Consider the ions ${{\rm{K}}^ + },{\rm{\;}}{{\rm{S}}^{2 - }},{\rm{C}}{{\rm{l}}^ - }{\rm{\;and\;C}}{{\rm{a}}^{2 + }}$. The radii of these ionic species follow the order

${{\rm{S}}^{2 - }} > C{{\rm{l}}^ - } > {{\rm{K}}^ + } > {\rm{C}}{{\rm{a}}^{2 + }}{\rm{\;}}$

${\rm{C}}{{\rm{a}}^{2 + }} > {{\rm{K}}^ + } > C{{\rm{l}}^ - } > {{\rm{S}}^{2 - }}$

${\rm{C}}{{\rm{l}}^ - } > {{\rm{S}}^{2 - }} > {{\rm{K}}^ + } > {\rm{C}}{{\rm{a}}^{2 + }}$

${\rm{C}}{{\rm{a}}^{2 + }} > C{{\rm{l}}^ - } > K > {{\rm{S}}^{2 - }}$

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

1 Easy

1 of 171Question permalink

Understanding +4 | -1

Which of the following is the most electropositive element?

View answer and explanation

Correct option: C

Mg

Option C (Mg) is correct because electropositivity increases down and to the left of the periodic table, and Mg is more to the left and down compared to P, S, and Al.

Solution approach

Correct Answer + Core Reason: Mg. Electropositivity increases down and to the left of the periodic table.

Step-by-Step Solution:
Step 1: Recall the trend of electropositivity in the periodic table. Electropositivity (metallic character) increases as you move down a group and to the left across a period.
Step 2: Locate the elements P, S, Mg, and Al in the periodic table. P and S are in the same period (Period 3) to the right, Al is also in Period 3 but to the left, and Mg is in Period 3, Group 2, which is further to the left than Al.
Step 3: Apply the trend. Mg is the most electropositive element among the given options because it is located furthest to the left in the periodic table compared to the other elements.

Why other options are wrong:
- Option A: P: Incorrect, P is a nonmetal and is not electropositive.
- Option B: S: Incorrect, S is a nonmetal and is not electropositive.
- Option D: Al: Incorrect, Al is less electropositive than Mg because Mg is further to the left in the periodic table.

Key Concepts:
- Electropositivity (metallic character)
- Periodic trends

Memory Aids:
- Electropositivity increases down and to the left.

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

2 Easy

2 of 171Question permalink

Analyzing +4 | -1

Identify the least stable ion among the following

View answer and explanation

Correct option: B

B{e^ - }

The correct answer is B (

B{e^ - }

) because Beryllium (

Be

) has a stable electronic configuration (2s²) and adding an electron to form

Be^-

is energetically unfavorable.

Solution approach

Correct Answer + Core Reason: $B{e^ - }$ . Beryllium has a relatively stable electronic configuration (2s²) and adding an electron is energetically unfavorable.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the neutral atoms:
- Li: 1s² 2s¹
- Be: 1s² 2s²
- B: 1s² 2s² 2p¹
- C: 1s² 2s² 2p²

Step 2: Write the electronic configurations of the anions:
- $L{i^ - }$ : 1s² 2s² (stable)
- $B{e^ - }$ : 1s² 2s² 2p¹ (less stable than Be)
- ${B^ - }$ : 1s² 2s² 2p² (more stable than Be-)
- ${C^ - }$ : 1s² 2s² 2p³ (more stable than Be-)

Step 3: Beryllium has a filled 2s subshell. Adding an electron requires adding it to the higher energy 2p subshell, making it less stable compared to the other options.

Why other options are wrong:
- Option A: $L{i^ - }$ achieves a noble gas configuration (1s² 2s²), making it relatively stable.
- Option C: ${B^ - }$ has a partially filled 2p subshell, which is more stable than adding an electron to Be.
- Option D: ${C^ - }$ has a more filled 2p subshell than B-, making it more stable.

Key Concepts:
- Electronic Configuration
- Stability of Ions
- Octet Rule

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

Quick Tips:
- Consider the electronic configurations of the ions when determining their stability.

3 Easy

3 of 171Question permalink

Application +4 | -1

An element has E.C.

1{S^2}\,\,2{S^2}\,\,2{P^6}\,\,3{S^2}\,\,3{P^3}

predict their period, group and block

View answer and explanation

Correct option: A

period = 3, block = p, group = 15

Option A (period = 3, block = p, group = 15) is correct because the element's electronic configuration ends in 3p3, indicating the 3rd period, p-block, and group 15.

Solution approach

Correct Answer + Core Reason: period = 3, block = p, group = 15. The electronic configuration determines the period, block, and group of the element.

Step-by-Step Solution:

Step 1: Identify the period. The period is determined by the highest principal quantum number (n) in the electronic configuration. In this case, the highest n is 3, so the period is 3.

Step 2: Identify the block. The block is determined by the subshell where the last electron enters. Here, the last electron enters the p-subshell, so the block is p.

Step 3: Identify the group. For p-block elements, the group number is calculated as 10 + the number of electrons in the outermost p-subshell. In this case, there are 3 electrons in the 3p subshell, so the group number is 10 + 5 = 15. (Note: The number of electrons in the outermost p-subshell is 3, plus the 2 electrons in the s-subshell of the same shell, which gives 5. So, the group number is 10 + 5 = 15.)

Step 4: Final Answer: Period = 3, Block = p, Group = 15

Why other options are wrong:
- Option B: This is incorrect because the period is determined by the highest principal quantum number, which is 3, not 5. Also, the block is p, not s.
- Option C: This is incorrect because while the period and block are correctly identified, the group number is miscalculated. The group number for p-block elements is 10 + number of valence electrons in p-orbital which is 10+5=15, not 10.
- Option D: This is incorrect because the period is 3, not 4, and the block is p, not d.

Key Concepts:
- Electronic configuration
- Period, block, and group in the periodic table
- Principal quantum number
- Valence electrons

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:
- The period number is the highest principal quantum number (n) in the electronic configuration.
- The block is determined by the last subshell filled (s, p, d, or f).
- For s-block elements, the group number is the number of valence electrons.
- For p-block elements, the group number is 10 + the number of valence electrons in p-orbital.

