Chemistry Mock Test - 5

Concept:

Imperfections in crystal structure involving either a single atom or a few numbers of atoms are called as point imperfections/defects. 

Types of point defects or imperfections are:

• Vacancy defects
• Interstitial Defects
• Substitutional defects
• Schottky defects
• Frenkel defects

Difference between Schottky and Frenkel is explained with the help of the following diagram:

Explanation-

• The particles present in a solid are closely packed together in a set arrangement which means that their location is fixed. The force between the adjacent particles is also high which makes it all the more effective for the heat to transfer and pass by during collision. 
• Solids tend to have a great electrical conductivity, which is why they’re also known as conductors. They allow both heats as well as electricity to pass through them with utter ease. This is mainly due to the fact that their conductivity is broadly dependent on their atoms’ number of valence electrons. 
• At low temperatures, semiconductors tend to act as insulators and at higher temperatures, they act as conductors.
• This is because the electrons that gather around the atoms of the particular semiconductor are able to break away from their covalent bonds with a rise in temperature.
• That in turn lets the electrons move about freely in the lattice. And at room temperature, there are enough free electrons in a semiconductor to let it conduct some kind of current.
• All metals are good conductors of electricity because they contain free or mobile electrons. These free electrons conduct electric currents.

The correct order of conductivity in solids is K_metals >> K_insulators < K_{semiconductors}

Explanation:

Given Concentration of the molecule is the same (i.e. 0.01 M)

→ The process of conversion of a substance from its liquid form to its gaseous form at a particular temperature by absorbing heat is called boiling.

Experiments have shown that for dilute solutions the elevation of boiling point (∆Tb ) is directly proportional to the molal concentration of the solute in a solution.

Thus ∆Tb ∝ m  or ∆Tb = Kb × m × i

"i" is Vanthoff factor.

∆T_b ∝ i

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality,

K_b is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).

The unit of Kb is K kg mol-1.

→ Glucose is a covalent molecule and does not undergo dissociation (i = 1) therefore the concentration of solute particles in solutions is 0.01 M

→  Potassium chloride is an ionic compound and phenol forms phenoxide ion hence it shows a greater boiling point than glucose.

→ KCl → K⁺ + Cl^-  (i = 2)

→ So the correct order will be Potassium chloride solution > Phenol solution > Glucose solution

Explanation:

Given,

→ 18.25 grams of HCl (Solute)

⇒ Number of moles = 18.25/36.5 = 0.5 mole

→ Solvent (Water) = 500 g

We know, molality is given by:

\({\rm{Molality}} = \frac{{{\rm{Moles\;of\;Solute}}}}{{{\rm{Kilogram\;of\;solvent}}}} \)

\(\therefore {\rm{Molality}} = \frac{{0.5\; \times {\rm{\;}}1000}}{{500 }}\) = 1 m

Explanation:

Normality-

→ Normality is described as the number of gram or mole equivalents of solute present in one liter of a solution.

Normality:

• It is defined as the number of grams equivalent per liter of solution.
• Also known as equivalent concentration.

​\({\rm{Normality}} = \frac{{{\rm{Gram\;equivalents}}}}{{{\rm{volume\;of\;solution\;in\;liter}}}}\)

Option 1: 8 g of KOH per 100 ml

⇒ KOH (Gram equivalent weight = 56)

N = (8/56) × (1000/100) = 1.428 N

Option 2: 1 N phosphoric acid

Option 3: 6 g of NaOH per 100 mL

⇒ NaOH gram equivalent weight = 40g

N = (6/40) × (1000/100) = 1.5 N

Option 4: 0.5 M H2SO4

Converting Molarity into Normality 

As H₂SO₄  has basisity of 2 (2 H⁺ + SO₄^2-) 

→ Normality = 2 × 0.5 = 1 N

∴ Normality of 6 g of NaOH per 100 mL is high (i.e. 1.5 N)

The correct answer is isotonic solutions.

Concept:

• Colligative properties are the properties that depend upon the number of solute particles present in the solution. 
  They are : 
• Lowering of Vapour pressure:
  • The vapour pressure exerted by solute molecules on the surface of the solution decreases as solute particles are added to the solution.
  • The relative lowering vapour pressure as given by Raoult's law is:

​\(\Delta p = p_0 × x_2\)

• Elevation of the boiling point:
• The boiling point of a solution increases as we add solute particles to the pure solvent.
• The elevation in boiling point is directly proportional to the molality of the solution.

​ΔT_b = k_b × m; where m = molality of the solution and kb = molal elevation constant.

• Depression of freezing point:
  • The freezing point of a solution decreases as we add solute particles to the pure solvent.
  • The depression of the freezing point is also proportional to the molality of the solution.

​​ΔTf = k_f × m; where m = molality of the solution and k_f = molal depression constant

• Osmotic pressure:
  • ​When a solution and a pure solvent are separated by a semi-permeable membrane, due to differences in concentration, the solvent particles start moving towards the solution via the membrane. This phenomenon is called osmosis. However, the diffusion can be stopped by applying pressure over the membrane in the solution.
  • The osmotic pressure of a solution is the pressure required to stop osmosis when the solution is separated from pure solvent by a semi-permeable membrane.

