronic configuration and belongs to a later period, so it doesn't fit the $3d^2 4s^2$ configuration.

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

Memory Aids:
- Remember the order of filling orbitals: s, p, d, f.

Quick Tips:
- Knowing the atomic numbers and general electronic configurations of the first few d-block elements helps in quickly identifying the correct element.

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Analyzing +4 | -1

Which of the following species show maximum ionisation energy

View answer and explanation

Correct option: B

N{e^ + }

Option B (

N{e^ + }

) exhibits the maximum ionization energy because it has already lost one electron, resulting in a higher effective nuclear charge and making it more difficult to remove another electron.

Solution approach

Correct Answer + Core Reason: $Ne^+$ . Removing an electron from a positive ion requires more energy than removing an electron from a neutral atom due to increased effective nuclear charge.

Step-by-Step Solution:

Step 1: Define ionization energy.\nIonization energy is the energy required to remove an electron from a gaseous atom or ion.

Step 2: Consider the effective nuclear charge.\nThe effective nuclear charge experienced by an electron in $Ne^+$ is greater than that in $Ne$ , $F$ , or $F^-$ because it has already lost an electron.

Step 3: Analyze the given species.\nNe: Neutral neon atom.\nNe $^+$ : Neon ion with a +1 charge.\nF: Neutral fluorine atom.\nF $^-$ : Fluoride ion with a -1 charge.

Step 4: Removing an electron from a positive ion requires more energy than removing an electron from a neutral atom or a negative ion.

Why other options are wrong:
- Option A: $Ne$ is incorrect because it is a neutral atom, and removing an electron from a positive ion requires more energy.
- Option C: $F$ is incorrect because it is a neutral atom, and removing an electron from a positive ion requires more energy.
- Option D: ${F^ - }$ is incorrect because it is a negative ion, and removing an electron from a positive ion requires more energy.

Key Concepts:
- Ionization Energy
- Effective Nuclear Charge

Memory Aids:
- Positive ions have higher ionization energies than neutral atoms or negative ions.

Quick Tips:
- Consider the charge of the species when determining ionization energy.

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Remembering +4 | -1

Decreasing electro negativities of

F,Cl,Br

and I is

View answer and explanation

Correct option: B

F > Cl > Br > I

Option B (

F > Cl > Br > I

) is correct because electronegativity decreases down the halogen group.

Solution approach

Correct Answer + Core Reason: $F > Cl > Br > I$ . Electronegativity decreases down the halogen group.

Step-by-Step Solution:

Step 1: Recall the electronegativity trend in the halogen group.\nElectronegativity decreases down a group.

Step 2: Consider the positions of the halogens in the periodic table.
- F, Cl, Br, and I are all in group 17 (halogens).
- Fluorine is at the top of the group, followed by Chlorine, Bromine, and Iodine.

Step 3: Conclusion.\nTherefore, the decreasing order of electronegativity is $F > Cl > Br > I$ .

Why other options are wrong:
- Option A: $F < Cl < Br < I$ is incorrect because electronegativity decreases down the group.
- Option C: $Br < I < Cl < F$ is incorrect because electronegativity decreases down the group.
- Option D: $I < Br < Cl < F$ is incorrect because electronegativity decreases down the group.

Key Concepts:
- Electronegativity
- Halogens
- Periodic trends

Memory Aids:
- Electronegativity decreases down a group.

Quick Tips:
- Remember the electronegativity trend within the halogen group.

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Understanding +4 | -1

The screening effect of d-electrons is:

View answer and explanation

Correct option: D

Less than p-electrons

Option D (Less than p-electrons) is correct because d-electrons have a more diffused shape and are less effective at shielding the nuclear charge compared to p-electrons.

Solution approach

Correct Answer + Core Reason: Less than p-electrons. d-electrons have a more diffused shape, leading to less effective shielding.

Step-by-Step Solution:
Step 1: Understand the concept of the screening effect. The screening effect refers to the ability of inner electrons to shield outer electrons from the full nuclear charge.
Step 2: Consider the shapes of electron orbitals. s-electrons are most effective at shielding, followed by p-electrons, then d-electrons, and finally f-electrons.
Step 3: Apply the concept to the question. Since d-electrons are more diffused than p-electrons, they are less effective at shielding the nuclear charge.

Why other options are wrong:
- Option A: Equal to the p-electrons: This is incorrect because p-electrons are more effective at shielding than d-electrons.
- Option B: Much more than p-elecrons: This is incorrect because p-electrons are more effective at shielding than d-electrons.
- Option C: Same as f-electrons: This is incorrect because f-electrons are the least effective at shielding.
- Option D: Less than p-electrons: This is the correct answer.

Key Concepts:
- Screening effect (shielding effect)
- Atomic orbitals (s, p, d, f)
- Effective nuclear charge

Memory Aids:
- Shielding effectiveness: s > p > d > f

Quick Tips:
- Remember the order of shielding effectiveness based on the shape of the orbitals.

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Remembering +4 | -1

Which is the general outer electronic configuration of the coinage metals?

View answer and explanation

Correct option: B

\left( {n - 1} \right){d^{10}}n{s^1}

Option B is correct because coinage metals (Cu, Ag, Au) have a general outer electronic configuration of (n-1)d¹⁰ns¹ due to the stability of the fully filled d-orbital.

Solution approach

Correct Answer + Core Reason: $\left( {n - 1} \right){d^{10}}n{s^1}$ . Coinage metals have a completely filled d-orbital and one s-electron in their outermost shell.

Step-by-Step Solution:

Step 1: Identify the coinage metals.\nThe coinage metals are copper (Cu), silver (Ag), and gold (Au).

