Aryl halides are extremely less reactive towards Nucleophilic Substitution reactions. Answer: Aryl halides are extremely less reactive towards Nucleophilic Substitution reactions. because C—X bond

acquires a partial double bond character due to resonance. As a result, the bond cleavage in

haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic

substitution reaction.(draw the resonating structures of chlorobenzene)

2. 3. Haloarenes are less reactive than haloalkanes and haloalkenes.

C-X bond length in halobenzene is smaller than C-X bond length in CH3-X Answer: because C—X

bond in halobenzene acquires a partial double bond character due to resonance

4. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides

as main product. Answer: Haloalkanes react with KCN to form alkyl cyanides as main product while

AgCN forms isocyanides as main product.since KCN is predominantly ionic and provides cyanide ions

in solution. Although both carbon and Nitrogen can donate electron pair but carbon donates electron

pair instead of Nitrogen to form more stable C-C bond. How ever, AgCN is mainly covalent in nature

and Nitrogen is free to donate electron pair forming isocyanide as the main product.

5. Allyl chloride is more reactive than n - propyl chloride towards nucleophilic substitution reaction.

Answer: Allyl chloride is more reactive than n - propyl chloride towards nucleophilic substitution

reaction.because allyl carbocation is stabilized by resonance whereas n propyl carbocation is

stabilized by +I effect only.

6. 7. Allyl chloride is hydrolysed more readily than n-propyl chloride.

Tert-Butylbromide reacts with aq. NaOH by SN1 mechanism while n-butylbromide reacts by SN2

mechanism

8. Ethyl iodide undergoes SN2 undergoes reaction faster than ethyl bromide. Answer:Since I- ion is

better leaving group than Br- ion, Ethyl iodide reacts faster than ethyl bromide in SN2.

9. Benzyl chloride undergoes SN1 reaction faster than cyclohexyl methyl chloride. Answer: because in

case of benzyl chloride the carbocation formed is stablised by resonance.

10. p - nitro chlorobenzene undergoes nucleophilic substitution faster than chlorobenzene Answer: p -

nitro chlorobenzene undergoes nucleophilic substitution faster than chlorobenzene,due to presence

of electron withdrawing groups such as nitro group in p - nitro chlorobenzene.,it forms more stable

carbanion

11. The presence of nitro group (–NO2 ) at ortho and para postions in haloarenes increases the

reactivity of haloarenes towards Nucleophilic substitution reaction. Answer: The presence of nitro

group (–NO2 ) at ortho and para postions in haloarenes helps in stabilization of resulting carbanion by

–R and –I effects and hence increases the reactivity of haloarenes towards Nucleophilic substitution

reaction.

12. Electrophilic aromatic substitution reactions in haloarenes occur slowly. Answer:Chlorine

withdraws electrons through inductive effect and releases electrons through resonance. The

inductive effect is stronger than resonance and causes net electron withdrawal as a result

Electrophilic aromatic substitution reactions in haloarenes occur slowly

13. Although chlorine is an electron withdrawing group, yet it is ortho-

, para- directing in electrophilic

aromatic substitution reactions. Answer: As the weaker resonance (+R) effect of Chlorine which

stabilize the carbocation formed tends to oppose the inductive effect of Chlorine which destabilize

the carbocation at ortho and para postions and makes deactivation less for ortho and para postion.

14. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. Answer: The dipole

moment of chlorobenzene is lower than that of cyclohexyl chloride.because in chlorobenzene

electron withdrawing inductive effect is opposed by electron releasing resonance effect there fore it

is relatively less polar. On the other hand in cyclohexyl chloride there is only electron with drawing is

inductive effect of -Cl atom due to which more polar.CBSE 2021

15. Grignard reagents should be prepared under anhydrous conditions. Answer: Grignard reagents

should be prepared under anhydrous conditions.because Grignard reagent reacts with water and get

decomposed and form alkanes RMgX + H2O R-H + Mg (OH) X.

16. The treatment of alkyl chlorides with aq.KOH leads to the formation of alcohols but in the presence

of alc.KOH alkenes are major products. Answer: The treatment of alkyl chlorides with aq.KOH leads

to the formation of alcohols but in the presence of alc.KOH alkenes are major products.because in

presence of aq.KOH(more polar), nucleophilic substitution reaction takes place ,thus alcohols are

formed while in presence alc.KOH(less polar),elimination reaction takes place.thus alkenes are major

products.

