Chemistry NCERT Exemplar Solutions Class 12th Chapter Thirteen: Overview, Questions, Preparation

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Ans: (B) Triethylamine 

The structure of the given amines are shown below.

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Ans: Correct answer: (D) 

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Ans:  (C)

Because of the electron-releasing nature of the alkyl group, it (R) pushes electrons towards nitrogen, making the unshared electron pair more available for sharing with the acid's proton. Furthermore, the +I effect of the alkyl group stabilizes the substituted ammonium ion formed from the amine by dispersing the positive charge.

As a result, alkylamines are more powerful bases than ammonia.

As a result, the basic nature of aliphatic amines should increase as the number of alkyl groups increases.

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Ans:  (A) 

The NH₂ group is directly attached to the benzene ring in aniline and other aryl amines. As a result, the unshared electron pair on the nitrogen atom is conjugated with the benzene ring, making it less available for protonation.

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Ans: (C) 

S_N¹ reaction: A nucleophilic reaction that occurs in two steps, first is the bond-breaking step and the second is the production of the carbocation. The stability of carbocation formed in the second step determines the rate of reactivity of reactant toward S_N¹ reaction.  Here, C₆H₅CH₂Br,  

In the process of ionization, removal of bromine, a stable Benzyl carbocation is produced. Therefore, it is best suited for reaction through the S_N¹ mechanism. 

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Ans: (B)

Lithium aluminum hydride in ether is a strong reducing agent that donates its H? hydride ion to any C=O containing a functional group. In addition to LiAlH₄ to aryl nitro compounds, no reaction will be observed, the desired products of amines will not be produced, rather it will form diazobenzene products.

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Ans: (C) 

Potassium cyanide, as cyanide on reduction with sodium metal in alcohol, produces amine with increased carbon atoms. 

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Ans: (D)

Gabriel synthesis reaction:  

Gabriel’s synthesis is a method for producing primary amines. When phthalimide is treated with ethanolic potassium hydroxide, it forms a potassium salt of phthalimide, which when heated with an alkyl halide and then alkaline hydrolyzed yields the corresponding primary amine. This method cannot produce aromatic primary amines because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

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Ans: ( C)

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Ans: (D)

Lithium aluminium hydride in ether is a strong reducing agent that donates its hydride ion to any C=O containing functional groups into corresponding reduced compounds. Here, the amide group is converted to an amine functional group.

LiAlH₄ in ether is the best reagent for converting 2-phenylpropanamide to 2-phenylpropanamine.

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Ans: (B) 

Hoffmann bromamide reaction- it is also called degradation reaction as in this reaction primary amide group is treated with halogen first (Br) then the halogen-substituted amide product is converted to a primary amine with the release of carbon dioxide gas.

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Ans: (B) 

Hoffmann invented a method for producing primary amines by treating an amide with bromine in an aqueous or ethanolic sodium hydroxide solution. An alkyl or aryl group migrates from the amide's carbonyl carbon to the nitrogen atom during this degradation reaction.

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Ans: (D) 

The greater the electron density towards the ring, the greater its basic strength.

The electron withdrawing group reduces basic strength, whereas the electron donating group increases basic strength.

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Ans: (C)

Methyl amine reacts with HNO₃ form methanol with release of nitrogen gas and water as side products.

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Ans: (B) 

Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts, which are unstable and release nitrogen gas and alcohol quantitatively.

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Ans: (C)

In the process of nitration of benzene, firstly H₂SO₄ dissociates into H⁺ and HSO₄−. The proton produced further reacts with HNO₃ to give a positively charged complex that is unstable and dissociates into nitronium ions and water as products. The nitronium ion is an electron-deficient species i.e. electrophile that attacks the electron cloud of the benzene ring.

H₂SO₄  (conc.) ? H⁺+HSO₄⁻

H⁺+HNO₃? H₂NO₃⁺

H₂NO₃⁺? NO₂⁺ + H2O

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Ans: (C) 

Using active metals like iron in acidic conditions can be efficiently used for the reduction of nitro compounds. 

Under the acidic conditions, the intermediate compounds that may form are readily reduced to primary amine and pure product of primary amine is obtained. 

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Ans:  (B) 

Greater will be the basic character, greater will be its reactivity towards dilute hydrochloric acid. As, (B), secondary amine, it will be the most basic among all and hence the most reactive amine towards acid. 

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Ans: (A) 

On Reaction of acid anhydrides with primary amines, amides are formed as the primary product along with carboxylic acid. 