​Explanation:

Isotonic solutions:

• Diffusion of solvent molecules takes place when there is a difference of chemical potential (or simply concentration) between two solutions connected by a semi-permeable membrane.
• The diffusion can be stopped by applying a pressure π over the solution. This is the osmotic pressure.
• The osmotic pressure is independent of the nature of the membrane but depends on the following factors:
• The temperature remains constant, the osmotic pressure of a solution is directly proportional to its concentration.

​π = kC ; where C = concentration and k = proportionality constant

• Concentration remaining constant, the osmotic pressure is directly proportional to absolute temperature.

π = kT; where T = temperature

• Combining the two laws, we get

π = CRT; where R = universal gas constant

• The osmosis takes place until and unless the chemical potential of both the solutions becomes the same. This is the state of equilibrium. at this point, both solutions have the same osmotic pressure.
• When two solutions of equimolar concentration having the same osmolarity, are separated by a semi-permeable membrane, no net osmosis will take place, then the solution is called isotonic solutions. This is because the osmotic pressure on both sides is the same.

​Hypotonic solutions:

• When the concentration of solutes in a solution is less as compared to the other solution, it is called a hypotonic solution.
• In a hypotonic solution, diffusion takes place in an inward direction.

​Hypertonic solutions:​

• When the concentration of solutes in a solution is more as compared to the other solution, it is called a hypertonic solution.
• In a hypotonic solution, diffusion takes place in an outward direction.

Hence, two solutions with equal osmotic pressure are called isotonic.

Explanation:

Reduction potential according to electro chemical series

Zn E⁰ = -0.76

Fe E⁰ = -0.44

zinc will get oxidized first due to the high negative reduction potential,

But Fe will not get oxidized as it has low negative reduction potential compared to Zn.

Hence option 1 is correct.

Explanation:

Given:

1^st Part = Zn|ZnSo₄(0.01M) || CuSo₄(1.0M)Cu emf = E₁ and then changed to

2^nd Part = ZnSo₄(1.0M) || CuSo₄(0.01M) emf =E₂

RT/F = 0.059

So oxidation and reduction take place at anode and cathode,

Anode: Zn  →  Zn²⁺ +2^e-(oxidation)

Cathode: Cu²⁺ +2^e- →  Cu(reduction)

So n = 2 (No.of electrons transferred)

we will use the Nernst equation,

1^st Part

\(E_{cell}= E^\circ_{cell}-\frac{Rt}{nf}log\frac{[Product1]}{[Product2]}\)

\(E_{cell}= E^\circ_{cell}-\frac{0.059}{2}log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)

\(E_{cell}= E^\circ_{cell}-\frac{0.059}{2}log\frac{[0.01M]}{[1.0M]}\)

\(E_{cell}= E^\circ_{cell}-\frac{0.059}{2}(-2)\)

\(E_{cell}= E^\circ_{cell}+0.059 =E_1 \)

2^nd Part 

\(E_{cell}= E^\circ_{cell}-\frac{0.059}{2}log\frac{[1.0M]}{[0.01M]} \)

\(E_{cell}= E^\circ_{cell}-\frac{0.059}{2}(2) \)

\(E_{cell}= E^\circ_{cell}-0.059 =E_2 \)

Hence E₁ > E₂

Concept:

• An electrochemical cell is composed of an electrolyte and two electrodes, i.e., one anode and one cathode electrode.
• The Anode electrode is responsible for oxidation whereas, the cathode electrode is responsible for the reduction.
• For any chemical reaction, the cell reaction is always represented as,
• Oxidation (anode)||Reduction (cathode)

Explanation:

For the given chemical reaction, the half cell reaction is written as,

Oxidation (anode):  Cu → Cu⁺² + 2e^-

Reduction (cathode): Ag⁺ + e^- → Ag 

The cell reaction can be represented as

Cu(s)|Cu⁺²(aq)||Ag⁺(aq)|Ag(s)

Hence, the correct option is (1).

Explanation:

Molar conductance can be defined as the net conductance of all the ions produced when one mole of an electrolyte undergoes complete dissociation into its constituent ions. The formula for molar conductivity is quite similar to that of equivalent conductance, but it is denoted by the symbol ‘μ’.

The formula that can be used to calculate molar conductance is:

 μ = kV

It is important to note that equivalent conductance can be calculated as:

Equivalent conductance = (molar conductance) /n

Where, n = (molecular mass)/(equivalent mass)

Given data and Calculation-

l/a (cell constant) = 0.5cm^-1, R(resistance) = 50

N (normality) = 1.0

So the formula for equivalent conductance is,

 \(λ_{eq}=K\times\frac{1000}{N}\)

K= kappa, we will find K,

\(K=G\times\frac{l}{a}\)

here G = conductance which is the inverse of R means(1/R)

so \(K=\frac{l}{R}\times\frac{l}{a}\)

\(K=\frac{1}{50}\times0.5\)

K = 0.01

so now we got K let's put it into the formula of equivalent conductance which is

\(λ_{eq}=K\times\frac{1000}{N}\)

\(λ_{eq}=0.01\times\frac{1000}{1.0}\)

λ_eq=10 ohm^-1 cm² g eq^-1

Hence the correct option is 1

Explanation:

The relation between activation energy and temperature and the rate of reaction is given by the Arrhenius equation:

The Arrhenius equation is,

\(k = A{e^{ - {E_a}/RT}}\)

Where,

k = rate constant, A = Arrhenius constant

Ea = activation energy, T = temperature in K

​Given:

• The rate of reaction doubles with a 100 C rise in temperature.
• The temperature is raised by 500 C.