16. Alkyl halides, though polar, are immiscible with water. Answer: Alkyl halides, though polar, are

immiscible with water.because alkyl haides are unable to form H-bond with water as well as unable to

break the existing H-bonds among water molecules.

17. The solubility of haloalkanes in water very low.

18. Haloalkanes easily dissolve in organic solvents. Answer: because the new intermolecular attractions

between haloalkanes and organic solvents have much the same strength as ones being broken in the

separate haloalkanes and solvent molecules.

19. Halogen compounds used in industry as solvents are alkyl chlorides rather than bromides and

iodides. Answer: because alkyl chlorides are more stable and more volatile than bromides and

iodides.

20. Haloalkanes have higher boiling points as compared to those of corresponding alkanes.

Answer:Haloalkanes have higher boiling points as compared to those of corresponding alkanes.Due

to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon,

21. n- butyl bromide has higher point than tert.- butyl bromide. Answer: n- butyl bromide being a

straight chain alkyl halide has larger surface area than tert. butyl bromide. Larger the surface area

,larger the magnitude of the vanderwaal’s forces and higher is the boiling point.

22. The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. Answer: The boiling

points of alkyl halides decrease in the order: RI > RBr > RCl > RF.This is because with the increase in

size and mass of halogen atom, the magnitude of van der Waal forces increases

23. .P-Dichlorobenzene has higher m.p and solubility than those of o- and m- isomers. Answer: P-

Dichlorobenzene has higher m.p and solubility than those of o- and m- isomers. Since p-

dichlorobenzene has more symmetrical structure than other two isomers, hence its molecules fit

more closely in the crystal lattice and consequently stronger intermolecular attractive forces.

24. Racemic mixture is optically inactive. Answer: A Racemic mixture contains two enantiomers in d & l

equal proportions. As the rotation due to one enatiomer is cancelled by equal and opposite rotation

of another isomer. Therefore it is optically inactive..

25. 1-Bromobutane optically inactive but 2-Bromobutane is optically active. Answer: 2-Bromobutane

has four different groups attached to the tetrahedral carbon and has chiral carbon, therefore it is

optically active

22. Preparation of alkyl chloride from alcohols by treating it with Thionyl chloride SOCl2 is the best

method. Answer: Because it gives almost pure alkyl chloride since the by products of the reaction i.e.

SO2 and HCl are in gaseous phase.

23. Alkyl halides are generally not prepared in laboratory by free radical halogenations of alkanes.

Answer: Alkyl halides are generally not prepared in laboratory by free radical halogenations of

alkanes.since Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric

mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield

of any one compound is low.

24. Sulphuric acid is not used during the reaction of alcohols with KI. Answer: H2SO4 cannot be used

along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI

and then oxidizes it to I2

*Examples of solid organic compounds containing Nitrogen (N), Sulfur (S), Halogens (Cl, Br, I), and Phosphorus (P).used in element detection through Laissaigne method*

Compounds Containing Nitrogen (N):

1. Urea: CO(NH₂)₂

2. Benzamide: C₆H₅CONH₂

3. Aniline hydrochloride: C₆H₅NH₂·HCl

Compounds Containing Sulfur (S):

1. Thiourea: SC(NH₂)₂

2. Benzothiazole: C₇H₅NS

3. L-Cysteine: HSCH₂CH(NH₂)COOH

Compounds Containing Halogens (Cl, Br, I):

1. Chloroform (solid form): CHCl₃ (below 63°C)

2. Iodoform: CHI₃

3. Hexachlorobenzene: C₆Cl₆

Compounds Containing Phosphorus (P):

1. Triphenylphosphine oxide: (C₆H₅)₃PO

2. Phosphoric acid esters: (RO)₃PO (e.g., triethyl phosphate, where R = CH₂CH₃)

3. Aminomethylphosphonic acid: H₂NCH₂PO₃H₂

Compounds Containing Multiple Elements:

1. Nitrogen + Sulfur: Sulfanilamide (C₆H₄(SO₂NH₂)(NH₂))

2. Nitrogen + Halogen: Chloramine-T (C₇H₇ClNNaO₂S)

3. Sulfur + Phosphorus: Phosphorothioates (e.g., Malathion: C₁₀H₁₉O₆PS₂)

4. Nitrogen + Phosphorus: Ammonium phosphate (NH₄)₃PO₄

Chemistry class XI question paper based on redox reactions and chemical equilibrium

         SUB: CHEMISTRY                                CLASS:  11th           MM:  40

General instructions

Question no.01 to 10 carry one marks.