Step 2: Determine the electronic configurations of the coinage metals.
- Copper (Cu): [Ar] 3d¹⁰4s¹
- Silver (Ag): [Kr] 4d¹⁰5s¹
- Gold (Au): [Xe] 4f¹⁴5d¹⁰6s¹

Step 3: Generalize the outer electronic configuration.\nThe general outer electronic configuration of the coinage metals is (n-1)d¹⁰ns¹, where n is the principal quantum number of the outermost shell.

Why other options are wrong:
- Option A: Incorrect because $n{s^2}n{p^6}$ represents the electronic configuration of noble gases.
- Option B: Correct because $\left( {n - 1} \right){d^{10}}n{s^1}$ represents the electronic configuration of coinage metals.
- Option C: Incorrect because $\left( {n - 1} \right){d^{10}}n{s^2}$ does not represent the electronic configuration of coinage metals.
- Option D: Incorrect because $\left( {n - 1} \right){d^9}n{s^2}$ does not represent the electronic configuration of coinage metals.

Key Concepts:
- Electronic configuration
- Coinage metals

Memory Aids:
- Remember that coinage metals have a completely filled d-orbital and one s-electron.

Quick Tips:
- Review the electronic configurations of transition metals and their exceptions.

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Remembering +4 | -1

An element with atomic number 21 is a

View answer and explanation

Correct option: C

Transition element

Option C (Transition element) is correct because elements with atomic number 21 (Scandium) belong to the transition metal series.

Solution approach

Correct Answer + Core Reason: Transition element. Elements with atomic number 21 are transition metals.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 21: The element with atomic number 21 is Scandium (Sc).

Step 2: Determine the element's classification: Scandium is a transition metal because it has a partially filled d-orbital.

Step 3: Conclusion: The correct answer is Transition element.

Why other options are wrong:
- Option A: Halogen: Halogens are in Group 17 of the periodic table.
- Option B: Representative element: Representative elements are in Groups 1, 2, and 13-17, excluding transition metals.
- Option D: Alkali metal: Alkali metals are in Group 1 of the periodic table.

Key Concepts:
- Periodic Table classification
- Transition metals

Memory Aids:
- Remember that transition metals are located in the d-block of the periodic table.

Quick Tips:
- Know the general locations of different types of elements (alkali, alkaline earth, transition, halogens, noble gases) in the periodic table.

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Application +4 | -1

Which of the following combinations contain only iso electronic ions

View answer and explanation

Correct option: C

{P^{3 - }},{S^{2 - }},C{l^ - },Ar

Option C (

{P^{3 - }},{S^{2 - }},C{l^ - },Ar

) is correct because all the ions and the atom have 18 electrons each, making them isoelectronic.

Solution approach

Correct Answer + Core Reason: ${P^{3 - }},{S^{2 - }},C{l^ - },Ar$ . All species have 18 electrons.

Step-by-Step Solution:

Step 1: Calculate the number of electrons for each species in each option:
- A. ${N^{3 - }},{O^{2 - }},C{l^ - },Ne$ : 7+3=10, 8+2=10, 17+1=18, 10. Not isoelectronic.
- B. ${F^ - },Ar,{S^{2 - }},C{l^ - }$ : 9+1=10, 18, 16+2=18, 17+1=18. Not isoelectronic.
- C. ${P^{3 - }},{S^{2 - }},C{l^ - },Ar$ : 15+3=18, 16+2=18, 17+1=18, 18. Isoelectronic.
- D. ${N^{3 - }},{F^ - },{O^{2 - }},Ar$ : 7+3=10, 9+1=10, 8+2=10, 18. Not isoelectronic.

Step 2: Identify the combination with the same number of electrons for all species.

Why other options are wrong:
- Option A: ${N^{3 - }},{O^{2 - }},C{l^ - },Ne$ : Not all species have the same number of electrons.
- Option B: ${F^ - },Ar,{S^{2 - }},C{l^ - }$ : Not all species have the same number of electrons.
- Option D: ${N^{3 - }},{F^ - },{O^{2 - }},Ar$ : Not all species have the same number of electrons.

Key Concepts:
- Isoelectronic species.
- Calculation of number of electrons in ions.

Memory Aids:
- Anions gain electrons, cations lose electrons.

Quick Tips:
- Add the number of electrons gained to the atomic number for anions.
- Subtract the number of electrons lost from the atomic number for cations.

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Remembering +4 | -1

Which of the following element is metalloid?

View answer and explanation

Correct option: C

Ge

Option C is correct. Germanium (Ge) is a metalloid, possessing properties of both metals and nonmetals.

Solution approach

Correct Answer + Core Reason: Ge. Germanium is a metalloid, exhibiting properties intermediate between metals and nonmetals.

Step-by-Step Solution:

Step 1: Recall the definition of metalloids: Metalloids (also known as semi-metals) are elements that have properties of both metals and nonmetals.

Step 2: Identify the metalloids in the periodic table: Common metalloids include Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), and Polonium (Po).

Step 3: Check the given options:\nA. Bi - Bismuth is a metal.\nB. Sn - Tin is a metal.\nC. Ge - Germanium is a metalloid.\nD. C - Carbon is a nonmetal.

Why other options are wrong:
- Option A: Bismuth (Bi) is a metal.
- Option B: Tin (Sn) is a metal.
- Option D: Carbon (C) is a nonmetal.

Key Concepts:
- Metalloids (Semi-metals)
- Periodic Table

Memory Aids:
- Remember the common metalloids: B, Si, Ge, As, Sb, Te, Po.

Quick Tips:
- Know the location of metalloids in the periodic table (along the staircase line).

69 Easy

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Understanding +4 | -1

An element with atomic number of 50 belongs to

View answer and explanation

Correct option: B

p – block

Option B (p – block) is correct because an element with atomic number 50 is Tin, which is a p-block element.

