APPGCET Model Chemistry Question Paper-1

APPGCET Chemical Sciences Sample Paper-1 with Answers

We have prepared a model question paper-1 specifically for the Andhra Pradesh Post Graduate Common Entrance Exam. It is designed to help you practice and enhance your skills for the entrance test. By thoroughly practicing this question paper, you can aim to achieve a high rank and excel in the exam. Make sure to dedicate sufficient time and effort to master the content and format of the test.

What else we need to know about APPGCET exam?

APPGCET is a computer-based examination comprising 100 multiple-choice questions, each carrying one mark, resulting in a total of 100 marks. 

The test duration is 90 minutes, translating to 54 seconds per question. This time constraint poses a challenge, requiring quick comprehension and accurate answers. 

Notably, there is no negative marking for incorrect responses or unanswered questions. 

To qualify, candidates must get a minimum of 35% of the maximum marks, equivalent to 35 marks. 

The test includes various types of objective questions, such as analogies, classification, matching, and comprehension, drawn from the syllabus.

 

APPGCET Chemical Sciences Sample Paper-1 

1. Which of the following cycloalkanes is most stable?

1. Cyclobutane
2. Cyclohexane
3. Cyclopropane
4. Cyclopentane

Answer: Cyclohexane
Explanation: According to Baeyer's strain theory, angle strain values were calculated for cyclopropane, cyclobutane, cyclopentane, and cyclohexane, resulting in values of 24⁰.44’, 9⁰.44’, 0⁰.44’, and -5⁰.16’, respectively. These values represent the internal strain experienced by the carbon bonds in each cycloalkane. Baeyer initially assumed that cyclopentane, with its close-to-zero angle strain value, would be the most stable among them. However, experimental evidence later demonstrated that cyclohexane, which is strain-free and as stable as a chain alkane, is the most stable cycloalkane.

2. Camphor is ____________

1. A polymer
2. A sulfa drug
3. An amino acid
4. A bicyclic monoterpene

Answer: A bicyclic monoterpene
Explanation: Camphor has a molecular formula of C₁₀H₁₆O, which corresponds to (C₅H₈)₂O. This indicates that camphor consists of two isoprene units, making it a monoterpene. Additionally, the molecular formula of the corresponding saturated hydrocarbon is C₁₀H₁₈ (i.e. C_nH_2n-2), which suggests that camphor is bicyclic and contains two cyclic rings in its structure. In summary, camphor is a bicyclic monoterpene ketone.

3. Which of the given reactions exhibits Walden Inversion?

1. Bromination of alkene
2. Condensation reaction
3. SN² reaction
4. SN¹ reaction

Answer: SN² reaction

Explanation: Walden Inversion refers to the phenomenon where the configuration of a molecule undergoes a reversal during a chemical reaction. As a molecule typically exists in two enantiomeric forms around a chiral center, Walden inversion transforms it from one enantiomer to the other.

This reversal of configuration commonly occurs during nucleophilic substitution reactions, particularly through an SN² mechanism. In such reactions, the nucleophile can approach the stereocenter from either the front or the back direction

4. Nicotine, a toxic compound found in tobacco, falls within a category of organic compounds known as ______________

1. Narcotics
2. Carbohydrates
3. Alkaloids
4. Terpenoids

Answer: Alkaloids
Explanation:  Alkaloids are organic compounds of plant origin that contain basic nitrogen. The primary source of Nicotine is the tobacco plant, from which it is extracted as a salt of malic acid or citric acid. By further acidification and treatment with ether, the alkaloid nicotine is obtained. It has a molecular formula of C₁₀H₁₄N₂ and its IUPAC name is 3-(1-methyl-2-pyrrolidinyl) pyridine.

5. The reaction of benzene with nitric acid to form nitrobenzene is known as _______________________.

1. Nucleophilic substitution
2. Nucleophilic addition
3. Electrophilic substitution
4. Electrophilic addition

Answer: Electrophilic substitution
Explanation: Upon heating benzene with a mixture of concentrated nitric acid and concentrated sulfuric acid at 60°C, a process known as nitration occurs, leading to the formation of nitrobenzene. This reaction follows a specific mechanism, involving the following steps:
Step-1: The generation of a nitronium ion from concentrated nitric acid in the presence of concentrated sulfuric acid.
Step-2: The attack of the electrophilic nitronium ion on the benzene ring, resulting in the formation of a resonance-stabilized carbonium ion intermediate.
Step-3: The extraction of a proton from the carbonium intermediate, leading to the formation of nitrobenzene.
This substitution reaction is classified as an electrophilic substitution reaction due to the involvement of an electrophile (i.e.Nitronium ion).

6. Which of the following reaction is benzene known to readily undergo?

1. Electrophilic substitution reaction
2. Nucleophilic substitution reaction
3. Nucleophilic addition reaction
4. Electrophilic addition reaction

Answer: Electrophilic substitution reaction
Explanation: Benzene is stable due to its 6π-electron cloud. This makes benzene less reactive towards addition reactions as it would disrupt the delocalization of electrons and compromise its stability. Instead, benzene is prone to electrophilic attacks due to the presence of the delocalized electrons.

7. Diels Alder reaction is ______________

1. An electrophilic addition
2. A free radical addition
3. A nucleophilic addition
4. A cycloaddition

Answer: A cycloaddition
Explanation: The Diels-Alder reaction is a cycloaddition reaction that involves the condensation of a diene and a dienophile to form a six-membered compound. The presence of electron-withdrawing groups in the dienophile and electron-releasing groups in the conjugated diene promotes the cycloaddition reaction.

8. Which of the following is a rearrangement reaction?

1. Sandmeyer's reaction
2. Claisen reaction
3. Cannizzarro's reaction
4. Pinacole-pinacolone reaction

Answer: Pinacole-pinacolone reaction
Explanation: The pinacole-pinacolone rearrangement is a reaction that occurs in 1,2-glycols. It involves the shifting of an alkyl or aryl group from one of the glycol carbon to the other.