Thus, the reaction-coefficient = 2.

The number of times the reaction rate becomes double = 50/10 = 5

Hence, the rate of reaction becomes = 25 = 32 times.

Additional InformationThe relationship between reaction rate and temperature:

• According to Arrhenius, the molecules in a reaction must possess a certain amount of energy in order to undergo a chemical reaction.
• This means that before the reaction occurs, the energy of the molecules must be raised in order to bring any transformation.
• This energy-rich state is called the activated state of the molecules.
• The extra energy required to attain this minimum energy has to be supplied externally and is known as the activation energy
• Reactions with low requirements of activation energy have faster rates.
• The higher the activation energy, the slower is the reaction rate.

Explanation:

The temperature dependence of rate of chemical reaction is expressed by Arrhenius equation as, 

⇒ \(k = A{e^{ - {E_a}/RT}}\) 

where, A = Arrhenius factor or frequency or pre-exponential factor,

R = gas constant,

Ea = Activation energy.

Taking log on both sides of the Eq(i), the equation becomes

→  ln k = ln A - \(\frac{{{E_a}}}{{RT}}\).

On comparing with equation of straight line (y = mx + c), the nature of the plot of lnk vs 1/RT will be:

(i) Intercept = C = ln A

(ii) Slope/gradient = m = -Ea = -y ⇒ Ea = y.

So, the energy required to activate the reactant, (activation energy of the reaction, Ea is = y.

Explanation:

According to the adsorption theory of catalysis, the speed of the reaction increases because adsorption lowers the activation energy of the reaction

→ The increase in the concentration of the reactants on the surface increases the rate of reaction. Adsorption is an exothermic process, the heat of adsorption is utilized in enhancing the rate of the reaction. The catalytic activity is localized on the surface of the catalyst.

→ The mechanism involves five steps:

(i) Diffusion of reactants to the surface of the catalyst.

(ii) Adsorption of reactant molecules on the surface of the catalyst.

(iii) Occurrence of chemical reaction on the catalyst’s surface through the formation of an intermediate.

(iv) Desorption of reaction products from the catalyst surface, thereby, making the surface available again for more reactions to occur.

(v) Diffusion of reaction products away from the catalyst’s surface.  The surface of the catalyst unlike the inner part of the bulk has free valencies which provide the seat for chemical forces of attraction. When the gas comes in contact with such a surface, its molecules are held up there due to a loose chemical combination.

If different molecules are adsorbed side by side, they may react with each other resulting in the formation of new molecules. Thus, formed molecules may evaporate leaving the surface for the fresh reactant molecules

Explanation-

• Adsorption is defined as the deposition of molecular species onto the surface.
• The molecular species that gets adsorbed on the surface is known as adsorbate and the surface on which adsorption occurs is known as adsorbent.
• Common examples of adsorbents are the clay, silica gel, colloids, metals, etc.
• Adsorption is a surface phenomenon.
• The process of removal of adsorbent from the surface of adsorbate is known as desorption

Given data and Analysis:

• The extent of adsorption of a gas on a solid surface increase with the pressure of the gas at a constant temperature.
• Initially, the adsorption increases rapidly, however, as the pressure increase by gas molecules. Now, the adsorption becomes independent of the pressure of the gas.

→ Thus, the adsorption of a gas on a solid surface varies with the pressure of the gas as Fast → slow → independent of the pressure

Explanation-

• Adsorption is defined as the deposition of molecular species onto the surface.
• The molecular species that gets adsorbed on the surface is known as adsorbate and the surface on which adsorption occurs is known as adsorbent.
• Common examples of adsorbents are the clay, silica gel, colloids, metals, etc.
• Adsorption is a surface phenomenon.
• The process of removal of adsorbent from the surface of adsorbate is known as desorption.

As the amount of adsorbate in the solution increases the interaction of adsorbate with the adsorbent increase which led to an increase in the extent of adsorption.

So, from the above explanation, the correct answer is option 1.

Explanation:

• The stability of the colloidal sol is defined by some properties. The electrical charge on the colloidal particle is one of them.
• The electrical charge on the colloidal particle is explained by the process of preferential adsorption of ions from the solution.
• During the preparation of colloidal sol, an ionic colloidal adsorb ions common to its lattice.

It can be explained by the colloidal sol AgI is prepared by adding "KI" solution to the solution till "KI" is in slight excess, iodide ion (I⁻) will be absorbed in the surface of AgI particles thereby giving them a negative charge:

→ AgI + I⁻→ AgI:I⁻

→ The reaction that takes place when AgNO₃ and KI react together is: AgNO₃ + KI → AgI + KNO₃

• So for surrounding the silver iodide with iodide ion, the concentration of "KI" must be slightly more.
• AgNO₃, when dissolved in water, dissociates into silver cation and nitrate anion, and "KI", when dissolved in water, dissociates into potassium ion and iodide ion,
• When they are mixed AgI will be formed. If silver iodide has to be surrounded by iodide ion then the concentration of KI must be more

So, from the above options given, the option that has a concentration of "KI"$$slightly more than AgNO₃ is 50ml of 1M AgNO3 + 50mL of 1.5M "KI"

So option 2 is the correct answer.