Solution approach

Correct Answer + Core Reason: p – block. An element with atomic number 50 is Tin (Sn), which is a p-block element.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 50. The element with atomic number 50 is Tin (Sn).

Step 2: Determine the electronic configuration of Tin. The electronic configuration of Tin is [Kr] 4d10 5s2 5p2.

Step 3: Identify the block to which Tin belongs. Since the last electron enters the p-orbital, Tin belongs to the p-block.

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

Key Concepts:
- Electronic configuration
- p-block elements

Memory Aids:
- Remember the location of the p-block in the periodic table.

Quick Tips:
- Know the general electronic configurations of elements in each block to quickly identify the block to which an element belongs.

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Remembering +4 | -1

Ionization potential is lowest for:

View answer and explanation

Correct option: D

Alkali metals

Option D (Alkali metals) is correct because alkali metals have the lowest ionization potential due to their electronic configuration.

Solution approach

Correct Answer + Core Reason: Alkali metals. Alkali metals have the lowest ionization potential due to their electronic configuration of $ns^1$ .

Step-by-Step Solution:

Step 1: Recall the definition of ionization potential (IP). Ionization potential is the energy required to remove an electron from an isolated gaseous atom or ion.

Step 2: Consider the electronic configurations of the given options:
- Halogens: $ns^2np^5$ (high effective nuclear charge, strong attraction for electrons)
- Inert gases: $ns^2np^6$ (stable, completely filled valence shell)
- Alkaline earth metals: $ns^2$ (relatively stable, but less than inert gases)
- Alkali metals: $ns^1$ (easily lose one electron to achieve a stable noble gas configuration)

Step 3: Explain why alkali metals have the lowest IP. Alkali metals have only one electron in their outermost shell ( $ns^1$ ). This electron is weakly held by the nucleus due to the shielding effect of the inner electrons. Therefore, it is relatively easy to remove this electron, resulting in a low ionization potential.

Why other options are wrong:
- Option A: Halogens is incorrect because halogens have high ionization potentials due to their high effective nuclear charge and strong attraction for electrons.
- Option B: Inert gases is incorrect because inert gases have very high ionization potentials due to their stable, completely filled valence shell.
- Option C: Alkaline earth metals is incorrect because alkaline earth metals have higher ionization potentials than alkali metals due to their $ns^2$ configuration and stronger nuclear attraction.

Key Concepts:
- Ionization potential
- Electronic configuration
- Effective nuclear charge
- Shielding effect

Memory Aids:
- Alkali metals easily lose one electron.

Quick Tips:
- Remember that ionization potential generally increases across a period and decreases down a group.

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Understanding +4 | -1

The element with atomic number 36 belongs to …block in the Periodic Table.

View answer and explanation

Correct option: A

p

Option A (p) is correct because elements with atomic number 36 (Krypton) belongs to the p-block of the periodic table.

Solution approach

Correct Answer + Core Reason: p. Elements with atomic number 36 (Krypton) belong to the p-block.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 36. It is Krypton (Kr).

Step 2: Determine the electronic configuration of Krypton. The electronic configuration is $[Ar] 4s^2 3d^{10} 4p^6$ .

Step 3: Identify the last electron entering the orbital. The last electron enters the p-orbital.

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

Why other options are wrong:
- Option B: s-block elements have their last electron entering the s-orbital. Krypton's last electron enters the p-orbital.
- Option C: f-block elements have their last electron entering the f-orbital. Krypton's last electron enters the p-orbital.
- Option D: d-block elements have their last electron entering the d-orbital. Krypton's last electron enters the p-orbital.

Key Concepts:
- Electronic configuration
- Periodic table blocks (s, p, d, f)

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 electronic configuration helps determine the block to which an element belongs.

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Analyzing +4 | -1

Among the following elements, the one having the highest ionisation energy is

View answer and explanation

Correct option: C

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

Option C is correct because half-filled p orbitals (

p^3

) have extra stability, leading to a higher ionization energy.

Solution approach

Correct Answer + Core Reason: $\left[ {Ne} \right]\,3{s^2}\,3{p^3}$ . Half-filled p orbitals have extra stability, requiring more energy to remove an electron.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations:
A. $\left[ {Ar} \right]\,3{d^{10}}\,4{s^2}\,4{p^3}$ - This has a half-filled $4p$ subshell.
B. $\left[ {Ne} \right]\,3{s^2}\,3{p^1}$ - This has one electron in the $3p$ subshell.
C. $\left[ {Ne} \right]\,3{s^2}\,3{p^3}$ - This has a half-filled $3p$ subshell.
D. $\left[ {Ne} \right]\,3{s^2}\,3{p^2}$ - This has two electrons in the $3p$ subshell.

Step 2: Identify the most stable configuration: Half-filled and fully filled electronic configurations are more stable and require more energy to remove an electron.

Step 3: Compare the options: Option C has a half-filled $3p$ subshell, making it more stable than options B and D. Option A has a half-filled 4p subshell, but the question asks among the *following* elements, implying a focus on the Ne core options.

Step 4: Conclusion: Therefore, the element with the highest ionization energy is the one with the half-filled $3p$ subshell.

Why other options are wrong:
- Option A: $\left[ {Ar} \right]\,3{d^{10}}\,4{s^2}\,4{p^3}$ : While it has a half-filled p orbital, it's not among the elements with the Neon core, which the question seems to be implying.
- Option B: $\left[ {Ne} \right]\,3{s^2}\,3{p^1}$ : This has only one electron in the p orbital, making it easier to remove compared to a half-filled orbital.
- Option D: $\left[ {Ne} \right]\,3{s^2}\,3{p^2}$ : This has two electrons in the p orbital, making it easier to remove compared to a half-filled orbital.

