Chemistry Mock Test - 2

Explanation-

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.

The direction is common over a macroscopic volume which we term as a domain. The domain has a net magnetization and each domain directs itself, which results in its strong magnetization.

Explanation-

⇒ The structure of an ionic compound depends on the relative sizes of the cations and anions. Ionic compounds include salts, oxides, hydroxides, sulphides, and the majority of inorganic compounds. Ionic solids are held together by the electrostatic attraction between the positive and negative ions.

⇒ These ionic bonds between the charged particles result in a giant structure of ions.

⇒ Because the ions are held together tightly in these giant structures it takes a lot of energy to break all the bonds. As a result, ionic compounds have high melting points and boiling points.

⇒ Bigger ions form a close-packed structure.

⇒ Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size.

⇒ Occupation of all the voids is not necessary.

So statement given in option 4 is not true about ionic solids.

Explanation-

Packing Efficiency-

• A crystal lattice is made up of a very large number of unit cells where every lattice point is occupied by one constituent particle. The unit cell can be seen as a three-dimension structure containing one or more atoms. 

The percentage of spaces filled by the particles in the unit cell is known as the packing fraction of the unit cell.

Primitive cubic lattice-

• In the primitive cubic unit cell (PC), the atoms are present only at the corners. Every atom at the corner is shared among 8 adjacent unit cells.
• There are 4 unit cells in the same layer and 4 in the upper (or lower) layer. Therefore, a particular unit cell has only 1/8th of an atom.

The packing efficiency of PC lattice is 52.4%.

Body-centered cubic (bcc)-

• In body-centered cubic unit cell (BCC) has atoms at each corner of the cube and an atom at the center of the structure.

The packing efficiency of the BCC lattice is 68%.

Face-centered cubic (fcc)-

• In face-centered unit cell (FCC) contains atoms at all the corners of the crystal lattice and at the center of all the faces of the cube. The atom present at the face center is shared between 2 adjacent unit cells and only 1/2 of each atom belongs to an individual cell.

The packing efficiency of FCC lattice is 74%.

So the correct order of packing efficiency is fcc > bcc > pc.

Explanation:

The normality equation is 

⇒ N₁ V₁ + N₂ V₂ = N₃ (V₁ + V₂)

Suppose the volume of solution A mixed is "x" litre.

Then the volume of solution B will be "(2 - x)" litre.

Given,

⇒ Concentration of Solution A (N₁) = 0.5 N, Concentration of Solution B (N₂) = 0.1 N

⇒ Concentration of Mixed Solution (A & B) (N₃) = 0.2 N

⇒ Volume of mixed solution = 2 litre

Applying the normality equation,

⇒ N₁ V₁ + N₂ V₂ = N₃ (V₁ + V₂)

⇒ 0.5x + 0.1 (2 - x) = 0.2 × 2

⇒ 0.5x + 0.2 - 0.1x = 0.4

⇒ x = 0.5

So, the volume of A will be 0.5 lit 

Volume of B will be = 2 - 0.5 = 1.5 lit

Explanation:

If P⁰ and P are the vapour pressure of the pure solvent and solution and n₁ and n₂ are the moles of solute and solvent respectively in the solution,

As we know,

⇒ \(\frac{{{P^0} - P}}{{{P^0}}} = \frac{{{n_1}}}{{{n_1} + {n_2}}}\)

⇒ \(1 - \frac{P}{{{P^0}}} = \frac{{{n_1}}}{{{n_1} + {n_2}}}\)

⇒ \(\frac{P}{{{P^0}}} = 1 - \frac{{{n_1}}}{{{n_1} + {n_2}}}\)

⇒ \(\frac{P}{{{P^0}}} = \frac{{{n_2}}}{{{n_1} + {n_2}}}\)

⇒ \(P = {P^0}\left[ {\frac{{{n_2}}}{{{n_1} + {n_2}}}} \right]\)

∴ The correct relation between P and P0 is P = P0 \(\left[ {\frac{{{n_2}}}{{{n_1} + {n_2}}}} \right].\)

Additional Information Vapour pressure: 

• Vapour pressure of a liquid at any temperature may be defined as the pressure exerted by the vapour pressure above the liquid in equilibrium with the liquid at that temperature.
• The Vapour pressure of the solution at a given temperature is lower than the vapour pressure of the pure solvent at the same temperature.
• According to Raoult's law, the vapour pressure of a solution containing non-volatile solute is given by,

PA = PAo xA   

Where PA= vapour pressure of solute (A), PAo = vapour pressure in pure state, xA = mole fraction of solute = \(\frac{{{n_A}}}{{{n_B} + {n_A}}}\)

Factors affecting vapour pressure:

1. Nature of the liquid

• If the intermolecular forces of attraction in the liquid are weak, the molecules can easily leave the liquid and come into the vapour phase and hence the vapour pressure is higher.
• If the intermolecular forces of attraction in the liquid are strong, vapour pressure is weaker.