9. The reaction that involves the conversion of an aldehyde or ketone into a β-hydroxy ester by reacting it with an α-bromo ester is commonly referred to as ____________________

1. Wurtz reaction
2. Reformatsky reaction
3. Michael reaction
4. Diels Alder reaction

Answer: Reformatsky reaction
Explanation: It is a reaction between an α-bromo ester and a carbonyl compound (aliphatic/aromatic aldehyde or ketone) in the presence of zinc to form β-hydroxy ester

10. The reaction in which acetophenone is obtained from benzene by acylation in presence of anhydrous aluminium chloride is known as ____________________

1. Friedel Craft reaction
2. Reformatsky reaction
3. Michael reaction
4. Diels Alder reaction

Answer: Friedel Craft reaction
Explanation:  Here is the Fridel Craft reaction

                                                 AlCl₃

  C₆H₆  +  CH₃COCl -----------------→ C₆H₅COCH₃+HCl

Benzene   Acetyl chloride                           Acetophenone

11. The addition of HBr to propene in the presence of an organic peroxide leads to the formation of _______________

1. 2-Bromo propane
2. 1-Bromo propane
3. Both 1- and 2-bromo propanes
4. 1,2-Dibromo propane

Answer: 1-Bromo propane
Explanation: Khrasch's discovery revealed that the addition of HBr to unsymmetrical alkenes, such as propene, in the presence of an organic peroxide follows a different pattern than Markovnikov's rule. In this case, the negative part (Br^-) of the reagent (HBr) attaches to the carbon atom of the double bond that has more hydrogen atoms attached to it.

                                           R-O-O-R

 CH₃-CH=CH₂  + HBr -----------------→  CH₃-CH₂-CH₂Br

     1-Propene Hydrogen Bromide         1-Bromopropane(major product)

12. Dehydrohalogenation is a _____________________reaction

1. Addition
2. Condensation
3. Elimination
4. Substitution

Answer: Elimination
Explanation: When alkyl halides are reacted with an alcoholic solution of KOH, they undergo a process where a molecule of halogen acid is eliminated, resulting in the formation of alkenes. This reaction is classified as an elimination reaction since it involves the removal of HX (where X is halogen) from the alkyl halide.

 CH₃-CH₂Br  + KOH(alc.)---------→  CH₂=CH₂ + KBr + H₂O

   Ethyl Bromide                                       Ethylene

13. Which of the following compounds form only a mono-substituted derivative on bromination?

1. Propane
2. n-butane
3. 2-methyl butane
4. 2,2-dimethyl propane

Answer: Propane
Explanation: The bromination of propane yield a mixture of 1-bromo propane and 2-bromopropane. Both are mono-substituted bromo derivatives.

                                    h𝜈 

CH₃-CH₂-CH₃ + Br₂ ----------→ CH₃-CH₂-CH₂-Br + CH₃-CH(Br)-CH₃

  Propane                                      1-Bromopropane              2-Bromopropane

14. The formation of 2-butene from 2-butanol is an example of _________________

1. Hund's rule
2. Huckel's rule
3. Markonikov's rule
4. Seytziff's rule

Answer: Seytziff's rule
Explanation An alcohol when heated in the presence of Con. H₂SO₄, a molecule of water is eliminated and an alekene is formed. According to Satzeff's rule, the hydrogen atom is preferentially eliminated from the carbon atom with fewer number of hydrogen atoms.

                                         Conc. H₂SO₄/Δ

CH₃-CH₂-CH(OH)-CH₃ ----------------→ CH₃-CH=CH-CH₃

     2-Butanol                                                      2-Butene

15. Hydroboration of CH₃-CH=CH₂ gives___________________

1. Propan-1-ol
2. Propan-2-ol
3. Both 1- and 2- propanols
4. 1,2-propane diol

Answer: Propan-1-ol
Explanation When Propene reacts with Diborane, it produces Tri-n-propyl borane. Subsequently, upon oxidation with alkaline hydrogen peroxide, it yields 1-propanol.

16. The number of possible stereoisomers for 2,3-dibromobutane are_________________

1. 2
2. 3
3. 4
4. 5

Answer: 3
Explanation One stereoisomer of 2,3-dibromobutane is superimposable on its mirror image,resulting in a total of three possible stereoisomers instead of four.Three stereoisomers are formed by 2,3-dibromobutane, namely (S,S), (R,R), and (R,S)-Meso.

17. Isomeric monosaccharides which differ in their stereochemistry at α-carbon atom are known as_________________

1. Enantiomers
2. Diastereomers
3. Epimers
4. Anomers

Answer: Epimers
Explanation D-glucose and D-galactose serve as excellent examples of epimers, which are stereoisomers that differ in configuration at only one chiral carbon atom.

18. The reagent used to convert acetophenone to ethyl benzene is_______________________

1. CH₃CH₂MgX
2. LiAlH₄
3. H₂N-NH₂
4. Zn(Hg). HCl

Answer: Zn(Hg).HCl
Explanation The reduction of ketones or aldehydes to alkanes using zinc amalgam and concentrated hydrochloric acid is commonly referred to as Clemmensen's reduction.

                                       Zn/Hg

C₆H₅COCH₃ + 4H --------------------------→  C₆H₅CH₂CH₃

                                   Conc. HCl               

 Acetophenone                                                  Ethyl Benzene

19. Which of the following do not respond to 'Iodoform' test?

1. Acetaldehyde
2. Acetophenone
3. Methylethyl ketone
4. Benzaldehyde

Answer: Benzaldehyde

Explanation: Benzaldehyde does not give a positive result in the iodoform test due to the absence of an acetyl (CH₃-C=O) group.

20. p-nitrophenol is more soluble in water than o-nitrophenol because

1. p-nitrophenol is less volatile than o-nitrophenol
2. Of the presence of intermolecular hydrogen bonding in p-nitrophenol
3. Of the presence of intermolecular hydrogen bonding between molecules of p-nitrophenol and water
4. Of the presence of intramolecular hydrogen bonding in p-nitrophenol

Answer: Of the presence of intermolecular hydrogen bonding between molecules of p-nitrophenol and water

Explanation: The solubility of both ortho- and para-nitrophenols in water is attributed to their ability to form hydrogen bonds with water molecules.

In ortho-nitrophenol, the nitro group is positioned in a way that allows for intramolecular hydrogen bonding. On the other hand, in para-nitrophenol, the nitro group is situated in a manner that facilitates intermolecular hydrogen bonding.

