Question 1 5 / -1
The unit cell of a certain type of crystal is defined by three vectors a, b and c and the angle between a and b is α, b and c is β and a and c is γ. Now if α = β = γ = 90° and a = b ≠ c. The crystal structure is
Solution
Explanation:
If the atoms or atom groups in the solid are represented by points and the points are connected, the resulting lattice will consist of an orderly stacking of blocks or unit cells.
The orthorhombic unit cell is distinguished by three lines called axes of twofold symmetry about which the cell can be rotated by 180° without changing its appearance. This characteristic requires that the angles between any two edges of the unit cell be right angles but the edges may be any length.
There are 7 types of crystal systems:
Crystal System
Angles between Axis
Unit Cell Dimensions
Cubic
α = β = γ = 90°
a = b = c
Tetragonal
α = β = γ = 90°
a = b ≠ c
Orthorhombic
α = β = γ = 90°
a ≠ b ≠ c
Rhombohedral
α = β = γ ≠ 90°
a = b = c
Hexagonal
α = β = 90°, γ = 120°
a = b ≠ c
Monoclinic
α = γ = 90°, β ≠ 90°
a ≠ b ≠ c
Triclinic
α ≠ β ≠ γ
a ≠ b ≠ c
Question 2 5 / -1
Zn and Mg have the following crystal structure
Solution
Explanation:
Hexagonal closed pack structure:
In a hexagonal closed-packed structure, the third layer has the same arrangement of spheres as the first layer. Since the structure repeats itself after every two layers, the stacking for hcp may be described as "a-b-a-b-a-b". The atoms in a hexagonal closed packed structure efficiently occupy 74% of space while 26% is empty space. Ex Mg, Zn
Coordination Number and Number of Atoms Per Unit Cell:
The hexagonal closed packed (hcp) structure has a coordination number of 12 and contains 6 atoms per unit cell. The face-centered cubic (fcc) structure has a coordination number of 12 and contains 4 atoms per unit cell. The body-centered cubic (bcc) structure has a coordination number of 8 and contains 2 atoms per unit cell. The simple cubic structure has a coordination number of 6 and contains 1 atom per unit cell.
Crystal structure of Material
FCC:- Ni, Cu, Ag, Pt, Au, Pb, Al, Austenite or Ƴ-iron
BCC:- V, Mo, Ta, W, Ferrite or α-iron, δ-ferrite or δ-iron
HCP:- Mg, Zn
Cobalt:- HCP < 420°C, FCC > 420°C
Chromium:- HCP < 20°C, BCC > 20°C
Glass:- Amorphous
Important Points
In the crystal structure, the atomic packing factor (APF) or packing efficiency or packing fraction is the fraction of volume in a crystal structure that is occupied by constituent particles. It is a dimensionless quantity and always less than unity.
The atomic packing factor of different crystal structures is given in the table below:
Structure
Atomic packing factor
BCC
0.68
HCP
0.74
FCC
0.74
Diamond cubic
0.34
SC
0.52
Question 3 5 / -1
Which of the following statements is not true
Solution
Explanation:
Diamagnetic substance-
Diamagnetic substances are those substances that have a tendency to move from the stronger part to the weaker part of the external magnetic field. The diamagnetic substances get repelled by a magnet. The pairing of electrons cancels their magnetic moment in the diamagnetic substances. Paramagnetic substance-
Paramagnetic substances are those substances that get weakly magnetized in the presence of an external magnetic field. In the presence of an external magnetic field, these substances tend to move from a region of a weak to a strong magnetic field. In a paramagnetic material, the individual atoms possess a dipole moment, which when placed in a magnetic field, interact with one another, and get spontaneously aligned in a common direction, which results in its magnetization. Ferromagnetic substance-
Ferromagnetic substances are those substances that when it’s placed in an external magnetic field, get strongly magnetized. Also, they tend to move from a region of weak to the region of a strong magnetic field and get strongly attracted to a magnet. In a ferromagnetic material, the individual atoms possess a dipole moment, similar to a paramagnetic material. When placed in a magnetic field, the atoms interact with one another and get spontaneously aligned in a common direction. It can be magnetized permanently.
Antiferromagnetic substance-
The materials that exhibit antiferromagnetism are known as antiferromagnetic materials. When these materials are kept in the presence of a strong magnetic field, they get magnetized weakly in the direction of the magnetic field. This is known as antiferromagnetism. Antiferromagnetic materials are commonly found among the transition metal compounds. The domains in antiferromagnetic substances are oppositely oriented with respect to each other. So option 2 is the correct answer.
Question 4 5 / -1
Iron at 20°C is BCC with atoms of atomic radius 0.124 nm. What is the lattice constant ‘a’ for the cube edge of the iron unit cell ?
Solution
Concept :
Body-centered cubic (BCC) structure:
In ΔABC:
(BC)2 = (AB)2 + (AC)2
\({\left( {4r} \right)^2} = {\left( {\sqrt 2 a} \right)^2} + {\left( a \right)^2}\)
(4r)2 = 3a2
\(4r = \sqrt 3 \;a\)
\(\therefore r = \frac{{\sqrt 3 }}{4}a\)
Where, lattice parameters of BCC are:
r = atomic radius
a = edge length of the unit cell
Calculation :
Given:
Atomic radius (r) = 0.124 nm
Body-centered cubic (BCC) structure :
\( r = \frac{{\sqrt 3 }}{4}a\)
\( a = \frac{{4}}{\sqrt 3 }r\)
\( a = \frac{{4}}{\sqrt 3 }\times0.124=0.2864\;nm\)
Question 5 5 / -1
A saturated solution of sugar is prepared by dissolving 62.5 g of it in 250 g of water at 298 K. What is the solubility of sugar in water at this temperature?
Solution
The correct answer is 0.25 g per gram of water.
Concept:
Solution: Solute: The substance that is dissolved in the solvent. Solvent: The substance in which solute is dissolved. For example, in a salt solution, salt is solute and water is solvent.