Explanation-

Markovnikov’s rule-

When a protic acid (HX) is added to an asymmetric alkene, the acidic hydrogen attaches itself to the carbon having a greater number of hydrogen substituents whereas the halide group attaches itself to the carbon atom which has a greater number of alkyl substituents.

It can be simplified as -"Hydrogen is added to the carbon with the most hydrogens and the halide is added to the carbon with the least hydrogens."

Given data and analysis-

The final product as given is .

Double bonds are the Nucleophile sites.

Cl has more electronegativity so in HCl, Cl will have a partial negative charge and H will have a partial positive charge. 

Similarly in option B

So both options A and B are correct.

Concept:

• A colloid is a mixture in which one substance microscopically dispersed insoluble particles are suspended throughout other substances.
• The dispersed phase is the phase that is scattered or present in the form of colloidal particles.
• The dispersion medium is the medium in which the colloidal particles are dispersed.

Properties of colloidal solution:

• A colloid is a heterogeneous mixture.
• The size of the particles of a colloid is too small to be individually seen by naked eyes.
• Colloids are big enough to scatter a beam of light passing through them and make their path visible.
• The colloidal particles do not settle down when left undisturbed.
• They do not separate from the mixture by the process of filtration.
• But, a special technique of separation called centrifugation can be used to separate the particles.

Explanation:

Sol:

• Foam is a colloidal solution in which the particles are solid and the medium is a liquid.
• Examples: Paint

Emulsion:

• An emulsion is a mixture of two or more liquids that are normally immiscible owing to liquid-liquid phase separation.
• Examples: Milk

Aerosol:

• Foam is a colloidal solution in which the particles are solid and the medium is a gas.
• Examples: Smoke

Gel:

• A gel is a complex fluid consisting of two or more phases composed of a solid dispersed during a liquid.
• Examples: Curd

Whereas, Mist is an example of Liquid aerosol.

Therefore, the correct answer is A-III, B-I, C-II, D-IV.

Additional Information

Classification of colloids: 

Explanation-

Crystal field Splitting

• Crystal field splitting is the difference in energy between d orbitals of ligands.
• Crystal field splitting number is denoted by the capital Greek letter Δ.
• Crystal field splitting explains the difference in color between two similar metal-ligand complexes.
• Δ tends to increase with oxidation number and increases down a group on the periodic table.

The wavelength of the energy absorbed by the coordination compound is inversely proportional to crystal field stabilization energy (CFSE).

CFSE depends on four main properties:

(i) Geometry like octahedral or tetrahedral.

(ii) Charge of the compound in case of metals.

(iii) Nature of ligand as like strong or weak.

(iv) Size of the orbital like 3d, 4d, or 5d, etc.

Given data and analysis:

• The correct order of absorption of the wavelength of light in the visible region will be [Co(H₂​O)₆​]³⁺ > [Co(NH₃​)₆​]³⁺ > [Co(CN)₆​]^3−.
• The order of ligand field strength is CN⁻>NH_3​>H₂​O.
• The stronger the ligand, the greater the crystal field splitting, the greater the energy absorbed, and the smaller the wavelength absorbed.

Explanation:

No bond exists between two sulphur atoms.  

Additional Information 

A double bond exists between two Sulphur atoms. 

No bond exists between two sulphur atoms.  

A single bond exists between two sulphur atoms.     

A single bond exists between two sulphur atoms.  

 No bond exists between two sulphur atoms.  

Explanation:

• When he heat ammonium dichromate, (NH4)2Cr2O7 it decomposes to give the following products:

(NH4)2Cr2O7 → 4H2O + N2 + Cr2O3

• When Barium azide decomposes, it gives :

Ba(N3)2 → Ba + 3N2

• Both Ammonium dichromate and Barium azide decompose to give Nitrogen.

Hence, on heating ammonium dichromate and barium azide separately we get N2 in both cases.

Important Points

•  N2O is produced via the decomposition of ammonium nitrate.

NH4NO3 → N2O + 2H2O

• NO2 is produced by heating nitrates of heavy metals such as lead.

​Pb(NO3)2 → NO2 + PbO + O2

Additional InformationDecomposition reaction:

• Those reactions in which a compound splits up into two or more simpler substances are known as decomposition reactions.
• The decomposition reactions are carried out by applying heat, light or electricity.
• Heat, light, or electricity provide energy that breaks a compound into two or more simpler compounds. 
• A decomposition reaction is the opposite of a combination reaction. 

The types of decomposition reaction are:

Thermal decomposition:

• When a decomposition reaction is carried out by heating, it is called thermal decomposition.

Example: CaCO3 → CaO + CO2

• When limestone (CaCO3 ) is heated then it changed to Quick lime (CaO) and carbon dioxide gas.      

Electrolytic decomposition:

• When current passed through an aqueous solution of a substance it breaks down into ions.

