Here's a Periodic Assessment (PA) 1 exam paper for Class 12 Chemistry, covering the specified chapters, designed for Jyoti Central High School .
Jyoti Central High School,
Periodic Assessment - 1
Class: XII
Subject: Chemistry
Chapters Covered:
* Chemical Kinetics
* d- and f-Block Elements
* Coordination Compounds
* Haloalkanes and Haloarenes
* Alcohols, Phenols, and Ethers
* Aldehydes, Ketones, and Carboxylic Acids
Maximum Marks: 70
Time Allotted: 3 Hours
General Instructions:
* All questions are compulsory.
* Marks for each question are indicated against it.
* Use log tables if necessary.
* Calculators are NOT permitted.
* Draw neat diagrams and write balanced chemical equations wherever required.
Section A: Very Short Answer Type Questions (1 mark each)
* For a reaction, the rate constant is k = 3.2 \times 10^{-4} mol L$^{-1}$ s$^{-1}$. What is the order of the reaction?
* Why are most transition metal ions paramagnetic?
* What is the coordination number of Ni in [Ni(CO)_4]?
* Write the IUPAC name of the following compound: CH_3CH(Cl)CH_2CH_3.
* Which is a stronger acid: phenol or ethanol? Give reason.
* Give the common name for CH_3CHO.
* Out of CH_3COOH and HCOOH, which is a stronger acid and why?
Section B: Short Answer Type Questions - I (2 marks each)
* What is molecularity of a reaction? Can it be zero or fractional? Explain.
* Give two reasons why transition metals show variable oxidation states.
* Write the formula for the following coordination compound: Potassium trioxalatochromate(III).
* Predict the major product formed when 2-bromopentane reacts with alcoholic KOH. Name the rule involved.
* How will you distinguish between ethanol and propan-1-ol using a suitable chemical test?
Section C: Short Answer Type Questions - II (3 marks each)
* The decomposition of a substance follows first order kinetics. If its concentration is reduced to 1/4th of its initial value in 50 minutes, calculate the rate constant of the reaction. (\log 2 = 0.3010)
* Account for the following:
(i) Transition metals and their compounds are good catalysts.
(ii) Cr^{2+} is a strong reducing agent while Mn^{3+} is a strong oxidizing agent.
* Using Valence Bond Theory, determine the hybridization, magnetic character, and geometry of [CoF_6]^{3-} (Atomic number of Co = 27).
* Give reasons for the following:
(i) Although chlorine is an electron-withdrawing group, it is o,p-directing in aryl halides.
(ii) Haloalkanes are less reactive than haloarenes towards nucleophilic substitution reactions.
* How will you convert:
(i) Propene to Propan-2-ol
(ii) Benzene to Phenol
* Give one chemical test to distinguish between:
(i) Propanal and Propanone
(ii) Benzoic acid and Phenol
* Write the reactions involved in the following:
(i) Rosenmund reduction
(ii) Cannizzaro reaction
* Arrange the following compounds in increasing order of their boiling points and give reasons:
Butanal, Butan-1-ol, Butane.
* Illustrate the following reactions with an example each:
(i) Wurtz-Fittig reaction
(ii) Williamson's synthesis
Section D: Case Study / Application Based Questions (4 marks each)
* Read the following passage and answer the questions that follow:
Transition elements form a large number of coordination compounds due to their small size, high ionic charges, and availability of d-orbitals for bond formation. These compounds exhibit various types of isomerism. Werner's theory was the first successful attempt to explain the nature of bonding in coordination compounds. The magnetic properties and color of these compounds are explained by crystal field theory. Ligands play a crucial role in determining the properties of coordination complexes.
(i) What is meant by ligand in coordination compounds? Give one example of a bidentate ligand.
(ii) [NiCl_4]^{2-} is paramagnetic while [Ni(CO)_4] is diamagnetic, although both are tetrahedral. Why?
(iii) Write the IUPAC name for [Co(NH_3)_5Cl]Cl_2.
(iv) Why do transition metals form colored compounds?
* Read the following passage and answer the questions that follow:
Alcohols, phenols, and ethers are organic compounds containing oxygen. Alcohols and phenols contain -OH groups, while ethers contain the -O- linkage. The presence of the -OH group in alcohols and phenols influences their physical and chemical properties significantly. The acidic nature of alcohols and phenols varies, with phenols being more acidic than alcohols due to resonance stabilization of the phenoxide ion. Ethers are relatively inert but can be cleaved by strong acids.
(i) Give a chemical test to distinguish between primary, secondary, and tertiary alcohols.