4 Easy

4 of 171Question permalink

Understanding +4 | -1

The correct order according to size is

View answer and explanation

Correct option: C

{O^{2 - }} > {O^ - } > O

Option C (

{O^{2 - }} > {O^ - } > O

) is correct because as the negative charge increases, the electron-electron repulsion also increases, leading to an increase in size.

Solution approach

Correct Answer + Core Reason: ${O^{2 - }} > {O^ - } > O$ . Increasing negative charge increases ionic size.

Step-by-Step Solution:

Step 1: Consider the effect of adding electrons to an atom.

Step 2: Adding electrons increases electron-electron repulsion, which causes the electron cloud to expand.

Step 3: Therefore, the more negative the ion, the larger its size.

Step 4: Comparing O, O⁻, and O²⁻, the order of increasing size is O < O⁻ < O²⁻.

Why other options are wrong:
- Option A: This order is incorrect because it suggests that the neutral atom is larger than its negative ions.
- Option B: This order is incorrect because it suggests that O⁻ is larger than O²⁻, which is not true.
- Option D: This order is incorrect because it suggests that the neutral atom is larger than its negative ions, and that O²⁻ is smaller than O⁻.

Key Concepts:
- Ionic size
- Electron-electron repulsion
- Effect of charge on size

Memory Aids:
- Negative charge = bigger size.

Quick Tips:
- Remember that adding electrons increases size, and removing electrons decreases size.

5 Easy

5 of 171Question permalink

Application +4 | -1

Element with electronic configuration

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

3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^3

belongs to the following group of the periodic table

View answer and explanation

Correct option: B

$5 th$

Option B ($5 th$) is correct because the electronic configuration ends with

5p^3

, which indicates that the element belongs to the 5th group (Group 15) of the p-block elements.

Solution approach

Correct Answer + Core Reason: $5 th$. The electronic configuration ends with $5p^3$ , indicating group 15 (5th group of p-block).

Step-by-Step Solution:

Step 1: Analyze the electronic configuration: $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^3$ .

Step 2: Identify the valence shell: The valence shell is the 5th shell ( $5s^2 5p^3$ ).

Step 3: Determine the number of valence electrons: The number of valence electrons is 2 + 3 = 5.

Step 4: Determine the group number: Since the last electron enters the $p$ -orbital, it's a $p$ -block element. The group number is 10 + number of electrons in the $p$ -orbital = 10 + 5 = 15. The question asks for the group of p-block, so it is the 5th group of p-block elements.

Why other options are wrong:
- Option A: $2 {nd}$ : Incorrect group number.
- Option C: $3 {rd}$ : Incorrect group number.
- Option D: $7 {th}$ : Incorrect group number.

Key Concepts:
- Electronic configuration and its relation to the periodic table.
- Group number determination.

Memory Aids:
- For $p$ -block elements, group number = 10 + number of valence electrons in $p$ -orbital.

Quick Tips:
- Identify the valence shell and the number of valence electrons to determine the group number.

6 Easy

6 of 171Question permalink

Analyzing +4 | -1

On which of the following Dobereiner's Triad law is not applicable?

View answer and explanation

Correct option: C

{F}, {Cl}, {Br}

The correct answer is C (

{F}, {Cl}, {Br}

) because the atomic mass of Chlorine is not approximately the average of Fluorine and Bromine.

Solution approach

Correct Answer + Core Reason: ${F}, {Cl}, {Br}$ . The atomic mass of Cl is not approximately the average of F and Br.

Step-by-Step Solution:

Step 1: Recall Dobereiner's Triad law, which states that the atomic weight of the middle element is approximately the average of the atomic weights of the other two elements.

Step 2: Check each option:
- ${Cl}, {Br}, {I}$ : (35.5 + 127)/2 = 81.25 (close to 79.9 for Br)
- ${Ca}, {Sr}, {B}$ : (40 + 137)/2 = 88.5 (not close to 87.6 for Sr - this is a bad triad)
- ${F}, {Cl}, {Br}$ : (19 + 79.9)/2 = 49.45 (not close to 35.5 for Cl)
- ${Li}, {Na}, {K}$ : (7 + 39)/2 = 23 (close to 23 for Na)

Step 3: The triad law is not applicable for ${F}, {Cl}, {Br}$ because the atomic mass of Cl (35.5) is not approximately the average of F (19) and Br (79.9).

Why other options are wrong:
- Option A: ${Cl}, {Br}, {I}$ approximately follows the triad law.
- Option B: ${Ca}, {Sr}, {B}$ approximately follows the triad law.
- Option D: ${Li}, {Na}, {K}$ approximately follows the triad law.

Key Concepts:
- Dobereiner's Triad Law
- Atomic Weights

Memory Aids:
- Remember the definition of Dobereiner's Triad Law.

Quick Tips:
- Check if the atomic weight of the middle element is approximately the average of the other two.

7 Easy

7 of 171Question permalink

Application +4 | -1

The electronegativity of A and B are 0.7 and 4.0 respectively. The bond formed in AB and

{B_2}

molecule will be

View answer and explanation

Correct option: A

Ionic and covalent resp

Option A (Ionic and covalent resp) is correct because the large electronegativity difference between A and B leads to ionic bonding in AB, while identical atoms in B2 form a covalent bond.

Solution approach

Correct Answer + Core Reason: Ionic and covalent resp. Large electronegativity difference leads to ionic bond, similar electronegativity leads to covalent bond.

Step-by-Step Solution:

Step 1: Identify the electronegativity difference between A and B. The electronegativity difference is 4.0 - 0.7 = 3.3.

Step 2: Determine the type of bond formed between A and B. Since the electronegativity difference is greater than 1.7, the bond is considered ionic.

Step 3: Determine the type of bond formed between B and B in the B2 molecule. Since the electronegativity difference is 0, the bond is purely covalent.

Step 4: Conclude the types of bonds formed in AB and B2.

Why other options are wrong:
- Option B: Covalent & ionic resp: This is incorrect because the bond in AB is ionic, not covalent.
- Option C: Ionic and dative resp: Dative bonds are coordinate covalent bonds, not typically formed between elements with such a large electronegativity difference.
- Option D: Metallic & covalent resp: Metallic bonds are formed between metal atoms, not between A and B.

Key Concepts:
- Electronegativity
- Ionic bond
- Covalent bond
- Electronegativity difference and bond type

Memory Aids:
- Large electronegativity difference -> Ionic bond
- Small electronegativity difference -> Covalent bond

Quick Tips:
- Remember the electronegativity difference threshold for ionic vs. covalent bonds (approximately 1.7).

