Question
Read the passage given below and answer the following questions:
Iron forms many complexes in its +2 and +3 oxidation states such as [Fe(H2O)6]2+ (A); [Fe(CN)6]4- (B); [Fe(H2O)6]3+ (C); [Fe(CN)6]3- (D), etc., They exhibit, different magnetic properties and undergo different hybridisation of iron.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following statements is correct?
  1. (B) is paramagnetic while (C) is diamagnetic.
  2. Both (B) and (D) are outer orbital complexe.
  3. Both (A) and (C) are paramagnetic.
  4. (A) is outer orbital complex and (C) is inner orbital complex.
  1. The complex having maximum magnetic moment is:
  1. (A)
  2. (B)
  3. (C)
  4. (D)
  1. Which of the following does not represent correct configuration of the d-orbitals in the given complexes?
  1. $\text{(A)}:\text{t}^4_{2\text{g}}\text{e}^2_\text{g}$
  2. $\text{(B)}:\text{t}^6_{2\text{g}}\text{e}^0_\text{g}$
  3. $\text{(C)}:\text{t}^4_{2\text{g}}\text{e}^1_\text{g}$
  4. $\text{(D)}:\text{t}^5_{2\text{g}}\text{e}^0_\text{g}$
  1. The spin only magnetic moment of complexes (A), (B), (C) and (D) are respectively (in BM).
  1. $2\sqrt{6},0,\sqrt{35},\sqrt{3}$
  2. $0,2\sqrt{6},\sqrt{35},\sqrt{3}$
  3. $\sqrt{15,}2\sqrt{6},\sqrt{3},0$
  4. $\sqrt{3},\sqrt{8},0,\sqrt{15}$
  1. Which of the given complexes are outer orbital complexes?
  1. (A) and (B) only
  2. (B) and (C) only
  3. (A) and (C) only
  4. (B) and (D) only

Answer

  1. (c) Both (A) and (C) are paramagnetic.

Explanation:

(A): sp3d2 hybridisation (outer orbital)

No. of unpaired electrons = 4

(B): d2sp3 hybridisation (inner orbital)

No. of unpaired electrons = 0

(C): sp3d2 hybridisation (outer orbital)

No. of unpaired electrons = 5

(D): d2sp3 hybridisation (inner orbital)

No. of unpaired electron = 1

  1. (c) (C)

Explanation:

It has 5 unpaired electrons.

  1. (c) $\text{(C)}:\text{t}^4_{2\text{g}}\text{e}^1_\text{g}$

Explanation:

As H2O is a weak ligand so, it should be $\text{t}^4_{2\text{g}}\text{e}^1_\text{g}.$

  1. (a) $2\sqrt{6},0,\sqrt{35},\sqrt{3}$

Explanation:

Magnetic moments of (A), (B), (C) and (D) are respectively.

$\sqrt{4(4+2)},0,\sqrt{5(5+2)},\sqrt{1(1+2)}$

  1. (c) (A) and (C) only

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The electrochemical cell shown below is concentration cell.

M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm-3) |M The emfof the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059V.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. The solubility product (Ksp' mol3 dm-9) of MX2 at 298 K based on the information available for the given concentration cell is $(\text{take }2.303 \times \text{R}\times \frac{298}{\text{F}} = 0.059)$
  1. 2 × 10-15
  2. 4 × 10-15
  3. 3 × 10-12
  4. 1 × 10-12
  1. The value of $\triangle\text{G}$ (in kJ mol-1) for the given cell is (take 1 F = 96500 C mol-1)

  1. 3.7
  2. -3.7
  3. 10.5
  4. -11.4
  1. The equilibrium constant for the foUowing reaction is:

$\text{Fe}^{2+}+\text{Ce}^{4+}\rightleftharpoons\text{Ce}^{3+}+\text{Fe}^{3+}$

(Given, $\text{E}^\circ_\frac{\text{Ce}^{4+}}{\text{Ce}^{3+}}=1.44\text{V}$ and $\text{E}^\circ_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}=0.68\text{V}$)