2. Effect of temperature

• As the temperature of a liquid is increased, the vapour pressure of the liquid increases.

Explanation:

Empirical formula-

• The empirical formula is the simplest formula for a compound which is defined as the ratio of subscripts of the smallest possible whole number of the elements present in the formula. It is also known as the simplest formula.

A cube has 8 corners and 6 faces

⇒ Each Atom at the corner contributes 1/8 of itself to a particular cubic crystal lattice.

8 atoms × 1/8 = 1 Atom ( at corners)

⇒ Each Atom at a face center contributes 1/2 of itself to a particular cubic crystal lattice.

6 atoms × 1/2 = 3 atoms

Given data and Calculation-

• "A" atoms at the corner of the cube.
• The effective number of atoms of A = 1/8 × 8 = 1 
• if all the atoms present on the body diagonal are replaced by atom C.
• So two atoms of A present at the corner will be replaced by C.
• So the effective number of A = 6/8 = 3/4 
• B ions on the centers of the faces of the cube.
• For B, 1 atom contributes to two cells.
• the effective number of atoms of B = 1/2 × 6 = 3
• Atom C is present at diagonal that is 2 at corners.

So effective number of C = 1/8× 2 =  1/4

So the formula will be A3B12C.

Explanation-

Van't Hoff factor-

• The Van’t Hoff factor offers insight into the effect of solutes on the colligative properties of solutions. It is denoted by the symbol ‘i’.
• The Van’t Hoff factor can be defined as the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass.
• The extent to which a  substance associates or dissociates in a solution is described by the Van’t Hoff factor.
• For example, when a non-electrolytic substance is dissolved in water, the value of i is generally 1.
• However, when an ionic compound forms a solution in water, the value of i is equal to the total number of ions present in one formula unit of the substance.

Given data and Analysis-

• 0.2 molal aqueous solution of urea is present.
• Urea is nonelectrolyte, so it neither undergoes association nor dissociation, Hence van't Hoff factor is = 1.

Explanation-

The relation between the number of moles of electrons and the charge is as follows.

Number of moles of electrons = Q/F

Here Q = Charge, F = Faraday = 96500 C

Given data and Calculation:

⇒ 9650 coulombs of charge pass through the electroplating bath.

⇒ number of moles of electrons = \(\frac{Q}{F} = \frac{{9650}}{{96500}} = 0.1\)

We know that one mole of silver nitrate is going to produce one mole of monovalent silver ion when silver nitrate is dissociated.

∴ The number of electrons produced will be 1 mole.

⇒ So, 0.1 moles of silver are going to be produced by 0.1 moles of silver nitrate.

∴ Mass of the silver = (number of moles) (molar mass)

= (0.1) × (108) g = 10.8 g

Explanation-

Activation energy-

• Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into a product.
• It can also be described as the minimum amount of energy needed to activate or energize molecules or atoms so that they can undergo a chemical reaction or transformation.

Endothermic reaction-

• Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products.
• These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. Physical processes can be endothermic as well – Ice cubes absorb heat energy from their surroundings and melt to form liquid water (no chemical bonds are broken or formed).

Given data and Analysis:

⇒ X⟶Y is endothermic
⇒ For an endothermic reaction, ΔH is positive
⇒Again, ΔH= Ef − Eb
∵ ΔH is positive, thus Eb < Ef.

Explanation:-

Secondary Cell:-

• A secondary cell or battery is one that may be electrically recharged to its original pre-discharge condition after use by running current through the circuit in the opposite direction as when it was discharged.
• Primary battery cells include the voltaic cell, the Daniel cell, the Lachance, the Bunsen cell, and the fuel cell.
• Secondary cells include the lead-acid cell and the Edison alkaline cell, which are two well-known examples.
• In contrast to a disposable or main battery, which is delivered fully charged and discarded after use, a rechargeable battery, storage battery, or secondary cell (technically a sort of energy accumulator) can be charged, discharged into a load, and recharged several times.
• One or more electrochemical cells make up this device.
• Because it gathers and stores energy through a reversible electrochemical reaction, the word "accumulator" is employed.
• From button cells to megawatt systems connected to stabilise an electrical distribution network, rechargeable batteries come in a variety of shapes and sizes. 
• Lead-acid, zinc-air, nickel-cadmium (NiCd), nickel-metal hydride (NiMH), lithium-ion (Li-ion), lithium iron phosphate (LiFePO4), and lithium-ion polymer are among the electrode materials and electrolytes utilised.

Explanation:-

Standard Reduction Potential:-

• The standard reduction potential of a chemical species is a measurement of its tendency to be reduced.
• The standard oxidation potential is a measurement of a chemical species' tendency to be oxidized rather than reduced.
• The gas fluorine is strongly electronegative and seeks electrons to complete its outer shell.
• Fluorine has a comparatively high reduction potential of +2.87V as a result of this, indicating that it is more likely to acquire electrons and become reduced.

