Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the correct answer from the options given below:

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release H^+ ions. Since the Te-H bond is weaker than the Se-H bond, H2Te releases protons much more readily than H2Se, making it a stronger acid: **Thus, Statement I is false.** For binary hydrides down any group (like Group 15, 16, or 17), atomic size increase weakens the covalent bond. A weaker bond releases protons more effectively, meaning that both **acidic strength and reducing character increase down the group**, while thermal stability decreases. Class 12 Chemistry: p-Block Elements

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the cor

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the correct answer from the options given below:

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release H^+ ions. Since the Te-H bond is weaker than the Se-H bond, H2Te releases protons much more readily than H2Se, making it a stronger acid: **Thus, Statement I is false.** For binary hydrides down any group (like Group 15, 16, or 17), atomic size increase weakens the covalent bond. A weaker bond releases protons more effectively, meaning that both **acidic strength and reducing character increase down the group**, while thermal stability decreases. Class 12 Chemistry: p-Block Elements

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the cor

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy…

Year	2026	2025	2024	Total
Questions	16	13	13	42

More p-Block Elements Previous-Year Questions — Page 5

Q44 jee_main_2025_29_jan_evening Group 15 Elements Trends

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family: (1)

1012 kJ mol⁻¹

(2)

1402 kJ mol⁻¹

(3)

834 kJ mol⁻¹

(4)

947 kJ mol⁻¹

A. $1012 kJ mol⁻¹$ $1012\text{ kJ mol}^{-1}$
B. $1402 kJ mol⁻¹$ $1402\text{ kJ mol}^{-1}$
C. $834 kJ mol⁻¹$ $834\text{ kJ mol}^{-1}$
D. $947 kJ mol⁻¹$ $947\text{ kJ mol}^{-1}$

Solution

Core Logic

The main element of the apatite mineral family (e.g., fluorapatite $Ca₅(PO₄)₃F$ $Ca_{5}(PO_{4})_{3}F$) is Phosphorus ( $P$ $P$).

The first four elements of Group 15 are $N$ $N$, $P$ $P$, $As$ $As$, $Sb$ $Sb$. First ionization enthalpy decreases regularly down the group:

IE₁(N) > IE₁(P) > IE₁(As) > IE₁(Sb)

Sorting the given enthalpy data values in decreasing order:

1402 > 1012 > 947 > 834

Assigning these to the elements:

$N = 1402 kJ mol⁻¹$ $N = 1402\text{ kJ mol}^{-1}$
$P = 1012 kJ mol⁻¹$ $P = 1012\text{ kJ mol}^{-1}$
$As = 947 kJ mol⁻¹$ $As = 947\text{ kJ mol}^{-1}$
$Sb = 834 kJ mol⁻¹$ $Sb = 834\text{ kJ mol}^{-1}$

Pattern Recognition

Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array.

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Class 12 Chemistry: p-Block Elements

Q41 jee_main_2025_28_jan_morning Inert Pair Effect and Ionization Enthalpy

Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

A. $+2 and +3$ $+2 \text{ and } +3$
B. $+4 and +3$ $+4 \text{ and } +3$
C. $+4 and +1$ $+4 \text{ and } +1$
D. $+1 and +4$ $+1 \text{ and } +4$

Solution

Core Logic

Let us check the trends for the provided main group elements ( $Al, In, Tl$ $\mathrm{Al, In, Tl}$ from Group 13 and $Ge, Sn, Pb$ $\mathrm{Ge, Sn, Pb}$ from Group 14):

Highest First Ionization Enthalpy ( $IE₁$ $\mathrm{IE}_1$): Out of these options, Germanium ( $Ge$ $\mathrm{Ge}$) sits highest and further right along its period layout, demonstrating the highest $IE₁$ $\mathrm{IE}_1$ value among this set. Its most stable oxidation state is $+4$ $+4$.
Lowest First Ionization Enthalpy ( $IE₁$ $\mathrm{IE}_1$): Indium ( $In$ $\mathrm{In}$) lies lowest leftward among these relative coordinates, maintaining the lowest $IE₁$ $\mathrm{IE}_1$. Its most stable group oxidation state is $+3$ $+3$ (as the inert pair effect is much more pronounced for the heavier element $Tl$ $\mathrm{Tl}$ which prefers $+1$ $+1$).

Pattern Recognition

Sees: $IE₁$ $\mathrm{IE}_1$ extrema vs stable oxidation state profiles. Trap: Forgetting that inert pair shifts display max stability values at $+1$ $+1$ for $Tl$ $\mathrm{Tl}$ and $+2$ $+2$ for $Pb$ $\mathrm{Pb}$, while lighter counterparts like $In$ $\mathrm{In}$ favor $+3$ $+3$ and $Ge$ $\mathrm{Ge}$ favors $+4$ $+4$.

