NEET Chemistry - MCQ Practice Questions

Understanding:

We need to explain why the $+1$ oxidation state becomes more stable relative to the $+3$ state as we descend Group 13 (B → Tl).

Step 1: Recall the inert pair effect.

In heavier elements of Groups 13–15, the $n{s}^2$ electron pair becomes increasingly reluctant to participate in bond formation. This is called the inert pair effect. As the principal quantum number ($n$) increases, the $ns$ electrons penetrate the core more effectively but are also poorly shielded by $d$ and $f$ electrons, making them more tightly held and less available for bonding.

Step 2: Consequence in Group 13.

Step 3: Evaluate other options.

Answer:

The increasing stability of the $+1$ oxidation state down Group 13 is due to the inert pair effect, where the $n{s}^2$ electrons become increasingly resistant to participation in bonding.

Quick Tip:

The inert pair effect is most pronounced in the 6th period. In Group 14, ${\text{Pb}}^{2+}$ is more stable than ${\text{Pb}}^{4+}$; in Group 15, ${\text{Bi}}^{3+}$ is more stable than ${\text{Bi}}^{5+}$ — same principle.

Understanding:

We need to identify which nitrate salt, on heating, gives both ${\text{NO}}_2$ and ${\text{O}}_2$ as products.

Step 1: Recall the thermal decomposition pattern of metal nitrates.

The products depend on the activity/reactivity of the metal:

Step 2: Distinguish between options A and D.

Both ${\text{Cu(NO}}_3{)}_2$ and ${\text{Pb(NO}}_3{)}_2$ give ${\text{NO}}_2+{\text{O}}_2$ on heating. Among these, ${\text{Pb(NO}}_3{)}_2$ is the most commonly cited NEET example for this reaction and is a standard laboratory preparation of ${\text{NO}}_2$.

The question asks which compound decomposes to give ${\text{NO}}_2$ and ${\text{O}}_2$ — ${\text{Pb(NO}}_3{)}_2$ is the textbook answer for this specific decomposition in NCERT.

Answer:

${\text{Pb(NO}}_3{)}_2$ decomposes on heating to give $\text{PbO}$, ${\text{NO}}_2$, and ${\text{O}}_2$:

Quick Tip:

This reaction is the standard NCERT laboratory method for preparing ${\text{NO}}_2$ gas. Remember: metal nitrates of metals between Mg and Cu in the activity series → metal oxide + ${\text{NO}}_2$ + ${\text{O}}_2$.

Understanding:

We need to give the correct IUPAC name of ${\text{K}}_2[{\text{PtCl}}_4]$.

Step 1: Identify the components of the complex.

Step 2: Determine the oxidation state of Pt.

So platinum is in the $+2$ oxidation state: $\text{Pt(II)}$.

Step 3: Apply IUPAC 2013 naming rules.

Step 4: Check other options.

Answer:

The correct IUPAC name of ${\text{K}}_2[{\text{PtCl}}_4]$ is potassium tetrachloridoplatinate(II).

Quick Tip:

For anionic complexes, the metal name takes the suffix -ate (e.g., platinate, ferrate, cuprate). The 2013 IUPAC rules use -ido for anionic ligands: fluorido, chlorido, bromido, cyanido, hydroxido.