CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Medium, Page 5

121 Medium

Understanding +4 | -1

The correct order of ionic radii is:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because in each case, the neutral atom is larger than its positive ions, and the negative ion is larger than the neutral atom.

Solution approach

Correct Answer + Core Reason: All of these. Removing electrons decreases size, adding electrons increases size.

Step-by-Step Solution:

Step 1: Analyze each option:
- A: $\mathrm{Fe} > F{\mathrm{e}^{2 + }} > F{\mathrm{e}^{3 + }}$ - Correct, as positive charge increases, size decreases.
- B: ${\mathrm{O}^{2 - }} > {\mathrm{O}^ - } > {\mathrm{O}^ + }$ - Incorrect, should be O+ > O > O-. However, the question states O2- > O- > O+ which is correct since adding electrons increases size and removing decreases size.
- C: ${\mathrm{I}^ - } > I > {\mathrm{I}^ + }$ - Correct, adding an electron increases size, removing decreases.

Step 2: General trend: Anions are larger than their parent atoms, and cations are smaller than their parent atoms.

Step 3: Conclusion: All the given orders are correct.

Why other options are wrong:
- Option A: While A is correct, it's not the only correct option.
- Option B: While B is correct, it's not the only correct option.
- Option C: While C is correct, it's not the only correct option.

Key Concepts:
- Ionic size
- Effect of charge on size

Memory Aids:
- Anions are bigger, cations are smaller.

Quick Tips:
- Remember the general trend of ionic size with charge.

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Remembering +4 | -1

The IUPAC name of an element with atomic number 119 is

View answer and explanation

Correct option: B

ununennium

Option B (ununennium) is correct because it follows the IUPAC nomenclature rules for naming elements with atomic number 119.

Solution approach

Correct Answer + Core Reason: ununennium. This name is derived from the digits of the atomic number 119 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 119.

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 119:
- 1: un
- 1: un
- 9: enn
- Add the suffix -ium
- Therefore, the name is ununennium.

Why other options are wrong:
- Option A: ununoctium is the name for element 118.
- Option C: unnilennium is not a valid IUPAC name.
- Option D: unununium is the name for element 111.

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.

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Analyzing +4 | -1

Maximum I.E is for

View answer and explanation

Correct option: B

L{i^ + }

Option B (

Li^+

) is correct because removing an electron from

Li^+

(electronic configuration

1s^2

) requires breaking a stable noble gas configuration, which needs maximum energy.

Solution approach

Correct Answer + Core Reason: $Li^+$ . Removing an electron from $Li^+$ (electronic configuration $1s^2$ ) requires breaking a stable noble gas configuration, which needs maximum energy.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the ions.
$Be^+$ : $1s^22s^1$
$Li^+$ : $1s^2$
$B^+$ : $1s^22s^2$
$Ne^+$ : $1s^22s^22p^5$

Step 2: Analyze the ionization process for each ion.
$Be^+ \rightarrow Be^{2+} + e^-$
$Li^+ \rightarrow Li^{2+} + e^-$
$B^+ \rightarrow B^{2+} + e^-$
$Ne^+ \rightarrow Ne^{2+} + e^-$

Step 3: Determine which ionization requires the most energy.\nRemoving an electron from $Li^+$ requires breaking the stable $1s^2$ (noble gas) configuration, which requires a very high amount of energy. Removing an electron from $Ne^+$ is also high, but it's not as stable as the completely filled 1s orbital of $Li^+$ .

Step 4: Final Answer: $Li^+$ has the maximum ionization energy.

Why other options are wrong:
- Option A: $Be^+$ : Removing an electron from $Be^+$ is easier because it only requires removing a 2s electron.
- Option C: $B^+$ : Removing an electron from $B^+$ is easier because it only requires removing a 2s electron.
- Option D: $Ne^+$ : Removing an electron from $Ne^+$ requires removing a 2p electron, but it is not as stable as the completely filled 1s orbital of $Li^+$ .

Key Concepts:
- Ionization energy
- Electronic configuration
- Noble gas configuration

Memory Aids:
- Removing an electron from a noble gas configuration requires a large amount of energy.

Quick Tips:
- Consider the electronic configuration of the ion before ionization.

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Remembering +4 | -1

Which of the following metals exhibits more than one oxidation state?

View answer and explanation

Correct option: D

Fe

Option D (Fe) is correct because Iron can exhibit multiple oxidation states such as +2 and +3.

Solution approach

Correct Answer + Core Reason: Fe. Iron exhibits more than one oxidation state.

Step-by-Step Solution:

Step 1: Recall the common oxidation states of the given metals.
- Na typically exhibits a +1 oxidation state.
- Mg typically exhibits a +2 oxidation state.
- Al typically exhibits a +3 oxidation state.
- Fe can exhibit +2 and +3 oxidation states.

Step 2: Identify the metal with multiple oxidation states.\nIron (Fe) is known to exhibit multiple oxidation states.

Step 3: Conclusion.\nTherefore, Iron exhibits more than one oxidation state.

Why other options are wrong:
- Option A: Sodium typically exhibits only a +1 oxidation state.
- Option B: Magnesium typically exhibits only a +2 oxidation state.
- Option C: Aluminum typically exhibits only a +3 oxidation state.

Key Concepts:
- Oxidation states
- Transition metals

Memory Aids:
- Iron can be in +2 or +3 oxidation state.

Quick Tips:
- Remember that transition metals often exhibit multiple oxidation states.

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Understanding +4 | -1

The periodic table of elements does not

View answer and explanation

Correct option: B

Tell us about the arrangement of atoms in a molecule

Option B (Tell us about the arrangement of atoms in a molecule) is correct because the periodic table primarily deals with elements and their properties, not molecular structure.

Solution approach

Correct Answer + Core Reason: Tell us about the arrangement of atoms in a molecule. The periodic table is about elements, not molecules.

Step-by-Step Solution:

Step 1: Recall the purpose of the periodic table.

Step 2: Understand that it organizes elements based on their properties and electronic configurations.

Step 3: Recognize that it does not directly provide information about the arrangement of atoms in molecules.

Why other options are wrong:
- Option A (Include the inert gases): The periodic table does include inert gases.
- Option C (Allow us to make accurate guess of the properties of undiscovered elements): The periodic table allows for prediction of properties.
- Option D (Reveal regularities in the occurance of elements with similar properties): The periodic table reveals periodic trends.

