Choose the correct statements about the hydrides of group 15 elements. A. The stability of the hydrides decreases in the order NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃$\mathrm{NH}_3 > \mathrm{PH}_3 > \mathrm{AsH}_3 > \mathrm{SbH}_3 > \mathrm{BiH}_3$ B. The reducing ability of the hydrides increases in the order NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3$ C. Among the hydrides, NH₃$\mathrm{NH}_3$ is strong reducing agent while BiH₃$\mathrm{BiH}_3$ is mild reducing agent. D. The basicity of the hydrides increases in the order NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3$ Choose the most appropriate from the option given below:

Core Logic

The precipitation and complex formation reactions of Lead are:

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).

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 12 Chemistry: d and f Block Elements

Core Logic

Evaluating Statement I:

• SiO₂$SiO_2$, CO₂$CO_2$, GeO$GeO$, GeO₂$GeO_2$ are acidic in nature.
• SnO$SnO$, SnO₂$SnO_2$, PbO$PbO$, PbO₂$PbO_2$ are amphoteric in nature.

Therefore, the pairs [SnO, SnO₂]$[SnO, SnO_2]$ and [PbO, PbO₂]$[PbO, PbO_2]$ contain oxides that are both amphoteric. Number of such pairs = 2. Statement I is True.

Evaluating Statement II:

• BF₃$BF_3$ 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 NH₃$NH_3$ to form an adduct.
• In BF₃$BF_3$, Boron is sp²$sp^2$ 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. BF₃$BF_3$ is the quintessential Lewis acid.

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

This is the classic Chromyl Chloride test:

In chromyl chloride (CrO₂Cl₂$\text{CrO}_2\text{Cl}_2$), chromium is in the +6$+6$ oxidation state.

Step 1: Final Conclusion

The gas is CrO₂Cl₂$\text{CrO}_2\text{Cl}_2$ and the oxidation state of Cr is +6$+6$, corresponding to option (2).

Pattern Recognition

Sees: qualitative analysis test for chloride ions (chromyl chloride test). Trap: Confusing oxidation state of chromium in dichromate versus chromyl chloride.

Chapter Mix

Class 12 Chemistry: p-Block Elements

Related Formula

Core Logic

The reagent K₂HgI₄$K_2HgI_4$ in alkaline medium is Nessler's reagent. It gives a brown precipitate (iodide of Millon's base) with ammonia (NH₃$NH_3$). Therefore, the compound EH₃$EH_3$ is NH₃$NH_3$, and the element (E) is Nitrogen (N). The binary cation is the ammonium ion (NH₄^+)$(NH_4^+)$.

We need to find the first ionization enthalpy of Nitrogen among the first elements of groups 13 (B), 14 (C), 15 (N), and 16 (O). Due to its stable half-filled 2p³$2p^3$ configuration, Nitrogen has a remarkably high first ionization energy, higher than Oxygen. The order is: B < C < O < N$B < C < O < N$. Among the given values (801, 1086, 1312, 1402), 1402 kJ mol⁻¹$kJ\text{ mol}^{-1}$ is the highest and corresponds to Nitrogen.

Step 1: Final Conclusion

Element E is Nitrogen, and its correct first ionization enthalpy is 1402 kJ mol⁻¹$kJ\text{ mol}^{-1}$.

Pattern Recognition

Always remember the ionization energy exception across period 2: N > O and Be > B due to stable half-filled and fully-filled orbitals.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: p-Block Elements

Core Logic

Evaluate Statement I: The bond length of HX increases down the group (HF < HCl < HBr < HI$HF < HCl < HBr < HI$) due to the increasing atomic radius of the halogen. Therefore, the halogen making the longest bond is Iodine (I). However, Iodine has the largest covalent radius in the group, not the smallest. Thus, Statement I is false.

Evaluate Statement II: In group 15 hydrides, the boiling point order is PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃$PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$. The hydride with the lowest boiling point is PH₃$PH_3$ (Phosphine). The maximum covalency of Phosphorus is 6 (as seen in PF₆^-$PF_6^-$) because it has empty d-orbitals, not 4. Only Nitrogen has a maximum covalency of 4. Since the element is P, statement II is also false.

Step 1: Final Conclusion

Both Statement I and Statement II are false.

Pattern Recognition

Standard p-block trends: Boiling points of hydrides show anomalies due to hydrogen bonding. NH₃$NH_3$, H₂O$H_2O$, and HF jump out of the trend. P, S, Cl hydrides mark the lowest points.

Chapter Mix

Class 12 Chemistry: p-Block Elements