As a result, p-nitrophenol exhibits higher solubility in water compared to o-nitrophenol.

21. The conversion of D-Glucose into D-Arabinose is known as _____________ process

1. Kiliani-Fischer synthesis
2. Epimerisation
3. Ruff's degradation
4. Fischer synthesis

Answer: Ruff's degradation

Explanation: In 1898, Otto Ruff made a significant discovery in synthetic chemistry that involved the reduction of carbon chains in monosaccharides. This method, known as Ruff's degradation, was named after him.

A notable example of this degradation method is the conversion of aldohexose (D-Glucose) to aldo-pentose (D-Arabinose). This transformation is achieved by oxidizing the aldohexose with Bromine water and subsequently treating it with ferric acetate and 30% of hydrogen peroxide.

                       1.Br₂/H₂O  2.  Ca(OH)₂

Aldohexose -----------------------------------→ Aldo-pentose

                      1.Fe(CH3COO)₃  2.  30% H₂O₂     

 (n-carbon atoms)                                           (n-1) carbon atoms

22. Which of the following phenols is a strongest acid?

1. Phenol
2. 2-Nitrophenol
3. 2,4-Dinitrophenol
4. 2,4,6-Trinitrophenol

Answer: 2,4,6-Trinitrophenol
Explanation: The acidity of picric acid is higher than that of phenol due to the presence of electron-withdrawing nitro groups in both the ortho and para positions. Greater acidic character is observed when there are more electron-withdrawing nitro groups present.

23. The IUPAC name of  CH₃-CH₂-CH-CH₂-CH₂-CH₃

                                                          ⃒

                                                         CH₃

1. 3-Methyl hexane
2. 2-Methyl hexane
3. 2-Ethyl pentane
4. 2-Propyl butane

Answer: 3-Methyl hexane
Explanation:  According to the IUPAC nomenclature, the long continuous carbon chain is numbered by starting from the end closest to the substituent.

24. The IUPAC name of  CH₃-CH-CH₂-CH₂-Cl

                                                 ⃒

                                                 CH₃

1. 1-Chloro 3-methyl butane
2. 2-Methyl 4-chloro butane
3. 3-Methyl 1-chloro butane
4. 1-Chloro pentane

Answer: 1-Chloro 3-methyl butane
Explanation: According to the IUPAC nomenclature, the long continuous carbon chain should be numbered in a way that assigns the lowest possible number to the carbon atom carrying the halogen.

25. The reaction of acetaldehyde with sodiumbisulhite is an example of _________________

1. Electrophillic substitution
2. Nucleophillic substitution
3. Nucleophillic addition
4. Condensation reaction

Answer: Nucleophillic addition
Explanation:  The higher electronegativity of oxygen compared to carbon results in a partial positive charge on the carbon atom of the C=O group in aldehydes and ketones. This partial positive charge makes them highly reactive towards nucleophilic addition reactions.

26. Which statement accurately describes the C-H bond energies in ethane, ethylene, and acetylene?

1. Greatest in acetylene
2. Greatest in etahne
3. Equal in all
4. Least in ethane

Answer: Greatest in acetylene

Explanation:

Bond energy tends to rise with increasing bond multiplicity and decreasing bond length. Comparing acetylene, ethylene, and ethane, acetylene has the highest bond order (3), followed by ethylene (2), and ethane (1). Additionally, the C-H bond length in acetylene is the shortest which is 105.7 pm. As a result, acetylene possesses the highest bond energy than the other. Whereas, Ethane's C-H bond length is 109.3 pm, and ethylene's is 108.7 pm.

27. Which of the following compounds is the most acidic?

1. Chloromethane
2. Methanol
3. Ethene
4. Ethyne

Answer: Methanol

Explanation:

The compound's acidity is determined by the electronegativity of the atom to which the hydrogen atom is bonded. In the case of methanol, the C-O bond is the most polar because oxygen exhibits a higher electronegativity compared to chlorine in methyl chloride. A more polar bond corresponds to a stronger acid.

28. Which of the following compounds exhibit geometrical isomerism?

1. Butene
2. Ethane
3. Propane
4. Pentane

Answer: Butene

Explanation:

Geometrical isomerism in a molecule becomes possible when it contains a double bond, and the atoms or groups connected to each carbon of the double bond are different. Among the provided options, butene fulfills all the essential criteria required for displaying geometrical isomerism.

29. The homolytic cleavage of the C-C bond gives ___________

1. Carbonium ion
2. Carbanion
3. A free radical
4. A neutral carbon atom

Answer: A free radical

Explanation:

When a shared pair of electrons is evenly split between two bonded atoms, the bond cleavage is termed homolytic bond fission. This process results in the creation of two neutral particles, each carrying one unpaired electron, and they are referred to as free radicals.

30. Which of the following is a lyophobic colloid?

1. Proteins
2. Rubber
3. Gelatin
4. Gold

Answer: Gold

Explanation:

In this type of colloidal system, the dispersed phase does not have an affinity for the dispersion medium so the colloid is not readily formed. Examples of lyophobic colloids are metals and their insoluble compounds such as sulphides.

31. Determine the osmotic pressure in atmospheres (atm) for a 5% glucose solution at 18 degrees Celsius.

1. 6.636
2. 5.231
3. 8.327
4. 8.415

Answer: 6.636

Explanation:

The formula to determine the osmotic pressure of solution (π) = nRT

Where, n = total number of moles of the substance

R= Gas constant

T = Absolute temperature in Kelvin

We have, n = ^W⁄_mv

The Weight of Glucose = 5 gm

Molecular weight of Glucose = 180

(C₆H₁₂O₆ = 12x6+1x12+6x16 =180)

Volume of Glucose in the solution = 0.1 lit

Number of moles of Glucose in the solution = ⁵⁄_{180 X 0.1}

n = 0.278

π = 0.278 gm/lit X 0.0821 lit atm degree^-1mole^-1 X 291K

[ R = 0.0821 lit atm degree^-1mole^-1
T = 273+18 = 291 Kelvin]

π = 6.636 atm

32. Which of the following is a macromolecule?

1. Decane (C₁₀H₂₂)
2. Copper Sulphate (CuSO₄)
3. Silica (SiO₂)
4. Poly vinyl chloride (H₂C-CHCl)_n

Answer: Poly vinyl chloride

Explanation:

Among all the given options, Polyvinyl chloride (PVC) is a polymer. It means n number of small-sized molecules of Vinyl chloride polymerizes in the presence of organic peroxide to form a giant PVC. Hence, it is a macromolecule.