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Class 11 Chemistry: The p-Block Elements Class 12 Chemistry: The p-Block Elements

Q jee_main_2025_04_april_evening Group 13 and 14 Periodic Trends

The elements of Group 13 with highest and lowest first ionisation enthalpies are respectively:

A. B & Ga
B. B & TL
C. TL & B
D. B & In

Solution

Related Formula

Group 13 Ionisation Enthalpy Order: B > Tl > Ga > Al > In

Core Logic

The first ionisation enthalpy trend for Group 13 elements is irregular due to poor shielding by d and f electrons:

Boron (B) is the smallest atom in the group with no d-orbital shielding issues, so it has the highest ionisation enthalpy.
As we move down, poor shielding by $3d$ $3d$ electrons causes a slight increase at Gallium ( $Ga$ $Ga$). Poor shielding by $4f$ $4f$ electrons causes a sharp increase at Thallium ( $Tl$ $Tl$).
Indium (In) ends up with the weakest effective attraction for its outermost valence electron, giving it the lowest first ionisation enthalpy.

Therefore, the highest and lowest elements are B & In respectively.

Pattern Recognition

Group 13 does not follow a linear downward trend. Remember the characteristic 'W' shape or zig-zag pattern of its ionisation energies. Indium sits at the absolute minimum point of this curve.

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Class 11 Chemistry: The p-Block Elements

Q39 jee_main_2025_04_april_evening Group 13 and 14 Periodic Trends

Given below are two statements : Statement (I): The first ionisation enthalpy of group 14 elements is higher than the corresponding elements of group 13. Statement (II) : Melting points and boiling points of group 13 elements are in general much higher than those the corresponding elements of group 14. In the light of the above statements, choose the most appropriate answer from the options given below:

A. Statement I is correct but Statement II is incorrect
B. Statement I is incorrect but Statement II is correct
C. Both Statement I and Statement II are incorrect
D. Both Statement I and Statement II are correct

Solution

Core Logic

Statement I is correct: Moving left-to-right across a period from Group 13 to Group 14 increases the effective nuclear charge ( $Zeff$ $Z_{\text{eff}}$) and decreases the atomic radius. Consequently, more energy is required to extract an electron, so Group 14 elements exhibit higher first ionisation enthalpies.
Statement II is incorrect: Group 14 elements (like Carbon, Silicon, Germanium) build robust, highly stable three-dimensional covalent network crystal structures. As a result, the melting and boiling points of Group 14 elements are generally much higher than those of the corresponding Group 13 elements.

Pattern Recognition

Covalent network solids (Group 14) create huge upward steps in phase transition energy compared to Group 13 frameworks (e.g., Gallium, which melts at nearly room temperature).

Chapter Mix

Class 11 Chemistry: The p-Block Elements

Q37 jee_main_2025_04_april_morning Group 15 Elements

Given below are two statements: Statement I: Nitrogen forms oxides with

+1

+5

oxidation states due to the formation of

pπ - pπ

bond with oxygen. Statement II: Nitrogen does not form halides with

+5

oxidation state due to the absence of d-orbital in it. In the light of given statements, choose the correct answer from the options given below:

A. $Statement I is true but Statement II is false$
B. $Both Statement I and Statement II are false.$
C. $Statement I is false but Statement II is true.$
D. $Both Statement I and Statement II are true.$

Solution

Core Logic

Statement I is true: Nitrogen has a small atomic size and high electronegativity, allowing it to form strong multiple $pπ - pπ$ $p\pi - p\pi$ bonds with oxygen atoms. This enables stable oxide forms across a wide range of oxidation numbers spanning $+1$ $+1$ to $+5$ $+5$ (e.g., $N₂O₅$ $N_2O_5$).
Statement II is true: Nitrogen is a second-period element with a valence shell configuration of $2s² 2p³$ $2s^2 2p^3$. It completely lacks vacant $2d$ $2d$ orbitals. As a result, it cannot expand its octet beyond a covalency of 4, preventing the synthesis of pentahalides like $NX₅$ $NX_5$.

Pattern Recognition

Second-period elements are structurally bounded by a maximum covalency of 4 due to the complete absence of d-orbitals. This explains why $NF₅$ $NF_5$ is unstable/non-existent while $PF₅$ $PF_5$ is easily synthesized.

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Class 12 Chemistry: The p-Block Elements

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