(RCO)₂O + RNH₂? RCONH₂ + RCOOH

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Ans:  (B) 

Diazonium salts react with copper powder and halogen acid to form aryl halide.

Gattermann reaction is a Sandmeyer reaction variant in which Cu powder replaces CuCl.

This substitution produces aryl halide more easily and under milder conditions than the Sandmeyer reaction.

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Ans:  (B) 

The reaction of transformation of primary alkyl halides to primary amines using potassium phthalimide is the Gabriel phthalimide reaction. 

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Ans: (D)

Weak electrophile such as Diazonium cation readily reacts with electron-rich compounds which are having electron-rich compounds such as hydroxyl group, amino group. They don't react with electron-withdrawing groups like nitro groups. Therefore, nitrobenzene will not undergo an azo coupling reaction with benzene diazonium chloride. 

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Ans:  (D) 

Since phenol is the strongest acid among the four options listed above, it is the weakest Brönsted base. The stronger the acid, the weaker the conjugate base.

Amines have a strong tendency to accept electrons thus they are a strong bronsted base while phenol is the strongest acid among all, therefore as per the relation of conjugative strong acids and weak bases, phenol is the weakest base. 

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Ans: (D) 

Pyrrolidine is the strongest of two bases because the lone pair of nitrogen does not involve sin resonance, and the presence of two alkyl basic compounds increases the basic strength among the given four compounds.

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Ans: (A)  

The basic strength of an atom is determined by its electron-donating capacity; in this case, the amide is the most basic due to the presence of a negative charge and two lone pairs of electrons on the nitrogen atom.

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Ans: (B)

Among primary, secondary, and tertiary amines, tertiary amines are the most volatile compounds as they don't have any strong intermolecular H-bonding between N-H like the primary and secondary amines have. 

So due to weak dipole-dipole interactions, (B) will be the most volatile. 

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Ans: (D)

Aliphatic and arylalkyl primary amines can be prepared by the reduction of the corresponding nitriles with LiAlH₄

Heating alkyl halide with primary, secondary and tertiary amine can be prepared by reduction of LiAlH4 followed by treatment with water.

Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis produces primary amine. This process is known as Gabriel phthalimide reaction. The number of carbon atoms in the chain of amines of product is same as reactant.

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Ans: (C and D) 

Chlorobenzene and bromobenzene are prepared by the sandmeyer's reaction. 

While fluorobenzene and iodobenzene are prepared by simple heating of diazonium salt with aqueous KI Solution.

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Ans: (A, B & C) 

Nitro compounds are reduced to amines by passing hydrogen gas through an acidic medium containing finely divided nickel, palladium, or platinum, as well as by reduction with metals in an acidic medium.

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Ans: (A and B)

When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, they form isocyanides or carbylamines, which have a foul odor. This reaction does not occur in secondary or tertiary amines.

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Ans: option (B and C) 

Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanol, respectively.

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Ans: Option a and b

NHCOCH₃ is an ortho- para directing group so ortho-para products are formed. 

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Ans: A, B & C  

As the NH₂ group is Ortho para directing group, so Br⁺ attacks only at the o and p positions thus the products formed corresponds to the option A, B and C.

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Ans: (a and b) 

Gabriel synthesis is a method for producing primary amines. When phthalimide is treated with ethanolic potassium hydroxide, it forms a potassium salt of phthalimide, which when heated with an alkyl halide and then alkaline hydrolyzed yields the corresponding primary amine.

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Ans: option a and c

(a) Primary alkyl halides react with ammonia to give primary amines. Correct reaction.

(B) Elimination process doesn't occur with aq. KOH, hydrolysis process takes place. (B) is incorrect. 

(c) Dehydrohalogenation I.e. elimination of HCl occurs which produces alkene as the product takes place with alc. KOH . (c) is correct.

(d) In this reaction, aliphatic primary amines, on treatment with nitrous acid, produce primary alcohol as a product.

Therefore (d) is incorrect. 

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Ans: a and b  

In addition to the nitro derivatives, direct nitration of aniline produces tarry oxidation products. Furthermore, aniline is protonated in the strongly acidic medium to form the anilinium ion, which is meta directing. As a result, in addition to the ortho and para derivatives, a significant amount of meta derivative is formed.

However, by protecting the? NH₂ group with an acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative obtained as the main product.

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Ans: Option (A and B)  

Benzene diazonium chloride reacts with phenol to form p-hydroxy azobenzene by coupling the phenol molecule at its para position with the diazonium salt. This is referred to as a coupling reaction. This is an illustration of an electrophilic substitution reaction. 

At room temperature, aniline reacts with bromine water to form a white precipitate of 2,4,6-tribromoaniline.