(ii) Arrange the following in increasing order of acidity: Ethanol, Phenol, Water.
(iii) Write the equation for the reaction of phenol with bromine water. What is the observed change?
(iv) How can you prepare diethyl ether using Williamson's synthesis?
Section E: Long Answer Type Questions (5 marks each)
* (a) A first-order reaction has a rate constant of 5.5 \times 10^{-14} s$^{-1}$. Calculate its half-life period.
(b) Define activation energy. How is it affected by the addition of a catalyst?
(c) What is the effect of temperature on the rate of reaction?
(d) For a reaction, A+B \rightarrow C, the rate law is given as Rate = k[A][B]^{1/2}. What is the overall order of the reaction?
* (a) How are potassium permanganate (KMnO_4) prepared from MnO_2? Write balanced chemical equations.
(b) Why do Ce^{3+} ions form paramagnetic compounds? (Atomic number of Ce = 58).
(c) What is lanthanoid contraction? State its two consequences.
* (a) How will you convert:
(i) 1-bromopropane to 2-bromopropane
(ii) Chlorobenzene to Aniline
(iii) Ethyl bromide to Propanoic acid
(b) Give reasons for the following:
(i) Butan-1-ol has a higher boiling point than Butanal.
(ii) Carboxylic acids do not give characteristic reactions of aldehyde or ketone group.
* (a) Give the main product(s) formed in the following reactions:
(i) CH_3COOH \xrightarrow{PCl_5}
(ii) CH_3COCH_3 \xrightarrow{NaOCl}
(iii) C_6H_5CHO \xrightarrow{Conc. \ NaOH}
(iv) CH_3OH \xrightarrow{CuO, 573 K}
(v) Anisole \xrightarrow{HI (conc.)}
(b) Write the mechanism of hydration of ethene to form ethanol.
Answer Key / Hints (for Teacher's reference - DO NOT distribute to students)
Section A
* Zero order
* Presence of unpaired electrons in d-orbitals.
* 4
* 2-Chlorobutane
* Phenol; due to resonance stabilization of phenoxide ion.
* Acetaldehyde
* HCOOH is stronger; +I effect of CH_3 in CH_3COOH destabilizes carboxylate ion more.
Section B
8. No. of reacting species in an elementary reaction. Cannot be zero/fractional as it involves actual participation of molecules.
9. (i) Small energy difference between (n-1)d and ns orbitals. (ii) Involvement of both (n-1)d and ns electrons.
10. K_3[Cr(C_2O_4)_3]
11. Pent-2-ene; Saytzeff's Rule (more substituted alkene is major product).
12. Iodoform test: Ethanol gives yellow ppt of iodoform with I_2/NaOH. Propan-1-ol does not.
CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3 \downarrow + HCOONa + 5NaI + 5H_2O
Section C
13. First order: k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}
[R]_0 = x, [R] = x/4, t=50 min.
k = \frac{2.303}{50} \log \frac{x}{x/4} = \frac{2.303}{50} \log 4 = \frac{2.303}{50} \times 2 \log 2
k = \frac{2.303 \times 2 \times 0.3010}{50} = \frac{1.386}{50} = 0.02772 \text{ min}^{-1}.
14. (i) Due to variable oxidation states and large surface area, they provide suitable reaction sites and activation energy.
(ii) Cr^{2+} (d^4) can lose one electron to form Cr^{3+} (d^3, stable half-filled t_{2g}), so it's a reducing agent. Mn^{3+} (d^4) can gain one electron to form Mn^{2+} (d^5, stable half-filled d-subshell), so it's an oxidizing agent.
15. Co is +3 in [CoF_6]^{3-}. Co$^{3+}$ (d^6). F^- is a weak field ligand. No pairing.
Electronic configuration of Co$^{3+}: $t_{2g}^4 e_g^2$ (high spin)
Hybridization: $sp^3d^2$ (outer orbital complex)
Geometry: Octahedral
Magnetic character: Paramagnetic (4 unpaired electrons)
16. (i) Although -Cl is electron-withdrawing by inductive effect, it has lone pairs that can donate electrons to the ring via resonance, leading to *o,p*-directing effect. Resonance effect dominates over inductive effect for directing nature.
(ii) In haloarenes, C-X bond has partial double bond character due to resonance, making it shorter and stronger. The phenyl cation formed by X- removal is unstable. Nucleophilic attack is hindered by electron density of the ring.