Given,

Fe2+ + Sn → Fe + Sn2+

⇒ The sum of the electric potential differences (PDs) caused by a separation of charges (electrons or ions) at each phase boundary (or interface) in the cell is the cell's emf.

⇒ The chemical composition of the two interacting phases determines the amplitude of each PD.

⇒ Because the Fe ion is undergoing reduction, Fe serves as the cathode and Sn serves as the anode in the cell reaction. Hence,

E^º_cell  = E^º_cathode  - E^º_anode

⇒  Eºcell  = - 0.44 - ( - 0.14) = - 0.30 V.

The negative EMF indicates that the process proceeds in the opposite direction spontaneously.

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.

Difference between Absorption and Adsorption

Explanation:-

EMF of Electromotive Cell:

• The maximum potential difference between two electrodes of a cell is known as the electromotive force of a cell, or EMF of a cell.
• It's also known as the net voltage between the half-reactions of oxidation and reduction.
  An electrochemical cell's EMF is primarily used to identify whether it is galvanic or not.
• We can calculate the resulting energy and the amount of charge flowing through the cell if we know the resultant energy and the amount of charge passing through the cell.
• It is the most straightforward method of calculating the EMF.
• We might instead use a definition similar to Ohm's law (V = IR).

The maximum potential difference between two electrodes is called the EMF.

E_cell = E_cathode - E_anode

⇒ \(E = E^o - \frac{0.059}{n} \times log\frac{[M]}{[M^{n+}]}\)

Explanation:

According to the given equation, the rate of disappearance of reactant A is twice the rate of disappearance of reactant B

Given 

⇒ Reaction 2A + B → A2B

• For expressing the rate of a reaction where stoichiometric coefficients of reactants or products are not equal to one, the rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients.
• Since the rate of consumption of A is twice the rate of formation of A₂B, to make them equal, the term [A] is divided by 2.

⇒ The rate of this reaction is given by 

\( ⇒ \frac{{ - 1}}{2}\frac{{d\left[ A \right]}}{{dt}} = \frac{{ - d\left[ B \right]}}{{dt}} = \frac{{d\left[ {{A_2}B} \right]}}{{dt}}\)

Explanation-

Pseudo-first-order reaction -

⇒ A pseudo-first-order reaction can be defined as a second-order or bimolecular reaction that is made to behave like a first-order reaction.

⇒ This reaction occurs when one reacting material is present in great excess or is maintained at a constant concentration compared with the other substance.

⇒ For example, a second-order of the reaction is given by the equation,

A + B → C + D

⇒ This reaction is dependent upon the concentrations of both A and B but one of the components is present in large excess and thus its concentration hardly changes as the reaction proceeds.

⇒ So, if component B is in large excess and the concentration of B is very high as compared to that of A, the reaction is considered to be a pseudo-first-order reaction with respect to A and if component A is in large excess and the concentration of A is very high as compared to that of B, the reaction is considered to be pseudo-first order with respect to B.

⇒ The value of the rate constant of pseudo-first-order reaction depends on the concentration of reactants present in excess amount.

⇒ This reaction occurs when one reacting material is present in great excess or is maintained at a constant concentration compared with the other substance.

Explanation-

Activation energy-

• Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into a product.
• It can also be described as the minimum amount of energy needed to activate or energize molecules or atoms so that they can undergo a chemical reaction or transformation.

Endothermic reaction-

• Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products.
• These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. Physical processes can be endothermic as well – Ice cubes absorb heat energy from their surroundings and melt to form liquid water (no chemical bonds are broken or formed).

Given data and Calculation:

⇒ An endothermic reaction A → B has an activation energy of 15 kcal/mole and energy of reaction of 5 kcal/mole

⇒ For an endothermic reaction, the activation energy for the backward reaction is equal to the difference between the activation energy of the forward reaction and the enthalpy change of the reaction.

Thus, the activation energy of the reaction B → A is 15 − 5 = 10 kcal/mole

Explanation:

A + 2B → Products 

Given,

→@ t = 0, [dA[/dt = 2.6 × 10-2 mol L-1s-1

From rate law 

\(\frac{{ - d\left[ A \right]}}{{dt}} = - \frac{1}{2}.\frac{{d\left[ B \right]}}{{dt}}\)

\(\therefore \frac{{d\left[ B \right]}}{{dt}} = 2.\frac{{d\left[ A \right]}}{{dt}}\)

= 2 × 2.6 × 10^-2

= 5.2 × 10^-2 mol L^-1 s^-1

Additional InformationRate Law:

• Rate law states that the rate of a reaction is proportional to the concentration of the reactants raised to the power of the order of the reaction.

In mathematical terms, we can say that

Rate = - dc/dt ; the rate of decrease of reactants

Rate = - dx/dt; the rate of increase of reactants

→ Rate = k [C]ⁿ; 

where 'k' = rate constant and 'n' = order of a reaction.