Explanation:
Solubility:
The maximum amount of a solute that can be dissolved in 100 g of a solvent at a specified temperature is known as the solubility of that solute in that solvent (at that temperature).
Given: Mass of sugar (solute) = 62.5 g
Mass of water (solvent) = 250g
Solubility = (Mass of solute / mass of solvent) × 100
= \(\frac{62.5}{250} × 100\)
= 25 % or 25/100 = 0.25 g per gram of water
Question 6 5 / -1
For a non-volatile solute
Solution
Explanation:
→ The non-volatile solute is simply a substance that does not easily vaporize. Furthermore, when dissolved in a solvent, it does not contribute to the vapor pressure of the solution
hence vapour pressure of solute is zero
Question 7 5 / -1
Which has minimum osmotic pressure
Solution
Explanation:
Osmotic pressure can be defined as the minimum pressure that must be applied to a solution to halt the flow of solvent molecules through a semipermeable membrane. It is a colligative property and is dependent on the concentration of solute particles in the solution.
It can be calculated with the formula π = iCRT
Where, π = the osmotic pressure
→ i = Van’t Hoff Factor (value = number of dissociated ions)
→ C = molar concentration of the solute in the solution
→ R = universal gas constant, T = temperature
In the above question, RT will remain constant and whatever the difference will be because of i × C
Option 1: 200 mL of 2 M NaCl solution ( i = 2, for NaCl)
π = 2 × concentration of NaCl × RT
π = 2 × 2 × RT → π = 4 RT
Option 2: 200 mL of 1 M glucose solution ( i = 1 as it is non electrolyte)
→ π = 1 × 1 × RT
→ π = 1 × 1 × RT → π = RT
Option 3: 200 mL of 2 M urea solution ( i = 1 as it is non electrolyte)
→ π = 2M × 1 × RT
→ π = 2 × 1 × RT → π = 2 × RT
∴ Glucose solution has minimum osmotic pressure.
Question 8 5 / -1
A plot of log x/m, against log P is straight line inclines at an angle 45°, when the pressure is 0.4 atm, and Freundlich parameter, k is 100, the amount of solute adsorbed per gram of the adsorbent will be:
Solution
Concept:
In adsorption isotherms, the fraction of moles of adsorbate (x) by the grams of adsorbent(m) versus P is plotted. The Freundlich isotherm gives us the empirical relationship between the amount of gas adsorbed and its equilibrium pressure P . It is seen that the amount of adsorption increases with an increase of pressure when the pressure is in the lower range. As the pressure increases , the rate of adsorption increases but reaches saturation after some point . After this point, the amount of adsorption doesn't change even when pressure is increased. The equation of Freundlich isotherm is given as follows: \(\frac{{\rm{x}}}{{\rm{m}}} \propto {\rm{\;}}{{\rm{P}}^{\frac{1}{{\rm{n}}}}}\) where k and n are constant for a particular adsorbent and adsorbate at a fixed temperature .
The graph is represented as follows.
Calculation:
Given:
Freundlich parameter, k = 100
Pressure P = .4 atm
Freundlich adsorption isotherm:
Freundlich gave an empirical relationship between the quantity of gas adsorbed by a unit mass of solid adsorbent and pressure at a particular temperature. \(\frac{{\rm{x}}}{{\rm{m}}} = {\rm{K}}.{{\rm{p}}^{1/{\rm{n}}}}\)
Taking log on both sides, we get,
\(\log \frac{{\rm{x}}}{{\rm{m}}} = \log {\rm{K}} + \frac{1}{{\rm{n}}}\log {\rm{p}}\)
We know that the formula of the slope is: \({\rm{m}} = \frac{{\rm{y}}}{{\rm{x}}}\)
In the formula, the slope is \(\frac{1}{{\rm{n}}}\) Therefore,
\(\Rightarrow \frac{{\rm{y}}}{{\rm{x}}} = \frac{1}{{\rm{n}}}\)
It is given that the angle θ = 45° and thus slope tanθ = tan 45° = 1
\(1= \frac{1}{{\rm{n}}}\)
Substituting the values in the isotherm equation, we get:
\(\log \frac{{\rm{x}}}{{\rm{m}}} = \log {\rm{100}} + \frac{1}{{\rm{1}}}\log {\rm{0.4}}\)
\(\log \frac{{\rm{x}}}{{\rm{m}}} =2 + \log {\rm{0.4}}\)
\(\log \frac{{\rm{x}}}{{\rm{m}}} =2 - 0.397\)
\({x\over m}=40g\)
Hence, the amount of solute adsorbed per gram of the adsorbent is 40g.
Question 9 5 / -1
The catalyst used for the reaction SO2 + O2 → SO3 in Contact's Process is:
Solution
Concept:
A catalyst is a substance that can be added to a reaction to enhance the reaction rate without getting consumed in the process. Catalysts typically speed up a reaction by lowering the activation energy or changing the reaction mechanism.Enzymes are proteins that act as catalysts in biochemical reactions. A stoichiometric amount of one reactant can mean the amount that will react completely with a given amount of another reactant, leaving no excess of either one. Catalyst does not take part in balancing the chemical reaction hence catalysts are not required in Stoichiometric amounts. The most common catalysts are - Aluminosilicates, Iron, Vanadium, Nickel, Platinum + Alumina.
Explanation: Contacts Process
Sulphuric acid is prepared via the contact's process which involves the following steps: Preparation of sulphur dioxide: In this step. sulphur burns in air or is oxidized to produce sulphur dioxide gas by the reaction: S + O2 → SO2
Oxidation of sulphur dioxide to sulphur trioxide: In the second step the following reaction takes place: SO2 + O2 → SO3
Oxidation of SO2 to SO3 is an exothermic process and releases heat. The process occurs with the help of catalyst vanadium pentoxide V2 O5 . Formation of oleum: Sulphur trioxide is then dissolved in 94% H2 SO4 solution to prepare oleum H2 S3 O7 which is also called fuming sulphuric acid. The last step is the dilution of oleum to give sulphuric acid. The catalyst thus used for the reaction SO2 + O2 → SO3 in Contact's Process is Vanadium pentoxide.