Key Concepts:
- Periodic Table
- Elements vs. Molecules
- Periodic Trends

Memory Aids:
- The periodic table is primarily about the properties of individual elements.

Quick Tips:
- Distinguish between the information provided by the periodic table and other chemical concepts.

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Analyzing +4 | -1

The first ionization energy of magnesium is smaller as compared to that of elements

X

and

Y

, but higher than that of

Z

. The elements

X

,

Y

and

Z

, respectively, are

View answer and explanation

Correct option: C

argon, chlorine and sodium

Option C is correct because ionization energy generally increases across a period and decreases down a group. Argon and chlorine have higher ionization energies than magnesium, while sodium has a lower ionization energy.

Solution approach

Correct Answer + Core Reason: argon, chlorine and sodium. Ionization energy increases across a period and decreases down a group.

Step-by-Step Solution:

Step 1: Understand the trend of ionization energy in the periodic table.\nIonization energy generally increases across a period (from left to right) and decreases down a group (from top to bottom).

Step 2: Analyze the given information.\nThe first ionization energy of magnesium (Mg) is smaller than that of elements X and Y, but higher than that of Z. This means X and Y are located to the right of Mg or above Mg in the periodic table, and Z is located to the left of Mg or below Mg.

Step 3: Evaluate the options.
- Option A: chlorine, lithium and sodium. Cl > Mg, Li < Mg, Na < Mg. Incorrect.
- Option B: argon, lithium and sodium. Ar > Mg, Li < Mg, Na < Mg. Incorrect.
- Option C: argon, chlorine and sodium. Ar > Mg, Cl > Mg, Na < Mg. Correct.
- Option D: neon, sodium and chlorine. Ne > Mg, Na < Mg, Cl > Mg. Incorrect.

Step 4: Conclusion.\nOption C (argon, chlorine and sodium) satisfies the given conditions.

Why other options are wrong:
- Option A: Lithium has a lower ionization energy than magnesium, so it cannot be X or Y.
- Option B: Lithium has a lower ionization energy than magnesium, so it cannot be X or Y.
- Option C: Argon and chlorine have higher ionization energies than magnesium, and sodium has a lower ionization energy.
- Option D: Sodium has a lower ionization energy than magnesium, so it cannot be X or Y.

Key Concepts:
- Ionization energy
- Periodic trends

Memory Aids:
- Ionization energy increases across a period (left to right).
- Ionization energy decreases down a group (top to bottom).

Quick Tips:
- Remember the general trends of ionization energy, electronegativity, atomic size, and metallic character in the periodic table.

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Remembering +4 | -1

In Mendleev's periodic table some pairs of elements do not follow increasing order of atomic weight known as anomolous pair. The correct example of anomolous pair is

View answer and explanation

Correct option: B

Th, Pa

Option B (

Th, Pa

) is the correct anomalous pair because Thorium and Protactinium's positions were not strictly based on increasing atomic weight in Mendeleev's table.

Solution approach

Correct Answer + Core Reason: $Th, Pa$ . This pair represents an exception in Mendeleev's periodic table where elements were not strictly arranged by increasing atomic weight.

Step-by-Step Solution:

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

Step 2: Understand that some element pairs were placed out of strict atomic weight order to maintain similar chemical properties within groups.

Step 3: Identify $Th$ and $Pa$ as one such pair where the placement deviates from strict atomic weight order.

Step 4: Conclude that $Th, Pa$ is an anomalous pair.

Why other options are wrong:
- **Option A ( $Cl, Ar$ ): Chlorine and Argon generally follow the increasing order of atomic weight in the modern context, although Argon was placed before Potassium by Mendeleev based on properties.
- Option C ( $Te, In$ ): Tellurium and Indium do not represent a classic anomalous pair in Mendeleev's original table. Tellurium and Iodine are the classic example.
- Option D ( $Co, Nb$ ): Cobalt and Niobium do not constitute a well-known anomalous pair in Mendeleev's periodic table. Cobalt and Nickel are the classic example.

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

Memory Aids:
- Remember the classic anomalous pairs like Te-I and Ar-K.

Quick Tips:**
- Focus on the historical context of Mendeleev's table and the exceptions to the atomic weight arrangement.

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Remembering +4 | -1

According to the Periodic Law of elements, the variation in properties of elements is related to their

View answer and explanation

Correct option: C

atomic numbers

Option C (atomic numbers) is correct because the modern periodic law states that the properties of elements are periodic functions of their atomic numbers.

Solution approach

Correct Answer + Core Reason: atomic numbers. The Periodic Law states that element properties vary with atomic number.

Step-by-Step Solution:

Step 1: Recall the Periodic Law: The modern Periodic Law states that the chemical and physical properties of elements are periodic functions of their atomic numbers.

Step 2: Conclusion: The correct answer is atomic numbers.

Why other options are wrong:
- Option A: atomic masses: The original periodic law was based on atomic masses, but this has been superseded by the modern law based on atomic numbers.
- Option B: nuclear masses: Nuclear mass is related to atomic mass, but the periodic law is based on atomic number.
- Option D: nuclear neutron-proton number ratios: This is not the basis of the periodic law.

Key Concepts:
- Periodic Law
- Atomic number

Memory Aids:
- Remember that the modern periodic table is organized by atomic number.

Quick Tips:
- Know the difference between the original and modern periodic laws.

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Understanding +4 | -1

Which one of the following has maximum ionisation potential?

View answer and explanation

Correct option: B

Be

Option B (Be) is correct because Be has a smaller atomic size and a higher effective nuclear charge compared to the other elements.

Solution approach

Correct Answer + Core Reason: Be. Beryllium has a smaller size and higher effective nuclear charge compared to the others.

Step-by-Step Solution:

Step 1: Recall the factors affecting ionization potential.

Step 2: Ionization potential is affected by factors such as nuclear charge, atomic size, and electronic configuration. Smaller atomic size and higher nuclear charge generally lead to higher ionization potential.

Step 3: Locate the elements K, Be, Na, and Mg on the periodic table. K and Na are in Group 1, while Be and Mg are in Group 2. Be is above Mg.

Step 4: Elements in Group 2 generally have higher ionization potentials than elements in Group 1 in the same period. Be is above Mg, so Be will have a higher ionization potential than Mg. K and Na will have lower ionization potentials than Be and Mg. Be has the smallest atomic size among these elements, and a relatively high effective nuclear charge, leading to the highest ionization potential.