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

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

Quick Tips:
- Look for half-filled or fully filled electronic configurations when comparing ionization energies.

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Remembering +4 | -1

Which group of periodic table contains no metal?

View answer and explanation

Correct option: C

VIIA

Option C (VIIA) is correct because Group VIIA (Halogens) consists entirely of non-metals.

Solution approach

Correct Answer + Core Reason: VIIA. Group VIIA (halogens) contains only non-metals.

Step-by-Step Solution:

Step 1: Analyze the groups:
- IA (Alkali metals): Contains metals (except Hydrogen).
- IIIA: Contains metals, metalloids, and non-metals.
- VIIA (Halogens): Contains only non-metals (Fluorine, Chlorine, Bromine, Iodine, Astatine).
- VIII (Noble gases): Contains non-metals.

Step 2: Identify the group with no metals: Group VIIA (Halogens) has no metals.

Why other options are wrong:
- Option A (IA): Group IA (Alkali metals) contains metals.
- Option B (IIIA): Group IIIA contains metals, metalloids, and non-metals.
- Option D (VIII): Group VIII (Noble gases) contains non-metals, but the question asks for a group that *contains no metal*, and VIIA is a better answer as it contains only non-metals.

Key Concepts:
- Periodic Table
- Metals, Non-metals, and Metalloids
- Group Properties

Memory Aids:
- Halogens are all non-metals.

Quick Tips:
- Know the general properties of different groups in the periodic table.

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Remembering +4 | -1

What would be the IUPAC name for the element with atomic number 115

View answer and explanation

Correct option: C

UuP

Option C (UuP) is correct because the IUPAC symbol for element 115 (Ununpentium) is Uup, which is often shortened to UuP.

Solution approach

Correct Answer + Core Reason: UuP. The IUPAC symbol for element 115 is Uup, often shortened to UuP.

Step-by-Step Solution:

Step 1: Determine the IUPAC name for element 115: The element with atomic number 115 is Ununpentium.

Step 2: Determine the IUPAC symbol: The IUPAC symbol is formed from the first letters of the numerical roots: Un-un-pentium, so Uup.

Step 3: Common abbreviation: The symbol Uup is sometimes abbreviated as UuP.

Step 4: Conclusion: The correct answer is UuP.

Why other options are wrong:
- Option A: Uun: This is the symbol for element 110 (Ununnillium).
- Option B: Uub: This is the symbol for element 112 (Ununbium).
- Option D: Uus: This is the symbol for element 117 (Ununseptium).

Key Concepts:
- IUPAC nomenclature
- Element symbols

Memory Aids:
- Remember the numerical roots used in IUPAC nomenclature: 0-nil, 1-un, 2-bi, 3-tri, 4-quad, 5-pent, 6-hex, 7-sept, 8-oct, 9-enn.

Quick Tips:
- Memorize the IUPAC nomenclature rules for naming elements with atomic numbers greater than 100.

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Remembering +4 | -1

Atomic radii of inert gases are expressed in terms of

View answer and explanation

Correct option: B

Vanderwaal’s radius

Option B (Vanderwaal’s radius) is correct because inert gases do not form covalent or metallic bonds, so their atomic radii are measured using van der Waals radii.

Solution approach

Correct Answer + Core Reason: Vanderwaal’s radius. Inert gases do not form covalent or metallic bonds, so their radii are measured using van der Waals forces.

Step-by-Step Solution:

Step 1: Understand the different types of atomic radii.

Step 2: Covalent radius: Half the distance between the nuclei of two covalently bonded atoms.

Step 3: Metallic radius: Half the distance between the nuclei of two adjacent atoms in a metallic crystal.

Step 4: Van der Waals radius: Half the distance between the nuclei of two non-bonded atoms in solid.

Step 5: Inert gases do not form covalent or metallic bonds.

Step 6: Therefore, their atomic radii are expressed in terms of van der Waals radius.

Why other options are wrong:
- Option A: Covalent radius: Inert gases do not form covalent bonds.
- Option C: Metallic radius: Inert gases do not form metallic bonds.
- Option D: All of the above: Only van der Waals radius is applicable to inert gases.

Key Concepts:
- Atomic radius
- Covalent radius
- Metallic radius
- Van der Waals radius
- Inert gases

Memory Aids:
- Inert gases use van der Waals radius.

Quick Tips:
- Remember that inert gases do not form chemical bonds, so their radii are measured using van der Waals forces.

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Remembering +4 | -1

Maximum oxidation state (+8) is exhibited by

View answer and explanation

Correct option: B

Ru and OS

Option B (Ru and OS) is correct because Ruthenium (Ru) and Osmium (Os) can exhibit a +8 oxidation state.

Solution approach

Correct Answer + Core Reason: Ru and OS. Ruthenium and Osmium are known to exhibit a +8 oxidation state.

Step-by-Step Solution:

Step 1: Oxidation state refers to the charge an atom would have if all bonds were ionic.

Step 2: Osmium (Os) and Ruthenium (Ru) are transition metals that can achieve a +8 oxidation state due to their electronic configurations and ability to form multiple bonds with oxygen and fluorine.

Step 3: Other transition metals and nonmetals generally do not reach such a high oxidation state.

Why other options are wrong:
- Option A (Co and Ni): Cobalt (Co) and Nickel (Ni) typically exhibit oxidation states up to +2 or +3, but not +8.
- Option C (Cl and I): Chlorine (Cl) and Iodine (I) are halogens that can have oxidation states up to +7, but not +8.
- Option D (Te and I): Tellurium (Te) and Iodine (I) do not reach +8 oxidation state.