By acetylation with acetic anhydride to protect the −NH₂ group, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.

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Ans: HNO₃ acts as a base in the nitrating mixture (HNO₃ + H₂SO₄) and provides the electrophile.

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Ans:

 Direct nitration of aniline is not possible on account of oxidation of -NH₂ group. However, nitration can be carried after protecting the -NH₂ group by acetylation to give acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines.
The acetyl group being electron withdrawing attracts the lone pair of electrons of the N-atom towards carbonyl group.

As a result, the activating effect -NH₂ group is reduced i.e., the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of —NHCOCH₃ group is less than that of —NH₂ group.

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Ans: When Benzyl amine is treated with nitrous acid, firstly BDC is formed which is unstable and decomposes to Benzyl alcohol. 

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Ans: The best reagents for the conversion of nitrile to primary amine are LiAlH4 and Sodium/Alcohol. By reduction, the nitriles can be converted into a corresponding primary amine.

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Ans: Hinsberg's reagent is also known as benzenesulphonyl chloride (C₆H₅SO₂Cl). When it reacts with primary and secondary amines, it produces sulphonamides. Allowing secondary and tertiary amines to react with Hinsberg's reagent allows them to be distinguished (benzenesulphonyl chloride C₆H₅SO₂Cl). Secondary amines react with Hinsberg's reagent to form an alkali-insoluble product. N, N-diethylamine, for example, reacts with Hinsberg's reagent to form N, N-diethylbenzenesulphonamide, which is insoluble in alkalis. Tertiary amines, on the other hand, are unaffected by Hinsberg's reagent.

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Ans: The reaction of aniline with nitrous acid at 273-278 K produces benzene diazonium chloride. The reaction of sodium nitrite with hydrochloric acid produces nitrous acid in the reaction mixture. Diazotisation is the process of converting primary aromatic amines into diazonium salts. Because of its instability, the diazonium salt is generally not stored or used immediately after preparation.

The crystalline solid benzene diazonium chloride is colorless.

It is easily soluble in water and stable at room temperature, but it reacts with water when warmed.

In the dry state, it decomposes easily.

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Ans: As −NH₂ is a strong activating group, the aniline will readily undergo electrophilic substitution reaction, and it is difficult to cease reaction at the mono substitution stage. 

Therefore, the activating group −NH₂ is protected by an acetylation process.

The acetylated complex formed utilizes the lone pair of nitrogen and are less available for donation, this helps to carry out the nitration reaction easily. 

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Ans: As the electronegativity of oxygen is more than the electronegativity of a nitrogen atom, the O−H bond is more polar than the N−H bond, therefore MeOH is stronger acid than MeNH₂ or MeNH₂ is stronger base than MeOH.

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Ans: Amides are the byproducts of the acylation reaction. The reaction is carried out in the presence of a stronger base than the amine, such as pyridine, which removes the formed HCl and shifts the equilibrium to the right.

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Ans: The azo products have an extended conjugate system with both aromatic rings linked by the –N=N- bond. These compounds are frequently colored and used as dyes. Benzene diazonium chloride reacts with phenol to form p-hydroxy azobenzene by coupling the phenol molecule in its para position with the diazonium salt. This is referred to as a coupling reaction.

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Ans: CS₂ is a non-polar solvent which decreases the activating effect of? NH₂. As a result, mono substitution occurs only at o- and p- positions giving a mixture of 2-bromoaniline (minor) and 4-bromoaniline (major) as products. 

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Ans:  CH₃CH₂CH₃ < CH₃CH₂NH₂ < CH₃CH₂OH

As oxygen is more electronegative than nitrogen therefore, the O-H bond is more polar than the N-H bond. So ethanol has more dipole moments than ethylamine. Propane is non- polar in nature hence, had the least among all. 

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Ans:

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Ans: N, N-dimethylbenzenanmine 

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Ans: Z is an aliphatic amine that produces a solid that is insoluble in base. This means that the reaction with C₆H₅SO₂Cl must produce a product with no replaceable hydrogen attached to nitrogen.

To put it another way, the amine must be a secondary amine.

Z stands for ethylmethylamine.

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Ans: Carbylamine reaction is shown by primary amines only, primary amines react with chloroform in the presence of alcoholic KOH to form isocyanides which upon catalytic reduction gives secondary amines as the final product. 

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Ans: On reaction with hydrochloric acid, aniline forms anilinium chloride ion which is water soluble. Therefore, the aniline solubility in aqueous HCl solution. 