Why other options are wrong:
- Option A: K has a large atomic size and low nuclear charge, resulting in a low ionization potential.
- Option C: Na has a large atomic size and low nuclear charge, resulting in a low ionization potential.
- Option D: Mg has a larger atomic size than Be, resulting in a lower ionization potential than Be.

Key Concepts:
- Periodic trends: Ionization potential
- Atomic size and nuclear charge

Memory Aids:
- Higher ionization potential means it's harder to remove an electron.

Quick Tips:
- Use the periodic table to compare the relative positions of the elements and consider the factors affecting ionization potential.

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Analyzing +4 | -1

Among the statements (I-IV), the correct ones are:
I)

Be

has smaller atomic radius compared to

Mg

II)

Be

has higher ionisation enthalpy than

Al

III) Charge/radius ratio of

Be

is greater than that of

Al

IV) Both

Be

and

Al

form mainly covalent compounds.

View answer and explanation

Correct option: B

(I), (II) and (IV)

Option B ((I), (II) and (IV)) is correct because Be has a smaller atomic radius than Mg, higher ionization enthalpy than Al, and both form mainly covalent compounds.

Solution approach

Correct Answer + Core Reason: (I), (II) and (IV). Be has smaller atomic radius compared to Mg, higher ionization enthalpy than Al, and both Be and Al form mainly covalent compounds.

Step-by-Step Solution:

Step 1: Analyze statement I: $Be$ has smaller atomic radius compared to $Mg$ . This is true because atomic radius increases down a group.

Step 2: Analyze statement II: $Be$ has higher ionisation enthalpy than $Al$ . This is true. Ionization enthalpy generally decreases down a group, but Be has a higher ionization enthalpy than Al due to its smaller size and higher effective nuclear charge. Also, Be has a completely filled s-orbital which makes it harder to remove an electron.

Step 3: Analyze statement III: Charge/radius ratio of $Be$ is greater than that of $Al$ . This is true. Be has a smaller size and higher charge density than Al, so its charge/radius ratio is greater.

Step 4: Analyze statement IV: Both $Be$ and $Al$ form mainly covalent compounds. This is true due to their high polarizing power and tendency to form coordinate covalent bonds.

Step 5: Identify the correct combination of statements: Statements I, II, and IV are correct.

Why other options are wrong:
- Option A: (II), (III) and (IV) is incorrect because statement I is also correct.
- Option C: (I), (III) and (IV) is incorrect because statement II is also correct.
- Option D: (I), (II) and (III) is incorrect because statement IV is also correct.

Key Concepts:
- Atomic radius trend
- Ionization enthalpy trend
- Charge/radius ratio
- Covalent character
- Diagonal relationship

Memory Aids:
- Down a group: atomic radius increases, ionization enthalpy decreases.
- Diagonal relationship: Be resembles Al.

Quick Tips:
- Remember the general trends in the periodic table, but be aware of exceptions like the Be-Al relationship.

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Analyzing +4 | -1

The electronic configuration of four elements are

(i)

[\mathrm{Xe}] 5 s^1

(ii)

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

(iii)

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

(iv)

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

Select the incorrect statement about these elements.

View answer and explanation

Correct option: B

(ii) is a

d

-block element

Option B is the incorrect statement. (ii) is an f-block element because the last electron enters the 4f orbital.

Solution approach

Correct Answer + Core Reason: (ii) is a $d$ -block element. The electronic configuration shows the last electron entering the 4f orbital, making it an f-block element, not a d-block element.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations:
(i) $[\mathrm{Xe}] 5 s^1$ - s-block element
(ii) $[\mathrm{Xe}] 4 f^{14}, 5 d^1, 6 s^2$ - f-block element (last electron enters the 4f orbital)
(iii) $[\mathrm{Ar}] 4 s^2 4 p^5$ - p-block element
(iv) $[\mathrm{Ar}] 3 d^7, 4 s^2$ - d-block element

Step 2: Evaluate each statement:\nA. (i) is a strong reducing agent - Alkali metals are strong reducing agents, so this is correct.\nB. (ii) is a $d$ -block element - Incorrect, it's an f-block element.\nC. (iii) has high magnitude of $\Delta e_g H$ - Halogens have high electron gain enthalpy, so this is correct.\nD. (iv) exhibits variable oxidation states - Transition metals exhibit variable oxidation states, so this is correct.

Why other options are wrong:
- Option A: (i) is indeed a strong reducing agent as it readily loses its single electron.
- Option C: (iii) is a halogen and has a high electron gain enthalpy.
- Option D: (iv) is a transition metal and exhibits variable oxidation states.

Key Concepts:
- Electronic Configuration
- Block Elements (s, p, d, f)
- Reducing Agent
- Electron Gain Enthalpy
- Variable Oxidation States

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

Quick Tips:
- Identify the block of an element by looking at the last electron's orbital.

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Analyzing +4 | -1

Electronic configurations of four elements A, B,

\mathrm{C}

and

\mathrm{D}

are given below :

(i)

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

(ii)

1 s^2 2 s^2 2 p^4

(iii)

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

(iv)

1 s^2 2 s^2 2 p^5

Which of the following is the correct order of increasing tendency to gain electron?

View answer and explanation

Correct option: A

(i)

<

(iii)

<

(ii)

<

(iv)

Option A ((i)

<

(iii)

<

(ii)

<

(iv)) is correct because the electron gain enthalpy becomes more negative as we move towards a more stable electronic configuration.

Solution approach

Correct Answer + Core Reason: (i) $<$ (iii) $<$ (ii) $<$ (iv). Electron gain enthalpy becomes more negative as the tendency to gain an electron increases.

Step-by-Step Solution:

Step 1: Analyze the electronic configurations.

Step 2: Determine the stability of each configuration.

Step 3: Relate stability to the tendency to gain an electron.

Step 4: (i) $1s^2 2s^2 2p^6$ is a noble gas configuration (Ne), very stable, least tendency to gain an electron.

Step 5: (ii) $1s^2 2s^2 2p^4$ needs 2 more electrons to achieve noble gas configuration.

Step 6: (iii) $1s^2 2s^2 2p^6 3s^1$ needs 7 more electrons to achieve noble gas configuration. It will lose one electron more easily.

Step 7: (iv) $1s^2 2s^2 2p^5$ needs 1 more electron to achieve noble gas configuration. Highest tendency to gain an electron.

Step 8: Therefore, the order of increasing tendency to gain electron is (i) $<$ (iii) $<$ (ii) $<$ (iv).

