Chemistry Marathon: From Polarity to p-Block (JEE/NEET PYQs)

Chemistry Marathon: From Polarity to p-Block (JEE/NEET PYQs)

Test your knowledge on five diverse and crucial chemistry concepts.

Question 1: Polarity of Molecules

Choose the polar molecule from the following:

(1) $CCl_4$ (2) $CO_2$ (3) $CH_2=CH_2$ (4) $CHCl_3$

Solution 1: Polarity

Key Concept: A molecule is non-polar if its net **dipole moment ($\mu$) is zero**. This usually occurs when the molecule is highly symmetrical, canceling out all bond dipoles.

• $CCl_4$ (Tetrahedral), $CO_2$ (Linear), and $CH_2=CH_2$ (Planar) are highly symmetrical, so $\mu = 0$.
• $CHCl_3$ (Trichloromethane) has a **tetrahedral geometry**, but the central carbon is bonded to one H atom and three Cl atoms. The bond dipoles are unequal and **do not cancel out** due to the difference in the polarity of the C-H and C-Cl bonds, resulting in a net dipole moment ($\mu \ne 0$).

**Correct Option: (4)** $CHCl_3$.

Question 2: Spin-Only Magnetic Moment

Consider the following complex ions: $P = [FeF_6]^{3-}$, $Q = [V(H_2O)_6]^{2+}$, $R = [Fe(H_2O)_6]^{2+}$. The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is:

(1) $R < Q < P$ (2) $R < P < Q$ (3) $Q < R < P$ (4) $Q < P < R$

Solution 2: Magnetic Moment ($\mu$)

The magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$, where $n$ is the number of **unpaired electrons**. Since all ligands are **weak field** (F$^-$ and $H_2O$), no pairing occurs beyond the necessary $d$-orbital filling.

• **P = $[FeF_6]^{3-}$:** Fe is in the $+3$ state ($3d^5$). $n=5$. $\mu = \sqrt{5(7)} = \sqrt{35}$ BM $\approx 5.92$ BM.
• **Q = $[V(H_2O)_6]^{2+}$:** V is in the $+2$ state ($3d^3$). $n=3$. $\mu = \sqrt{3(5)} = \sqrt{15}$ BM $\approx 3.87$ BM.
• **R = $[Fe(H_2O)_6]^{2+}$:** Fe is in the $+2$ state ($3d^6$). $n=4$. $\mu = \sqrt{4(6)} = \sqrt{24}$ BM $\approx 4.90$ BM.

Comparing values: $\sqrt{15} < \sqrt{24} < \sqrt{35}$.

Order: $\mathbf{Q < R < P}$. **Correct Option: (3)**.

Question 3: Acidity and Qualitative Analysis

Given below are two statements:

Statement (I): $p$-nitrophenol is more acidic than $m$-nitrophenol and $o$-nitrophenol.

Statement (II): Ethanol will give immediate turbidity with Lucas reagent.

(1) Statement I is true but Statement II is false (2) Both Statement I and Statement II are true (3) Both Statement I and Statement II are false (4) Statement I is false but Statement II is true

Solution 3: Acidity and Alcohols

Statement (I) Acidity: The $\text{-NO}_2$ group is electron-withdrawing ($\text{-M}$ and $\text{-I}$ effects), stabilizing the phenoxide ion and increasing acidity. $p$-nitrophenol is the most acidic because $o$-nitrophenol forms a stable **intramolecular H-bond** , which hinders proton release, and $m$-nitrophenol lacks the strong $\text{-M}$ effect at that position. Hence, **Statement (I) is True**.

Statement (II) Lucas Test: The Lucas test distinguishes between primary, secondary, and tertiary alcohols. Primary alcohols (like Ethanol) react the slowest, only giving turbidity upon **heating or after a long time** (hours). Tertiary alcohols react immediately. Hence, **Statement (II) is False**.

**Correct Option: (1)** (Statement I is true but Statement II is false).

Question 4: Periodic Table Classification

Given below are two statements:

Statement (I): The 4f and 5f - series of elements are placed separately in the Periodic table to preserve the principle of classification.

Statement (II): S-block elements can be found in pure form in nature.

(1) Statement I is false but Statement II is true (2) Both Statement I and Statement II are true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are false

Solution 4: Periodicity

Statement (I): The 4f (Lanthanides) and 5f (Actinides) series are placed separately at the bottom of the table primarily to **preserve the structure of the Periodic Table** and avoid making it excessively wide. This is done to maintain the principle of classification based on groups and periods. Hence, **Statement (I) is True**.

Statement (II): S-block elements (Alkali and Alkaline Earth metals, except Be and Mg) are highly **electropositive** and react readily with air and moisture. They are never found in pure or free form in nature, only in the combined state (salts, minerals). Hence, **Statement (II) is False**.

**Correct Option: (3)** (Statement I is true but Statement II is false).

Question 5: Assertion and Reason (Boron)

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Melting point of Boron (2453 K) is unusually high in group 13 elements.

Reason (R): Solid Boron has a very strong crystalline lattice.

(1) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (2) Both (A) and (R) are correct and (R) is the correct explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true

Solution 5: Boron's Melting Point

Assertion (A) is True: Boron has the highest melting point in Group 13 due to its small size and high ionization energy, which leads to strong bonding.

Reason (R) is True: Boron exhibits unique allotropic forms containing $\mathbf{B_{12}}$ icosahedral units that are linked together by strong **covalent bonds** in a complex, rigid crystalline lattice . Breaking this structure requires a very large amount of energy.

Since the exceptionally strong lattice (R) directly explains the unusually high melting point (A), **(R) is the correct explanation of (A)**.

**Correct Option: (2)**.

Key Takeaways

• **Polarity:** Look for asymmetry (e.g., $CHCl_3$ vs. $CCl_4$).
• **Magnetic Moment:** The value depends only on the number of unpaired electrons ($n$). Weak field ligands do not cause pairing in $d^4$ to $d^7$ ions.
• **Acidity:** $\mathbf{p-NO_2}$ is generally the most acidic because it avoids $o$-nitrophenol's intramolecular H-bonding.
• **S-Block:** These elements are highly reactive and found only in **combined states** in nature.