  1. 7.6 × 1012
  2. 6.5 × 1010
  3. 5.2 × 109
  4. 3.4 × 1012
  1. The solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the emf of the cell

Ag|Ag+ (satd. Ag2CrO4 soln) || Ag+ (0.1 M) | Ag

is 0.164V at 298 K, is:

  1. 3.359 × 10-12 mol3 L-3
  2. 2.287 × 10-12 mol3 L-3
  3. 1.158 × 10-12 mol3 L-3
  4. 4.135 × 10-12 mol3 L-3
  1. To calculate the emf of the cell, which of the foUowing options is correct?
  1. emf = Ecathode - Eanode
  2. emf = Eanode - Ecathode
  3. emf = Eanode + Ecathode
  4. None of these.
Read the passage given below and answer the following questions:
When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures remains intact. Examples of denaturation of protein are coagulation of egg white on boiling, curdling of milk, formation of cheese when an acid is added to milk.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Mark the wrong statement about denaturation of proteins.
  1. The primary structure of the protein does not change.
  2. Globular proteins are converted into fibrous proteins.
  3. Fibrous proteins are converted into globular proteins.
  4. The biological activity of the protein is destroyed.
  1. Which structure(s) of proteins remains(s) intact during denaturation process?
  1. Both secondary and tertiary structures.
  2. Primary structure only.
  3. Secondary structure only.
  4. Tertiary structure only.
  1. $\alpha$-helix and $\beta$-pleated structures of proteins are classified as:
  1. Primary structure.
  2. Secondary structure.
  3. Tertiary structure.
  4. Quaternary structure.
  1. Cheese is a:
  1. Globular protein.
  2. Conjugated protein.
  3. Denatured protein.
  4. Derived protein.
  1. Secondary structure of protein refers to:
  1. Mainly denatured proteins and structure of prosthetic groups.
  2. Three-dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain.
  3. Linear sequence of amino acid residues in the polypeptide chain.
  4. Regular folding patterns of continuous portions of the polypeptide chain.
Describe detailed information on classification of carbohydrates.
Explain the D and L notation method of spatial arrangement with respect to glucose.
Read the passage given below and answer the following questions:

Aniline activates the benzene ring by increasing electron density at ortho- and para-positions. Hence, it is o-, p-directing. -NH2 group strongly activates the ring therefore it is difficult to stop the reaction at monosubstitution stage. Among electrophilic substitution reaction, direct nitration of aniline is not done to get o- and p-nitroaniline because lone pair of electrons present at nitrogen atom will accept proton from nitrating mixture to give anilinium ion which is meta-directing.

Aniline with NaNO2 and HCI forms benzene diazonium chloride at very low temperature. Aromatic amines react with nitrous acid to form a yellow oily liquid known as N-nitrosoamines.

A statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Nitrating mixture used for carrying out nitration of benzene consists of cone. HNO3 + cone. H2SO4.

Reason: In presence of H2SO4, HNO3 acts as a base and produces $\text{NO}^+_2$ ions.

  1. Assertion: Anilinium chloride is more acidic than ammonium chloride.

Reason: Anilinium ion is not resonance-stabilised.

  1. Assertion: Nitrobenzene can be prepared from benzene by using mixture of cone. HNO3 and cone. H2SO4.

Reason: In the mixture, H2SO4 act as a acid.

  1. Assertion: In strongly acidic solution, aniline becomes less reactive towards electrophilic reagents.

Reason: The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is no longer available for resonance.

  1. Assertion: Nitration of aniline can be done conveniently by protecting -NH2 group through acetylation.

Reason: Acetylation of aniline results in the increase of electron density in the benzene ring.

For a first order reaction, A → Products, $\text{k}=\frac{2.303}{\text{t}}\log\frac{\text{a}}{\text{a}-\text{x}},$ where a is the initial concentration of A and (a - x) is the concentration of A after time t. k is rate constant. Its value is constant at constant temperature for a reaction. The time in which half of the reactant is consumed is called half-life period. Half-life period of a first order reaction is constant. Its value is independent of initial concentration or any other external conditions.