Why other options are wrong:
- Option B: (i) $<$ (ii) $<$ (iii) $<$ (iv): Incorrect order.
- Option C: (iv) $<$ (ii) $<$ (iii) $<$ (i): Incorrect order.
- Option D: (iv) $<$ (i) $<$ (ii) $<$ (iii): Incorrect order.

Key Concepts:
- Electronic Configuration
- Electron Gain Enthalpy
- Stability of Noble Gas Configuration

Memory Aids:
- Noble gases have the least tendency to gain electrons.

Quick Tips:
- Elements closer to achieving a noble gas configuration have a higher tendency to gain electrons.

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Application +4 | -1

An element, which belongs to third period and group 16 in the periodic table has electronic configuration.

View answer and explanation

Correct option: B

1 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^4

Option B (

1 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^4

) is correct because it represents an element in the third period (3 energy levels) and group 16 (6 valence electrons).

Solution approach

Correct Answer + Core Reason: $1 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^4$ . This configuration has 3 energy levels (3rd period) and 6 valence electrons (Group 16).

Step-by-Step Solution:

Step 1: Understand the question: We need to find the electronic configuration of an element in the 3rd period and group 16.

Step 2: Determine the period: The 3rd period means the highest principal quantum number (n) is 3. So, the electronic configuration should include $3s$ and $3p$ orbitals.

Step 3: Determine the group: Group 16 means the element has 6 valence electrons. These valence electrons will be in the $3s$ and $3p$ orbitals.

Step 4: Write the electronic configuration: The general configuration for the 3rd period and group 16 is $1s^2 2s^2 2p^6 3s^2 3p^4$ . The $3s$ orbital has 2 electrons, and the $3p$ orbital has 4 electrons, totaling 6 valence electrons.

Why other options are wrong:
- **Option A ( $1 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^3$ ): This configuration has 5 valence electrons, so it belongs to group 15, not 16.
- Option C ( $1 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^5$ ): This configuration has 7 valence electrons, so it belongs to group 17, not 16.
- Option D ( $1 {~s}^2, 2 {~s}^2 2 {p}^4$ ): This configuration only has two energy levels (2nd period), not the 3rd period.

Key Concepts:
- Electronic configuration
- Periodic table trends
- Valence electrons

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

Quick Tips:**
- The period number is the highest principal quantum number (n).
- For p-block elements, the group number is 10 + number of valence electrons.

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Application +4 | -1

If ionization potential for hydrogen atom is 13.6 eV, then ionization potential for

L{i^{ + 2}}

will be:

View answer and explanation

Correct option: D

122.4 eV

Option D (122.4 eV) is correct because the ionization potential is proportional to Z² for hydrogen-like species, where Z is the atomic number.

Solution approach

Correct Answer + Core Reason: 122.4 eV. The ionization potential is proportional to $Z^2$ for hydrogen-like species.

Step-by-Step Solution:

Step 1: Recall the formula for ionization energy of hydrogen-like species.
$IE = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$ \nFor hydrogen, $Z = 1$ and $n = 1$ , so $IE_H = 13.6 \text{ eV}$

Step 2: For $Li^{+2}$ , $Z = 3$ and $n = 1$ .
$IE_{Li^{+2}} = 13.6 \times \frac{3^2}{1^2} = 13.6 \times 9 = 122.4 \text{ eV}$

Step 3: Final Answer
$122.4 \text{ eV}$

Why other options are wrong:
- Option A: 54.4 eV: This is incorrect; it might be calculated using an incorrect formula or an incorrect value for Z.
- Option B: 6.8 eV: This is incorrect; it might be calculated by dividing 13.6 by 2.
- Option C: 13.6 eV: This is incorrect; it is the ionization potential for hydrogen, not $Li^{+2}$ .

Key Concepts:
- Ionization Potential
- Hydrogen-like Species

Memory Aids:
- Remember that ionization energy is proportional to $Z^2$ .

Quick Tips:
- Use the formula $IE = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$ for hydrogen-like species.
- Identify the atomic number (Z) correctly.

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Analyzing +4 | -1

Assertion (A):

\mathrm{Be}

and

\mathrm{Al}

show some similar properties.
Reason (R): The metallic radius of

\mathrm{Be}

is less than the metallic radius of

\mathrm{Al}

.

View answer and explanation

Correct option: B

Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion

Option B is correct. Beryllium (Be) and Aluminum (Al) exhibit diagonal relationship due to similar polarizing power, but the reason given about metallic radii is true but not the correct explanation for the similarity in properties.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion. Be and Al show similar properties due to similar charge/size ratio, leading to similar polarizing power, not solely due to metallic radii.

Step-by-Step Solution:

Step 1: Understand the Assertion: Be and Al exhibit a diagonal relationship in the periodic table, leading to similar chemical properties.

Step 2: Understand the Reason: The metallic radius of Be is indeed less than that of Al.

Step 3: Evaluate the relationship: While both the assertion and reason are individually correct, the smaller metallic radius of Be compared to Al is not the primary reason for their similar properties. The diagonal relationship arises from the similarity in ionic potential (charge/radius ratio) or polarizing power.

Step 4: Conclusion: The reason is not the correct explanation for the assertion.

Why other options are wrong:
- Option A: The reason is not the correct explanation, even though both are correct statements.
- Option C: The reason is correct, so this is incorrect.
- Option D: The assertion is correct, so this is incorrect.

Key Concepts:
- Diagonal Relationship in the Periodic Table
- Metallic Radius
- Polarizing Power
- Ionic Potential

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

Quick Tips:
- Consider the underlying reasons for periodic trends and relationships, not just superficial observations.

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Analyzing +4 | -1

The ionisation energy will be maximum for the process:

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because removing two electrons from Be requires significantly more energy due to its small size and relatively high nuclear charge compared to removing electrons from Cs, Li, or Ba.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be} \to \mathrm{B}{\mathrm{e}^{2 + }}$ . Removing two electrons from a small atom with a relatively high nuclear charge requires the most energy.

Step-by-Step Solution:

Step 1: Understand ionization energy: It's the energy required to remove an electron from an atom or ion in the gaseous phase.

Step 2: Consider the size and nuclear charge of each element.

Step 3: Ba is a large atom with low ionization energy. $\mathrm{Ba} \to \mathrm{B}{\mathrm{a}^{2 + }}$ will require energy, but not the maximum.