Explanation:

Ethylene diamine tetra acetic acid (EDTA):

• Ethylene diamine tetra acetic acid or EDTA is a complexing agent.
• EDTA Was first synthesized by Ferdinand Munz in 1935. This EDTA is also called amino polycarboxylic acid.
• It is colorless and water-soluble in its salt form.
• The structure consists of the basic ethane group which is attached to the two -NH₂ groups at the terminal positions forming an ethylene diamine.
• The two hydrogens from each amine group situated at the terminal positions are replaced by the acetic acid group(-CH₂COO) There are a total of four acetic acid groups attached to the ethylene diamine.
• The structure is as follows.

• As EDTA is a polydentate ligand, it is also a chelating ligand, it contains donor atoms so arranged that they can occupy more than one position around the same metal atom simultaneously in the first sphere of coordination.
• The resulting complexes are called chelates
• EDTA  is a hexadentate ligand.
• It has four carboxylic acid groups and two amine groups with lone pairs of electrons on them. It is polyprotic acid.
• Along with the four carboxylic acid groups,  EDTA  can add two more hydrogens to the two nitrogens of the amine groups.

Explanation-

van der Waals forces-

Van der Waals forces are weak intermolecular forces that are dependent on the distance between atoms or molecules. These forces arise from the interactions between uncharged atoms/molecules.

• Covalent bonds and ionic bonds are significantly stronger than Van der Waals forces
• These forces are additive in nature, they are made up of several individual interactions
• These forces cannot be saturated
• No directional characteristic can be attributed to these force

⇒ Atoms of noble gases are stable and do not have electrostatic forces of attraction or repulsion between them.

⇒ The noble gases are monatomic under standard conditions because their outermost shell is completely filled due to which they are extremely stable in nature. Due to this stability, weak interatomic forces exist between them known as van der Waal forces.

Ion - dipole forces-

⇒ An ion-dipole interaction is the intermolecular force of attraction between a charge ion (cation or anion) and a molecule.

⇒ It is found commonly in the solution where ionic compounds dissolve in polar solvents.

⇒ Ion-dipole attractions become stronger as the charge on the ion increases or as the magnitude of the dipole of the polar molecule increases.

London dispersion forces-

⇒ London dispersion forces are the weakest intermolecular force. London’s dispersion forces can be defined as a temporary attractive force due to the formation of temporary dipoles in a nonpolar molecule.

Magnetic force-

⇒ It is a consequence of electromagnetic force and is caused due to the motion of charges.

Explanation-

Option 1: HF is a stronger acid than HCI.

⇒ 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.

⇒ Due to the higher electronegativity of F, it is strongly bonded in HF, so It is not easy to remove the H⁺ ion. So HF is not a stronger acid than HCl.

So option 1 is not true.

Option 2: Among halide ions, iodide is the most powerful reducing agent.

⇒ The iodide ion has the lowest value of standard reduction potential. Hence, it is the most powerful reducing agent among the halide ions.

Option 3: Fluorine is the only halogen that does not show a variable oxidation state.

⇒ Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d orbitals and therefore, can expand their octets and show + 1, + 3, + 5, and + 7 oxidation states also.

Option 4: HOCI is a stronger acid than HOBr.

⇒ The weaker the O−H bond, the stronger the acid. Chlorine is more electronegative than Bromine which reduces the strength of the O−H bond in HOCl.

Thus HOCl is a stronger acid than HOBr.

Explanation:

Fajan's rule-

Fajans’ rule predicts whether a chemical bond will be covalent or ionic. A few ionic bonds have partial covalent characteristics which were first discussed by Kazimierz Fajans in 1923. At that time with the help of X-ray crystallography, he was able to predict ionic or covalent bonding with the attributes like ionic and atomic radius.

• Size of the ion: Smaller the size of the cation, the larger the size of the anion, and the greater the covalent character of the ionic bond.
• The charge of Cation: Greater the charge of cation, the greater the covalent character of the ionic bond.

⇒ MCl_2​ is more ionic than MCl₄.​

Explanation:

Co-ordinate complexes are incompletely ionizable in the aqueous solutions. These give a complexion that does not show complete ionization.

• The coordination complex is classified into positive, negative, and neutral complex based on the charge on the coordination sphere.
• Positive complex - charge on coordination sphere is positive and it is mentioned as cationic complex.
• Negative complex - charge on coordination sphere is negative and it is mentioned as an anionic complex.
• Neutral complex - no charge on coordination sphere and it is mentioned as neutral complex.

Given data and Analysis:

→ [Pt(NH3)2CI2] - Charge on complex part = 0

→ [Co(NH3)6]CI3 = [Co(NH3)6]³⁺ + 3Cl^- Charge on complex part = +3

→ [Ni(NH3)6]CI2 = [Ni(NH3)6] ²+ + 2Cl- Charge on complex part = +2

→ K4[Fe(CN)6] = [Fe(CN)6] ^4- + 4K⁺ Charge on complex part = -4 

So [Pt(NH3)2CI2]  is the neutral complex.