Additional Information
Fe acts as a catalyst in Haber's Process and Mo acts as a Promoter.
Question 10 5 / -1
Which of the following empirical equations represents the Freundlich adsorption curve?
Solution
Concept:
Adsorption:
Adsorption is defined as the deposition of molecular species onto the surface . The molecular species that gets adsorbed on the surface is known as Adsorbent and the surface on which adsorption occurs is known as Adsorbate. It is a surface phenomenon and occurs due to the unsaturation of forces on the surface of an adsorbate molecule.
There are two types of adsorption:Physical or physisorption where there exist only weak van der Waals forces between adsorbent and adsorbate. Chemical or chemisorption where new bonds are formed between adsorbent and adsorbate. Explanation:
In adsorption isotherms, the fraction of moles of adsorbate (x) by the grams of adsorbent(m) versus P is plotted. The Freundlich isotherm gives us the empirical relationship between the amount of gas adsorbed and its equilibrium pressure P . It is seen that the amount of adsorption increases with an increase of pressure when the pressure is in the lower range. As the pressure increases , the rate of adsorption increases but reaches saturation after some point . After this point, the amount of adsorption doesn't change even when pressure is increased. The equation of Freundlich isotherm is given as follows: \(\frac{{\rm{x}}}{{\rm{m}}} \propto {\rm{\;}}{{\rm{P}}^{\frac{1}{{\rm{n}}}}}\) where k and n are constant for a particular adsorbent and adsorbate at a fixed temperature .
The graph is represented as follows.
When the temperature is fixed and pressure is low , the adsorption becomes directly proportional to the pressure of the gas. The relation reduces to x=kP .
This signifies, 1/n = 1 or n = 1.
At a higher range of pressure , the adsorption becomes independent of pressure and the equation reduces to x/m= k.
Hence, the equation \(\frac{{\rm{x}}}{{\rm{m}}} \propto {\rm{\;}}{{\rm{P}}^{\frac{1}{{\rm{n}}}}}\) represents the Freundlich adsorption curve .
Important Points
In Freundlich adsorption isotherm, the value of 1/n is 0 to 1. Adsorption theory explains heterogeneous catalysis.
Question 11 5 / -1
Zn2+ → Zn(s) ; Eo = -0.76 V
Cu2+ → Cu(s) ; Eo = -0.34 V
Which of the following is spontaneous
Solution
Explanation :-
Spontaneous Reaction:
A spontaneous reaction is one in which the production of products is favored under the conditions of the reaction. A roaring campfire is an example of a spontaneous reaction since it is exothermic, meaning the system's energy decreases as energy is released as heat to the surroundings. An increase in entropy characterizes a spontaneous process in a perfectly isolated system. The total energy of an isolated system is always constant, according to the First Law of Thermodynamics. The first law illustrates the relationship between the work done by the system and the heat absorbed by the system without putting any restrictions on the direction of heat flow. The reaction's free energy can determine the spontaneity of a reaction is the key is ∆ G, and all you need is the reaction's enthalpy change or ∆ H. You can compute it in a variety of methods, including using enthalpies of formation. For a spontaneous reaction, Eºcell must have a positive value.
Thus, here Cu 2+ + Zn → Cu + Zn2+ having the least negative value is a spontaneous reaction.
Question 12 5 / -1
Electrode potential of Zn2+ / Zn is -0.76 V and that of Cu2+ / Cu is + 0.34V. The EMF of the cell constructed between these two electrodes is
Solution
Concept:
Standard Cell Potential
The standard voltage of a cell is the sum of the standard potential for the two half-reactions.
Standard Cell Potential ( E0 cell E0 oxidation E0 reduction
Where, E0 oxidation opposite of the standard reduction potential .
Explanation:
For the cell
Zn |Zn2+ ||Cu2+ | Cu
Standard electrode potentials for Zn = -0.76 V
Standard electrode potentials for Cu = +0.34 V
Therefore,
E0 cell E0 cathode - E0 Anode
E0 cell = 0..34 - (-0.76) = 1.10 V
The value of standard potential of the cell in volts will be 1.10 V
Question 13 5 / -1
The potential of standard hydrogen electrode is zero. This implies that
Solution
Explanation
As we know hydrogen's standard electrode potential is declared to be zero volts at all temperatures.
Hence using the relation,
ΔG0 = -nFE0
where,
-n =number of electrons transferred, F =Faraday's constant (96500 ),
E 0 = voltage of the cell
Now as we know that E0 = 0,
ΔG0 = -n × F× E0
ΔG0 = 0
Hence ΔGo f (H+ , aq) = 0
Question 14 5 / -1
Limiting molar conductivity of NH4 OH (i.e. λm o (NH4 OH) is equal to
Solution
Explanation
The molar conductivity of a solution at infinite dilution is known as limiting molar conductivity, and Kohlrausch’s Law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
In general, if an electrolyte of dissociation gives v+ cations and v– anions then its limiting molar conductivity is given by,
\(\wedge_m^\infty=V_+\lambda_+^\circ+V_-\lambda_-^\circ \)
Here, \(\lambda_+^\circ\) and \(\lambda_-^\circ\) are the limiting molar conductivities of cations and anions respectively.