Step 4: Be is a smaller atom than Ba, Li, and Cs. Removing two electrons will require more energy than removing one electron from Cs or Li.

Step 5: Cs and Li are alkali metals with low ionization energies. Removing one electron is relatively easy.

Step 6: Therefore, $\mathrm{Be} \to \mathrm{B}{\mathrm{e}^{2 + }}$ requires the maximum ionization energy.

Why other options are wrong:
- Option A: $\mathrm{Ba} \to \mathrm{B}{\mathrm{a}^{2 + }}$ : Barium is a large atom, so its ionization energy is lower than that of Beryllium.
- Option C: $\mathrm{Cs} \to \mathrm{C}{\mathrm{s}^ + }$ : Cesium is an alkali metal and has a very low ionization energy.
- Option D: $\mathrm{Li} \to \mathrm{L}{\mathrm{i}^ + }$ : Lithium is an alkali metal and has a low ionization energy.

Key Concepts:
- Ionization energy
- Atomic size
- Nuclear charge

Memory Aids:
- Smaller atoms have higher ionization energies.

Quick Tips:
- Consider the size and nuclear charge of the atoms involved. Smaller atoms and higher nuclear charges lead to higher ionization energies.

137 Medium

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Analyzing +4 | -1

S-I: The second ionisation energy of '

O

' is greater than that of '

N

'.

S-II: The half filled p-orbitals cause greater stability.

View answer and explanation

Correct option: A

Both I and II are correct

Option A is correct because both statements are true: the second ionization energy of O is greater than N, and half-filled p-orbitals contribute to greater stability.

Solution approach

Correct Answer + Core Reason: Both I and II are correct. The second ionization energy of oxygen is greater than that of nitrogen, and half-filled p-orbitals cause greater stability.

Step-by-Step Solution:

Step 1: Analyze Statement I: The electronic configuration of O is $1s^22s^22p^4$ and N is $1s^22s^22p^3$ . After removing one electron, O+ becomes $1s^22s^22p^3$ (half-filled) and N+ becomes $1s^22s^22p^2$ . Removing an electron from a half-filled configuration requires more energy. However, the second ionization energy of O is greater than that of N because after the first ionization, O+ has a half-filled configuration, while N+ does not. Therefore, it's harder to remove an electron from O+ than N+.

Step 2: Analyze Statement II: Half-filled and fully filled orbitals have extra stability due to exchange energy. This is a well-known concept. Statement II is correct.

Step 3: The reason is not the correct explanation for the assertion. The higher second ionization energy of oxygen is due to the higher effective nuclear charge experienced by the electrons in oxygen compared to nitrogen.

Why other options are wrong:
- Option B: Both statements are not incorrect.
- Option C: Statement I is not incorrect.
- Option D: Statement II is not incorrect.

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:
- Consider the electronic configurations of the ions when comparing ionization energies.

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Analyzing +4 | -1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The ionic radii of

{O^{2 - }}

and

M{g^{2 + }}

are same.
Reason: Both

{O^{2 - }}

and

M{g^{2 + }}

are isoelectronic species.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: D

(A) is false but (R) is true.

Option D ((A) is false but (R) is true) is correct because while both ions are isoelectronic, their ionic radii are different.

Solution approach

Correct Answer + Core Reason: (A) is false but (R) is true. ${O^{2 - }}$ and $M{g^{2 + }}$ are isoelectronic, but their ionic radii are not the same.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): The ionic radii of ${O^{2 - }}$ and $M{g^{2 + }}$ are the same. This is false. Although they are isoelectronic, they have different nuclear charges.

Step 2: Analyze Reason (R): Both ${O^{2 - }}$ and $M{g^{2 + }}$ are isoelectronic species. This is true. Both have 10 electrons.

Step 3: Determine the relationship between Assertion and Reason. Since the assertion is false and the reason is true, the correct option is (A) is false but (R) is true.

Why other options are wrong:
- Option A: Both (A) and (R) are true and (R) is the correct explanation of (A). This is incorrect because (A) is false.
- Option B: Both (A) and (R) are true but (R) is not the correct explanation of (A). This is incorrect because (A) is false.
- Option C: (A) is true but (R) is false. This is incorrect because (A) is false and (R) is true.

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

Memory Aids:
- Isoelectronic species have the same number of electrons.
- Higher nuclear charge leads to smaller ionic radius for isoelectronic species.

Quick Tips:
- When comparing isoelectronic species, consider the number of protons. The ion with more protons will have a smaller radius.

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Understanding +4 | -1

Which of the following transitions involves maximum amount of energy?

View answer and explanation

Correct option: D

{M^{2 + }}\left( \mathrm{g} \right) \to {M^{3 + }}\left( \mathrm{g} \right)

Option D (

{M^{2 + }}\left( \mathrm{g} \right) \to {M^{3 + }}\left( \mathrm{g} \right)

) is correct because the ionization energy increases with each successive ionization due to increased nuclear attraction on the remaining electrons.

Solution approach

Correct Answer + Core Reason: ${M^{2 + }}\left( \mathrm{g} \right) \to {M^{3 + }}\left( \mathrm{g} \right)$ . Removing an electron from a more positively charged ion requires more energy.

Step-by-Step Solution:

Step 1: Understand the concept of ionization energy.

Step 2: Ionization energy is the energy required to remove an electron from an atom or ion in the gaseous phase.

Step 3: Successive ionization energies increase because as electrons are removed, the remaining electrons are held more tightly by the nucleus due to a higher effective nuclear charge.

Step 4: Compare the given transitions. Removing an electron from a 2+ ion to form a 3+ ion requires significantly more energy than removing an electron from a neutral atom, a negative ion, or a 1+ ion.

Step 5: Therefore, the transition ${M^{2 + }}\left( \mathrm{g} \right) \to {M^{3 + }}\left( \mathrm{g} \right)$ involves the maximum amount of energy.

Why other options are wrong:
- Option A: ${M^ - }\left( \mathrm{g} \right) \to M\left( \mathrm{g} \right)$ : This involves removing an electron from a negative ion, which requires less energy than removing an electron from a positive ion or a neutral atom.
- Option B: $M\left( \mathrm{g} \right) \to {M^ + }\left( \mathrm{g} \right)$ : This is the first ionization energy, which is less than subsequent ionization energies.
- Option C: ${M^ + }\left( \mathrm{g} \right) \to {M^{2 + }}\left( \mathrm{g} \right)$ : This is the second ionization energy, which is less than the third ionization energy.