8 Easy

8 of 171Question permalink

Understanding +4 | -1

Which one of the following combinations represents a metallic element?

View answer and explanation

Correct option: A

2, 8, 2

Option A is correct because the electronic configuration 2, 8, 2 represents an element with 2 valence electrons, which is characteristic of metallic elements in Group 2 (alkaline earth metals).

Solution approach

Correct Answer + Core Reason: A. 2, 8, 2. This electronic configuration represents a metallic element because it has 2 valence electrons.

Step-by-Step Solution:

Step 1: Analyze the electronic configuration 2, 8, 2. This means the element has 2 electrons in the first shell, 8 in the second, and 2 in the outermost shell.

Step 2: Elements with 1, 2, or 3 valence electrons tend to be metals.

Step 3: Elements with 5, 6, or 7 valence electrons tend to be non-metals.

Step 4: Elements with 4 valence electrons tend to be metalloids.

Step 5: Elements with 8 valence electrons (except Helium with 2) are noble gases.

Why other options are wrong:
- Option B: 2, 8, 4 could be a metalloid or non-metal.
- Option C: 2, 8, 7 is a halogen, which is a non-metal.
- Option D: 2, 8, 8 is a noble gas, which is a non-metal.

Key Concepts:
- Electronic configuration
- Valence electrons
- Metals, non-metals, and metalloids

Memory Aids:
- Remember the general trend: few valence electrons = metal, many valence electrons = non-metal.

Quick Tips:
- Determine the number of valence electrons to predict the metallic character.

9 Easy

9 of 171Question permalink

Remembering +4 | -1

IUPAC name of an element with atomic number 110 is

View answer and explanation

Correct option: A

Ununnillium

Option A (Ununnillium) is correct because it follows the IUPAC nomenclature rules for naming elements with atomic number 110.

Solution approach

Correct Answer + Core Reason: Ununnillium. This name is derived directly from the digits of the atomic number 110 using IUPAC nomenclature.

Step-by-Step Solution:

Step 1: Understand the question: The question asks for the IUPAC name of an element with atomic number 110.

Step 2: Recall IUPAC nomenclature rules: The IUPAC nomenclature for elements with atomic numbers greater than 100 uses the following roots for the digits:
- 0: nil
- 1: un
- 2: bi
- 3: tri
- 4: quad
- 5: pent
- 6: hex
- 7: sept
- 8: oct
- 9: enn

Step 3: Apply the rules to atomic number 110:
- 1: un
- 1: un
- 0: nil
- Add the suffix -ium
- Therefore, the name is Ununnillium.

Why other options are wrong:
- Option B: Unununium is the name for element 111.
- Option C: Unnilennium is not a valid IUPAC name.
- Option D: Unnilunium is not a valid IUPAC name.

Key Concepts:
- IUPAC nomenclature for elements with atomic numbers greater than 100.

Memory Aids:
- Remember the roots for each digit (0-9).

Quick Tips:
- Break down the atomic number into its individual digits and apply the IUPAC rules.

10 Easy

10 of 171Question permalink

Understanding +4 | -1

Which of the following element can exhibit more than one oxidation state

View answer and explanation

Correct option: A

Cr

The correct answer is A (Cr) because Chromium (Cr) can exhibit multiple oxidation states due to its electronic configuration.

Solution approach

Correct Answer + Core Reason: Cr. Chromium exhibits variable oxidation states due to its electronic configuration and ability to lose different numbers of electrons.

Step-by-Step Solution:

Step 1: Recall the electronic configurations of the given elements.

Step 2: Cr has an electronic configuration of [Ar] 3d⁵ 4s¹. It can lose electrons from both the 3d and 4s orbitals, leading to multiple oxidation states (e.g., +2, +3, +6).

Step 3: Cs, Sr, and Rb are alkali and alkaline earth metals, which typically exhibit only one oxidation state (+1 and +2, respectively).

Why other options are wrong:
- Option B: Cs (Cesium) is an alkali metal and typically exhibits only a +1 oxidation state.
- Option C: Sr (Strontium) is an alkaline earth metal and typically exhibits only a +2 oxidation state.
- Option D: Rb (Rubidium) is an alkali metal and typically exhibits only a +1 oxidation state.

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

Memory Aids:
- Remember that transition metals often exhibit multiple oxidation states.

Quick Tips:
- Know the common oxidation states of alkali metals, alkaline earth metals, and transition metals.

11 Easy

11 of 171Question permalink

Understanding +4 | -1

Consider the ions

{\mathrm{K}^ + },\mathrm{\;}{\mathrm{S}^{2 - }},\mathrm{C}{\mathrm{l}^ - }\mathrm{\;and\;C}{\mathrm{a}^{2 + }}

. The radii of these ionic species follow the order

View answer and explanation

Correct option: D

{\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > \mathrm{C}{\mathrm{a}^{2 + }}\mathrm{\;}

Option D (

{\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > {\mathrm{Ca}^{2 + }}\mathrm{\;}

) is correct because ionic radii decrease with increasing nuclear charge and increase with increasing negative charge.

Solution approach

Correct Answer + Core Reason: ${\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > {\mathrm{Ca}^{2 + }}\mathrm{\;}$ . Ionic radii are affected by nuclear charge and number of electrons.

Step-by-Step Solution:

Step 1: Identify the number of electrons for each ion:
- K+ (19 - 1 = 18 electrons)
- S2- (16 + 2 = 18 electrons)
- Cl- (17 + 1 = 18 electrons)
- Ca2+ (20 - 2 = 18 electrons)

Step 2: All ions are isoelectronic (same number of electrons). Therefore, the ionic radius is determined by the effective nuclear charge (Zeff).