So,
\(\wedge_m^\circ (NH_4OH)=\wedge_m^\circ(NH_4^+)+\wedge_m^\circ(OH^-)\)
\(=\wedge_m^\circ(NH^+_4)+\wedge_m^\circ(Cl^-)-\wedge_m^\circ(OH^-) \)
\(=\wedge_m^\circ(NH_4^+)-\wedge_m^\circ-(Cl^-)-\wedge_m^\circ(Na^+)\)
\(=\wedge_m^\circ(NH_4Cl)+\wedge_m^\circ(NaOH)-\wedge_m^\circ(NaCl)\)
Question 15 5 / -1
If doubling the concentration of a reactant 'A' increases the rate 4 times and tripling the concentration of 'A' increases the rate 9 times, the rate is proportional to
Solution
Explanation-
The expression for the rate of the reaction is r = k[A]n ----- ( i )
When the concentration of A is doubled, the rate increases 4 times.
r′ = 4r = [2A]n ---- (ii)
Dividing ( ii ) by ( i ) ,
\(\frac{{4r}}{r} = {\frac{{[2A]}}{{{{[A]}^n}}}^n}\)
So n = 2
Tripling the concentration of 'A' increases the rate 9 times.
r′ = 9r = K[3A]n ---- (ii)
Dividing ( ii ) by ( i ) ,
\(\frac{{9r}}{r} = {\frac{{[3A]}}{{{{[A]}^n}}}^n}\)
So n = 2
So the rate is proportional to the Square of the concentration of 'A'.
Question 16 5 / -1
T50 of first-order reaction is 10 min. Starting with 10 mol L-1 rate after 20 min is
Solution
Explanation-
First-order reaction-
A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant . In other words, a first-order reaction is a chemical reaction in which the rate varies based on the changes in the concentration of only one of the reactants. Thus, the order of these reactions is equal to 1. The rate of reaction is shown as \(\frac{{dx}}{{dt}} = k[A]\) The half-life of a chemical reaction (denoted by ‘T50 ’) is the time taken for the initial concentration of the reactant(s) to reach half of its original value. At t = T 50 ’ [A] = \(\frac{{\left[ {{A_0}} \right]}}{2}\) Where [A] denotes the concentration of the reactant and [A0 ] denotes the initial concentration of the reactant. \({T_{50}} = \frac{{0.693}}{K}\) Thus, the half-life of a first-order reaction is equal to 0.693/k (where ‘k’ denotes the rate constant, whose units are s-1 ).
Given data and calculation-
T 50 of first-order reaction = 10 min
K = \(\frac{0.693}{10}\) =0.0693 min-1
Concentration at starting = 10 mol L-1
After 20 mins it must be passed two half lives.
So concentration will be 2.5 mol L-1 .
So the rate will be 0.0693 × 2.5 mol L -1 min -1
Question 17 5 / -1
The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii) The respective orders of the reactions are
Solution
Explanation-
Zero-order reaction-
Zero-order reaction is a chemical reaction wherein the rate does not vary with the increase or decrease in the concentration of the reactants. Therefore, the rate of these reactions is always equal to the rate constant of the specific reactions. The integral form of zero-order reactions can be rewritten as [A] = -kt + [A0 ]
Comparing this equation with that of a straight line (y = mx + c), an [A] against t graph can be plotted to get a straight line with a slope equal to ‘-k’ and intercept equal to [A0 ] as shown below.
First-order reaction-
A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant.
In other words, a first-order reaction is a chemical reaction in which the rate varies based on the changes in the concentration of only one of the reactants. Thus, the order of these reactions is equal to 1.
→ The concentration v/s time graph for a first-order reaction is provided below.
→ For first-order reactions, the equation ln [A] = -kt + ln [A0 ] is similar to that of a straight line (y = mx + c) with slope -k. This line can be graphically plotted as follows.
So from the figure given in the question, it is clear that the respective orders of the reactions are 1 and 0.
Question 18 5 / -1
Which of the following characteristics is not correct for physical adoption
Solution
Explanation:
Adsorption-
Physical adsorption:
This type of adsorption is also known as physisorption. It is due to weak Van der Waals forces between adsorbate and adsorbent. For example, H2 and N2 gases adsorb on coconut charcoal. Characteristics of physical adsorption:This type of adsorption is caused by physical forces. Physisorption is a weak phenomenon. This adsorption is a multi-layered process . Physical adsorption is not specific and takes place all over the adsorbant. Surface area, temperature, pressure, and nature of adsorbate effects physisorption. Adsorption increases with an increase in temperature. Both enthalpy and entropy of adsorption are negative. Adsorption is spontaneous. Energy for activation is low (20 – 40 kg/mol).
Important Points Chemical adsorption:
This type of adsorption is also known as chemisorption. It is due to strong chemical forces of bonding type between adsorbate and adsorbent. This type of adsorption is caused by chemical forces. It is a very strong process. This type of adsorption is almost a single-layered phenomenon.
Question 19 5 / -1
Choose the correct statement
Solution
Explanation:
Option 1-
As there is no unpaired electron, so it is diamagnetic.
So option 1 is correct.
Option 2-
It is more favorable to form six bonds rather than four. The crystal field stabilization energy is usually greater for octahedral than tetrahedral complexes. So Tetrahedral complexes are less stable than octahedral complexes.
Option 3-
[Fe(CN)6 ]3− and [FeF6 ]3− both are stable but the former is more stable. So this option is wrong.
Option 4-
Cu(NH3 )4 ]2+ has square planar geometry and is paramagnetic. So this option is wrong.
Question 20 5 / -1
Which of the following facts about complex [Cr(NH3 )6 ]CI3 is wrong
Solution
Explanation:
→ In the presence of NH3 a strong ligand, the 3d electrons pair up leaving two d-orbitals empty. Hence, the hybridization is d2 sp3 and is octahedral in shape.
→ As it undergoes d2 sp3 hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
→ The complex forms an inner orbital complex
→ C hlorine is present Outside the coordination sphere of the complex which gives a white precipitate with a silver nitrate solution
Question 21 5 / -1
The complex ion which has no 'd' electrons in the central metal atom is
Solution
Explanation:
Option 1: [MnO 4 ] -
Mn (central atom)
→ Atomic number z = 25
→ Electronic configuration = 3d5 4s2
→ Oxidation state Mn+7
→ Electronic configuration 3d0 4s0
In MnO4 - there is no "d" electrons in central metal atom.