17. (i) Propene to Propan-2-ol: $CH_3CH=CH_2 \xrightarrow{H_2O/H^+}$ or $\xrightarrow{(BH_3)_2 / H_2O_2, OH^-}$ (Hydroboration-oxidation gives anti-Markovnikov, hydration gives Markovnikov).
(ii) Benzene to Phenol: Benzene $\xrightarrow{Conc. H_2SO_4, \Delta}$ Benzenesulphonic acid $\xrightarrow{NaOH, \Delta}$ Sodium phenoxide $\xrightarrow{H^+/H_2O}$ Phenol. (Or Cumene process, Dow process)
18. (i) **Tollens' test / Fehling's test:** Propanal (aldehyde) will give silver mirror (Tollens') or red precipitate (Fehling's). Propanone (ketone) will not.
(ii) **FeCl_3$ test:** Phenol gives violet/green/blue color with neutral FeCl$_3$ solution. Benzoic acid does not.
19. (i) Rosenmund reduction: RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO + HCl
(e.g., CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl)
(ii) Cannizzaro reaction: (For aldehydes lacking \alpha-H)
2HCHO \xrightarrow{Conc. NaOH} CH_3OH + HCOONa
2C_6H_5CHO \xrightarrow{Conc. NaOH} C_6H_5CH_2OH + C_6H_5COONa
20. Order of boiling points: Butane < Butanal < Butan-1-ol.
Reasons:
* Butane (non-polar): Only weak London dispersion forces.
* Butanal (polar): Has dipole-dipole interactions due to polar C=O group, stronger than London forces.
* Butan-1-ol (highly polar): Forms strong intermolecular hydrogen bonds due to -OH group, leading to significantly higher boiling point.
21. (i) Wurtz-Fittig reaction: Alkyl halide + Aryl halide + Na \xrightarrow{dry\ ether} Alkylarene.
e.g., C_6H_5Br + CH_3Br + 2Na \xrightarrow{dry\ ether} C_6H_5CH_3 + 2NaBr (Toluene)
(ii) Williamson's synthesis: Alkyl halide + Sodium alkoxide/phenoxide \rightarrow Ether.
e.g., CH_3CH_2Br + CH_3ONa \rightarrow CH_3CH_2OCH_3 + NaBr (Methoxyethane)
e.g., C_6H_5ONa + CH_3Br \rightarrow C_6H_5OCH_3 + NaBr (Anisole)
Section D
22. (i) Ligand: An ion or molecule capable of donating a pair of electrons to the central metal atom/ion to form a coordinate bond. Example: Oxalate ion (C_2O_4^{2-}).
(ii) In [Ni(CO)_4], CO is a strong field ligand, causing pairing of electrons in the d-orbitals of Ni (oxidation state 0, d^{10}), resulting in diamagnetism. In [NiCl_4]^{2-}, Cl^- is a weak field ligand, no pairing occurs for Ni (oxidation state +2, d^8), leaving two unpaired electrons, thus paramagnetic.
(iii) Pentaamminechlorocobalt(III) chloride
(iv) Transition metals have unpaired electrons in their d-orbitals. When visible light falls on them, these electrons absorb energy and undergo d-d transitions, exciting to higher energy d-orbitals. The remaining transmitted light appears colored.
23. (i) Lucas Test: Primary, secondary, and tertiary alcohols react with Lucas reagent (Conc. HCl + ZnCl_2).
* Tertiary: Immediate turbidity.
* Secondary: Turbidity after 5-10 minutes.
* Primary: No turbidity at room temperature.
(ii) Increasing order of acidity: Ethanol < Water < Phenol.
(iii) Phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol.
C_6H_5OH + 3Br_2 \xrightarrow{H_2O} C_6H_2Br_3OH \downarrow + 3HBr
(iv) Diethyl ether using Williamson's synthesis:
CH_3CH_2ONa + CH_3CH_2Br \rightarrow CH_3CH_2OCH_2CH_3 + NaBr
(Sodium ethoxide + Ethyl bromide \rightarrow Diethyl ether + Sodium bromide)
Section E
24. (a) Half-life for first order: t_{1/2} = \frac{0.693}{k}
t_{1/2} = \frac{0.693}{5.5 \times 10^{-14} s^{-1}} = 1.26 \times 10^{13} s.
(b) Activation Energy (E_a): Minimum amount of energy reactants must possess for a reaction to occur. A catalyst lowers the activation energy by providing an alternative reaction pathway, thus increasing the rate of reaction.
(c) Effect of temperature: Rate of reaction generally increases with increase in temperature. For many reactions, rate doubles for every 10$^\circC rise in temperature. This is because higher temperature increases kinetic energy of molecules, leading to more effective collisions and higher fraction of molecules with energy $\ge E_a$.