33. Which of the following is the unit of a first-order rate constant?

1. Sec^-1
2. MolesL^-1sec^-1
3. Moles^-2Lsec^-1
4. Moles^-1Lsec^-1

Answer: Sec^-1

Explanation:

Since the first-order rate constant (k₁) varies inversely with the time (t), it's unit is sec^-1

34. The reaction between H₂ and Cl₂ in presence of sunlight is a ____________________

1. First order reaction
2. Chain reaction
3. Zero order reaction
4. Second order reaction

Answer: Zero order reaction

Explanation:

The reaction between hydrogen and chlorine in the presence of sunlight is considered zero-order because the rate of the reaction remains constant and is not influenced by the concentrations of the reactants. Over time, the concentration of products steadily increases in a linear fashion.

35. In which of the following processes, the entropy will decrease?

1. Boiling of water
2. Freezing of a liquid
3. Expansion of a gas
4. Melting of ice

Answer: Freezing of a liquid

Explanation

Entropy is an extensive property that quantifies the degree of randomness in a system. When we examine the freezing of a liquid, we observe that the system releases heat energy to its surroundings, all while its temperature remains constant. And the liquid transforms into a more structured, solid state, which reduces disturbance in the system. As a result, the entropy decreases.

36. The equivalent conductance at infinite dilution is equal to________________

1. The sum of the ionic conductances of cation and anion
2. The sum of the transport numbers of the cation and anion
3. The sum of the specific conductivity and cell constant
4. The sum of equivalent and specific condctivities

Answer: The sum of the ionic conductances of cation and anion

Explanation

According to Kohlrausch's law, at infinite dilution, when the dissociation is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the other ion it's paired with. The value of equivalent conductance at infinite dilution for any electrolyte is equal to the sum of the ionic conductance of cation and anion.

37. From the values of bond energies it is possible to calculate__________________

1. Heat of solution
2. Heat of dilution
3. Heat of formation
4. Heat of neutralisation

Answer: Heat of formation

Explanation:

Bond energy represents the energy released when one mole of a specific bond forms between the corresponding atoms or radicals. 

For instance, the bond energy of an O-H bond is 33 Kcal/mole, signifying that the formation of one O-H bond releases 33 Kcal of energy. Consequently, during the formation of one mole of an H₂O molecule (which comprises two O-H bonds), 66 Kcal of energy is liberated. The standard enthalpy of formation of H₂O is found to be -68.4 Kcal/mole. 

This remarkable consistency between theoretical and experimental assessments demonstrates that bond energy values can be employed to compute the heat of formation of a substance from its constituent elements.

38. The definition of phosphorescence is ________________

1. S₀→S₁
2. S₁→S₀
3. T₁→S₀
4. S₁→T₁

Answer: T₁→S₀

Explanation:

Phosphorescence is characterized by a molecule transitioning from the triplet state (T₁) to the ground singlet state (S₀), resulting in the release of energy in the form of either heat or light radiation. The emission of energy during this transition occurs over a relatively extended period, ranging from 10^-3 to 10¹ seconds, making phosphorescence a notably slow process.

39. Which of the following electrolyte is used in the Weston standard cell?

1. CdSO₄.8H₂O
2. CuSO₄.5H₂O
3. ZnSO₄.7H₂O
4. Saturated KCl

Answer: CdSO₄.8H₂O

Explanation:

The Weston cell, also known as the Weston standard cell, is a wet-chemical cell known for its exceptional voltage stability.
It consists of an anode composed of a cadmium-mercury amalgam, and a cathode made of pure mercury covered with a paste containing Mercurous sulfate and additional mercury.
The electrolyte used is a saturated solution of cadmium sulfate (CdSO4.8H2O), while the depolarizer is a paste of Mercurous sulfate.
Electrical connections to the cadmium amalgam and mercury are established through platinum wires.

40. The variation of heat of reaction with temperature is called _______________

1. Classius-Clapeyron equation
2. Hesss's equation
3. Kirchhoff's equation
4. Carnot's theorem

Answer: Kirchhoff's equation

Explanation:

According to Kirchoff's equation, the change in enthalpy for a reaction at constant pressure per degree of temperature change is equal to the difference in the heat capacities between the products and the reactants at constant pressure. ΔC_p = ^{ΔH₂ -ΔH₁} ⁄ _T2 - T1

41. Which of the following expression gives the hydrolysis constant of the salt of a weak acid and strong base?

1. K_h = ^K_w⁄_Ka.Kb
2. K_h = ^K_w⁄_Ka
3. K_h = ^K_w⁄_Kb
4. K_h = ^K_w.K_a⁄_Kb

Answer: K_h = ^K_w⁄_Ka

Explanation:

K_h = ^K_w⁄_Ka
Where, K_h = Hydrolysis constant
K_w = Ionic product of water and its value is 1.008 x 10^-14 at 25^oC
K_a = Dissociation constant of acid
The inversely proportional relationship between K_h and K_a demonstrates that when an acid is weaker (with a small K_a value), it exhibits a higher degree of hydrolysis (with a higher K_h value).

42. The Joule-Thomson coefficient for an ideal gas is ___________________

1. Zero
2. Positive
3. Negative
4. Either positive or negative

Answer: Zero

Explanation:

Under the conditions where an ideal gas follows all gas laws regardless of temperature and pressure, with its molecules exerting no mutual attractive forces, the Joule-Thomson coefficient for an ideal gas is zero.

43. Which of the following statement best defines the inversion temperature?

1. The temperature at which a gas liquifies
2. The temperature to which a gas should be cooled before applying pressure
3. The temperature above which the liquid cannot exist
4. The tepmerature at which both gas and liquid can coexist

Answer: The temperature to which a gas should be cooled before applying pressure

Explanation:

The inversion temperature is the temperature at which the sign of the Joule-Thomson coefficient changes. At this temperature, the value of Joule-Thomson coefficient is zero.
It's important to note that for a gas to be cooled through Joule-Thomson expansion, the gas must initially be brought to a temperature below its inversion temperature and then expanded adiabatically through a porous plug from a higher to lower pressure. As the temperature decreases during the experiment, the extent of cooling increases.