Option 2: [Co(NH 3 ) 6 ] 3+
Co (central atom)
→ Atomic number = 27
→ Electronic configuration = 3d7 4s2
→ Oxidation state Co+3
→ Electronic configuration 3d6 4s0
Option 3: [Fe(CN) 6 ] 3-
Fe (central atom)
→ Atomic number = 26
→ Electronic configuration = 3d6 4s2
→ Oxidation state Fe+3
→ Electronic configuration 3d5 4s0
Option 4: [Cr(H 2 O) 6 ] 3+
Cr (central atom)
→ Atomic number = 24
→ Electronic configuration = 3d5 4s1
→ Oxidation state Cr+3
→ Electronic configuration 3d3 4s0
Question 22 5 / -1
Amongst H2 O, H2 S, H2 Se and H2 Te the one with the highest boiling point is:
Solution
Explanation-
Boiling Point-
The boiling point of a liquid is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure of the liquid’s environment. At this temperature, the liquid is converted into a vapor. In the absence of other intermolecular forces, the higher the molecular mass the greater the boiling point .
Hydrogen bonding-
In a molecule, when a hydrogen atom is linked to a highly electronegative atom, it attracts the shared pair of electrons more and so this end of the molecule becomes slightly negative while the other end becomes slightly positive. The negative end of one molecule attracts the positive end of the other and as a result, a weak bond is formed between them. This bond is called the hydrogen bond. The compounds having hydrogen bonding show abnormally high melting and boiling points. The high melting and boiling point of the compound containing hydrogen bonds is due to the fact that some extra energy is needed to break these bonds.
H2 O is a liquid whereas H2 S, H2 Se, and H2 Te are all gases at ordinary temperatures. In water, hydrogen bonding causes linkages in the water molecules which result in the boiling point of water being more than that of the other compounds.
So Option 1 is the correct answer.
Question 23 5 / -1
In the following reaction the catalyst used is
Solution
Explanation:
Catalysts-
catalysts are used to break or rebuild the chemical bonds between the atoms which are present in the molecules of different elements or compounds. In essence, catalysts encourage molecules to react and make the whole reaction process easier and more efficient.
Given data and analysis:
The reaction will be as follows:
→ The conversion will include treating the hexane molecule with the catalyst that is oxides of metals like vanadium oxide or chromium oxide.
→ This will remove hydrogen atoms from hexane and form a cyclic structure called cyclohexane. The reaction is:
→ This cyclohexane at a very high temperature loses more hydrogen atoms (3 hydrogen molecules) and undergoes aromatization to form aromatic benzene. The reaction is:
→ So the final reaction will be like this-
So here catalysts used are Cr 2 O 3 and Al 2 O 3.
Question 24 5 / -1
The acid having O - O bond is
Solution
Explanation-
Peroxide is any of a class of chemical compounds in which two oxygen atoms are linked together by a single covalent bond i.e. O−O bond. Peroxide linkage (−O−O−) is present in H2 S2 O8 as its name also suggests peroxide linkage H2 S2 O8 The molecular structure of oxoacids is shown in the figure below. So has H 2 S 2 O 8 O - O bond.
Question 25 5 / -1
The incorrect statement among the following is
Solution
Explanation-
Allotropes of Carbon-
Carbon due to its capability of having variable oxidation states or coordination numbers makes carbon one of the few elements to have multiple numbers of allotropic forms. Allotropes of carbon-Diamond, Lonsdaleite, Graphite, Q-carbon,C60 Allotropes of Oxygen-
The 2 allotropes of oxygen are diatomic oxygen and ozone. Allotropic forms of sulphur-
Rhombic, octahedral or α-sulphur Monoclinic, prismatic or β-sulphur Plastic or γ-sulphur Milk of sulphur colloidal or δ-sulphur. So S 8 is not the only allotropic form of sulphur, it has other forms of allotropes also.
Red Phosphorus has more atoms linked together in a network than white phosphorus does, which makes it much more stable . It is not quite as flammable, but given enough energy, it still reacts with air.
So option 3 is the correct answer.
Question 26 5 / -1
Which of the following statement is wrong
Solution
Explanation-
Option 1-
As we move down the group, the tendency to form a covalent bond with small H So this is wrong .
Option 2-
pπ-pπ-bonds are the bonds that are formed by the overlapping of p orbitals, perpendicular to the internuclear axis. Due to the small size of the Nitrogen atom and the availability of incompletely filled 2p orbitals nitrogen forms a Pπ - Pπ bond, but it can not form a dπ - Pπ bond. So this is correct.
Option 3-
N−N single bond is weaker than the P−P bond due to the smaller size of N as compared to P. Smaller size of N leads to a smaller N−N bond length. As a result, the lone pair of electrons on both the N atoms repel each other leading to instability or weakening of the N−N bond. Because of the larger size of the P atom, the P−P bond length is more and the lone pair-lone pair repulsion between P atoms is less which makes the P−P bond stronger than the N−N bond. So this is correct.
Option 4-
Four stable resonance structures are possible for N2 O4 . So this option is also wrong.
Here both options 1 and 4 are wrong.
Question 27 5 / -1
Chlorine acts as bleaching agent only in presence of
Solution
Explanation:
→ Cl2 + H2 O (from moisture) → HCl + HClO
→ HClO is not stable and decomposes to HCl + [O] (Nascent oxygen) due to oxidizing power of Nascent oxygen chlorine acts as a bleaching agent.
