p-Block Elements

S-1 is correct as gallium is used in manufacturing of thermometers. S-2 is incorrect

To determine which of the following materials is not a semiconductor, we must understand what a semiconductor is and then look at the properties of each material listed.

Semiconductors are materials that have a conductivity level between conductors (such as metals) and insulators (such as most ceramic materials).

They can conduct electricity under certain conditions or when doped with impurities.

The most common semiconductors are silicon and germanium, but some compound materials like gallium arsenide and silicone carbide also behave as semiconductors.

Now, analyzing each option given: Option A: Copper oxide - Copper oxide can behave as a semiconductor, particularly in some specific structures or when used in thin-film solar cells.

Thus, it can exhibit semiconductor properties.

Option B: Graphite - Graphite is a form of carbon where the carbon atoms are bonded together in a sheet-like form.

It is known for its excellent electrical conductivity due to the delocalization of pi electrons across the carbon layers.

Its behavior is more characteristic of a conductor rather than a semiconductor.

Option C: Silicon - Silicon is the most well-known and widely used semiconductor material in electronic devices, including transistors, diodes, and integrated circuits.

Its semiconducting properties are due to its band structure, which can be altered by doping with other elements.

Option D: Germanium - Germanium is another classic semiconductor material used in the early development of electronic devices.

Though not as commonly used as Silicon in modern electronics, due to its similar crystal structure and semiconducting properties, Germanium remains an important semiconductor material.

Based on the provided information, Option B: Graphite is not a semiconductor.

Its properties align more with those of a conductor rather than a semiconductor, distinguishing it from the other options listed which all have semiconducting capabilities under certain conditions.

Among the group 15 elements (N, P, As, Sb, Bi), the lightest two (N and P) are commonly considered nonmetals.

Of these, nitrogen (N) is known to have the weakest single bond to itself (N–N).

The N–N single‐bond enthalpy ($\approx 160$\,kJ/mol) is lower (weaker) than the P–P single‐bond enthalpy ($\approx 200$\,kJ/mol).

Therefore, the “nonmetallic group 15 element with the weakest E–E bond” is nitrogen.

Maximum Covalency of Nitrogen Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., $\mathrm{NH_3}$), it can expand to four bonds in certain cationic species such as ammonium $\mathrm{NH_4}^+$ or $\mathrm{NF_4}^+$.

In such species, nitrogen has a formal positive charge but is still forming four covalent bonds.

Hence, the maximum covalency of nitrogen is 4.

Answer: 4 (Option C).

Ammonia (NH₃) has a more available lone pair on nitrogen than PH₃ on phosphorus due to smaller size and higher electronegativity.

Thus, NH₃ is a stronger base and has a higher proton affinity than PH₃.

This statement is correct.

Sulfur dioxide (SO₂) is quite versatile in redox chemistry.

In many reactions, it actually behaves as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation state).

While under some conditions it may act as an oxidizing agent, it is well known and widely used for its reducing properties.

Hence, the claim that it “cannot act as a reducing agent” is incorrect.

Nitrogen dioxide (NO₂) is known to dimerise to form dinitrogen tetroxide (N₂O₄), especially at lower temperatures.

This statement is correct.

Phosphorus trifluoride (PF₃) is a known stable compound.

In contrast, a compound like nitrogen pentafluoride (NF₅) is not observed, largely due to the limitations of nitrogen’s size and bonding capabilities in forming such a structure.

This statement is correct.

Based on the analysis, the incorrect statement is:

Tellurium dioxide ($\mathrm{TeO}_2$) acts as an oxidizing agent.

This is because it can accept electrons and be reduced from its +4 oxidation state to a lower oxidation state.

Tellurium hydride ($\mathrm{TeH}_2$) is considered acidic.

This is due to its relatively low bond dissociation energy, which makes it prone to breaking, releasing protons (H$^+$).

single bond weaker than due to more $\ell \mathrm{p}-\ell \mathrm{p}$ repulsion.

Bond length $\Rightarrow d_{p-p}>d_{\mathrm{N}-\mathrm{N}}(\operatorname{size} \uparrow$, B.L.

$\uparrow)$ In group 15 elements only N \& P show disproportionation in +3 oxidation state, As, $\mathbf{S b} \&$ Bi have almost inert for disproportionation in +3 oxidation state.

So both statements are false.

(i) True: Tl1+ is more stable than Tl3+ due to the inert pair effect.

Therefore, Tl3+ acts as a powerful oxidizing agent. (ii) True: The standard reduction potential for Al3+/Al is -1.66 V, indicating that Al3+ is difficult to reduce and thus highly stable. (iii) False: Tl3+ is not stable in solution. (iv) True: Tl1+ is more stable than Tl3+. (v) True: Both Al3+ and Tl1+ are highly stable in their respective oxidation states.

Statement 1 is correct since left to right 1 E increases in general in periodic table.

Statement 2 is incorrect since M.P. of group 14 elements is more than group 13 elements.

B2O3 is an acidic oxide. Al2O3 and Ga2O3 are amphoteric oxide. In2O3 and Tl2O are basic oxide.

In oxide of nitrogen it can achieve +5 oxidation state because it can form $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen e.g.

$\mathrm{N}_2 \mathrm{O}_5$ Nitrogen cannot form halide in +5 oxidation state because it does not contain d-orbital. e.g. $\quad \mathrm{NX}_5$ does not exist X = halide