Q51
jee_main_2026_21_jan_morning
Reactions of Lead Compounds
Consider the following reactions.
PbCl_2 + K_2CrO_4 rightarrow A + 2KCl$PbCl_{2} + K_{2}CrO_{4} \rightarrow A + 2KCl$ (Hot solution)
A + NaOH rightleftharpoons B + Na_2CrO_4$A + NaOH \rightleftharpoons B + Na_{2}CrO_{4}$
PbSO_4 + 4CH_3COONH_4 rightarrow (NH_4)_2SO_4 + X$PbSO_{4} + 4CH_{3}COONH_{4} \rightarrow (NH_{4})_{2}SO_{4} + X$
In the above reactions, A, B and X are respectively
- A. mathrmNa_2[mathrmPb(OH)_2] , PbCrO_4text and (mathrmNH_4)_2[mathrmPb(mathrmCH_3mathrmCOO)_4]$\mathrm{Na}_{2}[\mathrm{Pb(OH)}_{2}] , PbCrO_{4}\text{ and }(\mathrm{NH}_{4})_{2}[\mathrm{Pb}(\mathrm{CH}_{3}\mathrm{COO})_{4}]$
- B. PbCrO_4 , mathrmNa_2[mathrmPb(OH)_4]text and [mathrmPb(mathrmNH_3)_4]mathrmSO_4$PbCrO_{4} , \mathrm{Na}_{2}[\mathrm{Pb(OH)}_{4}]\text{ and }[\mathrm{Pb}(\mathrm{NH}_{3})_{4}]\mathrm{SO}_{4}$
- C. mathrmNa_2[mathrmPb(OH)_2] , PbCrO_4text and [mathrmPb(mathrmNH_3)_4]mathrmSO_4$\mathrm{Na}_{2}[\mathrm{Pb(OH)}_{2}] , PbCrO_{4}\text{ and }[\mathrm{Pb}(\mathrm{NH}_{3})_{4}]\mathrm{SO}_{4}$
- D. PbCrO_4 , mathrmNa_2[mathrmPb(OH)_4]text and (mathrmNH_4)_2[mathrmPb(mathrmCH_3mathrmCOO)_4]$PbCrO_{4} , \mathrm{Na}_{2}[\mathrm{Pb(OH)}_{4}]\text{ and }(\mathrm{NH}_{4})_{2}[\mathrm{Pb}(\mathrm{CH}_{3}\mathrm{COO})_{4}]$
Solution
### Core Logic
The precipitation and complex formation reactions of Lead are:
mathrmPbCl_2 + mathrmK_2mathrmCrO_4 rightarrow mathrmPbCrO_4 + 2mathrmKCl quad (textHot solution) quad textso, A is mathrmPbCrO_4$$\mathrm{PbCl}_2 + \mathrm{K}_2\mathrm{CrO}_4 \rightarrow \mathrm{PbCrO}_4 + 2\mathrm{KCl} \quad (\text{Hot solution}) \quad \text{so, A is } \mathrm{PbCrO}_4$$
mathrmPbCrO_4 + 4mathrmNaOH \ (excess) rightarrow mathrmNa_2[mathrmPb(OH)_4] + mathrmNa_2mathrmCrO_4 quad textso, B is mathrmNa_2[mathrmPb(OH)_4]$$\mathrm{PbCrO}_4 + 4\mathrm{NaOH \ (excess)} \rightarrow \mathrm{Na}_2[\mathrm{Pb(OH)}_4] + \mathrm{Na}_2\mathrm{CrO}_4 \quad \text{so, B is } \mathrm{Na}_2[\mathrm{Pb(OH)}_4]$$
mathrmPbSO_4 + 4mathrmCH_3mathrmCOONH_4 rightarrow (mathrmNH_4)_2 [mathrmPb(CH_3COO)_4] + (mathrmNH_4)_2mathrmSO_4 quad textso, X is (mathrmNH_4)_2[mathrmPb(CH_3COO)_4]$$\mathrm{PbSO}_4 + 4\mathrm{CH}_3\mathrm{COONH}_4 \rightarrow (\mathrm{NH}_4)_2 [\mathrm{Pb(CH_3COO)}_4] + (\mathrm{NH}_4)_2\mathrm{SO}_4 \quad \text{so, X is } (\mathrm{NH}_4)_2[\mathrm{Pb(CH_3COO)}_4]$$
### Pattern Recognition
Lead forms a yellow precipitate of lead chromate (A), which is amphoteric and dissolves in excess NaOH to form soluble plumbate(II) complex (B). It also forms a stable soluble complex with ammonium acetate (X).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: p-Block Elements
Class 12 Chemistry: d and f Block Elements
Q55
jee_main_2026_21_jan_morning
Group 13 and 14 Compounds
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
### 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
Q40
jee_main_2025_02_april_evening
Group 16 Elements (Oxygen Family)
The nature of oxide (mathrmTeO_2)$(\mathrm{TeO}_2)$ and hydride (mathrmTeH_2)$(\mathrm{TeH}_2)$ formed by Te, respectively are:
- A. textOxidising and acidic$\text{Oxidising and acidic}$
- B. textReducing and basic$\text{Reducing and basic}$
- C. textReducing and acidic$\text{Reducing and acidic}$
- D. textOxidising and basic$\text{Oxidising and basic}$
Solution
### Related Formula
textBond Strength propto frac1textSize difference$$\text{Bond Strength} \propto \frac{1}{\text{Size difference}}$$
textAcidic Strength propto frac1textM-H Bond Dissociation Energy$$\text{Acidic Strength} \propto \frac{1}{\text{M-H Bond Dissociation Energy}}$$
### Core Logic
Let's analyze the properties of Tellurium compounds:
1. **Tellurium Dioxide** (mathrmTeO_2$\mathrm{TeO}_2$):
- Due to the **inert pair effect**, the +6$+6$ oxidation state of Tellurium is less stable, whereas its +4$+4$ state is relatively stable. However, in comparison to sulphur dioxide (which is a strong reducing agent), mathrmTeO_2$\mathrm{TeO}_2$ is oxidising because the lower oxidation states (like element Tellurium or +2$+2$) are chemically accessible. Thus, mathrmTeO_2$\mathrm{TeO}_2$ acts as an **oxidising agent**.