Question no.11 to 14 carry two marks.

  Question no. 15 to 17 carry three marks.

Question no.18 to 19 carry four marks.

Question no. 20 carry five marks.

 

SECTION A- (1M)

Q.1. We know about the relationship between Kc and Kp is 

Kp = Kc (RT)∆n

What will be the value of ∆n for the reaction

NH4Cl (s) → NH3 (g) + HCl (g)

(i) 1

(ii) 0.5

(iii) 1.5

(iv) 2

 

Q.2. In the case of the reaction

H2(g) + I2(g) → 2HI (g), 

the standard free energy is given as 

∆G > 0.

The equilibrium constant (K) will be ___

(i) K = 0

(ii) K > 1

(iii) K = 1

(iv) K < 1

 

Q.3.Which of the following is not the general feature of equilibria involving physical processes?

(i) Equilibrium is possible only in the closed system at the given temperature.

(ii) All measurable properties of the given system remain constant.

(iii) All the physical processes stop at equilibrium.

(iv) The opposing process occurs at the same rate, and there is the dynamic, however stable condition.

 

 

Q.4.PCl5, PCl3, and Cl2 are at equilibrium at 500K in the closed container, and their concentrations are given as 0.8 × 10–3mol L–1, 1.2 × 10–3 mol L–1, and 1.2 × 10–3 mol L–1 respectively. The value of Kc for the given reaction

PCl5 (g) PCl3 (g) + Cl2 (g) will be,

 

 

(i) 1.8 × 103mol L–1

(ii) 1.8 × 10–3

(iii) 1.8 × 10–3 L moL–1

(iv) 0.55 × 104

 

Q.5. The acidity of the compound BF3 could be explained based on which among the following concepts?

(i) Arrhenius’s concept

(ii) Bronsted Lowry’s concept

(iii) Lewis’s concept

(iv) Bronsted Lowry as well as Lewis’s concept.

 

Q.6. Which of the following is not a redox reaction?

(a) CaCO3 → CaO + CO2

(b) O2 + 2H2 → 2H2O

(c) Na + H2O → NaOH + 1/2H2

(d) MnCl3 → MnCl2 + 1/2 Cl2

 

Q.7.The most powerful oxidising agent among the following is:

(a) H2SO4

(b) H3BO3

(c) HPO3

(d) H3PO4

 

Q.8.The oxidation number of Cr in K2Cr2O7 is:

(a) -6

(b) +6

(c) +2

(d) -2

 

 

 

Q9. Consider the following reaction:

Zn + Cu2+ → Zn2+ + Cu

With reference to the above, which one of the following is the correct statement?

(a) Zn is reduced to Zn2+ ions.

(b) Zn is oxidised to Zn2+ ions.

(c) Zn2+ ions are oxidised to Zn.

(d) Cu2+ ions are oxidized to Cu.

 

 

Q.10. Reduction never involves:

(a) gain of electrons

(b) decrease in oxidation number

(c) loss of electrons

(d) decrease in valency of electropositive component

 

 

 

                  SECTION- B (2M EACH)

 

Q.11. The value of Kc for the given reaction 

    2HI (g) ⇋ H2 (g) + I2 (g) is 1 × 10-4

    At the given time, the composition of the reaction mixture is   given as follows;

 [HI] =2 × 10-5 mol, [H2] =1 × 10-5 mol as well as [I2] =1 × 10-5 mol.

 

 Q.12. Predict whether the solutions for the following salts are neutral, acidic or basic: NaCl, KBr, NH4NO3, and KF.

In which direction would the reaction proceed?

 

Q.13. Calculate the oxidation number of phosphorus in the following species.

