Given below are two statements :
Statement I : The number of pairs among [SiO_2, CO_2]$[SiO_{2}, CO_{2}]$, [SnO, SnO_2]$[SnO, SnO_{2}]$, [PbO, PbO_2]$[PbO, PbO_{2}]$ and [GeO, GeO_2]$[GeO, GeO_{2}]$, which contain oxides that are both amphoteric is 2.
Statement II : BF_3$BF_{3}$ is an electron deficient molecule can act as a lewis acid, forms adduct with NH_3$NH_{3}$ and has a trigonal planar geometry.
In the light of the above statement, choose the correct answer from the option given below.
A.textBoth Statement I and Statement II are true.$\text{Both Statement I and Statement II are true.}$
B.textBoth Statement I and Statement II are false.$\text{Both Statement I and Statement II are false.}$
C.textStatement I is true but Statement II is false.$\text{Statement I is true but Statement II is false.}$
D.textStatement I is false Statement II is true.$\text{Statement I is false Statement II is true.}$
Solution & Explanation
### Core Logic
Evaluating Statement I:
- SiO_2$SiO_2$, CO_2$CO_2$, GeO$GeO$, GeO_2$GeO_2$ are acidic in nature.
- SnO$SnO$, SnO_2$SnO_2$, PbO$PbO$, PbO_2$PbO_2$ are amphoteric in nature.
Therefore, the pairs [SnO, SnO_2]$[SnO, SnO_2]$ and [PbO, PbO_2]$[PbO, PbO_2]$ contain oxides that are both amphoteric. Number of such pairs = 2. Statement I is True.
Evaluating Statement II:
- BF_3$BF_3$ has 6 electrons in the outermost shell of the central Boron atom. It is electron-deficient and acts as a Lewis acid.
- It accepts a lone pair from Lewis bases like NH_3$NH_3$ to form an adduct.
- In BF_3$BF_3$, Boron is sp^2$sp^2$ hybridized, resulting in a trigonal planar geometry. Statement II is True.
### Step 1: Conclusion
Both statements are factually correct.
### Pattern Recognition
Oxides of heavier Group 14 elements (Sn, Pb) are typically amphoteric in both their +2 and +4 oxidation states. BF_3$BF_3$ is the quintessential Lewis acid.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: p-Block Elements
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Given below are two statements:
Statement I: textH_2textSe$\text{H}_2\text{Se}$ is more acidic than textH_2textTe$\text{H}_2\text{Te}$.
Statement II: textH_2textSe$\text{H}_2\text{Se}$ has higher bond enthalpy for dissociation than textH_2textTe$\text{H}_2\text{Te}$.
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.$\text{Both Statement I and Statement II are false.}$
B.textBoth Statement I and Statement II are true.$\text{Both Statement I and Statement II are true.}$
C.textStatement I is true but Statement II is false.$\text{Statement I is true but Statement II is false.}$
D.textStatement I is false but Statement II is true.$\text{Statement I is false but Statement II is true.}$
Solution
### Core Logic
Let us analyze the periodic properties of chalcogen hydrides (textGroup 16$\text{Group 16}$):
1. **Bond Dissociation Enthalpy (Delta_textdisH$\Delta_{\text{dis}}H$)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (r_textTe > r_textSe$r_{\text{Te}} > r_{\text{Se}}$). This increase in size leads to poorer orbital overlap with the small 1s$1s$ orbital of hydrogen, resulting in a longer and weaker textM-H$\text{M-H}$ bond. Consequently, the bond dissociation enthalpy decreases:
Delta_textdisH: textH_2textSe (276 text kJ mol^-1) > textH_2textTe (238 text kJ mol^-1)$$\Delta_{\text{dis}}H: \text{H}_2\text{Se } (276 \text{ kJ mol}^{-1}) > \text{H}_2\text{Te } (238 \text{ kJ mol}^{-1})$$
**Thus, Statement II is true.**
2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release textH^+$\text{H}^+$ ions. Since the textTe-H$\text{Te-H}$ bond is weaker than the textSe-H$\text{Se-H}$ bond, textH_2textTe$\text{H}_2\text{Te}$ releases protons much more readily than textH_2textSe$\text{H}_2\text{Se}$, making it a stronger acid:
textAcidic Strength: textH_2textSe < textH_2textTe$$\text{Acidic Strength: } \text{H}_2\text{Se} < \text{H}_2\text{Te}$$