Explanation:

Given compound is [Co(NH3)6]CI3.

⇒ The surrounding atoms, ions, and molecules around the central transition metal ion are known as Ligands.

⇒ Ligand is cobalt.

⇒ The oxidation state of ′CO′:

⇒ x + 6 (0)=3

IUPAC name will be Hexammine cobalt (III) chloride.

Additional Information Rules For Naming Coordination Compound-

• The ligands are always written before the central metal ion in the naming of complex coordination complexes.
• When the coordination center is bound to more than one ligand, the names of the ligands are written in an alphabetical order which is not affected by the numerical prefixes that must be applied to the ligands.
• When there are many monodentate ligands present in the coordination compound, the prefixes that give insight into the number of ligands are of the type: di-, tri-, tetra-, and so on.
• When there are many polydentate ligands attached to the central metal ion, the prefixes are of the form bis-, tris-, and so on.
• The names of the anions present in a coordination compound must end with the letter ‘o’, which generally replaces the letter ‘e’. Therefore, the sulfate anion must be written as ‘sulfato’ and the chloride anion must be written as ‘chlorido’.
• The following neutral ligands are assigned specific names in coordination compounds: NH3 (ammine), H2O (aqua or aquo), CO (carbonyl), and NO (nitrosyl).
• After the ligands are named, the name of the central metal atom is written. If the complex has an anionic charge associated with it, the suffix ‘-ate’ is applied.
• When writing the name of the central metallic atom in an anionic complex, priority is given to the Latin name of the metal if it exists (with the exception of mercury).
• The oxidation state of the central metal atom/ion must be specified with the help of roman numerals that are enclosed in a set of parentheses.
• If the coordination compound is accompanied by a counter ion, the cationic entity must be written before the anionic entity.

Concept:

• Phosphorus is a group 15 or group V A element.
• It has outer electronic configuration ns²np³.
• Phosphorus unlike Nitrogen exists in many forms such as red phosphorus, white phosphorus, etc, each of them having a different lattice structure.
• Phosphorus mainly exists as P₄, and is a tetrahedral molecule where each Phosphorus atom is linked to another P atom via single covalent bonds.
• This makes Phosphorus to be much more reactive than Nitrogen as single bonds are much easier to break.
• The breaking of the P-P bond requires much less energy.
• This makes Phosphorus quite reactive in nature.

Explanation:

• The element P has five electrons in the outermost shell and is short of three electrons to complete its octet.
• One case may be such that P can accept three electrons from other elements and be anion P^3-, which is not easy considering the energy of electron gain enthalpy.
• So, they make mostly covalent compounds by sharing three electrons in most cases and P^3- is formed in very rare cases.
• As Phosphorus has a vacant 3d orbital, which can accept electrons in it, it can expand its octet and sometimes exhibits a covalency of 5 in compounds such as PCl₅.
• In PCl₅, the oxidation state of P is +5, in that case, all 5 valence electrons take part in bonding.
• In PCl₃, the oxidation state of P is +3.
• In P₂O₅ also, the oxidation state of P is +5.
• Phosphorus shows the oxidation state of -3 in compounds such as phosphine, PH₃.

Hence, the three important oxidation states of Phosphorus are -3, +3 and +5.

Explanation:

Strong Reducing behavior of H₃PO₂ is due to the presence of one- OH group and two P-H bonds

H Shows reducing behavior

Explanation:

⇒ The strength of an acid is directly proportional to the stability of its conjugated bases.

The correct answer is It is an α-amino acid.

Key Points

• Glycine is an α-amino acid.
• The amino acid glycine has a single hydrogen atom in its side chain.
• It is one of the most basic amino acids found in nature.
• The molecular formula for glycine is C₂H₅NO₂.
• It contains 2 atoms of carbon, 5 atoms of hydrogen, 1 atom of nitrogen and 2 atoms of oxygen
• It is the smallest of all amino acids, and it has a side chain as well as a hydrogen molecule.
• Amino acids are required for all metabolic functions and human life processes.
• In the year 1820, glycine was extracted from a material known as gelatin.
• It was later found by Henri Braconnot, a French scientist and pharmacist.
• Glycine was created by heating gelatinous material with sulfuric acid.

Explanation-

⇒ Xenon is an inert gas. Its electronic configuration is [Kr]4d¹⁰5s²5p⁶. All orbitals that are filled have paired electrons.

⇒ When one, two, or three electrons are promoted from 5p (filled) to 5d (empty) orbitals, two, four, and six half-filled orbitals are formed. 

Explanation-

Optical Isomerism

• Compounds that exhibit optical isomerism feature similar bonds but different spatial arrangements of atoms forming non-superimposable mirror images.
• These optical isomers are also known as enantiomers.
• Enantiomers differ from each other in their optical activities.
• Dextro enantiomers rotate the plane of polarized light to the right whereas laevo enantiomers rotate it to the left, as illustrated below.
• 

Given data and Analysis-

Both [Co(en)3]CI3, and cis - [Co(en)2CI2]CI have optical isomers as mentioned in the image.