In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Rate of reaction doubles when concentration of reactant is doubled if it is a first order reaction.

Reason: Rate constant also doubles.

  1. Assertion: For the first order reaction, half-life period is expressed as $\text{t}_\frac{1}{2}=\frac{2.303}{\text{k}}\log2.$

Reason: The half-life time of a first order reaction is not always constant and it depends upon the initial concentration of reactants.

  1. Reason: The half-life time of a first order reaction is not always constant and it depends upon the initial concentration of reactants.

Reason: Acid only acts as a catalyst whereas alkali acts as one of the reactants.

  1. Assertion: For a first order reaction, the concentration of the reactant decreases exponentially with time.

Reason: Rate of reaction at any time depends upon the concentration of the reactant at that time.

  1. Assertion: Half-life period for a first order reaction is independent of initial concentration of the reactant.

Reason: For a first order reaction, $\text{t}_\frac{1}{2}=\frac{0.693}{\text{k}},$ where k is rate constant.

Read the passage given below and answer the following questions:
A compound (X) containing C, H and O is unreactive towards sodium. It also does not react with Schiff s reagent. On refluxing with an excess of hydroiodic acid, (X) yields only one organic product ( Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorus and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The compound (X) is an:
  1. Acid.
  2. Aldehyde.
  3. Alcohol.
  4. Ether.
  1. The IUPAC name of the acid formed is:
  1. Methanoic acid.
  2. Ethanoic acid.
  3. Propanoic acid.
  4. Butanoic acid.
  1. Compound (Y) is:
  1. Ethyl iodide.
  2. Methyl iodide.
  3. Propyl iodide.
  4. Mixture of (a) and (b).
  1. Compound (Z) is:
  1. Methanol.
  2. Ethanol.
  3. Propanol.
  4. Butanol.
  1. Compound (X) on treatment with excess of Cl2 in presence of tight gives:
  1. $\propto-$ Chlorodiethyl ether.
  2. $\propto,\propto'-$ Dichlorodiethyl ether.
  3. Perchlorodiethyl ether.
  4. None of these.
Read the passage given below and answer the following questions:

Transition metal oxides are compounds fanned by the reaction of metals with oxygen at high temperature. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5· V2O4 dissolves in acids to give VO2+ salts. Transition metal oxides are commonly utilized for their catalytic activity and semi conductive properties. Transition metal oxides are also frequently used as pigments in paints and plastic. Most notably titatnium dioxide. One of the earliest application of transition metal oxides to chemical industry involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid. Since then, many other applications have emerged, which include benzene oxidation to maleic anhydride on vandium oxides; cyclohexane oxidation to adipic acid on cobalt oxides. An important property of the catalyst material used in these processes is the ability of transition metals to change their oxidation state under a given chemical potential of reductants and oxidants.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which oxide of vanadium is most likely to be basic and ionic?
  1. VO
  2. V2O3
  3. VO2
  4. V2O5
  1. Vanadyl ion is:
  1. $\text{VO}^{2+}$
  2. $\text{VO}^{+}_2$
  3. $\text{V}_{2}\text{O}^+$
  4. $\text{VO}^{3-}_4$
  1. Which of the following statements is false?
  1. With fluorine vanadium can form VF5.
  2. With chlorine vanadium can form VCl5.
  3. Vanadium exhibits highest oxidation state in oxohalides VOCl3, VOBr3 and fluoride VF5.
  4. With iodine vanadium cannot form Vl5 due to oxidising power of V5+ and reducing nature of I-.
  1. The oxidation state of vanadium in V2O5 is:
  1. $\frac{+5}{2}$

  2. +7
  3. +5
  4. +6
  1. Identify the oxidising agent in the following reaction.

