Given below are two statements: Statement I: The N-N single bond is weaker and longer than that of P-P single bond Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo disproportionation reactions. In the light of above statements, choose the correct answer from the options given below:

Let us analyze both statements systematically: * **Statement I:** The N-N single bond is indeed weaker than the P-P single bond due to high inter-electronic repulsion between non-bonding lone pairs on the small nitrogen atoms. However, nitrogen has a smaller atomic size than phosphorus, making the N-N single bond shorter (140 pm) compared to the P-P single bond (221 pm). Thus, Statement I is false because it incorrectly claims it is longer. * **Statement II:** In Group 15, only nitrogen and phosphorus compounds in the +3 oxidation state readily undergo disproportionation. As we go down the group (As, Sb, Bi), the +3 oxidation state becomes increasingly stable due to the inert pair effect, meaning they do not readily undergo disproportionation. Hence, Statement II is false as a general trend across all group 15 elements. Since the N-N bond is shorter and heavier elements in the +3 state do not disproportionate readily, both statements are evaluated to be false. Sees: "N-N single bond longer" ightarrow Absolute error. Small atoms form short bonds, always. Inert pair effect stabilizes +3 lower down the group, preventing disproportionation reactions. Class 12 Chemistry: p-Block Elements

Given below are two statements: Statement I: The N-N single bond is weaker and longer than that of P-P single bond Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo d

Let us analyze both statements systematically: * **Statement I:** The N-N single bond is indeed weaker than the P-P single bond due to high inter-electronic repulsion between non-bonding lone pairs on the small nitrogen atoms. However, nitrogen has a smaller atomic size than phosphorus, making the N-N single bond shorter (140 pm) compared to the P-P single bond (221 pm). Thus, Statement I is false because it incorrectly claims it is longer. * **Statement II:** In Group 15, only nitrogen and phos

Given below are two statements: Statement I: The N-N single bond is weaker and longer than that of P-P single bond Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo disproportionation reactions. In the light of above statements, choose the correct answer from the options given below:

Let us analyze both statements systematically: * **Statement I:** The N-N single bond is indeed weaker than the P-P single bond due to high inter-electronic repulsion between non-bonding lone pairs on the small nitrogen atoms. However, nitrogen has a smaller atomic size than phosphorus, making the N-N single bond shorter (140 pm) compared to the P-P single bond (221 pm). Thus, Statement I is false because it incorrectly claims it is longer. * **Statement II:** In Group 15, only nitrogen and phosphorus compounds in the +3 oxidation state readily undergo disproportionation. As we go down the group (As, Sb, Bi), the +3 oxidation state becomes increasingly stable due to the inert pair effect, meaning they do not readily undergo disproportionation. Hence, Statement II is false as a general trend across all group 15 elements. Since the N-N bond is shorter and heavier elements in the +3 state do not disproportionate readily, both statements are evaluated to be false. Sees: "N-N single bond longer" ightarrow Absolute error. Small atoms form short bonds, always. Inert pair effect stabilizes +3 lower down the group, preventing disproportionation reactions. Class 12 Chemistry: p-Block Elements

Given below are two statements: Statement I: The N-N single bond is weaker and longer than that of P-P single bond Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo d

Let us analyze both statements systematically: * **Statement I:** The N-N single bond is indeed weaker than the P-P single bond due to high inter-electronic repulsion between non-bonding lone pairs on the small nitrogen atoms. However, nitrogen has a smaller atomic size than phosphorus, making the N-N single bond shorter (140 pm) compared to the P-P single bond (221 pm). Thus, Statement I is false because it incorrectly claims it is longer. * **Statement II:** In Group 15, only nitrogen and phos

Given below are two statements: Statement I: The N-N single bond is weaker and longer than that of P-P single bond…

More The p-Block Elements Previous-Year Questions — Page 2

Q45 jee_main_2025_08_april_evening Group 16 Hydrides

Given below are two statements: Statement I:

textH_2textSe

is more acidic than

textH_2textTe

. Statement II:

textH_2textSe

has higher bond enthalpy for dissociation than

textH_2textTe

. In the light of the above statements, choose the correct answer from the options given below:

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

Solution

### Core Logic Let us analyze the periodic properties of chalcogen hydrides (

textGroup 16

): 1. **Bond Dissociation Enthalpy (

Delta_textdisH

)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (

r_textTe > r_textSe

). This increase in size leads to poorer orbital overlap with the small

1s

orbital of hydrogen, resulting in a longer and weaker

textM-H

bond. Consequently, the bond dissociation enthalpy decreases:

Delta_textdisH: textH_2textSe (276 text kJ mol^-1) > textH_2textTe (238 text kJ mol^-1)

**Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release

textH^+

ions. Since the

textTe-H

bond is weaker than the

textSe-H

bond,

textH_2textTe

releases protons much more readily than

textH_2textSe

, making it a stronger acid:

textAcidic Strength: textH_2textSe < textH_2textTe

**Thus, Statement I is false.** ### Pattern Recognition 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements

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)

1012text kJ mol^-1

(2)

1402text kJ mol^-1

(3)

834text kJ mol^-1

(4)

947text kJ mol^-1

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

Solution

### Core Logic The main element of the apatite mineral family (e.g., fluorapatite

Ca_5(PO_4)_3F

) is Phosphorus (

P

). The first four elements of Group 15 are

N

P

As

Sb

. First ionization enthalpy decreases regularly down the group:

IE_1(N) > IE_1(P) > IE_1(As) > IE_1(Sb)

Sorting the given enthalpy data values in decreasing order:

1402 > 1012 > 947 > 834

Assigning these to the elements: *

N = 1402text kJ mol^-1

P = 1012text kJ mol^-1

As = 947text kJ mol^-1

Sb = 834text kJ mol^-1

### Pattern Recognition Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array. ### Evaluation Rubric / Model Answer null ### Chapter Mix 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. $+2 text and +3$ $+2 \text{ and } +3$
B. $+4 text and +3$ $+4 \text{ and } +3$
C. $+4 text and +1$ $+4 \text{ and } +1$
D. $+1 text and +4$ $+1 \text{ and } +4$

Solution

### Core Logic Let us check the trends for the provided main group elements (

mathrmAl, In, Tl

from Group 13 and

mathrmGe, Sn, Pb

from Group 14): - **Highest First Ionization Enthalpy (

mathrmIE_1

):** Out of these options, Germanium (

mathrmGe

) sits highest and further right along its period layout, demonstrating the highest

mathrmIE_1

value among this set. Its most stable oxidation state is **

+4

**. - **Lowest First Ionization Enthalpy (

mathrmIE_1

):** Indium (

mathrmIn

) lies lowest leftward among these relative coordinates, maintaining the lowest

mathrmIE_1

. Its most stable group oxidation state is **

+3

** (as the inert pair effect is much more pronounced for the heavier element

mathrmTl

which prefers

+1

). ### Pattern Recognition Sees:

mathrmIE_1

extrema vs stable oxidation state profiles. Trap: Forgetting that inert pair shifts display max stability values at

+1

for

mathrmTl

and

+2

for

mathrmPb

, while lighter counterparts like

mathrmIn

favor

+3

and

mathrmGe

favors

+4

. ### Evaluation Rubric / Model Answer null ### Chapter Mix 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

textGroup 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

electrons causes a slight increase at Gallium (

Ga

). Poor shielding by

4f

electrons causes a sharp increase at Thallium (

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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

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