Key Concepts:
- Ionization Energy
- Successive Ionization Energies
- Effective Nuclear Charge

Memory Aids:
- Remember that each successive ionization requires more energy: IE1 < IE2 < IE3, etc.

Quick Tips:
- When comparing ionization energies, consider the charge of the ion. The higher the positive charge, the greater the ionization energy.

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Application +4 | -1

If electronegativity of

x

be 3.2 and that of

y

be 2.2 , the percentage ionic character of

x y

is -

View answer and explanation

Correct option: A

19.5

Option A (19.5) is correct because the percentage ionic character can be estimated using the formula: % ionic character = 16|XA - XB| + 3.5|XA - XB|².

Solution approach

Correct Answer + Core Reason: 19.5. The percentage ionic character is estimated using the electronegativity difference.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference between x and y.\nElectronegativity difference = |3.2 - 2.2| = 1.0

Step 2: Use the formula to estimate the percentage ionic character.\nPercentage ionic character = 16|XA - XB| + 3.5|XA - XB|²\nPercentage ionic character = 16(1.0) + 3.5(1.0)² = 16 + 3.5 = 19.5\%

Why other options are wrong:
- Option B: 18.5: This is incorrect because the calculation is not accurate.
- Option C: 9.5: This is incorrect because the calculation is not accurate.
- Option D: 29.5: This is incorrect because the calculation is not accurate.

Key Concepts:
- Electronegativity
- Percentage ionic character

Memory Aids:
- Remember the formula to estimate the percentage ionic character.

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

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Application +4 | -1

Predict the correct formula of compounds, by the following pair of elements like aluminium and sulphur with the help of oxidation states,

View answer and explanation

Correct option: B

\mathrm{Al}_2 \mathrm{S}_3

Option B (

\mathrm{Al}_2 \mathrm{S}_3

) is correct because it represents the neutral compound formed by balancing the charges of Al

^{3+}

and S

^{2-}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Al}_2 \mathrm{S}_3$ . The correct formula is determined by balancing the oxidation states of aluminium (+3) and sulfur (-2).

Step-by-Step Solution:

Step 1: Identify the oxidation states of aluminium and sulfur.\nAluminium (Al) has an oxidation state of +3.\nSulfur (S) has an oxidation state of -2.

Step 2: Criss-cross the oxidation states to determine the subscripts in the formula.\nAl $^{3+}$ and S $^{2-}$ will combine to form Al $_2$ S $_3$ .

Step 3: Verify that the compound is neutral.
2(Al $^{3+}$ ) + 3(S $^{2-}$ ) = 2(+3) + 3(-2) = +6 - 6 = 0

Step 4: Final Answer: The correct formula is $\mathrm{Al}_2 \mathrm{S}_3$ .

Why other options are wrong:
- Option A: $\mathrm{Al}_2 \mathrm{S}_2$ : This formula implies that the oxidation states are not properly balanced. It suggests a 1:1 ratio, which is incorrect.
- Option C: $\mathrm{Al}_3 \mathrm{S}_2$ : This formula incorrectly assigns the subscripts, not balancing the charges properly.
- Option D: $\mathrm{Al}_4 \mathrm{S}_2$ : This formula also incorrectly assigns the subscripts, not balancing the charges properly.

Key Concepts:
- Oxidation states
- Chemical formula determination
- Balancing charges in compounds

Memory Aids:
- Criss-cross method for determining subscripts in chemical formulas.

Quick Tips:
- Always check if the resulting compound is neutral by summing the charges.

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Understanding +4 | -1

Positive electron gain enthalpy is highest for

View answer and explanation

Correct option: C

Ne

Option A (Xe) has the highest positive electron gain enthalpy because noble gases have stable, filled electron shells and resist adding an extra electron.

Solution approach

Correct Answer + Core Reason: Xe. Noble gases have stable, filled electron shells and resist adding an extra electron.

Step-by-Step Solution:

Step 1: Recall the definition of electron gain enthalpy. It is the energy change when an electron is added to a neutral gaseous atom.

Step 2: Consider the electronic configurations of the given elements.\nNe: $1{s^2}2{s^2}2{p^6}$ \nAr: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$ \nKr: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}$ \nXe: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}$

Step 3: Noble gases have completely filled electron shells, making them very stable. Adding an electron to a noble gas requires a significant amount of energy, resulting in a positive electron gain enthalpy.

Step 4: Xe has the largest atomic size among the given noble gases. While all have positive electron gain enthalpy, the larger the atom, the less the electron density, and the lower the repulsion for an incoming electron. However, the positive value is still the highest due to the stable configuration. The question is tricky, but since all are noble gases, the answer is the one with the highest positive value, which is Xe.

Why other options are wrong:
- Option B: Kr has a positive electron gain enthalpy, but it is lower than Xe.
- Option C: Ne has a positive electron gain enthalpy, but it is lower than Xe.
- Option D: Ar has a positive electron gain enthalpy, but it is lower than Xe.

Key Concepts:
- Electron Gain Enthalpy
- Electronic Configuration
- Stability of Noble Gas Configurations

Memory Aids:
- Noble gases have positive electron gain enthalpies.

Quick Tips:
- Remember that noble gases resist adding electrons due to their stable configurations.

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Understanding +4 | -1

In which one of the following pairs the radius of the second species is greater than that of the first?

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because for isoelectronic species, the species with fewer protons has a larger radius.

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}^{2 - }},{\mathrm{N}^{3 - }}$ . For isoelectronic species, more negative charge means larger size.

Step-by-Step Solution:

Step 1: Analyze each pair of ions.

Step 2: For option A (Na, Mg), both are neutral atoms. Mg has a higher nuclear charge and smaller size than Na.

Step 3: For option B ( ${\mathrm{O}^{2 - }},{\mathrm{N}^{3 - }}$ ), both are isoelectronic (same number of electrons). N^3- has a smaller nuclear charge (7) than O^2- (8). Therefore, N^3- has a larger radius.

Step 4: For option C ( $\mathrm{L}{\mathrm{i}^ + },\mathrm{\;B}{\mathrm{e}^{2 + }}$ ), both are isoelectronic. Be²⁺ has a higher nuclear charge and smaller size than Li⁺.

Step 5: For option D ( $\mathrm{B}{\mathrm{a}^{2 + }},\mathrm{\;S}{\mathrm{r}^{2 + }}$ ), Ba and Sr are in the same group. Ba is below Sr, so Ba²⁺ has a larger radius.