Key Concepts:
- Oxidation State
- Transition Metals
- Electronic Configuration

Memory Aids:
- Remember that Osmium and Ruthenium are exceptions and can exhibit a +8 oxidation state.

Quick Tips:
- Familiarize yourself with the common oxidation states of various elements, especially transition metals.

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Understanding +4 | -1

Which of the following sequence regarding ionisation potential of coinage metal is correct:

View answer and explanation

Correct option: C

\mathrm{Cu} > Ag < Au

Option C (

\mathrm{Cu} > Ag < Au

) is correct because ionization potential generally decreases down a group, but gold's relativistic effects cause it to have a higher ionization potential than silver.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cu} > Ag < Au$ . The ionization potential of Au is higher than Ag due to relativistic effects.

Step-by-Step Solution:

Step 1: Define ionization potential: Ionization potential is the energy required to remove an electron from a neutral gaseous atom.

Step 2: General trend: Ionization potential generally decreases down a group due to increasing atomic size and shielding effect.

Step 3: Coinage metals (Cu, Ag, Au) belong to Group 11 of the periodic table.

Step 4: Based on the general trend, we would expect the ionization potential to decrease from Cu to Ag to Au. However, relativistic effects in gold cause its outermost electrons to be more tightly bound to the nucleus, resulting in a higher ionization potential than silver.

Step 5: Therefore, the correct order is $\mathrm{Cu} > Ag < Au$ .

Why other options are wrong:
- Option A: $\mathrm{Cu} > Ag > Au$ is incorrect because Au has a higher ionization potential than Ag.
- Option B: $\mathrm{Cu} < Ag < Au$ is incorrect because Cu has a higher ionization potential than Ag, and Au has a higher ionization potential than Ag.
- Option C: $\mathrm{Cu} > Ag < Au$ is the correct answer.
- Option D: $\mathrm{Ag} > Cu < Au$ is incorrect because Cu has a higher ionization potential than Ag.

Key Concepts:
- Ionization Potential
- Periodic Trends
- Relativistic Effects

Memory Aids:
- Gold is special due to relativistic effects.

Quick Tips:
- Be aware of exceptions to periodic trends due to relativistic effects, especially for heavy elements.

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Understanding +4 | -1

Given below are two statement : One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: Size of

B{k^{3 + }}

ion is less than

N{p^{3 + }}

ion.
Reason R: The above is a consequence of the lanthanoid contraction.
In the light of the above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: D

Both A and R are true and R is the correct explanation of A

Option D (Both A and R are true and R is the correct explanation of A) is correct because the size of

B{k^{3 + }}

is indeed smaller than

N{p^{3 + }}

due to the actinide contraction, analogous to lanthanide contraction.

Solution approach

Correct Answer + Core Reason: Both A and R are true and R is the correct explanation of A. The size decreases due to actinide contraction which is similar to lanthanide contraction.

Step-by-Step Solution:

Step 1: Understand Assertion A: Size of $B{k^{3 + }}$ ion is less than $N{p^{3 + }}$ ion.

Step 2: Understand Reason R: The above is a consequence of the lanthanoid contraction.

Step 3: Analyze the positions of Bk and Np in the periodic table. Both are actinides.

Step 4: Recall the concept of lanthanide/actinide contraction. As we move across the lanthanide or actinide series, the effective nuclear charge increases due to poor shielding by the 4f or 5f electrons, respectively. This causes a contraction in size.

Step 5: Since Bk is further along the actinide series than Np, it experiences a greater actinide contraction, resulting in a smaller ionic size.

Step 6: Therefore, Assertion A is true, and Reason R correctly explains why.

Why other options are wrong:
- Option A: A is false but R is true: Assertion A is true, so this option is incorrect.
- Option B: Both A and R are true but R is not the correct explanation of A: Reason R is the correct explanation for Assertion A, so this option is incorrect.
- Option C: A is true but R is false: Reason R is true, so this option is incorrect.

Key Concepts:
- Lanthanide contraction
- Actinide contraction
- Effective nuclear charge
- Periodic trends

Memory Aids:
- Remember that lanthanide and actinide contraction leads to a decrease in size across the series.

Quick Tips:
- Understand the concept of effective nuclear charge and its effect on ionic size.

79 Easy

79 of 171Question permalink

Application +4 | -1

E. N of the element A is

{E_1}

and I.E is

{E_2}

. Hence, E. A will be

View answer and explanation

Correct option: A

2{E_1} - {E_2}

Option A (

2{E_1} - {E_2}

) is correct. Electron affinity (E.A) can be related to electronegativity (E.N) and ionization energy (I.E) using Mulliken's formula: E.N = (I.E + E.A)/2. Rearranging, E.A = 2*E.N - I.E.

Solution approach

Correct Answer + Core Reason: $2{E_1} - {E_2}$ . Electron affinity (E.A) can be related to electronegativity (E.N) and ionization energy (I.E) using Mulliken's formula.

Step-by-Step Solution:

Step 1: Recall Mulliken's formula for electronegativity:
$\chi = \frac{{I.E. + E.A.}}{2}$ \nWhere $\chi$ is electronegativity, I.E. is ionization energy, and E.A. is electron affinity.

Step 2: In this problem, E.N. = ${E_1}$ and I.E. = ${E_2}$ . We want to find E.A.

Step 3: Substitute the given values into Mulliken's formula:
${E_1} = \frac{{{E_2} + E.A.}}{2}$

Step 4: Solve for E.A.:
$2{E_1} = {E_2} + E.A.$
$E.A. = 2{E_1} - {E_2}$

Why other options are wrong:
- Option B: ${E_1} - {E_2}$ - Incorrect rearrangement of Mulliken's formula.
- Option C: ${E_1} - 2{E_2}$ - Incorrect rearrangement of Mulliken's formula.
- Option D: $\left( {{E_1} + {E_2}} \right)/2$ - This is the formula for electronegativity, not electron affinity.