Step 3: Higher the effective nuclear charge, smaller the ionic radius. The nuclear charges are:
- K+ (Z = 19)
- S2- (Z = 16)
- Cl- (Z = 17)
- Ca2+ (Z = 20)

Step 4: Arrange the ions in increasing order of nuclear charge: S2- < Cl- < K+ < Ca2+

Step 5: Since ionic radius decreases with increasing nuclear charge, the order of ionic radii is: S2- > Cl- > K+ > Ca2+

Why other options are wrong:
- Option A: $\mathrm{Ca}^{2 + } > \mathrm{K}^ + } > C\mathrm{l}^ - } > \mathrm{S}^{2 - }}$ : This is incorrect because it places the ion with the highest nuclear charge (Ca2+) as having the largest radius, which is the opposite of the correct trend.
- Option B: $\mathrm{Cl}^ - } > \mathrm{S}^{2 - } > \mathrm{K}^ + } > \mathrm{Ca}^{2 + }}$ : This is incorrect because it incorrectly orders S2- and Cl- relative to each other. S2- has a lower nuclear charge than Cl-, so it should have a larger radius.
- Option C: $\mathrm{Ca}^{2 + } > C\mathrm{l}^ - } > K > \mathrm{S}^{2 - }}$ : This is incorrect because it includes neutral K, which is not isoelectronic with the other ions, and also incorrectly orders the ionic radii based on nuclear charge.

Key Concepts:
- Isoelectronic species
- Effective nuclear charge (Zeff)
- Ionic radius trends

Memory Aids:
- For isoelectronic species, higher positive charge = smaller radius.
- Anions are larger than their neutral atoms; cations are smaller.

Quick Tips:
- When comparing isoelectronic species, focus on the nuclear charge. The ion with the greater nuclear charge will have a smaller radius.

12 Easy

12 of 171Question permalink

Remembering +4 | -1

The outer electronic configuration of transition element is

View answer and explanation

Correct option: A

\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}

Option A (

\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}

) is correct because it accurately represents the general outer electronic configuration of transition elements, allowing for variations in the d and s orbital filling.

Solution approach

Correct Answer + Core Reason: $\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}$ . This configuration allows for the filling of the (n-1)d orbitals with 1 to 10 electrons and the ns orbital with 0 to 2 electrons, which is characteristic of transition elements.

Step-by-Step Solution:

Step 1: Recall the definition of transition elements. Transition elements are those elements that have partially filled d orbitals in their elemental or most stable oxidation state.

Step 2: Consider the general electronic configuration of transition elements. The general electronic configuration is $(n-1)d^{1-10} ns^{0-2}$ , where n is the principal quantum number of the outermost shell.

Step 3: Understand the variations in the electronic configuration. The $ns$ orbital can have 0, 1, or 2 electrons due to the small energy difference between the $(n-1)d$ and $ns$ orbitals. This leads to exceptions like Cr and Cu.

Why other options are wrong:
- **Option B ( $n{s^2}\,\,n{p^{1 - 6}}$ ): This is the general electronic configuration of p-block elements, not transition elements.
- Option C ( $\left( {n - 2} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}$ ): The d-orbital being filled is (n-1)d, not (n-2)d. (n-2)f orbitals are filled in inner transition elements (lanthanides and actinides).
- Option D ( $\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^2}$ ): This configuration is not entirely correct as it doesn't account for the possibility of having 0 or 1 electrons in the ns orbital, as seen in elements like Chromium (Cr).

Key Concepts:
- Electronic configuration of transition elements
- Definition of transition elements
- Filling of d-orbitals

Memory Aids:
- Remember that transition elements involve the filling of the (n-1)d orbitals.

Quick Tips:**
- Pay attention to the (n-1) factor for d-orbitals in transition elements.

13 Easy

13 of 171Question permalink

Understanding +4 | -1

Amphoteric oxide combinations are in

View answer and explanation

Correct option: C

\mathrm{Sn}{\mathrm{O}_2},\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3},\mathrm{ZnO}

Option C (

\mathrm{Sn}{\mathrm{O}_2},\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3},\mathrm{ZnO}

) is correct because all three oxides are known to be amphoteric.

Solution approach

Correct Answer + Core Reason: $\mathrm{Sn}{\mathrm{O}_2},\mathrm{\;}{\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3},\mathrm{ZnO}$ . These oxides are amphoteric.

Step-by-Step Solution:

Step 1: Define amphoteric nature.\nAn amphoteric oxide can react with both acids and bases.

Step 2: Consider the nature of the given oxides.
- $SnO_2$ is amphoteric.
- $Al_2O_3$ is amphoteric.
- ZnO is amphoteric.

Step 3: Conclusion.\nTherefore, the combination of amphoteric oxides is $\mathrm{Sn}{\mathrm{O}_2},\mathrm{\;}{\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3},\mathrm{ZnO}$ .

Why other options are wrong:
- Option A: $\mathrm{ZnO},\mathrm{\;}{\mathrm{K}_2}\mathrm{O},\mathrm{S}{\mathrm{O}_3}$ is incorrect because $K_2O$ is basic and $SO_3$ is acidic.
- Option B: $\mathrm{ZnO},{\mathrm{P}_2}{\mathrm{O}_5},\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7}$ is incorrect because $P_2O_5$ and $Cl_2O_7$ are acidic.
- Option D: $\mathrm{Pb}{\mathrm{O}_2},\mathrm{Sn}{\mathrm{O}_2},\mathrm{S}{\mathrm{O}_3}$ is incorrect because $SO_3$ is acidic.

Key Concepts:
- Amphoteric oxides
- Acidic oxides
- Basic oxides

Memory Aids:
- Remember the common amphoteric oxides: $ZnO, Al_2O_3, SnO_2, PbO, PbO_2, BeO$ .

Quick Tips:
- Remember the common amphoteric oxides.

14 Easy

14 of 171Question permalink

Analyzing +4 | -1

Among the following second ionisation energy is maximum for

View answer and explanation

Correct option: C

Cr

Option C is correct because Cr+ has a stable half-filled d5 configuration, making it difficult to remove another electron.

Solution approach

Correct Answer + Core Reason: Cr. The second ionization energy is maximum for Cr because $Cr^+$ has a stable half-filled $d^5$ configuration.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the elements:
- V (Vanadium): [Ar] 3d³ 4s²
- Ti (Titanium): [Ar] 3d² 4s²
- Cr (Chromium): [Ar] 3d⁵ 4s¹
- Mn (Manganese): [Ar] 3d⁵ 4s²

Step 2: Write the electronic configurations of the corresponding ions after the first ionization:
- V⁺: [Ar] 3d³ 4s¹
- Ti⁺: [Ar] 3d² 4s¹
- Cr⁺: [Ar] 3d⁵
- Mn⁺: [Ar] 3d⁵ 4s¹

Step 3: Analyze the stability of the ions: Cr⁺ has a 3d⁵ configuration, which is a stable half-filled d-orbital. Removing an electron from a stable configuration requires more energy.