Step 6: The question asks for the pair where the second species is *greater* than the first. Only option B fits this condition, but the question is asking for the opposite. The correct interpretation is which option is the radius of the *second* species greater than the *first*. In option B, N3- has 10 electrons and 7 protons. O2- has 10 electrons and 8 protons. Since N3- has a smaller nuclear charge, it's radius is larger.

Why other options are wrong:
- Option A: Mg has a higher nuclear charge than Na, making it smaller.
- Option C: Be²⁺ has a higher nuclear charge than Li⁺, making it smaller.
- Option D: Ba²⁺ is larger than Sr²⁺ because it is located below Sr in the periodic table.

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

Memory Aids:
- For isoelectronic species, higher positive charge = smaller size, higher negative charge = larger size.

Quick Tips:
- When comparing isoelectronic species, focus on the number of protons (nuclear charge).

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Application +4 | -1

First and second ionisation energies of magnesium are 7.646 and

15.035 \mathrm{eV}

respectively. The amount of energy in

\mathrm{kJ}

needed to convert all the atoms of magnesium into

\mathrm{Mg}^{2+}

ions present in

12 \mathrm{mg}

of magnesium vapours is :

[Given :

1 \mathrm{eV}=96.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

]

View answer and explanation

Correct option: A

1.1

Option A (1.1) is correct because it calculates the total ionization energy for converting all Mg atoms in 12mg to Mg²⁺ ions, considering both first and second ionization energies.

Solution approach

Correct Answer + Core Reason: 1.1. The energy needed is calculated by summing the first and second ionization energies and scaling by the number of moles of Mg.

Step-by-Step Solution:

Step 1: Calculate the number of moles of Mg: $n = \frac{m}{M} = \frac{12 \times 10^{-3} \text{ g}}{24 \text{ g/mol}} = 0.5 \times 10^{-3} \text{ mol}$ .

Step 2: Calculate the total ionization energy in eV: $IE_{total} = IE_1 + IE_2 = 7.646 + 15.035 = 22.681 \text{ eV}$ .

Step 3: Convert the total ionization energy to kJ/mol: $IE_{total} = 22.681 \text{ eV} \times 96.5 \text{ kJ/mol/eV} = 2188.9 \text{ kJ/mol}$ .

Step 4: Calculate the total energy needed for 0.5 x 10⁻³ mol of Mg: $E = n \times IE_{total} = 0.5 \times 10^{-3} \text{ mol} \times 2188.9 \text{ kJ/mol} = 1.094 \text{ kJ} \approx 1.1 \text{ kJ}$ .

Why other options are wrong:
- Option B (1.5): This is incorrect. It likely arises from an error in the calculation or a misunderstanding of how to combine the ionization energies.
- Option C (2): This is incorrect. It likely arises from an error in the calculation or a misunderstanding of how to combine the ionization energies.
- Option D (0.5): This is incorrect. It likely arises from only considering one of the ionization energies or an error in the calculation.

Key Concepts:
- Ionization Energy
- Moles Calculation
- Unit Conversion

Memory Aids:
- Remember to convert units appropriately (eV to kJ/mol).

Quick Tips:
- Pay attention to units and conversions.

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Remembering +4 | -1

An element with atomic number 59 belongs to

View answer and explanation

Correct option: A

f

-block

Option A (

f

-block) is correct because an element with atomic number 59 is Praseodymium, which is a lanthanide and belongs to the f-block.

Solution approach

Correct Answer + Core Reason: $f$ -block. An element with atomic number 59 is Praseodymium (Pr), which is a lanthanide and belongs to the f-block.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 59. The element with atomic number 59 is Praseodymium (Pr).

Step 2: Determine the electronic configuration of Praseodymium. The electronic configuration of Praseodymium is [Xe] 4f3 6s2.

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

Why other options are wrong:
- **Option B ( $d$ -block): Elements in the d-block have their last electron entering a d-orbital.
- Option C ( $p$ -block): Elements in the p-block have their last electron entering a p-orbital.
- Option D ( $s$ -block): Elements in the s-block have their last electron entering an s-orbital.

Key Concepts:
- Electronic configuration
- f-block elements
- Lanthanides

Memory Aids:
- Remember that lanthanides and actinides belong to the f-block.

Quick Tips:**
- Know the location of the f-block in the periodic table to quickly identify f-block elements.

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Understanding +4 | -1

The element whose electronic configuration is

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

is:

View answer and explanation

Correct option: B

Metal

Option B is correct because the element with the electronic configuration

1s^22s^22p^63s^2

has 2 valence electrons in the 3s orbital, indicating it is a metal (specifically Magnesium).

Solution approach

Correct Answer + Core Reason: Metal. The electronic configuration indicates a metal because it has 2 valence electrons.

Step-by-Step Solution:

Step 1: Identify the electronic configuration: $1s^22s^22p^63s^2$ .

Step 2: Determine the number of valence electrons: The outermost shell is the 3rd shell, which has 2 electrons in the 3s orbital.

Step 3: Elements with 1, 2, or 3 valence electrons are generally metals.

Step 4: Elements with 5, 6, or 7 valence electrons are generally non-metals.

Step 5: Elements with 4 valence electrons are generally metalloids.

Step 6: Elements with 8 valence electrons are noble gases.

Why other options are wrong:
- Option A: Metalloids typically have 4 valence electrons.
- Option C: Noble gases have 8 valence electrons (except Helium with 2).
- Option D: Non-metals typically have 5, 6, or 7 valence electrons.

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.

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Application +4 | -1

Predict the period, group and block of an element having electronic configuration

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

.

View answer and explanation

Correct option: A

Period

=3^{\text {rd}},

block

= p,

group

=16

Option A (Period = 3rd, block = p, group = 16) is correct because the highest principal quantum number is 3 (period 3), the last electron enters the p-orbital (p-block), and the number of valence electrons is 6 (group 16).

Solution approach

Correct Answer + Core Reason: Period $=3^{\text {rd}},$ block $= p,$ group $=16$ . The electronic configuration indicates the element is in the 3rd period, p-block, and group 16.

Step-by-Step Solution:

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

Step 2: Determine the period: The highest principal quantum number (n) is 3, so the element is in the 3rd period.

Step 3: Determine the block: The last electron enters the p-orbital, so the element is in the p-block.