2. **Tellurium Hydride** (mathrmTeH_2$\mathrm{TeH}_2$):
- Tellurium is a very large atom. The orbital overlap between Tellurium and Hydrogen is extremely poor. Hence, the mathrmTe-H$\mathrm{Te-H}$ bond is very long and has very **low bond dissociation energy**.
- This allows mathrmTeH_2$\mathrm{TeH}_2$ to easily release mathrmH^+$\mathrm{H^+}$ in solution, making it highly **acidic**.
### Step 1: Final Verification
Therefore, the nature of mathrmTeO_2$\mathrm{TeO}_2$ is oxidising, and the nature of mathrmTeH_2$\mathrm{TeH}_2$ is acidic.
### Pattern Recognition
Periodic Trend: As we go down Group 16:
- Acidic strength of hydrides increases: mathrmH_2O < H_2S < H_2Se < H_2Te$\mathrm{H_2O < H_2S < H_2Se < H_2Te}$.
- Reducing character of hydrides also increases.
- Reducing power of dioxides decreases: mathrmSO_2$\mathrm{SO_2}$ (reducing) rightarrow mathrmTeO_2$\rightarrow \mathrm{TeO_2}$ (oxidising).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: p-Block Elements
Q38
jee_main_2025_07_april_morning
Properties of Group 14 Elements
The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715mathrm\ kJ\ mol^-1$715\mathrm{\ kJ\ mol}^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions mathrmA^2+$\mathrm{A}^{2+}$, mathrmB^4+$\mathrm{B}^{4+}$ respectively is:
- A. textboth reducing$\text{both reducing}$
- B. textboth oxidising$\text{both oxidising}$
- C. textreducing and oxidising$\text{reducing and oxidising}$
- D. textoxidising and reducing$\text{oxidising and reducing}$
Solution
### Core Logic
For Group 14 (textC, textSi, textGe, textSn, textPb$\text{C}, \text{Si}, \text{Ge}, \text{Sn}, \text{Pb}$):
- The ionisation energies generally decrease down the group, but there is an anomaly between textSn$\text{Sn}$ and textPb$\text{Pb}$ due to relativistic contraction / poor shielding of 4f electrons in textPb$\text{Pb}$.
- Thus, the first ionisation enthalpy of Tin (mathrmSn$\mathrm{Sn}$) is 708 text kJ mol^-1$708 \text{ kJ mol}^{-1}$ and Lead (mathrmPb$\mathrm{Pb}$) is 715 text kJ mol^-1$715 \text{ kJ mol}^{-1}$. These are indeed the lowest in the group.
- Hence, element **A** is mathrmSn$\mathrm{Sn}$ and **B** is mathrmPb$\mathrm{Pb}$.
Nature of their ions:
- mathrmA^2+ = mathrmSn^2+$\mathrm{A}^{2+} = \mathrm{Sn}^{2+}$: Since mathrmSn^4+$\mathrm{Sn}^{4+}$ is more stable than mathrmSn^2+$\mathrm{Sn}^{2+}$, mathrmSn^2+$\mathrm{Sn}^{2+}$ readily undergoes oxidation to +4$+4$, acting as a strong **reducing agent**.
- mathrmB^4+ = mathrmPb^4+$\mathrm{B}^{4+} = \mathrm{Pb}^{4+}$: Due to the strong **inert pair effect**, mathrmPb^2+$\mathrm{Pb}^{2+}$ is highly stable compared to mathrmPb^4+$\mathrm{Pb}^{4+}$. Thus, mathrmPb^4+$\mathrm{Pb}^{4+}$ is eager to reduce to +2$+2$, acting as a strong **oxidising agent**.
### Pattern Recognition
Inert pair effect becomes extremely prominent at the bottom of the group. Lead's most stable state is +2$+2$, making mathrmPb^4+$\mathrm{Pb}^{4+}$ oxidising. Tin's stable state is +4$+4$, making mathrmSn^2+$\mathrm{Sn}^{2+}$ reducing.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: p-Block Elements
Class 11 Chemistry: Periodic Classification of Elements