44. What do you mean by Gold number?

1. The number of gold atoms present in a standard gold sol
2. The minimum amount of an electrolyte which should be added to a standard gold sol which just prevents its coagulation
3. The minimum amount of gold chloride which should be dispersed in water to give a stable gold sol
4. The number of moles of a protective colloid which should be added to a standard gold sol to prevent its coagulation

Answer: The number of moles of a protective colloid which should be added to a standard gold sol to prevent its coagulation

Explanation:

The Gold number is the quantity of protective colloid, measured in milligrams, needed to prevent the coagulation of a 10 ml standard gold solution when 1 ml of a 10% sodium chloride solution is added to it.
In this context, coagulation of gold sol is observed through a color shift from red to blue as the particle size increases.
A higher Gold number signifies weaker protective abilities of the lyophilic colloid because it implies a larger amount is necessary to prevent coagulation.

45. A photovoltaic cell consists of which of the following semiconductor junctions that exposed to sunlight

1. A juction of two p-type semiconductors
2. A junction of a metal and a n-type semiconductor
3. A junction of a n-and p-type semiconductors
4. A junction of two n-type semiconductors

Answer: A junction of a n-and p-type semiconductors

Explanation: 

The photovoltaic effect refers to the phenomenon where voltage or electric current is generated in a photovoltaic cell upon exposure to sunlight. These cells are typically constructed using two distinct types of semiconductors—a p-type and an n-type—combined to form a p-n junction.

This junction creates an electric field, prompting electrons to migrate to the positive p-side and holes to the negative n-side. Consequently, negatively charged particles move in one direction while positively charged particles move in the opposite direction.

When photons of light, which are packets of electromagnetic radiation, strike these cells, they transfer energy to electrons within the semiconductor material.

This energy causes the electrons to transition to a higher energy state known as the conduction band. In this excited state, the electrons become mobile within the material, resulting in the generation of an electric current within the cell.

46. What is a space lattice in a crystal?

1. An arrangement of molecules/atoms/ions in a three dimensional space in a random manner
2. The repeating pattern of molecules/atoms/ions in a crystal structure
3. It is a three-dimensional collection of points (molecules/atoms/ions) which have identical environment
4. It is the simplest unit or a building block of entire crystal

Answer: It is a three-dimensional collection of points (molecules/atoms/ions) which have identical environment

Explanation:

A space lattice is an arrangement of an infinite set of points within a three-dimensional space that defines the positioning of fundamental units, such as atoms, ions, or molecules, within a crystal structure.

47. Which of the following shows the Bragg's equation?

1. nλ = 2 sinθ
2. λ = 2d sinθ
3. nλ = 2d sinθ
4. λ = ⁿ⁄_2d sinθ

Answer: nλ = 2d sinθ

Explanation:

Bragg's law states that when X-rays are incident on a crystal composed of regularly spaced atomic planes, they penetrate into the crystal and strike the atoms in different planes. X-rays are then reflected from these planes.
This principle led to the development of an equation, known as Bragg's equation, which establishes a relationship between the wavelengths of the X-rays (λ), the spacing between successive planes (d), and the angle of incidence (θ).

48. The defects caused by missing or misplacing atoms are called ________________

1. Interstitial defects
2. Point defects
3. Schotky defects
4. Frenkel defects

Answer: Point defects

Explanation:

Point defects within a crystal emerge as a result of the absence of certain ions at specific lattice sites or the occupation of vacant lattice sites or interstitial positions.
These defects are categorized into two types: (a) Stoichiometric defects and (b) Non-stoichiometric defects.

49. The two effects associated with the Debye-Huckel theory are______________

1. Relaxation effect and asymmetry effect
2. Relaxation effect and mesomeric effect
3. Relaxation effect and electrophoretic effect
4. Only mesomeric effect

Answer: Relaxation effect and electrophoretic effect

Explanation:

According to the Debye-Huckel-Onsager theory of strong electrolytes, two factors influence the mobility of ions:

Relaxation effect:

Each ion in an electrolyte solution is surrounded by an atmosphere of oppositely charged ions. When electric field is not applied, the ionic atmosphere is symmetrical.

On application of the electric field the central positive ion will move towards cathode, while its negatively charged ions move towards anode. As a result, the symmetry of the ionic atmosphere is destroyed and becomes asymmetrical ionic atmosphere.

In the rear (back) of the moving ion, there will be an excess of negatively charged ions. Therefore, the central positive ion which is moving experiences a backward pull. The effect thus decreases the mobility of ion and this effect is known as “Asymmetric effect”.

It is also called relaxation effect because it takes some time to form a new ionic atmosphere.

Electrophoretic Effect: 

The ionic atmosphere around a central ion is associated with solvent molecules. When an electric field is applied, a cation moves towards the cathode, and the negatively charged ionic atmosphere with solvent molecules moves towards the anode. This counter movement retards the central ion's movement, referred to as the "Electrophoretic Effect."

50. When a gas is subjected to adiabatic expansion it gets cooled because__________

1. Fall in temperature

2. Loss of kinetic energy

3. Decrease in velocity

4. Energy lost in doing work

Answer: Fall in temperature

Explanation:

In adiabatic process, expansion takes place at the expense of internal energy. Due to this, internal energy becomes less resulting in the fall in temperature

51. The light emitted by a firefly is an example of ________________

1. Fluorescence

2. Phosphorescence

3. Photosynthesis

4. Chemiluminescence

Answer: Chemiluminescence

Explanation: 
Luminescence is the occurrence of light emission by entities other than heat. When chemical reactions result in the emission of light, this phenomenon is termed chemiluminescence.

The light of firefly can occur through the oxidation of a protein, Luciferin, by atmospheric oxygen in the presence of the enzyme Luciferase. The excited state product of the reaction emits radiation of a specific wavelength, manifesting as a visible glow.