Step 4: Compare the second ionization energies: Removing an electron from Cr⁺ will require significantly more energy due to the stability of the half-filled d-orbital.

Step 5: Conclusion: Therefore, the second ionization energy is maximum for Cr.

Why other options are wrong:
- Option A: V: V⁺ does not have a stable electronic configuration, so its second ionization energy will not be as high as Cr.
- Option B: Ti: Ti⁺ does not have a stable electronic configuration, so its second ionization energy will not be as high as Cr.
- Option D: Mn: Mn⁺ loses its 4s¹ electron relatively easily, so its second ionization energy will not be as high as Cr.

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 and require more energy to remove an electron.

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

15 Easy

15 of 171Question permalink

Remembering +4 | -1

Modern periodic table states that the physical and chemical properties of the elements are periodic functions of thier atomic number. This is given by:

View answer and explanation

Correct option: B

Moseley

Option B (Moseley) is correct because Moseley discovered that the properties of elements are periodic functions of their atomic number.

Solution approach

Correct Answer + Core Reason: Moseley. Moseley's work established that the atomic number is the fundamental property determining the chemical behavior of an element.

Step-by-Step Solution:

Step 1: Recall the history of the periodic table:
- Mendeleev arranged elements by atomic weight.
- Moseley discovered the atomic number and rearranged the periodic table based on it.
- Hund's rule deals with electron configuration within orbitals.
- Bohr's model describes the structure of the atom.

Step 2: Identify the scientist responsible for the modern periodic law: Moseley.

Why other options are wrong:
- Option A (Mendeleev): Mendeleev arranged the periodic table based on atomic weight, not atomic number.
- Option C (Hund's): Hund's rule describes the filling of electrons in orbitals.
- Option D (Bohr): Bohr's model describes the structure of the atom but not the periodic law.

Key Concepts:
- Modern Periodic Law
- Atomic Number
- Mendeleev's Periodic Table

Memory Aids:
- Moseley = Atomic Number.

Quick Tips:
- Remember that the modern periodic table is based on atomic number, not atomic weight.

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Remembering +4 | -1

According to Dobereiner, Atomic weight of the middle element was the

View answer and explanation

Correct option: A

Arithmetic mean of the other two elements

Option A (Arithmetic mean of the other two elements) is correct because Dobereiner's triads were based on the principle that the atomic weight of the middle element is approximately the arithmetic mean of the other two.

Solution approach

Correct Answer + Core Reason: Arithmetic mean of the other two elements. Dobereiner's Law of Triads stated that the atomic weight of the middle element is approximately the arithmetic mean of the atomic weights of the other two elements in the triad.

Step-by-Step Solution:

Step 1: Recall Dobereiner's Law of Triads. Dobereiner observed that certain groups of three elements (triads) had similar chemical properties, and the atomic weight of the middle element was approximately the arithmetic mean of the other two.

Step 2: Understand the arithmetic mean. The arithmetic mean of two numbers is calculated by adding the numbers and dividing by 2.

Step 3: Apply the concept to Dobereiner's triads. For example, in the triad of Lithium (Li), Sodium (Na), and Potassium (K), the atomic weight of Sodium is approximately the average of the atomic weights of Lithium and Potassium.

Why other options are wrong:
- Option B (Twice to the sum of other two elements): This is incorrect; Dobereiner's law involves the arithmetic mean, not twice the sum.
- Option C (Thrice to the sum of other two elements): This is incorrect; Dobereiner's law involves the arithmetic mean, not thrice the sum.
- Option D (Equal to the sum of other two elements): This is incorrect; Dobereiner's law involves the arithmetic mean, not the sum.

Key Concepts:
- Dobereiner's Law of Triads
- Arithmetic mean

Memory Aids:
- Remember that Dobereiner's triads involved finding the average atomic weight of two elements to approximate the atomic weight of the middle element.

Quick Tips:
- Know the basic principles of early attempts to classify elements, such as Dobereiner's triads.

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Remembering +4 | -1

The elements present in the core of earth are collectively known as

View answer and explanation

Correct option: D

Siderophiles

Option D (Siderophiles) is correct because siderophiles are elements that tend to dissolve in iron, which is the primary component of the Earth's core.

Solution approach

Correct Answer + Core Reason: Siderophiles. These elements have a strong affinity for iron and are found in the Earth's core.

Step-by-Step Solution:

Step 1: Understand the question: The question asks about the elements present in the Earth's core and their collective name.

Step 2: Recall the composition of Earth's core: The Earth's core is primarily composed of iron and nickel.

Step 3: Define the terms:
- Lithophiles: Elements that prefer to be in silicate rocks (Earth's crust and mantle).
- Nucleophiles: A chemical term referring to species that are attracted to positive charges.
- Chalcophiles: Elements that prefer to combine with sulfur.
- Siderophiles: Elements that prefer to be in iron.

Step 4: Apply the definitions: Since the Earth's core is primarily iron, the elements present there are siderophiles.

Why other options are wrong:
- Option A: Lithophiles are found in the Earth's crust and mantle, not the core.
- Option B: Nucleophiles are a chemical concept and not related to the composition of the Earth's core.
- Option C: Chalcophiles are associated with sulfur-containing minerals, not the Earth's core.

Key Concepts:
- Composition of Earth's core
- Geochemical classification of elements

Memory Aids:
- Sidero- means iron (think of siderosis, iron overload).

Quick Tips:
- Remember the main elements in the Earth's core (iron and nickel) to identify siderophiles.

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Application +4 | -1

Oxidation state of oxygen in compounds

\mathrm{OF}_2

and

\mathrm{Na}_2 \mathrm{O}

respectively are

View answer and explanation

Correct option: A

+2,-2

Option A (+2, -2) is correct because in OF2, oxygen is bonded to the more electronegative fluorine, giving it a positive oxidation state, while in Na2O, oxygen is bonded to the less electronegative sodium, giving it a negative oxidation state.

Solution approach

Correct Answer + Core Reason: +2, -2. Oxygen has +2 oxidation state when bonded to Fluorine and -2 when bonded to Sodium.