Example: 2H2O → 2H2 + O2

Photo decomposition reaction:

• When a compound is broken down by photons or absorption of light, it is called a photodecomposition reaction.

Example: 2AgCl → 2Ag + Cl2

Explanation-

Option 1-

→ HNO₂ acts as an acid in an aqueous solution as follows:

→ In an aqueous solution, it dissociates to give an H⁺ and NO₂^⁻

→ The proton attaches with water to form a hydronium ion.

→ The reaction is given as:

HNO₂ + H₂O ⇄ NO₂^- + H₃O⁺

It is very stable in an aqueous solution. So option 1 is the correct answer.

Option 2, 3, and 4-

• Whether HNO₂ acts as a reducing or oxidizing agent can be found by examining the oxidation number of nitrogen in the compound.
• Hydrogen and Oxygen always have oxidation numbers of +1 and −2 respectively.
• Since the compound has no charge, the sum of oxidation numbers should add up to zero.

Taking the oxidation number of nitrogen as "x", we get: 

+1 + x + (−2 × 2) = 0 

⇒ x = 4 − 1 = 3

⇒ It is known that nitrogen can exhibit oxidation numbers in the range of -3 to +5.

⇒ Since the oxidation state of nitrogen in HNO₂ is +3, it can get reduced and lowered, meaning that the compound can get oxidized or reduced.

⇒ So it can act as both oxidizing and reducing agent.

⇒So options 2,3 and 4 are wrong.

Explanation:

→ A formal charge is used to predict the energy of any substance. The less the formal charge on atoms of a compound more it is stable. Stability is inversely proportional to energy.

→ Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally, the lowest energy structure is the one with the smallest formal charges on the atoms.

→ Formula to calculate formal charge = Number of valence electrons - Number of non-bonding electrons - 1/2 × Number of bonding electrons

The formal charge on the 'S' atom

= 6 - 0 - (1/2) × 12 = 0 

The formal charge on three 'O' atoms

= 6 - 4 - (1/2) × 4 = 0

Hence option 2 is correct.

Explanation:

If the intermolecular force is more then the boiling point will be more.

Hydrides of group 15

NH₃

PH₃

AsH₃

SbH₃

BiH₃

As we go from top to bottom the molecular weight increases.

→ Boiling points of the nitrogen family generally increase on moving down the group but the boiling point of NH₃ is higher than PH₃ and AsH₃ due to hydrogen bonding. Hence the order will be 

SbH3  >  NH3  > AsH3  > PH3

Explanation:

1. Bleaching powder has power because of its chlorine.
2. When bleaching powder is kept for a long time it loses its chlorine.
3. After the loss of chlorine, it changes into calcium chloride and calcium chlorate

→ 3Ca(OCl)₂ → 2CaCl₂ + Ca(ClO₃)₂

Explanation-

I₂ is a weak oxidizing agent.

I₂ is less reactive than Cl, Br, and F.

KCl → K⁺ + Cl^-

KF → K⁺ + F^-

KBr → K⁺ + Br^-

I₂ can not oxidize Cl^-, Br^-, F^- to Cl₂, Br_2, and F₂ as it is a weak oxidizing agent.

As the reactivity decreases when we move downward in the table, so upper halogen can replace the lower halogen. 

So I₂ can not replace  Cl-, Br-, F-  as it is less reactive.

'So None of the above will be correct.

Explanation-

⇒ An acid can be defined as a species that readily gives hydrogen ions in the solution. If the hydrogen atom is weakly bonded then it will be released easily, making the acid stronger in nature.

⇒Similarly, if the hydrogen atom is strongly bonded then it cannot be released easily, making the acid a weak acid.

Given data and Analysis-

Option 1: HI > HBr > HCI

⇒ HCl, HBr, and HI are binary acids within the same group. The atom bonded to hydrogen that is Cl, Br, and I respectively becomes larger as we proceed down the group, which tends to weaken the bond to H.

⇒ As a result, binary acids become more acidic as we move from top to bottom in the group. So, the correct order of acidic strength is: HI > HBr > HCl

Option 2: HIO4 > HBrO4 > HCIO4

⇒ These are oxyacids with the same number of oxygens but with different central atoms.

⇒ As the electronegativity of the central atom increases, the bond strength of O-H weakens, resulting in an increase in acidity. So, the correct order of acidic strength is: HIO4 < HBrO4 < HCIO4

So the order of acidic strength is not correct in option 2.

Option 3: HCIO4 > HCIO₃ > HCIO3

⇒ In the case of oxyacids, the H is always bonded to one of the oxygens.

⇒ In oxyacid series such as HCIO4, HCIO₃, and HCIO2 with the increase in the number of oxygens bonded to the central atom, the oxidation number of the central atom also increases.

⇒ This causes a weakening of the O-H bond strength. As a result, an increase in acidity. So, the correct order of acidic strength is:

HCIO4 > HCIO3 > HCIO3.

Option 4: HF > H2O > NH3

⇒ In hydrides, the acidity of the hydrides of the corresponding elements also increases with an increase in electronegativity of the element.

⇒ F is the most electronegativity followed by O, N. As a result, the acidity increases.

So, the correct order of acidic strength is: HF > H2O > NH3

Explanation:​

Hydrogen and helium are the most commonly used lift gases.