Key Concepts:
- Electronegativity
- Ionization Energy
- Electron Affinity
- Mulliken's Formula

Memory Aids:
- Remember Mulliken's formula: $\chi = \frac{{I.E. + E.A.}}{2}$

Quick Tips:
- Be familiar with the relationship between electronegativity, ionization energy, and electron affinity.

80 Easy

80 of 171Question permalink

Understanding +4 | -1

Which of the following cannot be isoelectronic

View answer and explanation

Correct option: D

Two different atoms

Option D (Two different atoms) is correct because isoelectronic species have the same number of electrons, and two different atoms will inherently have different numbers of protons and therefore different numbers of electrons in their neutral state.

Solution approach

Correct Answer + Core Reason: Two different atoms. Isoelectronic species have the same number of electrons, which is not possible for different neutral atoms.

Step-by-Step Solution:

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

Step 2: Consider the options:
- A. Two different cations: Can be isoelectronic (e.g., $Na^+$ and $Mg^{2+}$ ).
- B. Two different anions: Can be isoelectronic (e.g., $O^{2-}$ and $F^-$ ).
- C. Cation and anion: Can be isoelectronic (e.g., $Na^+$ and $F^-$ ).
- D. Two different atoms: Cannot be isoelectronic because neutral atoms have a number of electrons equal to their atomic number, and different atoms have different atomic numbers.

Why other options are wrong:
- Option A: Two different cations: Cations can lose different numbers of electrons to become isoelectronic.
- Option B: Two different anions: Anions can gain different numbers of electrons to become isoelectronic.
- Option C: Cation and anion: A cation and an anion can gain or lose electrons to have the same number of electrons.

Key Concepts:
- Definition of isoelectronic species.
- Cations and anions formation.

Memory Aids:
- Iso- means same, electronic means electrons.

Quick Tips:
- To determine if species are isoelectronic, calculate the number of electrons for each and compare.

81 Easy

81 of 171Question permalink

Understanding +4 | -1

Which is the property of non-metal?

View answer and explanation

Correct option: A

Electronegative

Option A (Electronegative) is correct because non-metals tend to gain electrons and are therefore electronegative.

Solution approach

Correct Answer + Core Reason: Electronegative. Non-metals tend to gain electrons to form negative ions.

Step-by-Step Solution:

Step 1: Understand the question. The question asks for a property of non-metals.

Step 2: Recall the properties of non-metals.
- Non-metals are generally electronegative, meaning they attract electrons.
- Non-metal oxides are acidic.
- Non-metals tend to be oxidizing agents, not reducing agents.
- Non-metals have high ionization potentials.

Step 3: Identify the correct property. Electronegativity is a property of non-metals.

Why other options are wrong:
- Option B: Basic nature of oxide: Metal oxides are generally basic, while non-metal oxides are acidic or neutral.
- Option C: Reducing property: Metals are generally reducing agents, while non-metals are oxidizing agents.
- Option D: Low ionisation potential: Metals have low ionization potentials, while non-metals have high ionization potentials.

Key Concepts:
- Electronegativity
- Properties of Metals and Non-metals
- Ionization Potential

Memory Aids:
- Non-metals are electron "attractors" (electronegative).

Quick Tips:
- Remember the general trends in electronegativity and ionization potential on the periodic table.

82 Easy

82 of 171Question permalink

Remembering +4 | -1

Which of the following belong to the category of transition metal

View answer and explanation

Correct option: C

Fe

Option C is correct. Iron (Fe) is a transition metal because it has a partially filled d-orbital.

Solution approach

Correct Answer + Core Reason: Fe. Iron (Fe) belongs to the category of transition metals because it has a partially filled d-orbital.

Step-by-Step Solution:

Step 1: Define Transition Metals: Transition metals are elements that have partially filled d orbitals in at least one of their common oxidation states.

Step 2: Analyze the electronic configurations of the given elements:\nA. K: $[\mathrm{Ar}] 4 s^1$ - Alkali metal\nB. Ra: $[\mathrm{Rn}] 7 s^2$ - Alkaline earth metal\nC. Fe: $[\mathrm{Ar}] 3 d^6 4 s^2$ - Transition metal\nD. Al: $[\mathrm{Ne}] 3 s^2 3 p^1$ - p-block element

Step 3: Conclusion: Iron (Fe) has a partially filled d-orbital and is therefore a transition metal.

Why other options are wrong:
- Option A: Potassium (K) is an alkali metal and does not have a partially filled d-orbital.
- Option B: Radium (Ra) is an alkaline earth metal and does not have a partially filled d-orbital.
- Option D: Aluminum (Al) is a p-block element and does not have a partially filled d-orbital.

Key Concepts:
- Transition Metals
- Electronic Configuration

Memory Aids:
- Transition metals are in the d-block of the periodic table.

Quick Tips:
- Know the location of transition metals in the periodic table.

83 Easy

83 of 171Question permalink

Remembering +4 | -1

Of the following elements, which one has the highest electronegativity?

View answer and explanation

Correct option: A

F

Option A (F) is correct because Fluorine has the highest electronegativity among the halogens.

Solution approach

Correct Answer + Core Reason: F. Fluorine has the highest electronegativity among the halogens.

Step-by-Step Solution:

Step 1: Recall the electronegativity trend in the halogen group.\nElectronegativity decreases down a group.

Step 2: Consider the positions of the halogens in the periodic table.
- F, Cl, Br, and I are all in group 17 (halogens).
- Fluorine is at the top of the group, followed by Chlorine, Bromine, and Iodine.