Step-by-Step Solution:

Step 1: Determine the oxidation state of oxygen in OF2.\nFluorine is more electronegative than oxygen. Fluorine always has an oxidation state of -1. Since there are two fluorine atoms, the total negative charge is -2. To balance this, oxygen must have an oxidation state of +2.

Step 2: Determine the oxidation state of oxygen in Na2O.\nSodium is less electronegative than oxygen. Sodium always has an oxidation state of +1. Since there are two sodium atoms, the total positive charge is +2. To balance this, oxygen must have an oxidation state of -2.

Why other options are wrong:
- Option B: -2, +2: This is incorrect because the oxidation state of oxygen is +2 in OF2 and -2 in Na2O.
- Option C: -2, -1: This is incorrect because the oxidation state of oxygen is -2 in Na2O, not -1.
- Option D: +1, -1: This is incorrect because the oxidation state of oxygen is +2 in OF2 and -2 in Na2O.

Key Concepts:
- Oxidation state
- Electronegativity

Memory Aids:
- Oxygen usually has an oxidation state of -2, except when bonded to fluorine or itself.

Quick Tips:
- Remember that fluorine is the most electronegative element and always has an oxidation state of -1.

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Remembering +4 | -1

The longest and shortest periods are

View answer and explanation

Correct option: C

6 & 1

Option C (6 & 1) is correct because the 6th period is the longest and the 1st period is the shortest in the periodic table.

Solution approach

Correct Answer + Core Reason: 6 & 1. The 6th period contains 32 elements, making it the longest, while the 1st period contains only 2 elements, making it the shortest.

Step-by-Step Solution:

Step 1: Recall the structure of the periodic table.

Step 2: Identify the number of elements in each period:
- 1st period: 2 elements (Hydrogen and Helium)
- 2nd period: 8 elements
- 3rd period: 8 elements
- 4th period: 18 elements
- 5th period: 18 elements
- 6th period: 32 elements
- 7th period: Incomplete, but contains 32 elements if completed

Step 3: Determine the longest and shortest periods. The 6th period is the longest, and the 1st period is the shortest.

Why other options are wrong:
- Option A (1 & 6): This is incorrect because it reverses the order; the 6th period is the longest, not the shortest.
- Option B (2 & 6): This is incorrect because the 2nd period is not the shortest; the 1st period is the shortest.
- Option D (1 & 7): This is incorrect because the 7th period is incomplete, and the 1st period is the shortest.

Key Concepts:
- Periodic table structure
- Number of elements in each period

Memory Aids:
- Remember that the 1st period is very short (2 elements) and the 6th period is the longest (32 elements).

Quick Tips:
- Know the number of elements in each period to quickly identify the longest and shortest periods.

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Analyzing +4 | -1

Which of the following pairs of atomic numbers represents elements belonging to the same group?

View answer and explanation

Correct option: C

13 and 31

Option C (13 and 31) is correct because elements with atomic numbers 13 and 31 belong to the same group (Group 13).

Solution approach

Correct Answer + Core Reason: 13 and 31. Elements with the same number of valence electrons belong to the same group.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each element.

Step 2: Identify the valence electrons for each element.

Step 3: Compare the number of valence electrons. Elements with the same number of valence electrons belong to the same group.

Step 4: Element with atomic number 13 (Aluminum) has electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^1$ . Valence electrons = 3.

Step 5: Element with atomic number 31 (Gallium) has electronic configuration $[Ar] 3d^{10} 4s^2 4p^1$ . Valence electrons = 3.

Step 6: Since both have 3 valence electrons, they belong to the same group (Group 13).

Why other options are wrong:
- Option A: 11 and 20: 11 (Na) is in Group 1, 20 (Ca) is in Group 2.
- Option B: 12 and 30: 12 (Mg) is in Group 2, 30 (Zn) is in Group 12.
- Option D: 14 and 33: 14 (Si) is in Group 14, 33 (As) is in Group 15.

Key Concepts:
- Electronic Configuration
- Valence Electrons
- Periodic Table Groups

Memory Aids:
- Elements in the same group have the same number of valence electrons.

Quick Tips:
- Knowing the electronic configuration helps determine the group number.

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Understanding +4 | -1

Which of the following oxides is most acidic in nature?

View answer and explanation

Correct option: A

BeO

Option A (BeO) is correct because BeO is the most acidic among the given oxides due to Be's high polarizing power.

Solution approach

Correct Answer + Core Reason: BeO. BeO is the most acidic in nature.

Step-by-Step Solution:

Step 1: Recall the trend of acidity of oxides in a group.\nAcidity of oxides decreases down a group for alkaline earth metals.

Step 2: Consider the positions of the elements in the periodic table.
- Be, Mg, Ca, and Ba are all in group 2 (alkaline earth metals).
- Be is at the top of the group, followed by Mg, Ca, and Ba.

Step 3: Conclusion.\nTherefore, BeO is the most acidic among the given oxides.

Why other options are wrong:
- Option B: MgO is less acidic than BeO.
- Option C: CaO is less acidic than BeO.
- Option D: BaO is less acidic than BeO.

Key Concepts:
- Acidity of oxides
- Periodic trends

Memory Aids:
- Acidity decreases down a group for alkaline earth metal oxides.

Quick Tips:
- Remember the acidity trend of oxides in the periodic table.

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Understanding +4 | -1

The second electron affinity is zero for

View answer and explanation

Correct option: B

Halogens

Option B (Halogens) is correct because after gaining one electron, halogens achieve a stable noble gas configuration, making it energetically unfavorable to add another electron.

Solution approach

Correct Answer + Core Reason: Halogens. After gaining one electron, halogens achieve a stable noble gas configuration, making the second electron affinity highly unfavorable (positive).

Step-by-Step Solution:

Step 1: Define second electron affinity: The energy change when an electron is added to a uninegative gaseous ion.

Step 2: Consider the electronic configuration of halogens: Halogens have 7 valence electrons and need only one more electron to achieve a stable noble gas configuration.

Step 3: After gaining one electron, a halogen becomes a negative ion with a stable electronic configuration. Adding another electron would require overcoming significant electron-electron repulsion and disrupting the stable configuration.

Step 4: Therefore, the second electron affinity for halogens is highly unfavorable and is considered to be positive (endothermic), effectively zero for the purpose of this question.

