JEE Main 2024 question paper pdf with solutions- Download Jan 27 Shift 2 Chemistry Question Paper pdf

Step 1: Understanding the Paschen series.

Paschen series corresponds to electronic transitions ending at \( n = 3 \).


Step 2: Longest wavelength = smallest energy difference.

This occurs when the electron falls from \( n = 4 \) to \( n = 3 \).

\[ \frac{1}{\lambda} = R\left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R\left( \frac{1}{9} - \frac{1}{16} \right) = R\left( \frac{7}{144} \right) \]

Thus,
\[ \lambda = \frac{144}{7R} \] Quick Tip: For longest wavelength in any series, consider the transition from the nearest higher level.

For a first-order reaction, the integrated rate law is:
\[ t = \frac{2.303}{k} \log\frac{a}{a-x} \]

Step 1: Time for 99.9% completion.

Here, \( a = 100 \), \( x = 99.9 \).
\[ t_1 = \frac{2.303}{k} \log\frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{2.303}{k} \times 3 \]

Step 2: Time for 50% completion (half-life).
\[ t_2 = \frac{2.303}{k} \log\frac{100}{50} = \frac{2.303}{k} \log 2 = \frac{2.303}{k} \times 0.3010 \]

Step 3: Ratio
\[ \frac{t_1}{t_2} = \frac{3}{0.3010} \approx 10 \] Quick Tip: For first-order reactions, high percentage completion times are many times larger than the half-life.

Step 1: Evaluating S1.

Ce\(^{4+}\) has the configuration of xenon (noble gas), so it is highly stable. Thus, statement I is correct.


Step 2: Evaluating S2.

Ce\(^{4+}\) readily gains electrons to get reduced to Ce\(^{3+}\), which means it acts as an \emph{oxidising agent, not a reducing agent. Hence statement II is incorrect.


Step 3: Final conclusion.

Only statement I is correct.
Quick Tip: Any ion that easily gains electrons acts as an oxidising agent; one that easily loses electrons acts as a reducing agent.

Step 1: Understanding SN1 mechanism.

SN1 reactions proceed through formation of a stable carbocation intermediate.


Step 2: Analysing the options.

Carbocations adjacent to double bonds (vinylic carbocations) are extremely unstable, as the positive charge is directly on the sp\(^2\) carbon.


Step 3: Conclusion.

C = C\(^{\oplus}\) does not undergo SN1 because vinylic carbocations are too unstable to form.
Quick Tip: SN1 is favoured by 3° > 2° > 1° carbocations; vinylic carbocations do not form.

Step 1: Conformations of ethane.

Ethane has infinite conformations due to free rotation around the C–C bond.


Step 2: Staggered vs Eclipsed.

Staggered conformation is most stable because repulsions between H atoms are minimized. Eclipsed conformations have maximum torsional strain and are least stable.


Step 3: Final conclusion.

Therefore, the incorrect statement is that eclipsed form is more stable.
Quick Tip: Staggered conformations are most stable due to minimum torsional strain.

\[ PbS + 4O_3 \rightarrow PbSO_4 + 4O_2 \]
Here, \( x = 4 \) and \( y = 4 \).

\[ x + y = 4 + 4 = 8 \] Quick Tip: Ozone often oxidises sulphides to sulphates while releasing oxygen.

Step 1: General formula.

Monocarboxylic acids follow the formula C\(_n\)H\(_{2n}\)O\(_2\).


Step 2: First and second members.

First member (n = 1): HCOOH

Second member (n = 2): CH\(_3\)COOH


Step 3: Conclusion.

Therefore, the second homologue is acetic acid (CH\(_3\)COOH).
Quick Tip: Homologues differ by –CH\(_2\)– group in any homologous series.

Step 1: Writing configurations.

Zn → 3d\(^{10}\)4s\(^{2}\)

Cu → 3d\(^{10}\)4s\(^{1}\)

Cd → 4d\(^{10}\)5s\(^{2}\)

Ag → 4d\(^{10}\)5s\(^{1}\)


Step 2: Evaluating correctness.

All four elements have completely filled d\(^{10}\) subshells in their ground state.


Step 3: Conclusion.

Thus, option (a) is correct since all listed elements possess a d\(^{10}\) configuration.
Quick Tip: Elements with filled d\(^{10}\) shells show high stability due to electronic symmetry.

Step 1: Hybridisation check.

[Co(NH\(_3\))\(_6\)]\(^{3+}\) → Inner orbital complex → d\(^{2}\)sp\(^{3}\)

[PtCl\(_6\)]\(^{2-}\) → Square planar or octahedral (uses d–orbitals) → d\(^{2}\)sp\(^{3}\)

SF\(_6\) → Uses sp\(^{3}\)d\(^{2}\) (equivalent to d\(^{2}\)sp\(^{3}\)) → octahedral

BrF\(_2\) → Does not have d\(^{2}\)sp\(^{3}\); uses sp\(^{3}\)d hybridisation.


Step 2: Conclusion.

Two of the species show d\(^{2}\)sp\(^{3}\) hybridisation.
Quick Tip: Octahedral geometry corresponds to d\(^{2}\)sp\(^{3}\) hybridisation in coordination compounds.

Step 1: Checking electronic configurations.

Cs\(^{+}\) → [Xe] (after losing 1 electron)

Sr\(^{2+}\) → [Kr] (after losing 2 electrons)


Step 2: Checking remaining ions.

Fe\(^{2+}\) → 3d\(^{6}\) (not noble gas configuration)

Pb\(^{2+}\) → [Xe] 4f\(^{14}\)5d\(^{10}\)6s\(^{2}\) (not noble gas configuration)


Step 3: Final conclusion.

Only Cs\(^{+}\) and Sr\(^{2+}\) attain noble gas configuration, so total = 2.
Quick Tip: Ions that lose electrons to reach s\(^{2}\)p\(^{6}\) configuration usually achieve noble gas stability.

The EMF of a hydrogen electrode is given by:
\[ E_{H^{+}/H_2} = -0.0591 \times pH \]

Step 1: Substitute pH = 3.
\[ E = -0.0591 \times 3 = -0.1773\ V \]

Step 2: Conclusion.

Thus, the EMF of the hydrogen electrode at pH 3 is \(-0.1773\) V.
Quick Tip: For every increase of 1 unit in pH, hydrogen electrode potential decreases by 0.0591 V.

Step 1: Definition of chiral carbon.

A carbon atom bonded to four different substituents is chiral.


Step 2: Analysis of compounds.

(i) The carbon adjacent to the carbonyl has identical substituents → not chiral.

(ii) Symmetrical diketone ring → no chiral centre.

(iii) CH\(_3\)–CH\(_2\)–CH(I)–CH\(_2\)–CH\(_3\): the carbon with I is attached to four different groups → chiral.

(iv) CH\(_3\)–CH\(_2\)–CH\(_2\)–C(OH)(H)–COOH: the central C attached to OH, H, COOH and propyl group → chiral.


Step 3: Conclusion.

Thus, the number of chiral compounds = 2.
Quick Tip: A quick way to detect chirality: look for tetrahedral carbon with all four different groups.

Step 1: Calculate moles of NaOH.
\[ Moles = \frac{84}{40} = 2.1 \]

Step 2: Use molarity formula.
\[ M \times V = moles \] \[ 3 \times V = 2.1 \] \[ V = 0.7\ L \]

Step 3: Convert to mL.
\[ 0.7\ L = 700\ mL \] Quick Tip: Always convert the final molarity volume to litres before converting to mL.