**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
Q44jee_main_2025_29_jan_eveningGroup 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$1012\text{ kJ mol}^{-1}$
(2) 1402text kJ mol^-1$1402\text{ kJ mol}^{-1}$
(3) 834text kJ mol^-1$834\text{ kJ mol}^{-1}$
(4) 947text kJ mol^-1$947\text{ kJ mol}^{-1}$
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$Ca_{5}(PO_{4})_{3}F$) 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:
IE_1(N) > IE_1(P) > IE_1(As) > IE_1(Sb)$$IE_1(N) > IE_1(P) > IE_1(As) > IE_1(Sb)$$
Sorting the given enthalpy data values in decreasing order:
1402 > 1012 > 947 > 834$$1402 > 1012 > 947 > 834$$
Assigning these to the elements:
* N = 1402text kJ mol^-1$N = 1402\text{ kJ mol}^{-1}$
* P = 1012text kJ mol^-1$P = 1012\text{ kJ mol}^{-1}$
* As = 947text kJ mol^-1$As = 947\text{ kJ mol}^{-1}$
* Sb = 834text 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: p-Block Elements
Q41jee_main_2025_28_jan_morningInert 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$\mathrm{Al, In, Tl}$ from Group 13 and mathrmGe, Sn, Pb$\mathrm{Ge, Sn, Pb}$ from Group 14):
- **Highest First Ionization Enthalpy (mathrmIE_1$\mathrm{IE}_1$):** Out of these options, Germanium (mathrmGe$\mathrm{Ge}$) sits highest and further right along its period layout, demonstrating the highest mathrmIE_1$\mathrm{IE}_1$ value among this set. Its most stable oxidation state is **+4$+4$**.
- **Lowest First Ionization Enthalpy (mathrmIE_1$\mathrm{IE}_1$):** Indium (mathrmIn$\mathrm{In}$) lies lowest leftward among these relative coordinates, maintaining the lowest mathrmIE_1$\mathrm{IE}_1$. Its most stable group oxidation state is **+3$+3$** (as the inert pair effect is much more pronounced for the heavier element mathrmTl$\mathrm{Tl}$ which prefers +1$+1$).
### Pattern Recognition
Sees: mathrmIE_1$\mathrm{IE}_1$ extrema vs stable oxidation state profiles.
Trap: Forgetting that inert pair shifts display max stability values at +1$+1$ for mathrmTl$\mathrm{Tl}$ and +2$+2$ for mathrmPb$\mathrm{Pb}$, while lighter counterparts like mathrmIn$\mathrm{In}$ favor +3$+3$ and mathrmGe$\mathrm{Ge}$ favors +4$+4$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: The p-Block Elements
Class 12 Chemistry: The p-Block Elements
Q27jee_main_2025_03_april_morningGroup 15 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 disproportionation reactions.
In the light of above 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
Let us analyze both statements systematically:
* **Statement I:** The textN-textN$\text{N}-\text{N}$ single bond is indeed weaker than the textP-textP$\text{P}-\text{P}$ 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 textN-textN$\text{N}-\text{N}$ single bond shorter (140text pm$140\text{ \pm}$) compared to the textP-textP$\text{P}-\text{P}$ single bond (221text pm$221\text{ \pm}$). 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$+3$ oxidation state readily undergo disproportionation. As we go down the group (As, Sb, Bi), the +3$+3$ 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.
### Step 1: Verification of Conclusions
Since the textN-textN$\text{N}-\text{N}$ bond is shorter and heavier elements in the +3$+3$ state do not disproportionate readily, both statements are evaluated to be false.
### Pattern Recognition
Sees: "textN-textN$\text{N}-\text{N}$ single bond longer"
ightarrow$
ightarrow$ Absolute error. Small atoms form short bonds, always. Inert pair effect stabilizes +3$+3$ lower down the group, preventing disproportionation reactions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: p-Block Elements
Q39jee_main_2025_04_april_eveningGroup 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 (Z_texteff$Z_{\text{eff}}$) 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).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: The p-Block Elements
More p-Block Elements Questions — jee_main_2026_21_jan_morning
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