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Ans:

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Ans:

This is an example of an electrophilic aromatic substitution reaction. phenol generates phenoxide ion in alkaline medium, which is more electron-rich than phenol and thus more reactive to electrophilic attack. In this reaction, the electrophile is an aryldiazonium cation.

The faster the reaction, the stronger the electrophile.

The cation p-nitrophenyldiazonium is a stronger electrophile than the cation p-toluene diazonium.

As a result, it preferentially couples with phenol.

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Ans:

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Ans: (i)- (d), (ii)- (c), (iii)- (a), (iv)- (b)

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 Ans: (i)- (b), (ii)- (a), (iii)- (d), (iv)- (c)

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Ans: (c)

In the acryl derivative, the delocalisation of electrons of the nitrogen atom occur over the carbonyl group, this decreases the electron density on the nitrogen atom that it no more perform as nucleophile and don't react with next acylating alkaline molecule. 

Therefore, Assertion statement is correct but reason is incorrect.

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Ans: (a)

In Hoffmann bromamide degradation reaction as in this reaction primary amide group is treated with halogen first bromine then the halogen substituted amide product is coverts to primary amine with the release of carbon dioxide gas. Both the statements assertion and reason are incorrect therefore the correct option is a. 

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Ans: (d)

Ethylbenzene sulphonamide is soluble in alkali because it has acidic hydrogen.

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Ans: (b)

N, N-die ethyl benzene sulphonamide, there is no acidic hydrogen present on the N-atom which can make it soluble. Therefore, it is insoluble in alkali. 

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Ans: (d)

Fe+2HCl→FeCl₂+2 [H] FeCl₂+H₂O (g)→FeO+2HCl

The nascent hydrogen formed act as reducing agent for the reduction of nitro compounds.

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Ans: (a)

Aromatic primary amines cannot be prepared using Gabriel phthalimide synthesis as aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide. Both the statement's assertion and reason are incorrect, hence the correct answer is a.

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Ans: (d) 

Acetanilide is less basic than aniline because electron density of nitrogen is lowered by acetyl group.

Number of Fe atoms = $\frac{0.34}{100}*\frac{3.3}{56}*6.022*10^{23}$

= 1.206 * 10²⁰

According to Fajan’s rule,

Covalent character $\propto$ size of anion

$\therefore Ca{F}_2<CaC{l}_2<CaB{r}_2<Ca{l}_2$

t_1/2 = 200 sec and it is 1st order reaction

$K=\frac{2.303log2}{200}=\frac{2.303}{t}log\frac{Co}{0.2Co}$

$\frac{log2}{200}=\frac{1}{t}log5$

$t=\frac{7}{3}\times 200=466.67sec=467sec$

the finest gold sol is red in colour, as the size of particle increases, it appears purple then blue and finally gold. The colour of colloidal solution depends on the wavelength of light scattered by the dispersed particles. The wavelength of light further depends on size and nature of the particles. Hence, both assertion and reason are true and reason is the correct explanation of assertion.

              Element                            Melting Point

              Al                                        933 K

              Ga                                      303 K

              ln                                        430 K

              Se                                       490 K

Al is extracted from Al₂O₃.2H₂O  i.e, Bauxite ore.

Kindly go through the solution 
$2Mn{O}_4^{-}+5H_2{O}_2+6H^{+}\to 2M{n}^{2+}+8H_{2}O+5O_2$

Due to high lattice energy, LiF is sparingly soluble in water. Li+ has high hydration energy among its group members due to the smallest size.

Kindly consider the following

              K₂ [Cu (CN)₄]

              Oxidation number of Cu is +1

              Cu⁺ = [Ar]3d¹⁰ ® Diamagnetic

              Match list-I with list-II

Correct order of priority is

$-S{O}_{3}H>-COCl>-CON{H}_2>-CN$

The structure consisting benzene ring is called benzenoidal structure and options (A) and (B) do not contain benzene ring, so they are non benzenoidal.

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution 

Vulcanization of rubber is carried out by heating a mixture of isoprene and sulphur.

Other name of animal starch is glycogen.

Phenolphtalein is a pH dependant indicator. It is a weak acid which is colourless in acidic medium but gives pink colour in basic medium. The pink colour is due to its conjugate form. Phenolphthalein dissociates in basic medium. Therefore, assertion is true but reason is false.