(a) HPO32- and

(b) PO43-

 

Q.14 Balance the following redox reactions by half reaction method   

   Cr2O7 2-- + Fe2+(aq.)  → Cr3+(aq.) + Fe3+ (aq.)  (in acidic solution) 

 

 

 

 

​                 SECTION- C (3M EACH)

Case Study 1:

The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:

 

 

Oxidation: An increase in the oxidation number of the element in the given substance.

Reduction: A decrease in the oxidation number of the element in the given substance.

Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.

Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.

Redox reactions: Reactions which involve change in oxidation number of the interacting species.

Types of Redox Reactions

1.) Combination reactions -A combination reaction may be denoted in the manner:

A + B → C

Either A and B or both A and B must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:

2.) Decomposition reactions- Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.

Examples of this class of reactions are:

It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction. This may also be noted here that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction.

3.) Displacement reactions- In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:

X + YZ → XZ + Y

Displacement reactions fit into two categories: metal displacement and non-metal displacement.

(a) Metal displacement: A metal in a compound can be displaced by another metal in the uncombined state. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores.

(b) Non-metal displacement: The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water. Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals.

4.) Disproportionation reactions – Disproportionation reactions are a special type of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction. The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation.

Here the oxygen of peroxide, which is present in –1 state, is converted to zero oxidation state in O2 and decreases to –2 oxidation state in H2O.

 

 

Q.15.

(i) In … an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element.

a) displacement reaction

b) decomposition reaction

c) disproportionation reaction

d) combination reaction

 

ii) … leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.

 a) displacement reaction

 b) decomposition reaction

 c) disproportionation reaction

 d) combination reaction

 

iii) In …. an element in one oxidation state is simultaneously oxidised and reduced.

a) displacement reaction

b) decomposition reaction

c) disproportionation reaction

d) combination reaction

 

 

Case Study 1: Le Chatelier’s principle is also known as the equilibrium law, used to predict the effect of change on a system at chemical equilibrium. This principle states that equilibrium adjusts the forward and backward reactions in such a way as to accept the change affecting the equilibrium condition. When factor-like concentration, pressure, temperature, and inert gas that affect equilibrium are changed, the equilibrium will shift in that direction where the effects caused by these changes are nullified. This principle is also used to manipulate reversible reactions in order to obtain suitable outcomes.

 

Q16.Which one of the following conditions will favour the maximum formation of the product in the reaction?

 

(i)

(a) Low temperature and high pressure
(b) Low temperature and low pressure
(c) High temperature and high pressure
(d) High temperature and low pressure

 

 

(ii)For the reversible reaction,

The equilibrium shifts in forwarding direction
(a) by increasing the concentration of NH3(g)
(b) by decreasing the pressure
(c) by decreasing the concentrations of N2(g) and H2(g)
(d) by increasing pressure and decreasing temperature.

 

(iii)Favourable conditions for manufacture of ammonia by the reaction

(a) low temperature, low pressure and catalyst
(b) low temperature, high pressure and catalyst
(c) high temperature, low pressure and catalyst
(d) high temperature, high pressure and catalyst

 

Q.17. Justify that the following reactions are redox reactions:
(a) CuO (s) + H2(g) —–> Cu(s) + H20(g)
(b) Fe2O3 (s) + 3CO(g) —-> 2Fe(s) + 3CO2(g)
(c) 4BCl3(g) +3LiAlH4(s) ——> 2B2H6(g) + 3LiCl(s) + 3AlCl3(s)

 

                     SECTION -D (4M EACH)

 

Q.18.Write formulas for the following compounds:
(a) Mercury (II) chloride, 

(b) Nickel (II) sulphate,

(c) Tin (IV) oxide, 

(d) Thallium (1) Sulphate

 

Q.19.What is the effect of:
(i) addition of H2 

(ii) addition of CH3OH
(iii) removal of CO

(iv) removal of CH3OH

 

 

 

 

 

 

                   SECTION -E (5 M)

Q.20. At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl5) is 8.3 x 10-3 . if decomposition proceeds as:

(a) Write an expression for Kc for the reaction
(b) What is the value of Kc for the reverse reaction at the same temperature.
(c) What would be the effect on Kc 

if
(i) More of PCl5is added 

(ii) Temperature is increased.