Why other options are wrong:
- Option A: Alkali metals readily lose one electron and do not have a significant electron affinity.
- Option B: Halogens is the correct answer.
- Option C: Noble gases already have a stable electronic configuration and have very low electron affinity.
- Option D: Transition metals have variable electron affinities depending on their electronic configuration.

Key Concepts:
- Electron Affinity
- Electronic Configuration
- Noble Gas Configuration

Memory Aids:
- Halogens want one electron, not two.

Quick Tips:
- Remember the electronic configurations of different groups in the periodic table.

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Understanding +4 | -1

Elements in the same vertical group of the periodic table have generally the same:

View answer and explanation

Correct option: D

Number of electrons in the outermost shell of their atoms

Option D (Number of electrons in the outermost shell of their atoms) is correct because elements in the same group have the same number of valence electrons, which determines their chemical properties.

Solution approach

Correct Answer + Core Reason: Number of electrons in the outermost shell of their atoms. Elements in the same group have the same number of valence electrons.

Step-by-Step Solution:

Step 1: Recall the definition of a group in the periodic table.

Step 2: Understand that elements in the same group have the same number of valence electrons.

Step 3: Recognize that valence electrons determine chemical properties.

Why other options are wrong:
- Option A (Atomic number): Elements in the same group have different atomic numbers.
- Option B (Electronic configuration): While the *outermost* electronic configuration is similar, the overall electronic configuration is different due to different numbers of electrons.
- Option C (Atomic mass): Elements in the same group have different atomic masses.

Key Concepts:
- Periodic Table Groups
- Valence Electrons
- Electronic Configuration
- Chemical Properties

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

Quick Tips:
- Focus on the relationship between valence electrons and chemical properties.

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Understanding +4 | -1

Which of the following set of elements cannot be a triad

View answer and explanation

Correct option: B

B, Al, Ga

Option B (B, Al, Ga) is correct because the atomic mass of Al is not the arithmetic mean of B and Ga, thus not forming a triad.

Solution approach

Correct Answer + Core Reason: B, Al, Ga. A triad is a group of three elements where the atomic mass of the middle element is approximately the arithmetic mean of the other two.

Step-by-Step Solution:

Step 1: Define what a triad is. A triad consists of three elements with similar chemical properties where the atomic weight of the middle element is approximately the average of the atomic weights of the first and third elements.

Step 2: Check each option:
- A: Li (7), Na (23), K (39). (7+39)/2 = 23. Na is approximately the average of Li and K.
- B: B (10.8), Al (27), Ga (69.7). (10.8+69.7)/2 = 40.25. Al is not approximately the average of B and Ga.
- C: Be (9), Mg (24.3), Ca (40.1). (9+40.1)/2 = 24.55. Mg is approximately the average of Be and Ca.
- D: Cl (35.5), Br (79.9), I (126.9). (35.5+126.9)/2 = 81.2. Br is approximately the average of Cl and I.

Step 3: Conclude that B, Al, and Ga do not form a triad.

Why other options are wrong:
- Option A: Li, Na, and K form a triad because the atomic weight of Na is approximately the average of Li and K.
- Option C: Be, Mg, and Ca form a triad because the atomic weight of Mg is approximately the average of Be and Ca.
- Option D: Cl, Br, and I form a triad because the atomic weight of Br is approximately the average of Cl and I.

Key Concepts:
- Dobereiner's Triads
- Atomic mass

Memory Aids:
- Remember the definition of a triad: the middle element's atomic mass is the average of the other two.

Quick Tips:
- Calculate the average atomic mass of the first and third elements in each set and compare it to the atomic mass of the middle element.

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Analyzing +4 | -1

The electronic configurations of four elements are :

(i)

[{Xe}] 6 s^1

(ii)

[{Xe}] 4 f^{14} 5 d^1 6 s^2

(iii)

[{Ar}] 4 s^2 4 p^5

(iv)

[{Ar}] 3 d^7 4 s^2

Which one of the following statements about these elements is not correct?

View answer and explanation

Correct option: B

(ii) is a d-block element

The correct answer is (ii) is a d-block element (B) because

[Xe] 4 f^{14} 5 d^1 6 s^2

is an f-block element (Lanthanide) since the last electron enters the f-orbital after filling the d-orbital.

Solution approach

Correct Answer + Core Reason: (ii) is a d-block element. $[{Xe}] 4 f^{14} 5 d^1 6 s^2$ is an f-block element.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations of the given elements.

Step 2: (i) $[{Xe}] 6 s^1$ is an s-block element (alkali metal) and a strong reducing agent.

Step 3: (ii) $[{Xe}] 4 f^{14} 5 d^1 6 s^2$ is an f-block element (Lanthanide) because the last electron enters the f-orbital.

Step 4: (iii) $[{Ar}] 4 s^2 4 p^5$ is a p-block element (halogen) and has high negative electron gain enthalpy.

Step 5: (iv) $[{Ar}] 3 d^7 4 s^2$ is a d-block element (transition metal) and shows variable oxidation states.

Step 6: Determine which statement is incorrect.

Step 7: Statement (ii) is incorrect because it states that the element is a d-block element, but it is an f-block element.

Why other options are wrong:
- Option A: (i) is a strong reducing agent is correct because alkali metals are strong reducing agents.
- Option C: (iii) has high negative electron gain enthalpy is correct because halogens have high negative electron gain enthalpy.
- Option D: (iv) shows variable oxidation state is correct because transition metals show variable oxidation states.

Key Concepts:
- Electronic configuration
- Periodic table blocks
- Reducing agents
- Electron gain enthalpy
- Oxidation states

Memory Aids:
- Remember the electronic configurations of elements in different blocks.

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

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Remembering +4 | -1

At present, how many elements are known

View answer and explanation

Correct option: D

118

Option D (118) is correct because there are currently 118 known elements in the periodic table.

Solution approach

Correct Answer + Core Reason: 118. Currently, there are 118 known elements.

Step-by-Step Solution:

Step 1: Recall the number of known elements: The periodic table has been expanded to include 118 elements.

Step 2: Consider the options: The correct number is 118.

Why other options are wrong:
- Option A (110): This is an outdated number of known elements.
- Option B (112): This is an outdated number of known elements.
- Option C (113): This is an outdated number of known elements.

Key Concepts:
- Periodic Table
- Number of Elements

Memory Aids:
- The periodic table is constantly being updated with new elements.

Quick Tips:
- Keep up-to-date with the latest discoveries in chemistry.