Question 28 5 / -1
The incorrect statement(s) about actinides is/are:
Solution
Concept:
Elements whose f-orbital getting filled up by electrons are called f-block elements. They are divided into lanthanoids and actinoids. The first series of elements are called lanthanides and include elements with atomic numbers beginning from 57 and ending at 71. These elements are non-radioactive. The second series of elements are called actinides and include elements with atomic numbers beginning from 89 and ending at 103. They are radioactive. Actinides are the elements whose atomic number is from 90 to 103 . The members of the series are Thorium, Protactinium, Uranium, Neptunium, Plutonium, Americium, Curium, Berkelium, Californium, Einsteinium, Fermium, Mendelevium, Nobelium, Lawrencium. Explanation:
All the actinide elements are radioactive , hence the statement All the actinides are radioactive is correct. In the actinoids, the last electron goes into the 5f orbital . Now, the 5f orbital due to its large size and diffused nature has a very poor shielding effect. As a result of the poor shielding effect, much of the nuclear charge leaks from the electron cloud, as a result, the outer electrons face more nuclear charge of the nucleus. As the atomic number increases, the nuclear charge increases and as a result, the ionic radii or the atomic radii decrease from Thorium to Lawrencium . This is known as Actinoid contraction. Thus, the statement Actinides do not exhibit actinide contraction is incorrect. The poor shielding effect of 4f electrons causes Lanthanide contraction is Lanthanides. Comparing it to 5f orbital, as 5f orbitals are larger in size, its shielding effect is much poor , as a result of which the electrons in 5forbitals are less tightly bound. Also, it is much farther than the nucleus compared to 4f orbitals. Hence, the statement: The 5f electrons of actinides are bound less tightly than the 4f electrons is also correct. The transuranium compounds are the elements with an atomic number greater than 92. The transuranium compounds are all radioactive and are man-made elements. Hence, the statement: The trans uranium elements are prepared artificially is also correct. Hence, the correct options are: 1,2 and 3.
Question 29 5 / -1
In qualitative analysis when H2 S is passed through an aqueous solution of salt acidified with dil. HCI, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives ______
Solution
Explanation-
→ In qualitative analysis when H2 S is passed through an aqueous solution of salt acidified with diluted HCl, a black precipitate is obtained.
→ On boiling the precipitate with diluted HNO3 it forms a solution of blue colour.
→ The addition of an excess of an aqueous solution of ammonia to this solution gives a deep blue solution of [Cu(NH3 )4 ]2+
Hence option 2 is correct.
The equations are as below-
→ CuSO4 + H2 S → CuS(black ppt) + H2 SO4
→ CuS + 2HNO3 → Cu(NO3 )2 (blue solution) + H2S
→ Cu(NO3 )2 + 4NH3 → Cu(NH3 )4]2+ (Deep Blue solution) + 2NO3 -
Question 30 5 / -1
Addition of methanol to 2 - methylpropene in the presence of conc. H2 SO4 gives
Solution
Explanation:
→ Addition of methanol to 2 - methylpropene in the presence of conc. H 2 SO 4 gives tert - butyl methyl ether
Question 31 5 / -1
Select the incorrect statement(s).
Solution
Explanation:
Option 1
→ Electrophilicity increased due to protonation
Option 2
Aldehydes are typically more reactive than ketones as
Aldehydes are less hindered than ketones The carbonyl carbon in aldehydes generally has a more partial positive charge than in ketones due to the electron-donating nature of alkyl groups. Aldehydes only have one e- donor group while ketones have two. Option 3
→ One carbonyl group is oriented with the oxygen atom pointing toward the water and is hydrogen-bonded to a water molecule. hence Carbonyl compounds can form hydrogen bonding with H 2 O
Option 4
→ Chloral hydrate is stable because of the formation of intramolecular hydrogen bonding between the hydrogen atom
Question 32 5 / -1
From which of the following tertiary butyl alcohol is obtained by the action of methyl magnesium bromide?
Solution
Concept:
Grignard Reagent:
The Grignard reagent is given as RMgX where X i s a halogen, and R is an alkyl or aryl (based on a benzene ring) group. They react with a variety of compounds such as acids, aldehydes, ketones, alkynes to form addition products. Grignard reagent usually attacks on nucleophilic center as:
The polarity of Grignard reagents is opposite to that expected because here the carbon atom bears a negative charge and acts as nucleophiles. This effect is called the Umpolang Effect or reversal of polarity. Explanation:
Methyl magnesium bromide is CH3 MgX and falls in the category of Grignard Reagents . The methyl group bears a partial negative charge and the metal bears a partial positive charge . Ketones react with a Grignard reagent to give tertiary alcohols as addition products. Aldehydes give secondary alcohol and formaldehyde gives primary alcohol. The reaction is a step-up reaction as the number of carbon atoms increases in the product. The reaction of acetone with methyl magnesium halide is given as follows:
The product of the reaction is tertiary butyl alcohol. Hence, tertiary butyl alcohol is obtained by the action of methyl magnesium bromide on acetone CH3 COCH3 .
Important Points
Reaction of carbon dioxide with methyl magnesium bromide gives acetic acid as product.
Question 33 5 / -1
Among the following acids least pKa value is of
Solution
Explanation-
The quantitative behavior of acids and bases in solution can be understood only if their pKa values are known.
In particular, the pH of a solution can be predicted when the analytical concentration and pKa values of all acids and bases are known; conversely, it is possible to calculate the equilibrium concentration of the acids and bases in solution when the pH is known.
pK a value-
The pKa value is one method used to indicate the strength of an acid. pKa is the negative log of the acid dissociation constant or Ka value . A lower pKa value indicates a stronger acid . That is, the lower value indicates the acid more fully dissociates in water.
The lowest pKa value is for oxalic acid as it gives the least number of hydrogen ions in the aqueous solution after its dissociation so it has the lowest dissociation constant value.
Question 34 5 / -1
When acetic acid reacts with P2 O5 , product formed
Solution
Explanation:
When acetic acid reacts with P 2 O 5 , the product formed is acetic anhydride
Question 35 5 / -1
Grignard reagent is a source of
Solution
Explanation:
Grignard Reagent:
The Grignard reagent is given as RMgX where X i s a halogen, and R is an alkyl or aryl (based on a benzene ring) group. They react with both aldehydes and ketones to form addition products. → Grignard reagents are the first source of carbanions (literally, "anions of carbon") . The Lewis structure of the CH3 - ion suggests that carbanions can be Lewis bases or electron-pair donors.