52. Which of the following prove the wave nature of electron?

1. Uncertainty principle

2. Photoelectric effect

3. Electron diffraction

4. Zeeman effect

Answer: Electron diffraction

Explanation: 
The experimental confirmation of the wave nature of electrons was conducted by Davisson and Germer. In their experiment, an electron beam from an incandescent tungsten filament was directed at a nickel crystal within an electric field, resulting in the formation of electron diffraction rings. The similarity between these diffraction rings and the pattern observed in X-ray diffraction substantiates the wave-like characteristics of electrons.

53. If the azimuthal quantum number of an electron is '1', its magnetic quantum numbers are____________

1. 0,1,2

2. -1,0,-2

3. -1,0,+1

4. 1,2,3

Answer: -1,0,+1

Explanation: 
When subjected to an external magnetic field, electrons within a subshell align themselves into various orientations known as orbitals. Each orbital is identified by a distinct m value, where the m values span from -l to 0 to +l. The maximum value for m is given by 2l+1, where l represents the azimuthal quantum number.
For instance, if the azimuthal quantum number (l) is 1, the maximum number of m values is 3, namely -1, 0, and +1.

54. Precipitation from a saturated solution occurs when___________

1. The ionic product and solubility product are equal

2. The ionic prosuct exceeds solubility product

3. The ionic product is less than the solubility product

4. The ionic product id much lower than the solubility product

Answer: The ionic prosuct exceeds solubility product

Explanation: 
The ionic product of an electrolyte must not surpass its solubility product value, regardless of the origin of the ions.
In cases where the ionic product momentarily exceeds the solubility product due to an increase in the concentration of one or both types of ions, the solution becomes supersaturated, leading to precipitation.

55. When Sodium chloride is added to a mixture of oil and caustic soda, the soap gets precipitated because_____________

1. Nacl is an ionic compound

2. Of an increase in concentration of Na⁺ ion

3. Of the insolubility of the soap in presence of Cl^- ions

4. Of decrease in the solubility product of NaCl in presence of the soap

Answer: Of an increase in concentration of Na⁺ ion

Explanation: 

Soaps are the sodium or potassium salts derived from higher fatty acids such as stearic acid, palmitic acid, oleic acid, and others.

To produce soap, fats and oils are placed in soap kettles, and a sodium hydroxide (NaOH) solution is introduced with stirring. The mixture is then heated. After the completion of the reaction, small portions of salt (NaCl) are added gradually with stirring, a process known as salting. Through salting, pure soap separates and rises to the surface. This salting process involves a common ion effect.

NaCl, being a strong electrolyte, ionizes almost completely, leading to a high concentration of sodium ions in the solution and the suppression of NaOH dissociation.

56. The temperature at which the boundary surface between liquid and its vapour disappears is called _____________

1. Boiling point

2. Critical point

3. Melting point

4. Freezing point

Answer: Critical point

Explanation: Several gases, including carbon dioxide, ammonia, and sulfur dioxide, undergo liquefaction under conditions of low temperature and high pressure. At the critical point of liquefaction, the distinction between the liquid and gas phases vanishes, and both phases exhibit identical characteristics.

57. For a gas the deviation from ideal behaviour is maximum at___________

1. 0^oC, 1 atm

2. -10^oC, 1 atm

3. 0^oC, 5 atm

4. -10^oC, 5 atm

Answer: -10^oC, 5 atm

Explanation: 

An ideal gas is one that follows the general gas equation and other gas laws under all temperature and pressure conditions, but it is a theoretical concept and doesn't exist in reality.

In practice, all known gases are real gases, and they exhibit deviations from ideal behavior, particularly at high pressures and low temperatures, as observed by scientists like Amagat and Andrew.

A convenient method to illustrate a gas's deviation from ideal behavior is by plotting the compressibility factor against pressure.

From the given options, the gases exhibit the maximum deviation from ideal behavior at the lowest temperature of -10 degrees Celsius and the highest pressure of 5 atmospheres.

58. The correct representation of Raoult's law is___________________________

1. ^Po-P⁄_P = ⁿ⁄_n+N

2. ^Po+P⁄_P = ⁿ⁄_n+N

3. ^P-Po⁄_P = ⁿ⁄_n+N

4. ^1-Po⁄_P = ⁿ⁄_n+N

Answer: 

^P-Po⁄_P = ⁿ⁄_n+N

Explanation: For a solution containing a non-volatile solute and volatile solvent, the law can be stated as "the relative lowering in vapor pressure for a solution is equal to the mole fraction of the solute.

^P-Po⁄_P = ⁿ⁄_n+N

Where, P= Vapor pressure of pure solvent and it is constant at a particular temperature
P^O= Vapor pressure of solution
N= number of moles of the solvent
n= number of moles of the solute

59. The degrees of freedom at eutectic point are ________________

1. 0

2. 1

3. 2

4. 3

Answer: 0

Explanation:  The formula to calculate the degree of freedom of a system is;

F=C-P+2

Where, F= Degrees of freedom
C = Number of components
P= Number of phases

Let us consider two components A and B which are completely miscible in the liquid state. They do not react to form any compound in the liquid state but on solidification , they form an easy melting intimate mixture called eutectic

From the above, at the eutectic, the number of components are 2 (both A and B).

And the number of phases for the component A at the eutectic are 2 ( both liquid and solid). Similarly, for the component B, the value of P=2. The total number of phases at the eutectic for both A and B are 2+2 =4

So, F= 2-4+2=0

60. Lanthanides can be separated rapidly and effectively by____________________

1. Complex formation method

2. Solvent extraction method

3. Valency change method

4. Ion exchange method

Answer: Ion exchange method

Explanation: Lanthanides can be efficiently separated through the ion-exchange process, a two-step procedure involving:

1. Addition of the solution of the mixture of lanthanides: Introducing a solution containing lanthanide ions over a column of synthetic ion exchange resin.
2. Use of eluents: Adding the eluent to the synthetic ion exchange resin column.

It causes the following changes:
Ammonium ions displace metal ions from the resin, and the eluted lanthanide ions form a complex with citrate ions.

As the hydrated Lu³⁺ ion is the largest and least firmly bound to the resin, its citrate complex is the first to emerge from the bottom of the column during elution.

Conversely, the hydrated La³⁺ ion, being the smallest and most firmly bound, emerges last from the bottom of the column.

