Model Question Paper-ii

Hexagonal close-packing can be arranged by two layers A and B placed one over another which can be represented by the following diagram:

Here, we can say that the second and third layers are not exactly aligned. So statement (iii) is not correct. Only the first and third layers are identical A-A or B-B. So options (i),(ii) and (iv) are correct.

Brine is a saturated solution of Sodium chloride in water. It contains a substantial amount of chloride(Cl^-) ions compared to hydroxyl (OH^-) ions.

Reduction potential value of Chlorine (Cl₂ + 2e → 2Cl^-) is 1.36V whereas that of Oxygen (O₂+ 2H₂O + 4e → 4OH-) is 0.40V.So O₂ gas should be oxidized at anode instead of Cl₂.

But since the amount of Cl^- ions are much more than OH- ions, so Cl^- ions are preferentially oxidized at the anode to liberate Cl₂ gas.

In qualitative analysis, when H₂S gas is passed through an aqueous solution of salt acidified with HCl, a black ppt of CuS is formed.

CuSO₄ + H₂S + dil HCl → CuS + H₂SO₄

(Black ppt)

When ppt of CuS is boiled with dilute HNO₃,it forms a blue coloured solution due to the following reactions given below:

3CuS + 8HNO₃→ 3Cu(NO₃)₂ + 2NO + 3S + 4H₂O

S + 2HNO₃→ H₂SO₄ + NO

2Cu²⁺ + SO₄^2- + 2NH₃ + 2H₂O → Cu(OH)₂ + CuSO₄ + 2NH₄OH

The given compound is a tertiary amine. There are two methyl groups and one n-butyl group attached to N atom.

Here the parent chain contains four carbon atoms. For two similar groups ‘di' is used.

In dilute sulphuric acid solution, hydrogen will be reduced at the cathode

and H₂O is oxidized at anode.

2H₂O → O₂ + 4H⁺ + 4e^-

while in a concentrated solution of sulphuric acid, SO₂^- ions are oxidized to tetrathionate ions.

Lyophobic sols are not stable and are easily precipitated by adding electrolytes. Lyophilic colloids like gums, soaps, gelatin etc are added to prevent coagulation of lyophobic sol. The process is known as ‘protection' and the lyophilic colloids are termed as protective colloids.

Lyophilic colloids cover-up of the particles of the lyophobic colloid by the formation of a protective film or layer.

Thus both the options (i) and (ii) are correct.

Emulsion is obtained when two immiscible or partially miscible liquids are shaken. Emulsions are of two types.

i) Oil dispersed in water(O/W type) and ii) Water dispersed in oil (W/O type).The O/W type on standing separate into two layers. So it is unstable. To stabilize the emulsion, a third component known as an emulsifying agent is added. The emulsifying agent helps to form an interfacial layer between suspended particles and the medium. So a stable emulsion is formed. For O/W and W/O type emulsifying agent used is gum and heavy metal salts respectively.

Zone refining process is very useful to get semi-conductor metals like germanium, Silicon etc. in a highly pure state. This method is based on the principle that impurities present in metals are more soluble in the molten state than in the solid-state of the metal.

Natural rubber is soft and sticky at a higher temperature and brittle at low temperature. Upon cross-linking the rubber becomes hard and tough with greater tensile strength. The vulcanized rubber is highly elastic, having low water absorption tendency and it is resistant to oxidation.

Aspartame is an artificial sweetener which is a non-carbohydrate compound. This is a methyl ester having dipeptide linkage between aminoacids Aspartic acid and Phenylalanine.

Aspartic acid Phenylalaninemethylester Aspartame

-NH₂ group of Aspartic acid joins with –COOH group of Phenylalaninemethylester to form dipeptide unit ( -CONH).

In semiconductors, electrons occupy different energy bands namely valence band and conduction band. There is an energy difference between the highest occupied valence band and lowest unoccupied conduction band. This is known as the bandgap which is quite small in semiconductors.

With the increase in temperature, electrons in the valance band get excited and jump to the conduction band, which increases ii's electrical conductivity.