Question 36 5 / -1
Which of the following is not formed by sandmeyer reaction
Solution
Explanation:
Sandmeyer's reaction-
→ Sandmeyer reaction is a type of substitution reaction that is widely used in the production of aryl halides from aryl diazonium salts. Copper salts like chloride, bromide or iodide ions are used as catalysts in this reaction.
→ Notably, the Sandmeyer reaction can be used to perform unique transformations on benzene. The transformations include hydroxylation, trifluoromethylation, cyanation, and halogenation .
In sandmeyer's reaction Iodo Benzene cannot be formed because of its large size.
Question 37 5 / -1
Which one is the Swartz reaction from the following
Solution
Explanation:
→ Alkyl fluorides are prepared by heating alkyl bromide or chloride in presence of metallic fluoride-like AgF, SbF3, or Hg2 F2 . This reaction is known as the Swartz reaction.
→ CH 3 Br + AgF → CH 3 F + AgBr is an example of the Swartz reaction.
In Swartz's reaction:
Higher alkanes are formed from lower alkyl halide. Alkanes with an even number of carbon atoms are formed. The reaction takes place in the presence of dry ether because sodium is a very reactive metal if it forms oxide with water. It is a simple nucleophilic substitution reaction.
Additional Information NaI in presence of dry acetone give alkyl iodide. This reaction is known as the Finkelstein reaction.
→ 2 CH 3 Cl + 2 Na \(\xrightarrow{Dryether}\) CH 3 .CH 3 + 2 NaCl, During this reaction Ethane is formed from methyl chloride. This is the Wurtz reaction.
Question 38 5 / -1
ortho and para - nitrophenols are separated by which of the follwoing method?
Solution
Explanation
Ortho nitrophenol will do intra molecular H bonding whereas para nitrophenol will do intermolecular H bonding. Hence the surface area of para nitrophenol is more than that of ortho nitrophenol and more the surface area more is the boiling point.
So ortho nitrophenol is more volatile than para nitrophenol and can be separated by heating. This process is called steam-distillation.
Question 39 5 / -1
Phenol reacts with excess of bromine water to produce
Solution
Explanation:-
The reaction of phenol with bromine in the presence of water:
When phenol is mixed with sufficient bromine water, a white precipitate of 2,4,6-tribromophenol forms. Phenol interacts with bromine water to produce 2,4,6-tribromophenol. Ionization is much easier in water. When phenol is ionized, it produces the phenoxide ion, which is much more ortho,para-directing. Bromine is also more easily ionized, resulting in a large number of bromonium ions. Reaction:
Question 40 5 / -1
Phenolic compound is formed in
Solution
Explanation:-
Phenolic Compound:
Phenolic substances are secondary metabolites created by phenylpropanoid metabolization in plants' shikimic acid and pentose phosphate. From simple phenolic molecules to highly polymerized compounds, they all comprise benzene rings with one or more hydroxyl substituents. Hydroxyphenols and dihydroxybenzenes are examples of simple substituted phenol compounds. Catechol (1,2-dihydroxybenzene), resorcinol (1,3-dihydroxybenzene), and hydroquinone are among examples (1,4-dihydroxybenzene) Reaction:
Diazonium salts are formed when an aromatic primary amine is treated with nitrous (NaNO2 + HCl) acid at 273–278 K. In nature, these diazonium salts are quite reactive. These diazonium salts finally hydrolyze to phenols when heated with water.
Question 41 5 / -1
Which of the following is Hofmann mustard oil reaction?
Solution
Explanation:
→ In the Hofmann mustard oil reaction of primary amines, the black precipitate is due to HgS. It is a test of primary amine.
→ Primary amine gives alkyl isothiocyanate having mustard oil like smell.
RNH2 + CS2 → S = C(SH) - NHR \(\xrightarrow {\rm HgCI_2}\) RNCS + HgS + 2HCI
Question 42 5 / -1
Maximum basic in gas phase is
Solution
Explanation:
In the Gaseous phase, Basicity depends on carbocation
the more stable the carbonation the more basic the compound is
⇒ Basic Strength ∝ + I Effect & Alkyl group (- R) show + I effect
So in (CH 3 CH 2 ) 3 N - (3 ° Amine), it has more alkyl group
⇒ Stability of 3 °> 2° carbonation > 1° > NH3
Given compounds,
⇒ NH 3 , CH 3 CH 2 NH 2 (1 ° Amine), (CH 3 CH 2 ) 2 NH (2 ° Amine), (CH 3 CH 2 ) 3 N (3 ° Amine)
∴ (CH 3 CH 2 ) 3 N is the most stable.
Question 43 5 / -1
Which of the following compounds have the lowest Boiling rate?
Solution
Concept:
Boiling point of organic compounds:
The temperature at which the vapor pressure of a liquid becomes equal to the outer atmospheric pressure is called the boiling point of the liquid. At the boiling point, the liquid starts to go to the gaseous phase. in doing so, it has overcome the intermolecular forces of attraction in between the molecules. When the strength of attraction is low , the liquids start boiling easily and have a lower boiling point. When the forces are high, the liquid has a high boiling point . The intermolecular forces depend upon:Van der Waals forces of attraction between the molecules.Van der Waals forces increase with the increase in molecular mass of the compounds.Van der Waals forces decrease as branching occurs in organic compounds. The strength of H-bonding > Van der Waals forces and thus Hydrogen bonding affects the boiling point . When there is intermolecular H- bonding present , it increases the boiling point as extra energy is required to break these bonds. When ion-dipole interactions are present, the boiling point also increases. Explanation:
The boiling point in amines:
Amines have higher boiling points than their corresponding alkanes this is because of the presence ofDipole-Dipole interaction Hydrogen bonding in Primary and secondary amines . Primary amines such as propylamine have H - Bonding present in them between the lone pair on the N atom of one molecule and the H atom of the neighbouring one. They also have strong dipole - dipole interaction between them. There are two H atoms with N atom of primary amines and thus they can form two H bonds . On the other hand, the N atom lies in the centre in secondary amines and that makes the permanent dipole magnitude lesser than that of primary amines of comparable molecular weights. This decreases the magnitude of dipole-dipole interactions and the strength of H bonding becomes less. Thus secondary amines have lesser boiling points than primary amines of comparable molecular masses. Hence, the boiling point of diethylamine is lesser than that of butylamine . Between propylamine and butylamine, propylamine has a lower molecular mass , hence, its boiling point is lesser than that of butylamine. Here Aniline has the highest molecular mass so it has the highest boiling point . The boiling point of propylamine is lesser than that of di ethyl amine because of its lower molecular weight . Here the interaction is dominated by Van der Waals interaction and not H - bonding. The decreasing order of boiling point is Aniline > butyl amine > di ethyl amine > propylamine. Hence, the compound with the lowest boiling rate is propylamine.