Explanation-

S_N2 reaction-

⇒ The S_N2 reaction mechanism involves the nucleophilic substitution reaction of the leaving group (which generally consists of halide groups or other electron-withdrawing groups) with a nucleophile in a given organic compound.

⇒The rate-determining step of this reaction depends on the interaction between the two species, namely the nucleophile and the organic compound.

⇒ There are two ways in which the nucleophile can attack the stereocenter of the substrate:

• A frontside attack where the nucleophile attacks from the same side where the leaving group is present, resulting in the retention of stereochemical configuration in the product.
• A backside attack where the nucleophile attacks the stereocenter from the opposite side of the carbon-leaving group bond, resulting in inversion of stereochemical configuration in the product.

⇒ Since purely S_N2 reactions show 100% inversion in stereochemical configuration, it is clear that these Reactions occur through a backside attack.

⇒ Thus, the nucleophile displaces the leaving group in the given substrates. It can be noted that primary and secondary substrates can take part in S_N2 reactions whereas tertiary substrates can not.

⇒ Reactivity in S_N​² reactions depends upon the steric hindrance.

⇒ Since the steric hindrance increases in the order: CH_3​X>1^∘>2^∘>3°, therefore, reactivity in SN² reactions decreases in the same order, i.e., option (c) is correct.

Explanation:

⇒ The reaction is known as the bromination reaction, which is an example of an electrophilic substitution reaction.

⇒ In this reaction, Br₂ and Fe react with each other and form Br⁺ (bromonium ion) which is an electrophile and CCl₃ is an electron-withdrawing group as well as an m-directing group so the bromine ion will be substituted on meta position.

Explanation:

⇒ Halogenation of Aromatic Hydrocarbons.

⇒ NaI + AgNO₃ → AgI + NaNO₃

Where AgI is (Yellow ppt)

⇒ CH₂I bond is weak so it will break easily and with ANO₃ solution NaI give a yellow precipitate of AgI

Explanation:

Let's know first the structure of m-cresol(meta cresol)

As the CH₃ group is attached to the meta position it is called m-cresol, if CH₃ is attached to the ortho position it is called ortho cresol.

For the IUPAC name, we have to number the carbons giving priority to the functional group

As the OH group is attached to the benzene ring we will call it phenol and at the 3^rd position we have methyl group, So it becomes,

3-methylphenol.

Hence option 1 is correct.

Explanation:-

Reimer-Tiemann Reaction:

• The Reimer-Tiemann reaction converts phenol to salicylic acid.
• When phenol is burned with carbon tetrachloride in the presence of aq. NaOH at 34)K, the product generated is salicylic acid, which may then be hydrolyzed with dil. HCl.
• The attack of dichlorocarbene on the benzene ring is involved.
• Salicyaldehyde is generated when phenol is treated with chloroform in an aqueous sodium hydroxide solution and then hydrolyzed with acid.

Reaction:

Explanation:

⇒ C₂H₅Br (ethyl bromide) \(\xrightarrow{Aqueous\ NaOH}\) CH₃CH₂OH (ethyl alcohol) [Compund - A]

⇒ CH3CH2OH \(\xrightarrow[-1/2H_2]{Na}\) CH₃CH₂ONa (sodium ethoxide) [Compund - B]

⇒ CH3CH2ONa \(\xrightarrow{CH_3l}\) CH₃-CH₂-O-CH₃ (ethylmethyl ether) [Compund - C]

• Hydrogen gas is evolved when ethanol reacts with sodium.
• 2Na + 2CH₃CH₂OH 🡪 2CH₃CH₂O–Na (Sodium ethoxide) + H₂ (g)

• Alcohols react with sodium leading to the evolution of hydrogen.
• With ethanol, the other product is sodium ethoxide.
• Ethanol:-
  • Ethanol is a liquid at room temperature.
  • Ethanol is commonly called alcohol and is the active ingredient of all alcoholic drinks.
  • In addition, because it is a good solvent, it is also used in medicines such as tincture iodine, cough syrups, and many tonics.
  • Ethanol is also soluble in water in all proportions.

Explanation:

⇒ When benzoic acid reacts with soda lime we get benzene and CO₂ is released, this reaction is known as the decarboxylation reaction and hence it does not give hydrocarbon.

Additional Information 

Explanation:

⇒ When salicylic acid is subjected to bromination using Br₂ water CO₂ is released and on ortho and para position of phenol will be replaced by Br forming tribromophenol.