Ring test is used to detect the presence of nitrate ions (NO₃^-) in chemistry qualitative analysis. The required chemical reactions involved for the Ring test' is:

i) NO₃^- + conc.H₂SO₄→ HSO₄^- + HNO₃

ii) 2HNO₃→ 2NO + H₂O + 3O

iii) 6FeSO₄ + 3O+ 3H₂SO ₄→ 3Fe₂(SO₄)₃ + 3H₂O

iv) 6FeSO₄ + 2HNO₃ + 3H₂SO₄ → 3Fe₂(SO₄)₃ + 2NO +4H₂O

v) FeSO₄ + NO + 5H₂O →

Brown ring is formed at the junction of the layers of the solution and sulphuric acid is obtained, which confirms the presence of nitrate ions in the solution.

Crystal field splitting energy Δ0 increases in the order:

[Cr(Cl)₆]^3– < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3–

Among the ligands Cl, NH₃ and CN, Cl is the weakest ligand and CN is the strongest one. Therefore, increasing order of crystal field splitting energy Δ0 is: [Cr(Cl)₆]^3– < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3–

The structure of allyl chloride and n-propyl chloride are the following:

Hydrolysis(addition of water molecule) involves the formation of a carbocation. Due to the presence of π-bond in allyl carbocation, this molecule can undergo resonance and hence gets stabilised.

But in n-propyl carbocation, no resonance is possible. So allyl chloride is hydrolysed more readily than n-propyl chloride.

Melamine and Formaldehyde are the two monomers for the synthesis of the above polymer Melamine.

The polymerisation occurs in the following way.

Antacids suppress the symptoms of acidity and not the cause. They work by neutralizing the acid produced in the stomach i.e. they do not control the cause of the production of more acid.

Antihistamines are the drugs that suppress the action of histamine which is the chemical responsible for the stimulation of secretion of acids (example pepsin and HCl) in the stomach.

Thus, antihistamines are used over antacids for better treatment of hyperacidity.

The vapour pressure above the liquid is inversely proportional to the boiling point of the liquid i.e. higher the vapour pressure above the liquid, lower is the boiling point of the liquid.

NaCl is a non-volatile substance, it decreases the vapour pressure above water. Thus, increasing the boiling point of the water.

Methyl alcohol is volatile in nature, so it increases the vapour pressure above water when added to it. Thus, it decreases the boiling point of the water.

The oxidation of water at anode requires over potential in the electrolysis of aqueous sodium chloride, so Cl^- ion is oxidized at the anode. The overpotential condition is due to kinetically slow oxidation of water.

The overpotential of water is greater than the standard potential for the oxidation of Cl^- ion.

Thus, chloride ion gets oxidized at the anode in electrolysis of aqueous NaCl.

The process of recovery of copper from low-grade ore involves two steps:

1. Leaching i.e. use of acids or suitable reagents that dissolve ore but not the impurities.

2. This leached solution is reacted with scrap iron (cheap in cost) or hydrogen to give copper.

Thus, the end product of the above reactions is copper.

Reaction of P₄O₆ in H₂O is:

The reaction of Phosphorous acid with water is:

The above two reactions can be written in combined form:

1 mole of is neutralized by 8 moles of NaOH

of P₄O₆ will be neutralized by moles of NaOH

Molarity of NaOH solution is 0.1 M i.e. 0.1 mole is present in 1 L of water

So, moles of NaOH will be present in =

The volume of 0.1 M NaOH required is 0.4 L or 400 mL.

The structure must be cis-octahedral in nature.

The trans octahedral form does not show the optical activity, so the structure can be cis octahedral in nature.

The example of such complex is .

The major product will be the 2-Chloro-2-Methyl propane

The reaction mechanism involves the electrophilic addition reaction according to the Markovnikov Rules.

The reaction mechanism is

Isobutylene breaks into the primary and secondary carbocation in the presence of H⁺ ion.

A secondary carbocation is much stable due to the +R effect of methyl group so it is more stable to undergo the electrophilic addition reaction.