Although helium is twice as heavy as (diatomic) hydrogen, they are both so much lighter than air that this difference is inconsequential.

Helium is the second lightest gas. For that reason, it is an attractive gas for lifting as well.

• A major advantage is that this gas is Non-combustible.
• Helium is inert and non-flammable
• Hydrogen is flammable when mixed with the oxygen of surrounding air.

Additional Information 

• Hydrogen:
  • Hydrogen gas consists of Hydrogen molecules.
  • The molecule consists of two atoms.
  • The atom has only one electron.
  • The nucleus of the hydrogen atom consists of only one proton.
• Helium:
  • Helium is the second lightest atom.
  • One nucleus of a Helium atom consists of two protons and two neutrons.
  • Helium atoms do not combine into molecules.
  • That is why it is called an inert gas, which consists of free atoms.

Explanation:

In the periodic table, the Actinoid element or the actinide element, is any of a series of 15 consecutive chemical elements from actinium to lawrencium (atomic numbers 89–103).

• As a group, Actinoid elements are significant largely because of their property of being radioactive.
• Curium (Cm), a synthetic chemical element having atomic number 96 comes under the Actinoid series of the periodic table.
• Actinoid elements are actinium, thorium (Th), protactinium, uranium, neptunium, plutonium (Pu), americium, curium, berkelium, californium, einsteinium, fermium, mendelevium, nobelium (No) and lawrencium.

Given elements

Uranium, Americium, Terbium, Thorium

Of this Uranium, Americium, & Thorium belong to the actinide series

→ Tb (Terbium) belongs to Lanthanide Series

→ Actinides are elements with atomic numbers from 90 to 103 following the element Actinium.

→ The symbol An is used while referring to any of the actinide series elements All actinide series elements are radioactive in nature, they release a large amount of energy on radioactive decay. In actinides, the valence electrons enter into the 5f orbital

1. The electronic configuration of uranium is [Rn] 5f3 6d1 7s2 as the last electron is entering in 5f orbital so, it is an actinide.
2. The electronic configuration of americium is [Rn] 5f7 7s2 as the last electron is entering in 5f orbital so, it is an actinide.
3. The electronic configuration of terbium is [Xe] 4f9 6s2  as here the last electron is entering the 4f orbital, it is not an actinide. it is lanthanide.
4. The electronic configuration of thorium is [Rn]7s2 6d2 5f0 and is an actinide.

Additional Information

• 58-71 atomic numbers belong to the first series of the f-block i.e., the lanthanides.
• Lanthanides start from Lanthanum and the last lanthanide is Lutetium.
• But, similar to Actinium, Lanthanum (atomic no. 57) is not included in the f-block and belongs to group 3.
• Hence, the f-block lanthanides are from atomic number 58-71.
• The Lanthanide series include La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb (Terbium), Dy, Ho, Er, Tm, Yb, and Lu.

Content:

• Benzene and other aromatic compounds show characteristic electrophilic substitution reactions.
• In this reaction, a hydrogen atom of the aromatic ring is substituted by an electrophile.
• The substitution takes place by addition- elimination mechanism.

Mechanism:

• In the first step, the benzene ring donates pi electrons to the electrophile.
• One of the carbon atoms forms a bond with the electrophile.
• In the second step, the complex formed then loses a proton from the saturated carbon atom with the help of a base.
• The aromatic ring is finally regenerated in the last step.
• The products formed in the reaction depend on the substituents called directing groups attached to it.
   

Explanation:

• Benzene reacts with Carbon monoxide and HCl gas to undergo formylation of benzene.
• The electrophile here is H-C(+)=O and the reaction is an electrophilic substitution reaction.
• The reaction is called the Gatterman Koch reaction and the reaction proceeds as follows:
• The first step is the generation of the electrophile.
• The second step is the donation of the pi-electron density of benzene to the electrophile.
• The aromatic ring is finally regenerated in the last step.

The net reaction is the formation of benzaldehyde.

Hence, Benzene reacts with CO and HCl under pressure to form C6H5CHO.

Explanation:

When ethyl alcohol is used as a solvent for methyl magnesium iodide it reacts and gives methane

C₂H₅OH + CH₃MgI → C₂H₅OMgI + CH₄(methane)

Hence ethyl alcohol can not be used as solvent

Explanation:

1. The main contaminants in the ether are mostly peroxide which is formed due to oxidation.
2. when KI is added to ether, the peroxide liberates iodine.
3. also, starch is used, which turns the solution blue which indicates impurity.

Explanation:

The source of nitrogen in Gabriel synthesis of amines is potassium phthalimide C6H4(CO2)N-K+

Explanation:

Reduction of nitro aryl compounds in presence of Fe and HCl gives aromatic primary amines.

Explanation:

⇒ The Gabriel synthesis is a chemical reaction that transforms primary alkyl halides into primary amines. Traditionally, the reaction uses potassium phthalimide. The name of the reaction comes from the German chemist Siegmund Gabriel.

Explanation-

→ Ethanal is more electrophilic than propanal. So, enolate of propanal will attack as nucleophile on ethanal.

→ The product formed is a β−hydroxy aldehyde.

So, Compound A is .