(d) Overall order = Sum of powers of concentrations in rate law = $1 + 1/2 = 1.5$ (or 3/2).
25. (a) **Preparation of Potassium Permanganate from MnO_2$:**
(i) Fusion of pyrolusite ore (MnO_2) with KOH in presence of air or an oxidizing agent like KNO_3 to form potassium manganate (K_2MnO_4, green).
2MnO_2 + 4KOH + O_2 \xrightarrow{heat} 2K_2MnO_4 + 2H_2O
(ii) Electrolytic oxidation of manganate to permanganate:
K_2MnO_4 \xrightarrow{Electrolytic\ oxidation\ in\ alkaline\ solution} KMnO_4
MnO_4^{2-} \xrightarrow{oxidation} MnO_4^- + e^- (at anode)
(b) Ce^{3+} (Atomic number of Ce = 58) has electronic configuration [Xe]4f^1 5d^0 6s^0. It has one unpaired electron in the 4f subshell, making it paramagnetic.
(c) Lanthanoid contraction: The steady decrease in atomic and ionic radii of the lanthanoids with increasing atomic number.
Consequences:
(i) Similarity in size of elements of 4d and 5d series (e.g., Zr and Hf).
(ii) Difficulty in separation of lanthanoids.
26. (a) Conversions:
(i) 1-bromopropane to 2-bromopropane:
CH_3CH_2CH_2Br \xrightarrow{Alc. KOH} CH_3CH=CH_2 (Propene)
CH_3CH=CH_2 \xrightarrow{HBr} CH_3CH(Br)CH_3 (2-bromopropane - Markovnikov's addition)
(ii) Chlorobenzene to Aniline:
C_6H_5Cl \xrightarrow{NH_3, Cu_2O, 473K, 60atm} C_6H_5NH_2 (Aniline) (or via reduction of nitrobenzene)
(iii) Ethyl bromide to Propanoic acid:
CH_3CH_2Br \xrightarrow{KCN} CH_3CH_2CN (Propanenitrile)
CH_3CH_2CN \xrightarrow{H_3O^+ / \Delta} CH_3CH_2COOH (Propanoic acid)
(b) Reasons:
(i) Butan-1-ol has a higher boiling point than Butanal: Butan-1-ol can form extensive intermolecular hydrogen bonding due to the presence of the -OH group, which requires more energy to break, leading to a higher boiling point. Butanal only has weaker dipole-dipole interactions.
(ii) Carboxylic acids do not give characteristic reactions of aldehyde or ketone group: In carboxylic acids, the carbonyl carbon is attached to an -OH group. The electron pair on the oxygen of the -OH group is involved in resonance with the carbonyl group, reducing the electrophilic character of the carbonyl carbon. Hence, they don't show reactions like nucleophilic addition, which is characteristic of aldehydes and ketones.
27. (a) Main product(s) formed:
(i) CH_3COOH \xrightarrow{PCl_5} CH_3COCl (Acetyl chloride)
(ii) CH_3COCH_3 \xrightarrow{NaOCl} CHI_3 \downarrow + CH_3COONa (Iodoform reaction, if NaOI is implied, yellow ppt)
(iii) C_6H_5CHO \xrightarrow{Conc. NaOH} C_6H_5CH_2OH + C_6H_5COONa (Benzyl alcohol and Sodium benzoate - Cannizzaro reaction)
(iv) CH_3OH \xrightarrow{CuO, 573 K} HCHO (Formaldehyde)
(v) Anisole \xrightarrow{HI (conc.)} C_6H_5OH + CH_3I (Phenol and Iodomethane)
(b) Mechanism of hydration of ethene to form ethanol:
Step 1: Protonation of alkene to form carbocation.
CH_2=CH_2 + H_3O^+ \rightarrow CH_3-CH_2^+ + H_2O
Step 2: Nucleophilic attack of water on carbocation.
CH_3-CH_2^+ + H_2O \rightarrow CH_3-CH_2-O^+H_2
Step 3: Deprotonation to form alcohol.
CH_3-CH_2-O^+H_2 + H_2O \rightarrow CH_3-CH_2-OH + H_3O^+
Note to Students:
* This question paper is designed to test your understanding of the concepts from the specified chapters.
* Practice writing balanced chemical equations and reaction mechanisms.
* Review distinguishing tests and named reactions thoroughly.
* Ensure you understand the reasons behind chemical properties and reactivity trends.
* All the best for your PA-1!
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