61. The strong field ligand is____________________

1. en

2. NH₃

3. NO₂^-

4. CN^-

Answer: CN^-

Explanation: Strong field ligands can cause electron pairing, resulting in the formation of low spin complexes that are either diamagnetic or weakly paramagnetic due to fewer availability of unpaired electrons.

The spectrochemical series provides the arrangement of ligands based on decreasing crystal field splitting energy.

CN^->NO₂^->en>NH₃.

The series indicates that CN^- is a strong field ligand.

62. The ^N⁄_P ratio of ₈O¹⁴ is____________________

1. 0.9

2. 0.75

3. 0.68

4. 0.5

Answer: 0.75

Explanation: ₈O¹⁴ has 8 protons and 6 neutrons.

The ^N⁄_P ratio of ₈O¹⁴ is ⁶⁄₈ = 0.75

63. The element belonging to (4n+2) radioactive series is _____________________

1. Th

2. U

3. Np

4. Ac

Answer: U

Explanation: The series of spontaneous changes that occur from the unstable nucleus is called radioactive disintegration series.

The known four disintegration series are given below;

1. Thorium series- (4n series)- It starts from Thorium-232 and ends at Lead-208
2. Neptunium series- (4n+1) series- It starts from Neptunium-237 and ends at Bismuth-209
3. Uranium series- (4n+2) series- It starts from Uranium-238 and ends at Lead-206
4. Actinium series- (4n+3) series- It starts from Actinium-235 and ends at Lead-207

64. The element that can have the highest oxidation number is__________________

1. N
2. O
3. Cl
4. C

Answer: Cl

Explanation:Oxidation number is defined as a number that represents an electric charge which an atom or ion has or appears to have when combined with other atoms.

With an atomic number of 7, nitrogen can exhibit oxidation states of +5, +4, +3, +2, +1, 0, -1, -2, and -3. Among these, +5 represents the highest oxidation state achievable for nitrogen.

With an atomic number of 8, oxygen can exhibit oxidation states of 2, 1, 0, -1, -2. Among these, 2 represents the highest oxidation state achievable for oxygen.

With an atomic number of 17, Chlorine can exhibit oxidation states of 6, 5, 4, 3, 2, 1, 0, -1, -2. Among these, +6 represents the highest oxidation state achievable for Chlorine.

With an atomic number of 6,carbon can exhibit oxidation states of 4, 3, 2, 1, -1, -2, -4. Among these, 4 represents the highest oxidation state achievable for carbon.

Hence, Chlorine can have the highest oxidation number from the given options above.

65. The most important air pollutants responsible for depletion of the Ozone layer are_______________

1. Oxides of nitrogen
2. Oxides of sulphur
3. Oxides of halogens
4. Chlorofluoro hydrocarbons

Answer: Chlorofluoro hydrocarbons

Explanation: The main cause of Ozone layer depletion is the wide spread of Chlorofluorocarbon. Chlorofluorocarbons (CFCs) remain stable on Earth, but as they ascend to the stratosphere, exposure to intense ultraviolet radiation causes the release of chlorine. This chlorine then interacts with ozone molecules, stripping them of an oxygen atom and converting them into nascent or atomic oxygen. This process repeats, and a single chlorine atom has the potential to destroy thousands of ozone molecules.

CF₃Cl→C^•F₃+Cl^•
O₃+Cl^•→ClO^•+O₂
ClO^•+O→Cl^•+O₂
C^•F₃+Cl^•→CF₃Cl
O₃+O→2O₂

66. Carbon monoxide, a common gaseous component of automobile exhausts is known to be a lethal poison for human beings, which causes breathlessness malfunctioning of heart, lungs etc because

1. It forms cyano complexes with biometals
2. It forms a stable carboxy haemoglobin that hinders transport of oxygen to heart and lungs
3. It directly enters the lungs and interferes with their normal functioning
4. It forms iron carbonyls with iron in the blood

Answer: It forms a stable carboxy haemoglobin that hinders transport of oxygen to heart and lungs

Explanation; Carbon monoxide is more dangerous because it combines with haemoglobin to form stable compound, called carboxy haemoglobin. Due to its formation, the transport of oxygen from lungs to the cells is restricted.

67. Which of the following ions is colourless?

1. Sc³⁺
2. Ti³⁺
3. Fe³⁺
4. V³⁺

Answer: Sc³⁺

Explanation: The color exhibited by transition elements results from d-d electron transitions that absorb energy from the visible region of the electromagnetic spectrum. Scandium, with an atomic number of 21 (1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹), achieves a stable Argon gas configuration by losing three electrons to form Sc³⁺ ions. However, as it lacks unpaired d-electrons in its valence shell, it cannot undergo d-d electron transitions, rendering it colorless.

68. The maximum magnetic moment is shown by the ion with electronic configuration

1. 3d⁸
2. 3d⁷
3. 3d⁶
4. 3d⁵

Answer: 3d⁵

Explanation: The paramagnetic strength of transition element or its ion is expressed in terms of magnetic moment μ.

μ=[n(n+2)]^1/2BM

Where, n= number of unpaired electrons
BM stands for Bohr magneton

The d⁸ electronic configuration has two unpaired electrons, results in a magnetic moment of 2.828 BM.
μ=[2(2+2)]^1/2= 2.828 BM

The d⁷ electronic configuration has three unpaired electrons, results in a magnetic moment of 3.872 BM.
μ=[3(3+2)]^1/2=3.872 BM

The d⁶electronic configuration has four unpaired electrons, results in a magnetic moment of 4.898 BM.
μ=[4(4+2)]^1/2=4.898 BM

The d⁵ electronic configuration has five unpaired electrons, results in a magnetic moment of 5.916 BM.
μ=[5(5+2)]^1/2=5.916 BM

By comparing the above magmentic moment data of various partially filled d-orbitals, we can conclude that 3d⁵ orbital possess the maximum magnetic moment.