Step 3: Conclusion.\nTherefore, Fluorine has the highest electronegativity among the given halogens.

Why other options are wrong:
- Option B: Chlorine is less electronegative than Fluorine.
- Option C: Bromine is less electronegative than Fluorine.
- Option D: Iodine is less electronegative than Fluorine.

Key Concepts:
- Electronegativity
- Halogens
- Periodic trends

Memory Aids:
- Electronegativity decreases down a group.

Quick Tips:
- Remember the electronegativity trend within the halogen group.

84 Easy

84 of 171Question permalink

Understanding +4 | -1

The element with electronic configuration

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

is

View answer and explanation

Correct option: C

an alkaline earth metal

Option C (an alkaline earth metal) is correct because the electronic configuration

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

indicates that the element has 2 valence electrons in the s-orbital of the outermost shell, which is characteristic of alkaline earth metals.

Solution approach

Correct Answer + Core Reason: an alkaline earth metal. The electronic configuration ends with $3s^2$ , indicating two valence electrons.

Step-by-Step Solution:

Step 1: Analyze the given electronic configuration: $1{s^2}2{s^2}2{p^6}3{s^2}$ .

Step 2: Identify the valence shell: The valence shell is the outermost shell, which is the 3rd shell in this case.

Step 3: Determine the number of valence electrons: The valence shell has 2 electrons in the $3s$ orbital.

Step 4: Relate the number of valence electrons to the group: Elements with 2 valence electrons in the s-orbital belong to the alkaline earth metals (Group 2).

Why other options are wrong:
- Option A: member of alkali family: Alkali metals have 1 valence electron.
- Option B: a noble gas: Noble gases have a completely filled valence shell (except He).
- Option D: Unpredictable: The electronic configuration clearly indicates the element's group.

Key Concepts:
- Electronic configuration and valence electrons.
- Group numbers and valence electrons.
- Alkaline earth metals.

Memory Aids:
- Group 1: Alkali metals (1 valence electron).
- Group 2: Alkaline earth metals (2 valence electrons).
- Group 17: Halogens (7 valence electrons).
- Group 18: Noble gases (8 valence electrons, except He which has 2).

Quick Tips:
- The number of valence electrons determines the group number for s- and p-block elements.
- Know the general electronic configurations for each group.

85 Easy

85 of 171Question permalink

Remembering +4 | -1

The property of attracting electrons by the halogen atoms in a molecule is called

View answer and explanation

Correct option: C

Electronegtivity

Option C (Electronegtivity) is correct because electronegativity is the measure of the ability of an atom to attract electrons in a chemical bond.

Solution approach

Correct Answer + Core Reason: Electronegtivity. Electronegativity is the measure of the ability of an atom to attract electrons in a chemical bond.

Step-by-Step Solution:

Step 1: Define the term.\nElectronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.

Step 2: Relate to halogens.\nHalogens are highly electronegative due to their high effective nuclear charge and small atomic size.

Step 3: Conclusion.\nThe property of attracting electrons by the halogen atoms in a molecule is called electronegativity.

Why other options are wrong:
- Option A: Ionization potential is the energy required to remove an electron from an isolated gaseous atom.
- Option B: Electron affinity is the change in energy when an electron is added to a neutral gaseous atom.
- Option D: Electronic attraction is a general term and not specific to the ability of an atom to attract electrons in a bond.

Key Concepts:
- Electronegativity
- Halogens
- Ionization potential
- Electron affinity

Memory Aids:
- Electronegativity: Electron-attracting ability.

Quick Tips:
- Remember the definitions of ionization potential, electron affinity, and electronegativity.

86 Easy

86 of 171Question permalink

Remembering +4 | -1

Which is not a scale of measuring electronegativity?

View answer and explanation

Correct option: A

Stevenson’s scale

Option A (Stevenson’s scale) is correct because Stevenson's scale is not a recognized scale for measuring electronegativity.

Solution approach

Correct Answer + Core Reason: Stevenson’s scale. Stevenson's scale is not a recognized scale for measuring electronegativity.

Step-by-Step Solution:

Step 1: Recall the common scales for measuring electronegativity.\nCommon scales include Mulliken's scale, Allred-Rochow's scale, and Pauling's scale.

Step 2: Identify the incorrect scale.\nStevenson's scale is not a recognized scale for measuring electronegativity.

Step 3: Conclusion.\nTherefore, Stevenson's scale is not a scale of measuring electronegativity.

Why other options are wrong:
- Option B: Mulliken’s scale is a valid scale for measuring electronegativity.
- Option C: Allred-Rochow’s scale is a valid scale for measuring electronegativity.
- Option D: Pauling scale is a valid scale for measuring electronegativity.

Key Concepts:
- Electronegativity scales
- Mulliken's scale
- Allred-Rochow's scale
- Pauling scale

Memory Aids:
- Remember the common electronegativity scales: Mulliken, Allred-Rochow, and Pauling.

Quick Tips:
- Familiarize yourself with the different electronegativity scales and their names.

87 Easy

87 of 171Question permalink

Understanding +4 | -1

Correct increasing order of first ionisation potential is

View answer and explanation

Correct option: A

Na < Mg > Al < Si

Option A (Na < Mg > Al < Si) is correct because there's an exception between Mg and Al due to the electronic configuration.

Solution approach

Correct Answer + Core Reason: Na < Mg > Al < Si. There is an exception between Mg and Al due to electronic configuration.