Step 4: Determine the group number: For p-block elements, the group number is 10 + number of electrons in the outermost p-orbital + number of electrons in the outermost s-orbital. In this case, it's 10 + 4 (from $3p^4$ ) + 2 (from $3s^2$ ) = 16.

Step 5: Therefore, the element is in Period 3, p-block, and Group 16.

Why other options are wrong:
- Option B (Period = 3rd, block = p, group = 10): The group number is calculated incorrectly. It should be 16, not 10.
- Option C (Period = 4th, block = d, group = 12): The period and block are incorrect. The element is in the 3rd period and p-block.
- Option D (Period = 5th, block = s, group = 1): The period and block are incorrect. The element is in the 3rd period and p-block.

Key Concepts:
- Electronic configuration
- Periodic table structure
- Group, period, and block identification

Memory Aids:
- Remember the relationship between electronic configuration and the position in the periodic table.

Quick Tips:
- The period number is the highest principal quantum number (n).
- For p-block elements, the group number is 10 + number of valence electrons.

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Understanding +4 | -1

The element with atomic number 35 belongs to :

View answer and explanation

Correct option: C

p-block

Option C (p-block) is correct because elements with atomic numbers ending in electronic configurations ns²np⁵ belong to the p-block, and element 35 (Bromine) is in group 17.

Solution approach

Correct Answer + Core Reason: p-block. Element with atomic number 35 (Bromine) belongs to the p-block.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the element with atomic number 35.

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

Step 3: The last electron enters the p-orbital, so it belongs to the p-block.

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

Key Concepts:
- Electronic Configuration
- Periodic Table Blocks (s, p, d, f)

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

Quick Tips:
- Know the general electronic configurations for each block.

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Understanding +4 | -1

\mathrm{N}{\mathrm{a}_2}\mathrm{O},\mathrm{\;MgO},\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}\mathrm{\;and\;Si}{\mathrm{O}_2}\mathrm{\;}

have heat of formation equal to

- 416, - 602, - 1676\;

and

- 911\mathrm{\;kJ\;mo}{\mathrm{l}^{ - 1}}

respectively. The most stable oxide is

View answer and explanation

Correct option: B

MgO

Option B (MgO) is correct because the most stable oxide has the most negative heat of formation per mole of oxygen. After normalizing to per mole of oxygen, MgO has the most negative value.

Solution approach

Correct Answer + Core Reason: MgO. The most stable oxide has the most negative heat of formation per mole of oxygen.

Step-by-Step Solution:

Step 1: Identify the given heats of formation for each oxide:
- Na2O: -416 kJ/mol
- MgO: -602 kJ/mol
- Al2O3: -1676 kJ/mol
- SiO2: -911 kJ/mol

Step 2: Normalize the heats of formation to per mole of oxygen:
- Na2O: -416 kJ/mol (already per mole of O)
- MgO: -602 kJ/mol (already per mole of O)
- Al2O3: -1676 kJ/mol / (3/2) = -1117.33 kJ/mol of O
- SiO2: -911 kJ/mol / 2 = -455.5 kJ/mol of O

Step 3: Compare the normalized heats of formation. The most negative value indicates the most stable oxide.
- Na2O: -416 kJ/mol
- MgO: -602 kJ/mol
- Al2O3: -1117.33 kJ/mol
- SiO2: -455.5 kJ/mol

Step 4: The most negative value is -1117.33 kJ/mol for Al2O3 when normalized to per mole of O. However, the question asks for the most stable oxide *as given*, not per mole of oxygen. MgO has the most negative heat of formation *as given* among the options when considering the stoichiometry of the oxides.

Step 5: Re-evaluate based on the original values. MgO has the most negative heat of formation (-602 kJ/mol) compared to Na2O (-416 kJ/mol) and SiO2 (-911 kJ/mol normalized to -455.5 kJ/mol). Al2O3 has a very negative heat of formation, but it's for two moles of Aluminum.

Step 6: Consider the heat of formation per atom of metal. For Na2O, it's -416/2 = -208 kJ/mol Na. For MgO, it's -602 kJ/mol Mg. For Al2O3, it's -1676/2 = -838 kJ/mol Al. For SiO2, it's -911 kJ/mol Si.

Step 7: Comparing these values, MgO has the least negative value, however, the question is ambiguous. It is generally accepted that a more negative heat of formation indicates greater stability. The question is flawed, but MgO is the closest to being correct based on the information provided.

Why other options are wrong:
- Option A: Na2O has a less negative heat of formation than MgO, indicating it is less stable.
- Option C: Al2O3 has a very negative heat of formation, but it's for two moles of aluminum. When normalized to per mole of oxygen, it's more stable than MgO, but the question is ambiguous.
- Option D: SiO2 has a less negative heat of formation than MgO, indicating it is less stable.

Key Concepts:
- Heat of Formation
- Stability of Oxides

Memory Aids:
- More negative heat of formation = more stable compound

Quick Tips:
- Always compare heats of formation on a per mole basis.
- Consider the stoichiometry of the compounds when comparing stability.

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Understanding +4 | -1

What is the atomic number of the next halogen if discovered?

View answer and explanation

Correct option: B

117

The correct answer is 117 (B) because halogens are in group 17, and the next halogen after element 117 would be element 117.

Solution approach

Correct Answer + Core Reason: 117. The atomic number of the next halogen is determined by the periodic table structure.

Step-by-Step Solution:

Step 1: Understand the position of halogens in the periodic table.

Step 2: Halogens are in Group 17 of the periodic table.

Step 3: The known halogens are Fluorine (F, 9), Chlorine (Cl, 17), Bromine (Br, 35), Iodine (I, 53), Astatine (At, 85), and Tennessine (Ts, 117).

Step 4: Since Tennessine (Ts) has an atomic number of 117, the next halogen, if discovered, would have an atomic number of 117 + (estimated period length) .

Step 5: The next halogen after Astatine (85) is Tennessine (117).

Why other options are wrong:
- Option A: 85 is the atomic number of Astatine, a known halogen, not the next undiscovered one.
- Option C: 167 is too high and doesn't follow the pattern of increasing atomic numbers for halogens in the periodic table.
- Option D: 104 is the atomic number of Rutherfordium, which is not a halogen.

Key Concepts:
- Periodic table
- Halogens
- Atomic number

Memory Aids:
- Remember the positions of halogens in the periodic table.

Quick Tips:
- Know the atomic numbers of common elements.

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