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Understanding +4 | -1

In

6^{\text {th}}

period of the modern periodic table, electronic energy levels are in the order:

View answer and explanation

Correct option: A

6 \mathrm{s}, 4 \mathrm{f}, 5 \mathrm{d}, 6 \mathrm{p}

The correct answer is A (

6 \mathrm{s}, 4 \mathrm{f}, 5 \mathrm{d}, 6 \mathrm{p}

) because this is the order in which the electronic energy levels are filled according to the Aufbau principle and Hund's rule.

Solution approach

Correct Answer + Core Reason: $6 \mathrm{s}, 4 \mathrm{f}, 5 \mathrm{d}, 6 \mathrm{p}$ . This is the correct order of filling orbitals based on the (n+l) rule and Hund's rule.

Step-by-Step Solution:

Step 1: Recall the Aufbau principle, which states that electrons first fill the lowest energy levels.

Step 2: Determine the (n+l) values for each subshell: 6s (6+0=6), 4f (4+3=7), 5d (5+2=7), 6p (6+1=7).

Step 3: When (n+l) values are the same, the subshell with the lower 'n' value is filled first. Therefore, 4f is filled before 5d and 6p.

Step 4: The order of filling is 6s, then 4f, then 5d, and finally 6p.

Why other options are wrong:
- Option B: $6 \mathrm{s}, 6 \mathrm{p}, 4 \mathrm{f}, 5 \mathrm{d}$ is incorrect because 4f is filled before 5d and 6p.
- Option C: $4 \mathrm{f}, 5 \mathrm{d}, 6 \mathrm{s}, 6 \mathrm{p}$ is incorrect because 6s is filled before 4f.
- Option D: None is incorrect because option A is the correct order.

Key Concepts:
- Aufbau Principle
- Hund's Rule
- (n+l) Rule
- Electronic Configuration

Memory Aids:
- Remember the Aufbau principle: electrons fill orbitals in order of increasing energy.

Quick Tips:
- Use the (n+l) rule to determine the order of filling orbitals.

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Remembering +4 | -1

According to Mendeleev’s periodic law, the physical and chemical properties of the elements are the periodic functions of their

View answer and explanation

Correct option: A

Atomic weights

Option A (Atomic weights) is correct because Mendeleev's periodic law stated that the physical and chemical properties of elements are periodic functions of their atomic weights.

Solution approach

Correct Answer + Core Reason: Atomic weights. Mendeleev's periodic law was based on atomic weights.

Step-by-Step Solution:

Step 1: Recall Mendeleev's periodic law.\nMendeleev's periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic weights.

Step 2: Consider the other options.\nAtomic volume and atomic number were not the basis of Mendeleev's periodic law.

Why other options are wrong:
- Option B: Atomic volume was not the basis of Mendeleev's periodic law.
- Option C: Atomic number was not known at the time of Mendeleev's work.
- Option D: Mendeleev's periodic law was based on atomic weights.

Key Concepts:
- Mendeleev's periodic law
- Atomic weights

Memory Aids:
- Remember that Mendeleev arranged elements by increasing atomic weight.

Quick Tips:
- Know the historical development of the periodic table.

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Application +4 | -1

An element has a face centred cubic (fcc) structure with a unit cell edge length of 200 pm. The atomic radius is

View answer and explanation

Correct option: A

\frac{{100}}{{\sqrt 2 }}\,\,\mathrm{pm}

Option A (

\frac{{100}}{{\sqrt 2 }}\,\mathrm{pm}

) is correct because in an fcc structure,

4r = a\sqrt{2}

, where a is the edge length and r is the atomic radius.

Solution approach

Correct Answer + Core Reason: $\frac{{100}}{{\sqrt 2 }}\,\mathrm{pm}$ . In an FCC structure, the atoms touch along the face diagonal, so $4r = a\sqrt{2}$ .

Step-by-Step Solution:

Step 1: Identify the type of unit cell: Face-centered cubic (fcc).

Step 2: Recall the relationship between the atomic radius (r) and the edge length (a) in an fcc unit cell: $4r = a\sqrt{2}$

Step 3: Given the edge length, a = 200 pm.

Step 4: Substitute the value of 'a' into the equation: $4r = 200\sqrt{2}$ pm

Step 5: Solve for r: $r = \frac{{200\sqrt{2}}}{4} = \frac{{200}}{{2\sqrt{2}}} = \frac{{100}}{{\sqrt{2}}}$ pm

Why other options are wrong:
- Option B: $\frac{{200}}{{\sqrt 2 }}\,\mathrm{pm}$ is incorrect. This is the length of the face diagonal, not 4 times the radius.
- Option C: $\frac{{100}}{{\sqrt 3 }}\,\mathrm{pm}$ is incorrect. This relationship is for a body-centered cubic (bcc) structure, not fcc.
- Option D: $\sqrt 3 \times 100\,\mathrm{pm}$ is incorrect. This is not the correct relationship for either fcc or bcc structures.

Key Concepts:
- Face-centered cubic (fcc) unit cell
- Relationship between atomic radius and edge length in fcc: $4r = a\sqrt{2}$

Memory Aids:
- FCC: Atoms touch along the face diagonal.
- BCC: Atoms touch along the body diagonal.

Quick Tips:
- Memorize the relationships between atomic radius and edge length for simple cubic, bcc, and fcc unit cells.
- Draw a diagram of the unit cell to visualize the relationship.

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Remembering +4 | -1

The number of elements which can be accommodated in the present set-up of the long form of the periodic table is:

View answer and explanation

Correct option: B

118

Option B (118) is correct because the current long form of the periodic table accommodates 118 elements.

Solution approach

Correct Answer + Core Reason: 118. The long form of the periodic table currently accommodates 118 elements.

Step-by-Step Solution:

Step 1: Recall the number of known elements.\nThe current periodic table is set up to accommodate 118 elements, from Hydrogen (H) to Oganesson (Og).

Why other options are wrong:
- Option A: 120 is not the correct number of elements in the current periodic table.
- Option C: 122 is not the correct number of elements in the current periodic table.
- Option D: 124 is not the correct number of elements in the current periodic table.

Key Concepts:
- Periodic table
- Number of elements

Memory Aids:
- Remember the number of elements in the current periodic table.

Quick Tips:
- Know the organization of the periodic table.

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