Explanation-

Ketone-

⇒ In ketones, the carbonyl group has 2 hydrocarbon groups attached to it. These can be either the ones containing benzene rings or alkyl groups. Ketone does not have a hydrogen atom attached to the carbonyl group.

Example -

⇒ Nucleophile is a reagent that is rich in electrons and ketones have a carbonyl group which is deficient in electrons at the carbon atom.

So, the Nucleophile attacks the carbon atom of carbonyl.

Explanation-

→ Carboxylate acids are easily ionized because there is resonance in carboxylate ion due to π⁻ electron shifting so H⁺ get ionized very easily.

So Carboxylic acid undergoes ionization due to resonance stabilization of carboxylate ion.

 Additional Information

Hydrogen bonding-

• Hydrogen bonding refers to the formation of Hydrogen bonds, which are a special class of attractive intermolecular forces that arise due to the dipole-dipole interaction between a hydrogen atom that is bonded to a highly electronegative atom and another highly electronegative atom that lies in the vicinity of the hydrogen atom.

α - hydrogen-

The hydrogen attached to the α carbon is called α hydrogen.

Concept -

Friedel craft reaction: 

• A Friedel-Crafts reaction is a type of organic coupling reaction that uses an electrophilic aromatic substitution to attach substituents to aromatic rings.
• Alkylation and acylation processes are the two most common Friedel-Crafts reactions.
• In the year 1877, French chemist Charles Friedel and American chemist James Crafts developed these reactions.
• Friedel Craft Acylation: 
  • An acyl group is added to an aromatic ring in the Friedel-Crafts acylation procedure.
  • Typically, an acid chloride (R-(C=O)-Cl) and a Lewis acid catalyst, such as AlCl₃, are used.
  • The aromatic ring is converted to a ketone via a Friedel-Crafts acylation process.
  • Under these conditions, the reaction between benzene and an acyl chloride is shown below.
  • AlCl₃ is a Lewis Acid, or, to put it another way, an "electron hungry species."
  • When it encounters Acetyl Chloride (CH3COCl), it removes the -Cl minus from it and transforms itself into AlCl₄ minus.
  • It now emits the electrophile -COCH₃ minus.
  • Benzene undergoes aromatic electrophilic substitution reactions.
  • As a result, the other electrophile binds to the benzene ring, giving rise to Acetophenone (C₆H₅COCH₃).
  • AlCl₃ acts as a catalyst in this reaction because it releases its Cl minus near the end of the reaction, resulting in the formation of HCl as a byproduct.

Explanation -

• An acyl group is added to an aromatic ring in the Friedel-Crafts acylation procedure.
• Typically, an acid chloride (R-(C=O)-Cl) and a Lewis acid catalyst, such as AlCl3, are used.
• AlCl₃ acts as a lewis acid as well as a catalyst for this reaction. 
• When it encounters Acetyl Chloride (CH₃COCl), it removes the -Cl minus from it and transforms itself into AlCl4 minus.

Hence, the correct answer to this question is Lewis acid. 

Additional Information

Freidel craft alkylation: 

• Friedel-Crafts Alkylation refers to the substitution of an alkyl group for an aromatic proton.
• With the help of a carbocation, an electrophilic attack on the aromatic ring is carried out.
• The Friedel-Crafts alkylation reaction employs alkyl halides as reactants to produce alkylbenzenes.

Explanation:

• Aromaticity is defined as a property of the conjugated cycloalkenes which enhances the stability of a molecule due to the delocalization of electrons present in the π-π orbitals.
• Aromatic molecules are said to be very stable, and they do not break so easily and also react with other types of substances. The organic compounds which are not said to be aromatic are known as aliphatic compounds. These might be in cyclic form, but only the aromatic rings have a special kind of stability.
• Aromaticity is the key factor in deciding maximum enolization in the given compound.
• Since aromatic compounds are more stable than any saturated /unsaturated(aliphatic). thus maximum enolization takes place in the aromatic compound.

So option 2 is the correct answer.

Explanation-

⇒ Carboxylic acids are the most acidic among the carbon compounds.

⇒ Acidity of water is determined by its reaction with sodium alkoxide which gives alcohol as one of the products.

⇒ This indicates that H₂O is more acidic than alcohol as only a stronger acid can displace a weaker acid from its salt. The reaction for the same is given below.

•  RONa + H₂O → ROH + NaOH
• Hence, water is stronger acid and alcohol is a weak acid.

⇒ Both water and alcohol form acetylene upon reaction with sodium acetylide. Hence, both water and alcohol are more acidic than acetylene.

• HC≡C–Na⁺ + HOH → HC≡CH + NaOH
• HC≡C–Na⁺ + ROH → HC≡CH + RONa

So correct order of acid strength RCOOH > HOH > ROH > CH ≡ CH.

Explanation:

→ In the first step, protonation takes place where O has a lone pair that acts as a donor and H- is the acceptor which gives protonated phenyl-methyl ether  +CH3I.

→ Here both SN₁ and SN₂ reaction take place but with SN₁ we get unstable phenyl carbocation so SN₂ reaction takes place and I- attack CH3 as a result of which bonded electrons will shift to O and we get Phenol + methyl iodide.

Explanation-

• Tertiary alcohol when react with copper at 573K dehydration takes place and alkene is formed
• Primary and secondary alcohol on reaction gives aldehyde and ketone by dehydrogenation. 