V2O+ 5Ca → 2V + 5CaO

  1. V2O5
  2. Ca
  3. V
  4. None of these.
Read the passage given below and answer the following questions:

According to Raoult's law, the partial pressure of two components of the solution maybe given as:

$\text{P}_\text{A}=\stackrel{{0}}{\hbox{PA }}\text{x}_\text{A}$ and $\text{P}_\text{B}=\stackrel{{0}}{\hbox{PB }}\text{x}_\text{B}$

For an ideal solution (obeys Raoult's law always)

$\Delta\text{H}_\text{mix}=0,\Delta\text{mix}=0$

All solutions do not obey Raoults law over entire range of concentration. These are known as non-ideal solutions.

For non-ideal solutions, $\text{P}_\text{A}\not=\stackrel{{0}}{\hbox{PA }}\text{x}_\text{A}$ or $\text{P}_\text{B}\not=\stackrel{{0}}{\hbox{PB }}\text{x}_\text{B}$

Positive deviation $\Rightarrow\text{P}_\text{A}>\stackrel{{0}}{\hbox{PA }}\text{x}_\text{A}$ and $\text{P}_\text{B}>\stackrel{{0}}{\hbox{PB }}\text{x}_\text{B}$

Negative deviation $\text{P}_\text{A}<\stackrel{{0}}{\hbox{PA }}\text{x}_\text{A}$ and $\text{P}_\text{B}\stackrel{{0}}{\hbox{PB }}\text{x}_\text{B}$

A statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: An ideal solution obeys Raoult's law.

Reason: In an ideal solution, solute-solute as well as solvent-solvent interactions are similar to solute-solvent interactions.

  1. Assertion: Acetone and aniline show negative deviations.

Reason: H-bonding between acetone and aniline is stronger than that between acetone-acetone and aniline-aniline.

  1. Assertion: Azeotropic mixtures are formed only by non-ideal solutions and they may have boiling points either greater than both the components or lesser than both the components.

Reason: The composition of the vapour phase is same as that of the liquid phase of an azeotropic mixture.

  1. Assertion: The solutions which show negative deviations from Raoult's law are called maximum boiling azeotropes.

Reason: 68% nitric acid and 32% water by mass fonn maximum boiling azeotrope.

  1. Assertion: $\Delta\text{H}_{\text{mix}}$ mix and $\Delta\text{V}_{\text{mix}}$ are positive for an ideal solution.

Reason: The interactions between the particles of the components of an ideal solution are almost identical as between particles in the liquids.

Read the passage given below and answer the following questions:

Fehling's reagent: Fehling's reagent is a mixture of two solutions. Fehllng's solution A is aqueous copper sulphate solution. Fehling's solution Bis alkaline sodium potassium tartarate (Rochelle salt).

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH(OH)COONa}\\\text{CuSo}_{4\text{(aq)}}+|\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH(OH)COOK}$

It is a mild oxidising agent. It is weaker than Tollens' reagent. It oxidises only aliphatic aldehydes to carboxylate ions and itself gets reduced to reddish brown precipitate of cuprous oxide. Aromatic aldehydes do not respond to Fehling's test. This reaction is used for the test of aliphatic aldehydes known as Fehling's reagent test.

In these questions (Q. No. l-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Fehling's solution can be used to distinguish between acetaldehyde and acetone.

Reason: Fehling's reagent is a mixture of two solutions.

  1. Assertion: Aromatic aldehydes can be distinguished from aliphatic aldehydes by Fehling's solution.

Reason: Aromatic aldehydes reduce Fehling's solution, but aliphatic aldehydes do not.

  1. Assertion: Fehling's solution oxidises acetaldehyde to acetic acid but not benzaldehyde to benzoic acid.

Reason: The C-H bond of -CHO group in benzaldehyde is stronger than in acetaldehyde.

  1. Assertion: CH3CHO and C6H5CH2CHO cannot be distinguished chemically by Fehling's solution.

Reason: CH3CHO and C6H5CH2CHO cannot be distinguished chemically by Fehling's solution.

  1. Assertion: Formaldehyde, when heated with Fehling's reagent produces a reddish brown ppt, of Cu.

Reason: Fehling's reagent oxidises fonnaldehyde to formate ion.