Thus, major product is 2-Chloro-2-Methyl propane.

The Lucas reagent is a solution of anhydrous zinc chloride in concentrated hydrochloric acid.

In Tertiary alcohols (3^o) the hydroxyl group is easily substituted by the halide group due to greater stability of the carbocation in comparison to primary carbocation and secondary carbocation formed during the reaction.

The greater stability is due to the more inductively donating alkyl groups. The hyperconjugation effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.

Thus, the correct order of the reaction is 3^o>2^o>1^o alcohols in increasing order.

Yes, we can convert CH₃NH₂ to a 2⁰ amine without getting any side products.

This reaction is called condensation reaction or amidation of alcohol.

This reaction involves the reaction between primary alcohol and a primary amine in the presence of a catalyst [RhH(PPh₃)₄] and yield of secondary amine is greater than 98%.

Amidation of alcohols.

The given disaccharide is Sucrose (table sugar). Its molecule is comprised of two monosaccharides i.e. fructose and glucose. The bond between the two monosaccharide units is called glycosidic linkage.

The sucrose has no anomeric hydroxyl groups (free ketone or free aldehydic group) so it is classified as a non-reducing sugar.

NaCl is a non-volatile substance when it is added to the water it decreases the melting point of the water simultaneously decreasing the vapor pressure of the water.

According to Raoult’s law, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction.

Thus, for water as a solvent in a solution

Where P_w is the partial pressure of water and x_W is a mole fraction of water in the solution.

Thus, on adding NaCl the mole fraction of water reduces.

Thus, vapor pressure also reduces.

Thus, both the assertion and reason are true and the reason is the correct explanation of the assertion.

In the ether, the bond angle is greater than the tetrahedral angle due to the fact that internal repulsion by the hydrocarbon part is greater than the external repulsion of the lone pair on oxygen.

Thus, both the assertion and reason are true and the reason is the correct explanation of the assertion.

The role of a catalyst is to provide an alternative path by forming an activated complex with lower activation energy without changing the enthalpy of reaction.

In the presence of a catalyst, the Gibbs free energy does not change as the energy difference between reactant and product remains the same.

A reference can be taken from the above graph that final product AB energy and the reactants (A and B) energies are the same in the absence and the presence of catalyst remains the same in the presence or absence of a catalyst.

OR

A pseudo-first-order reaction is a reaction that is truly second order but can be approximated to be first-order under special circumstances.

For Example, Hydrolysis of .01 mole of ethyl acetate in the presence of 10 moles of water, the reaction at first glance appears to be the first-order reaction but since the concentration of water does not change much in comparison to ethyl acetate the order of reaction becomes pseudo-first-order reaction.

For a pseudo-first-order reaction, the reactant with a large amount of concentration is ignored.

As the change in its concentration is negligible with comparison to another component in respect with time.

Hence, the rate only gets dependent upon the reactant in small quantity.

A is FeCr₂O₄, B is Na₂CrO₄, C is Na₂Cr₂O₇.2H₂O and D is K₂Cr₂O₇.

The reactions are

1.

2.

3.

Thus, A is chromite of iron which when reacts with the sodium carbonate gives a yellow solution (Na₂CrO₄), which in the presence of the H⁺ ion (acid) gives compound C (Na₂Cr₂O₇.2H₂O) which on treatment with KCl solution gives orange crystal of K₂Cr₂O₇.

OR

A is MnO₂, B is K₂MnO₄, C is KMnO₄ and D is KIO₃.

The reaction involved is

1.

2.

3.

Thus, the reaction involves these steps and their yields are respectively given.

The compound is Methyl Benzoate(A) which gives the 2,4 DNP test.

These are the compounds A, B, C with the following reactions.

OR

The reaction mechanism involves the following steps:

It is given that B gives negative tollents teat which means it is a Ketone.

The positive iodoform test suggest it has a structure similar to

Thus the compound B is propanone (common name acetone). It on the oxidation give acetic acid (compound C )and formic acid.