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: Consider the positions of Na, Mg, Al, and Si in the periodic table. They are all in the same period (Period 3), with Na being the leftmost and Si being the rightmost.
Step 3: Account for the exception between Mg and Al. Magnesium has a filled s-orbital ( $3s^2$ ), which is more stable than aluminum's ( $3s^23p^1$ ). Therefore, it requires more energy to remove an electron from Mg than from Al.
Step 4: Arrange the elements in increasing order of ionization potential, considering the exception. Na < Al < Mg < Si. However, since the question asks for the correct *increasing* order, we need to represent the exception correctly: Na < Mg > Al < Si

Why other options are wrong:
- Option B: Na < Mg < Al < Si: Incorrect, this does not account for the exception between Mg and Al.
- Option C: Na > Mg > Al > Si: Incorrect, this is the reverse order and does not account for the exception.
- Option D: Na < Mg < Al > Si: Incorrect, this incorrectly places Al higher than Mg.

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

Memory Aids:
- Remember the exception between Mg and Al: Mg > Al

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

88 Easy

88 of 171Question permalink

Understanding +4 | -1

Highest amount of energy will be required for the removal of the electron from

View answer and explanation

Correct option: A

s -

subshell

Option A (

s -

subshell) is correct because s-electrons are closest to the nucleus and thus experience the strongest effective nuclear charge, requiring the highest energy for removal.

Solution approach

Correct Answer + Core Reason: $s -$ subshell. Electrons in the s-subshell are closest to the nucleus and experience the highest effective nuclear charge.

Step-by-Step Solution:

Step 1: Ionization energy is the energy required to remove an electron from an atom or ion.

Step 2: The closer an electron is to the nucleus, the stronger the attraction and the higher the ionization energy.

Step 3: The order of penetration power of orbitals is $s > p > d > f$ . This means that s-electrons are, on average, closer to the nucleus than p-electrons, which are closer than d-electrons, and so on.

Step 4: Therefore, it requires more energy to remove an electron from the s-subshell compared to p, d, or f subshells.

Why other options are wrong:
- **Option B ( $p -$ subshell): p-electrons are farther from the nucleus than s-electrons, so they are easier to remove.
- Option C ( $d -$ subshell): d-electrons are farther from the nucleus than p-electrons, so they are even easier to remove.
- Option D ( $f -$ subshell): f-electrons are the farthest from the nucleus among the given options, so they are the easiest to remove.

Key Concepts:
- Ionization Energy
- Effective Nuclear Charge
- Orbital Penetration

Memory Aids:
- Remember the order of penetration power: s > p > d > f.

Quick Tips:**
- Consider the distance of the electron from the nucleus when comparing ionization energies.

89 Easy

89 of 171Question permalink

Analyzing +4 | -1

S-I: '

Be

' and '

Al

' have similar properties.
S-II: Cations of '

Be

' and '

Al

' have same polarizing power.

View answer and explanation

Correct option: A

Both I and II are correct

Option A (Both I and II are correct) is correct because Beryllium (Be) and Aluminum (Al) exhibit diagonal relationship due to similar polarizing power of their cations.

Solution approach

Correct Answer + Core Reason: Both I and II are correct. The diagonal relationship between Be and Al arises from the similarity in their charge/size ratio, leading to similar polarizing power.

Step-by-Step Solution:

Step 1: Identify the elements: Beryllium (Be) and Aluminum (Al).

Step 2: Recall the diagonal relationship in the periodic table. Elements diagonally adjacent often exhibit similar properties due to comparable charge/size ratios and polarizing power.

Step 3: Beryllium and Aluminum show a diagonal relationship.

Step 4: The polarizing power of an ion is directly proportional to its charge and inversely proportional to its size. $Polarizing\ Power \propto \frac{Charge}{Size}$

Step 5: $Be^{2+}$ and $Al^{3+}$ have similar polarizing power due to their charge and size.

Step 6: Similar polarizing power leads to similar chemical properties.

Why other options are wrong:
- Option B: Both I and II are incorrect: This is wrong because both statements are correct.
- Option C: I is correct and II is incorrect: This is wrong because both statements are correct.
- Option D: I is incorrect and II is correct: This is wrong because both statements are correct.

Key Concepts:
- Diagonal Relationship
- Polarizing Power
- Charge/Size Ratio

Memory Aids:
- Diagonal relationships occur due to similar charge/size ratios.

Quick Tips:
- Remember the diagonal relationships: Li-Mg, Be-Al, B-Si.

90 Easy

90 of 171Question permalink

Remembering +4 | -1

The elements in which electrons are progressively filled in 4f – orbital are called

View answer and explanation

Correct option: C

Lanthanoids

Option C is correct. Lanthanoids are the elements in which electrons are progressively filled in the 4f-orbital.

Solution approach

Correct Answer + Core Reason: Lanthanoids. Lanthanoids are the elements in which electrons are progressively filled in the 4f-orbital.

Step-by-Step Solution:

Step 1: Define Lanthanoids: Lanthanoids are the series of elements from Cerium (Ce) to Lutetium (Lu) in which the 4f orbitals are progressively filled.

Step 2: Define Actinoids: Actinoids are the series of elements from Thorium (Th) to Lawrencium (Lr) in which the 5f orbitals are progressively filled.

Step 3: Define Transition Elements: Transition elements are elements with partially filled d orbitals.

Step 4: Define Halogens: Halogens are the elements in Group 17.

Step 5: Conclusion: The elements in which electrons are progressively filled in 4f-orbital are called Lanthanoids.

Why other options are wrong:
- Option A: Actinoids fill the 5f orbitals.
- Option B: Transition elements fill the d orbitals.
- Option D: Halogens are Group 17 elements and do not involve filling f orbitals.

Key Concepts:
- Lanthanoids (4f series)
- Actinoids (5f series)
- Transition Elements (d-block)
- Halogens (Group 17)

Memory Aids:
- 4f orbitals = Lanthanoids, 5f orbitals = Actinoids.

Quick Tips:
- Know the location of Lanthanoids and Actinoids in the periodic table.

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