$\underset{\left(Colourless\right)}{HPh\left(aq.\right)}?H^{+}+\underset{\left(Pink\right)}{P{h}^{-}}$

PV = nmixRT

$n_{mix}=\frac{6*12.5}{0.083*300}?3$

Let moles of He is x

Moles of H₂ = 3 – x

4x + 2 (3 – x) = 10

x = 2 mol

Mass of He = 8 gm

 Number of neutrons = 26

Number of electrons = 25

% of extra electrons = $\frac{26-25}{25}\times 100=4\mathrm{\%}$

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Delta H=-59.6+1\times 8.314\times 300\times 10^{-3}=-57.10kJ/mol$

$\Delta T_b=i{K}_b{m}_1$

$\frac{\Delta T_b}{\Delta T_f}=\frac{K_b\times 1}{K_f\times 2}\Rightarrow \frac{3}{6}=\frac{1}{2}=\frac{K_b}{k_f}\times \frac{1}{2}$

$\frac{K_b}{K_f}=1\Rightarrow x=1$

$meqof{l}_2=meqofN{a}_2{S}_2{O}_3=20\times 0.002\times 1$

2 × mmol of l₂ = 0.4

mmol of l₂ = 0.2 mmol

mmol of Cu²⁺ = 0.2 × 2 × 10^-3

$\left[C{u}^{2+}\right]=\frac{0.4\times 10^{-3}}{10\times 10^{-3}}=0.04=4\times 10^{-2}M$

 $C_2{H}_{5}OH+PC{l}_3\to C_2{H}_{5}Cl+H_{3}P{O}_3$

$H_{3}P{O}_3+PC{l}_3\to H_4{P}_2{O}_5+HCl$

$\stackrel{+4}{Mn{F}_4}$ $\stackrel{+3}{Mn{F}_3}$

E.C = [Ar] 3d³                    [Ar]3d⁴

$\begin{array}{cc}\hfill \stackrel{+2}{Mn{F}_2}\hfill & \hfill \left[Ar\right]3d^5\hfill \end{array}\left[\begin{array}{l}{E}_{M{n}^{3+}|M{n}^{2+}}^0=1.57V\\ E_{M{n}^{4+}|M{n}^{2+}}^0=1.27V\end{array}\right]$

Moles of C₂H₆ =   $\frac{2.24\times 10^{-3}}{22.4}=10^{-4}$

Weight of C₂H₆ = 10^-4 × 30 = 3 mg

Ofloxacin is the only broad spectrum antibiotic given in the question.

Pencillin – G is narrow spectrum antibiotic.

Salvarsan is mainly active against spirochete, a bacteria that causes syphilis.

Terpineol is an antiseptic.

Meq of K₂Cr₂O₇ = Meq of Fe²⁺

(Molarity × Volume × nf) of K₂Cr₂O₇ = (molarity × volume × nf) of Fe²⁺

0.02 × 20 × 6 = M × 10 × 1

M = 0.24 M

Molarity = 24 × 10^-2 M

On decreasing pressure of NO by a factor of ‘2’ the rate of reaction decreases by a factor of ‘4’

$\therefore$ Order of reaction w.r.t ‘NO’ = 2

$K{O}_2,N{O}_2,Cl{O}_2,NO$ are paramagnetic.

$C_{P,m}=C_{v,m}+R$

$C_{v,m}=20.785-8.314=12.471J{K}^{-1}mo{l}^{-1}$

$n=\frac{5000}{12.471\times 200}=\frac{25}{12.471}\approx 2$

 $C{N}^{-},N{O}^{+}and{O}_2^{2+}$ have bond order = 3

Solubility of CaF₂ = S mol/L

$S=\frac{2.34\times 10^{-3}}{0.1\times 78}=\frac{2.34}{78}\times 10^{-2}=3\times 10^{-4}mol/L$

$K_{sp}\left(Ca{F}_2\right)=4S^3=4{\left(3\times 10^{-4}\right)}^3=108\times 10^{-12}$

= 0.0108 × 10^-8 (mol/L)³

Complex is $\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]Cl$

Primary valency = No. of ionisable species = 1

$2KMn{O}_4+5H_2{\stackrel{+3}{C}}_2{O}_4+3H_{2}S{O}_4\left(dil\right)\to K_{2}S{O}_4+2MnS{O}_4+10C{\stackrel{+4}{O}}_2+8H_{2}O$

Oxidation state of carbon changes from +3 to +4

$\therefore$ change in O. S is 1

 $\mathrm{\%}Opticalpurity=\frac{Observedrotationofmixture\times 100}{rotationofpureenautiomer}$

$=\frac{+12.6}{+30}\times 100=42\mathrm{\%}$

Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 × 0.5 n = 0.3 n

So, overall % age yield of complete reaction = $\frac{0.3n}{n}\times 100=30\mathrm{\%}$