Explanation:

• NO₂ on benzoic acid shows two effects -m and -I.
• On meta-position, the mesomeric (-m) effect has no significance, so on the ortho position nitro group shows -m,-I effect, on meta-position -I effect and on para position -m, -I effect.
• Ortho-nitro benzoic acid shows an ortho effect means NO₂ is present at the ortho position which is closer to COOH and is an electron-withdrawing group and will pull electrons towards itself, O^- has less charge so it will easily release H⁺ hence it increases its acidity. The Ortho effect forces the carbonyl group to change its polarity due to which resonance decreases and is less stable which increases acidity.
• Also para-Nitro benzoic acid has increased acidic strength as NO₂ at the para position will withdraw electrons but will be less than that of ortho nitro benzoic acid. 
• whereas meta benzoic acid shows only -I effect hence its acid strength is less than that of o-nitro and p-nitro.

So the order will be

• O-Nitro < P-Nitro < M-Nitro.

Concept:

• LiAlH₄ is used as a reducing agent in organic synthesis, especially for the reduction of esters, carboxylic acids, and amides.
• The solid is dangerously reactive toward water, releasing gaseous hydrogen (H₂).
• LiAlH₄ does not reduce double bond (C = C bond) but it reduces ester into alcohol.
• LiAlH₄ reduces a C-C double bond which is in conjugation i.e. resonance.
• LiAlH₄ reduces double bond in case of only when it is conjugation with phenyl ring but “cinnamic acid” is an exception because it has both carbonyl carbon as well as phenyl ring in its conjugation.

Here, it reduces an ester given above into alcohol.

\({\rm{C}}{{\rm{H}}_3}{\rm{CH}} = {\rm{CH}} - {\rm{C}}{{\rm{O}}_2}{\rm{C}}{{\rm{H}}_3} \xrightarrow[]{\text{LiAlH}_4}{\rm{C}}{{\rm{H}}_3}{\rm{CH}} = {\rm{CHC}}{{\rm{H}}_2}{\rm{OH}}\)

Explanation:

⇒ Alkoxymercuration-demercuration is a reaction in which an alkene (a molecule with a carbon-carbon double bond) reacts with an alcohol in the presence of mercuric acetate to form an alkoxymercury intermediate, which is then reduced with sodium borohydride to produce an ether.

⇒ Hence it does not involve carbocation as the intermediate.

Explanation:

By acetal formation

⇒ Acetals are commonly used to protect the carbonyl groups of aldehydes and ketones from basic, nucleophilic reagents. Once the protection is no longer needed, the acetal protecting group is easily removed, and the carbonyl group re-exposed, by treatment with dilute aqueous acid.

⇒ Aldehyde group can be protected by mercaptal formation against attack by acidic oxidising agents 

Additional Information

⇒ The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments.

⇒ As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general.

⇒ Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) Metallo-hydride, alkyl, and aryl reagents.

⇒ If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.

Explanation:

Cannizzaro reaction: 

→ The Cannizzaro reaction is a disproportionation process in which two molecules of an aldehyde combine with a hydroxide base to generate primary alcohol and a carboxylic acid.

Crossed Cannizzaro reaction :

→ When a mixture of formaldehyde and other aldehydes which has no α-hydrogen atom is treated with concentrated. alkali, formaldehyde is oxidized to carboxylic acid and another aldehyde is reduced to an alcohol. This reaction is crossed Cannizzaro's reaction.  

→ So the condition for crossed Cannizzaro reaction is that α-hydrogen must not be present. 

Option 1

Here we have acetyl acetaldehyde and formaldehyde but acetaldehyde possesses α-hydrogen hence crossed Cannizzaro's reaction is not possible.

Option 2

Here we have benzaldehyde and acetaldehyde where benzaldehyde does not have α- hydrogen but acetaldehyde has one hence crossed Cannizzaro's reaction is not possible.

Option 3

Here we have benzaldehyde and formaldehyde both are different aldehydes with no α-hydrogen and in presence of a strong base like conc NaOH forms sodium metheonic acid and benzyl alcohol hence it is a crossed Cannizzaro reaction.

Explanation:

Option 1

→ Cannizzaro reaction is a chemical reaction named after Stanislao Cannizzaro that involves the base-induced disproportionation of two molecules of a non-enolizable aldehyde to yield a carboxylic acid and a primary alcohol.

Option 2

→ A proton is exchanged by acid and alkoxide ions. When a base of high concentration is introduced, the aldehyde forms an anion which has a charge of 2. From this, a hydride ion is transferred to a second molecule of the aldehyde, forming carboxylate and alkoxide ions. The alkoxide ion also obtains a proton from the solvent for the reaction.

Option 3

→ Cannizzaro's reaction is not given by all the carbonyl compounds.

Option 4

→ A 50% aqueous or ethanolic solution is generally used in the Cannizaro reaction as it is used for aldehydes that do not have α-H. 

Concept:

Carbylamine reaction:

• Primary amines when treated with chloroform and ethanolic KOH gives carbylamines or isocyanides.
• The reaction occurs via carbene intermediate,
• Chloroform reacts with KOH to give chloro carbene: CCl₂.

The general form of the reaction is:

​

For example,

Explanation:

Step 1: Chloroform reacts with KOH to give chloro carbene: CCl2.