Aromatic amines have their lone pair of electrons on N atom delocalised into the ring which decreases their ability to participate in H bonding. Thus, aromatic amines have a lower boiling point than aliphatic amines of comparable molecular mass.
Question 44 5 / -1
Among the following which one does not act as an intermediate in Hofmann rearrangement?
Solution
Explanation:
Hoffmann bromamide reaction take pace as follows:
Hence RNC does not act as intermediate in Hoffmann rearrangement.
Question 45 5 / -1
Greatest amount of hydration is in
Solution
Explanation:
⇒ Hydration (Hydrolosis) → H2 O (H+ OH- )
⇒ OH- (Nucleophilic)( e- rich species)
The greatest amount of hydration will be in that compound in which carbonyl carbon has least electron density (i.e. attached to more electro negative groups) which is either option B or option C because of the presence of electron withdrawing Bromine atom.
But the attack on Aldehyde group is more than ketone group [ As aldehyde is more electrophilic than ketone group]
So Option 3 is the Corrrect.
Question 46 5 / -1
Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be ______
Solution
Explanation:
→ The primary structure of a protein refers to the linear sequence of amino acids in the polypeptide chain.
→ The primary structure is held together by covalent bonds such as peptide bonds, which are made during the process of protein biosynthesis or translation.
→ The primary structure of proteins tells us about the sequence of amino acids in which they are linked.
Additional Information
Polypeptide chain synthesis is signaled by two initiation codons AUG & GUG. AUG codes for methionine amino acid in eukaryotes and it codes for N-formyl methionine in prokaryotesSometimes GUG also functions as a start codon and codes for methionine amino acid otherwise, GUG codes for valine amino acid normally. So, it is clear that the first amino acid synthesized in polypeptide chain formation is methionine amino acid , hence the correct option will be 4 . Methionine, glycine, valine, lysine are the correct sequence of amino acids.
Question 47 5 / -1
Sucrose molecule is made up of
Solution
Explanation:
→ Sucrose molecule is made up of α−D− gluco pyranose and β−D− fructo furanose.
Additional Information
Fructose Also known as fruit sugar It's a simple ketonic monosaccharide found in many plants Fructose was discovered by Augustin-Pierre Dubrunfaut in 1847 Sucrose Sucrose is a common sugar Also called Table sugar For human consumption, sucrose is extracted and refined from either sugarcane or sugar beet Maltose Maltose is a disaccharide that is made up of two glucose units Also known as maltobiose or malt sugar Maltose was discovered by Augustin-Pierre Dubrunfaut
Question 48 5 / -1
RNA and DNA are chiral molecules, their chirality is due to
Solution
Explanation
Structure of ribose & deoxyribose:
→ When the sugar molecule is drawn in Fischer projection and the most oxidised carbon atom is at the top and the hydroxyl (−OH) on the bottom chiral centre points to the right side then it is known as D-sugar.
→ We know, ribose and deoxyribose sugars are D-sugars.→ Thus, RNA and DNA are chiral molecules, their chirality is due to D-sugar components.
Question 49 5 / -1
Which of the following statements is not correct?
Solution
Explanation-
Disinfectant-
A disinfectant can be defined as an antimicrobial agent that can be applied on the surface of some objects in order to destroy the microorganisms residing on them. Dilute solutions of some disinfectants can be used as antiseptics. Disinfectants are antimicrobial drugs. Antiseptics-
It can be defined as antimicrobial agents which can be applied to the body of living organisms to inhibit the action of microbes. They are not injected into the body like the antibiotics , rather they are applied on the surface of the skin to heal the living tissues in case of wounds and cuts.Dettol is the most commonly used antiseptic. It is a mixture of chloroxylenol and terpineol. Iodoform is also used as an antiseptic for wounds. Some antiseptics can be added to soaps.
Question 50 5 / -1
The drug is used as
Solution
Explanation-
Antacid-
The substances which neutralize the acids of the stomach are called antacids. Acidity in the stomach causes the sensation of heartburn, antacids are taken to relieve this sensation. They are taken orally to suppress the effects of acidity. Antacids are actually alkaline ions that directly neutralize the gastric acids of the stomach Analgesic-
Analgesics are the class of drugs that are used medically by patients to relieve pain without causing a loss of consciousness. Antimicrobial-
Infections and diseases may be caused by different types of organisms like bacteria, fungi, viruses, etc., in humans and animals. The drug used to prevent the pathogenicity of microorganisms is called an antimicrobial agent .
Antiseptic-
Antiseptics can be defined as antimicrobial agents which can be applied on the body of living organisms to inhibit the action of microbes. They are not injected into the body like antibiotics, rather they are applied on the surface of the skin to heal the living tissues in case of wounds and cuts. Dettol is the most commonly used antiseptic. It is a mixture of chloroxylenol and terpineol. Iodoform is also used as an antiseptic for wounds
It is Histamine & it is Antacid
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