Given data and Analysis-

Catalyst present is Cu.

Given compound is a tertiary alcohol so it will undergo dehydration and the following product will be formed. 

Explanation-

Grignard reagent-

• A Grignard reagent is an extremely strong nucleophile and can behave like carbonyl compounds with electrophiles.
• Grignard reagents are strong nucleophiles and can form carbon-carbon bonds, making them somewhat similar to organolithium reagents.
• When an amido group substituent is used instead of the alkyl substituent (amido magnesium halides are called Hauser Bases), the nucleophilicity of the reagent further increases.
• The carbonyl carbons of aldehydes and ketones are electrophilic in nature. In their Grignard reactions, the carbon-oxygen pi bond is cleaved and a new C-C bond is formed, resulting in the formation of an alkoxide.
• These alkoxides can be subjected to an acidic workup to yield alcohols.
• The reaction between a Grignard reagent and an ester proceeds in a manner similar to the Grignard reactions of aldehydes or ketones.
• The carbon-oxygen double bond is broken and a new carbon-carbon bond is formed. Alcohols are formed from the acidic workup of the resulting alkoxides.
• When there are alcohols on a carbon chain, the Grignard reagent will react to the alcohol with the hydrogen before it reacts with the carbonyl carbon.
• The same occurs when adding water to the reaction before the Grignard reagent has reacted with the aldehyde/ketone.

So here CH3CH2OH can not be converted into a carbonyl compound.

Explanation-

Terylene-

→ Ethylene glycol and terephthalic acid undergo condensation reactions to produce Terylene. The other name of Terylene is synthetic fiber.

• It is also a step-growth polymer.
• Terylene is extensively used in the textile industry to make hard wear.
• It is used for laundry usage as an automatic clothing vacuum packaging machine.
• Terylene is a very strong fiber and will suffer very little loss in strength when wet.
• It is elastic in nature and possesses the property of resisting creasing.

So option 2 is the correct answer.

Concept:

Terylene 

• Terylene is the chemical compound obtained by the polymerization of terephthalic acid and ethylene glycol.
• Zinc acetate and antimony trioxide at 420K - 460K are used as a catalyst for the condensation polymerization reaction.
• The chemical reaction is as follows,

Hence, the correct option is (3) Zinc acetate - antimony trioxide.

Explanation:

Hydrides-

• Compounds of hydrogen with less electronegative elements are known as hydrides.
• So when hydrogen reacts with any other element the product formed is considered to be a hydride.
• If we closely observe the periodic table hydride formation is not seen from VA group elements and this condition is known as the hydride gap.
• Hydrogen molecule usually reacts with many elements except noble gases to form hydrides.
• However, the properties may vary depending on the type of intermolecular force that exists between the elements, their molecular masses, temperature, and other factors.

2 negatively charged oxygen atoms make a very good hydride donor.

So option 2 is best hydride donor.

The correct answer is Both (A) and (R) are true but (R) is not the correct explanation of (A).

Key Points

• The resulting invert sugar syrup is sweeter than sucrose and is widely used in confectionery, bakery, and pastries, as it features enhanced moisture-preserving properties and is less prone to crystallization. Hence Assertion is correct.
• Invertase is used for the hydrolysis of sucrose into glucose and fructose. Hence Reason is also a correct but not the correct explanation of Assertion.

Additional Information

• Invert sugar is used in confectionary preparations (giving them added moisture) and in the preparation of sorbets and ice cream since it has the ability for controlling crystallization and creating a smoother mouthfeel.
• Inverted sugar is automatically produced when making jams since when by combining the sugar with the acid in the fruit and heating.
• Most of the sugar in honey is also inverted sugar.

Explanation-

Alkaline hydrolysis of Nitrile-

The nitrile is heated in a sodium hydroxide solution under reflux.

Instead of getting an ammonium salt as you would with a water-only reaction, you receive the sodium salt this time.

Ammonia gas is also produced.

Ammonia is produced by the interaction of ammonium ions with hydroxide ions.

In this type of hydrolysis, the final solution has to be acidified with a strong acid like dilute hydrochloric acid or dilute sulphuric acid to get the free carboxylic acid.

As previously stated, the ethanoate ion in sodium ethanoate will react with hydrogen ions.

CH₃CN + 2H₂0 → CH₃COO^- + NH₃

CH₃COO^- + H⁺ → CH₃COOH

So the end product is CH₃COOH (Ethanoic acid).

The correct answer is sucralose. 

Key Points

• Sucralose is a nonnutritive, zero-calorie artificial sweetener.
• It is a chlorinated sugar substitute that is about 600 times as sweet as sucrose.
• It is produced from sucrose when three chlorine atoms replace three hydroxyl groups.
• It is consumed as tablets (Blendy) by diabetic and obese patients. 
   

Additional Information

• Aspartame is an artificial non-saccharide sweetener 200 times sweeter than sucrose and is commonly used as a sugar substitute in foods and beverages.
• Saccharin is an artificial sweetener with effectively no food energy. It is about 550 times as sweet as sucrose but has a bitter or metallic aftertaste, especially at high concentrations.
• Alitame is an aspartic acid-containing dipeptide sweetener.