Step 2: The chlocarbene created in the reaction adds to the primary amines.

• Intramolecular hydride shift takes place.

Step:3 Two moles of HCl are eliminated to give the isocyanide.

1) 

2) 

3) 

Hence, the product of the reaction RNH2 + CHCl3 + KOH \(\mathop \to \limits^\Delta \)is RCN.

Important Points

• Only primary amines give this test.
• This test can be used to distinguish between primary and secondary amines. 

Concept:

• Pure amines are generally colourless, have a fishy smell, and are liquids in nature.
• The chemical reactivity of amines is mainly due to the presence of a lone pair of electrons on the N-atom due to which amines act as nucleophiles or bases.
• A nucleophile is a species that attacks an electron-deficient carbon and a base that attacks an electron-deficient atom.
• In aromatic amines, the lone pair of electrons on nitrogen activates the aromatic ring towards an electrophilic substitution reaction.
• Some reactions of amines are due to basic nature, alkylation, acylation, benzoylation, reaction with Hinsberg reagent, carbylamine reaction, etc.

Explanation:

Acylation of amines:

• Primary and secondary amines containing replaceable hydrogen atoms react with acid chlorides and acid anhydrides to form substituted amides.

• Tertiary amines do not contain replaceable hydrogen atoms so these do not undergo reaction with an acid chloride or acid anhydride.
• The acid generated during the reaction can protonate the amine and destroy its nucleophilic character. Hence, a base is used.
• Unlike in the case of the alkylation reaction of amines, the amide formed does not react further with organic halide or acid anhydride because the amide is non-basic and poor nucleophile due to the electron-withdrawing effect of the -CO-R group.
• Acyl chlorides are stronger acylating agents than anhydrides and esters.

Hence, when primary amines are treated with acid chlorides we get amides.

Additional Information

• Carboxylic acids do not form amides with amines. they only form salts.
• In the case of acylation, the presence of an H atom on Nitrogen is necessary in addition to the nucleophilic character of amines. 

Explanation-

Anomers-

An anomer is actually an epimer (also a cyclic saccharide) that differs in configuration, particularly at the acetal or hemiacetal carbon.

• An anomer is a kind of stereoisomer.
• The anomers are saccharides or glycosides that are epimers, which are distinct from each other in the configuration at C-2.
• If they are ketoses, or in the configuration of C-1, if they are aldoses.

Methyl α-D-glucoside and methyl β- D-glucoside are epimers and anomers because both differ in configuration only around C1​ and C1​ -carbon is called anomeric carbon or the glycosidic carbon.

 Important Points

Enantiomers-

⇒ Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other. 

Explanation-

Hemoglobin-

• It is a globular protein found in the RBCs, which are responsible for the transport of oxygen from the lungs to the entire body.
• It is a tetrameric protein made up of two alpha and two beta subunits. Each of the four chains is associated with an iron or heme group which binds with one oxygen molecule each.
• It carries oxygen from the lungs to the rest of the body.
• It shows a sigmoid curve in a graph of oxygen dissociation.
• It is a storage protein.
• Haemoglobin acts as an oxygen carrier in the blood because four Fe²⁺ ions of each haemoglobin can bind with four molecules of O_2​ and form oxyhaemoglobin.
• 4Hb + 4O₂​ → Hb_4​O₈ (Oxy-haemoglobin)​​

Explanation-

Peptide bonds-

• A peptide bond also sometimes called eupeptide bond is a chemical bond that is formed by joining the carboxyl group of one amino acid to the amino group of another. 
• A peptide bond is basically an amide-type of the covalent chemical bond. This bond links two consecutive alpha-amino acids from C1 (carbon number one) of one alpha-amino acid and N2 (nitrogen number two) of another.
• This linkage is found along a peptide or protein chain.

Given data and Analysis-

⇒ A peptide bond is a chemical bond formed between two molecules when a carboxylic group of one molecule reacts with an amino group of the other to form, thus releasing a water molecule. 

⇒ The CO-NH bond is called a peptide bond.

⇒ In the given peptide chain the number of peptide bonds is 2 (shown in the above image by a red mark).

Explanation-

• Starch is a polysaccharide produced by most of plants for the purpose of storage of energy.
• It is made up of several glucose units that are bound by glycosidic bonds.
• Starch consists of two types of molecules: branched amylopectin and helical amylose.
• Amylopectin comprises more mass of starch near about 75-80% and the rest belongs to amylose.
• To check for the presence of starch iodine solution is used as it instantly shows colour in the presence of iodine.
• Amylose present in starch is the reason for the change in its colour to blue.
• Iodine is not soluble in water so it is dissolved along with potassium iodide which forms a triiodide molecule which enters the helical structure of amylose and changes the colour of starch.
• Potassium iodide and Iodine solution are yellowish-orange and when it is dropped on any substance having starch then it converts to a blue-black solution indicating the presence of starch.