69. The correct set of quantum numbers for the valence electrons of Rb (Z =37) is ________________

1. 5,0,0,+¹⁄₂
2. 5,1,0,+¹⁄₂
3. 5,1,1,+¹⁄₂
4. 6,0,0,+¹⁄₂

Answer: 5,0,0,+¹⁄₂

Explanation:Rubidium with an atomic number of 37 has 1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s¹ electronic configuration. The last electron enters into the 5s orbital whose principal quantum number value is 5. Also, the azimuthal and magnetic quantum numbers values for the s-orbital are 0,0. Also the spin quantum number value for the electron rotating in the clockwise direction can be +¹⁄₂. Finally, the correct set of quantum numbers for the valence electrons of Rb (Z =37) is 5,0,0,+¹⁄₂

70. The isomerism shown by the pair of compounds [Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂ is

1. Bond or linkage isomerism
2. Ionisation isomers
3. Coordination isomerism
4. Geometrical isomers

Answer: Ionisation isomers

Explanation: The ionisation isomerism is shown by such compounds which have same composition but liberate different ions in solutions. In such isomers, the position of groups within or outside the coordination sphere differs.
[Pt(NH₃)₄Cl₂]Br₂ liberates Br^- ions in the solution.Whereas, [Pt(NH₃)₄Br₂]Cl₂ liberates Cl^- ions in the solution.

71. Tl with electrton configuration [Xe] 4f¹⁴5d¹⁰6s²6p¹ belongs to III group, yet its most stable oxidation state is +1 i.e. less than the group valence. This suggests that 6s² electrons show resistance for covalent bond formation. This is explained as due to the existence of____________

1. Ionic bonds in Tl compounds
2. An inert pair effect
3. Greater affinity of p-electron for bond formation
4. A greater tendency for Tl to form an ion of smaller radius

Answer: An inert pair effect

Explanation: The inert pair effect is the reluctance of outermost shell s-electrons to engage in bonding. Thallium, a member of the IIIA group in the periodic table, can display +1 and +3 oxidation states. However, its most stable oxidation state is +1, which is two units less than the group oxidation state of +3. This is attributed to the reluctance of valence shell s-electrons to partake in bond formation, where they remain as a paired electron pair—a phenomenon known as the inert pair effect.

72. The hybridisation of Cr in [Cr(NH₃)₆]³⁺ is_____________ and shape of the complex is ______________

1. dsp³, octahedral
2. d²sp³, tetrahedral
3. dsp², square planar
4. d²sp³, octahedral

Answer: d²sp³, octahedral

Explanation: In [Cr(NH₃)₆]³⁺ complex ion,chromium is in +3 oxidation state. The electronic configuration of Cr(Z=4) is 4s¹, 3d⁵. The configuration of Cr³⁺ ion is 3d³. Thus, the ion has six vacant i.e. two 3d, one 4s, and three 4p orbitals. These orbitals get hybridised to form six d²sp³hybrid orbitals. Clearly, six pairs of electrons, one from each NH₃ molecules occupy the six vacant hybrid orbitals. Hence, the geometry of the complex is octahedral.

73. The molarity of pure water is ____________________

1. 55.6
2. 5
3. 100
4. 18

Answer: 55.6

Explanation: Molarity is the number of moles of the solute dissolved per litre of the solution.

M=ⁿ⁄_v

M= Molarity of the given substance
n = number of moles of solute
v = volume of solution in litre

Density of water =^1g⁄_1ml

1 L = 1000 ml
1 ml = 1/1000 L

Density of water = ^1g⁄_¹⁄1000L

        =^{1 X 1000 g}⁄_1L

Density of water =^{mass of water}⁄_{volume of water}=1000g/L

The statement above confirms that the mass of pure water occupying 1 L volume equals 1000 g.
To find the number of moles of water, we divide the mass of water by its molecular mass.
The molecular mass of water (H₂O) is calculated as 2(1) + 1(16) = 18, where the atomic mass of hydrogen is 1 and that of oxygen is 16.
Therefore, the number of moles of water is ¹⁰⁰⁰⁄₁₈ =55.56 moles


Molarity of water = ^{number of moles of water}⁄_{volume of water}  =^{55.56 moles}⁄_{1 L}  =55.56 moles/litre=  55.56 M

74. Which of the following is an organometallic compound?

1. CH₃ONa
2. CH₃COONa
3. Pb(C₂H₅)₄
4. CH₃OMgX

Answer: Pb(C₂H₅)₄

Explanation: Organometallic compounds refer to those where a metal atom directly bonds to a carbon atom of a hydrocarbon molecule or radical. For instance, compounds like alkoxides, carbides, cyanides, and mercaptides don't fall under this category because the metal isn't linked to a carbon atom of an alkyl or aryl group. The same criterion applies to the mentioned options. With the exception of Pb(C₂H₅)₄, in all other compounds, the metal atom isn't directly linked to the carbon atom.

75. "Mass defect" is the difference between ___________ and the ________________ of an isotope

1. Neutrons, protons
2. Protons, electrons
3. Expected mass, actual mass
4. Actual mass, expected mass

Answer: Expected mass, actual mass

Explanation: Neutrons and protons reside within the nucleus, held together by strong nuclear attractive forces. Despite this, it's observed that the actual mass of the nucleus is consistently less than the combined mass of its constituent nucleons (neutrons and protons). This difference arises from the significant energy released during the process of nucleons binding together to achieve stability within the nucleus. Consequently, the variance between the actual mass of an isotope and the combined masses of its protons and neutrons is termed as mass defect.

76. The half life of Radon is 3.8 days. The time (in days) by which its ¹⁄₂₀ of the amount will be left behind is

1. 9
2. 16.45
3. 7.6
4. 1.9

Answer: 16.45

Explanation: 

The half-life of Radon = 3.8 days
The disintegration constant of Radon = λ = ^0.693⁄_t¹⁄2

λ = ^0.693⁄_3.8 days

The duration required for one-twentieth of a radon sample to remain is;

t = ^2.303⁄_λ  log ^N₀⁄_N

Where, N₀ is the number of Radon atoms present at t=0 which is 1 mole
N is the number of Radon atoms present after a certain time = ¹⁄₂₀  moles
On subtituting the above values, we get;

t = ^2.303⁄_^0.693⁄3.8 log ¹⁄_¹⁄20

t = ^{2.303 X 3.8}⁄_0.693 log 20

t =12.628 [log2 X log10]

t =12.628 [0.3010 + 1]

t =12.628 X 1.3010

t = 16.429 days

77. Which of the following ions is diamagnetic in nature?

1. Cr³⁺
2. Mn²⁺
3. Cu²⁺
4. Sc³⁺

Answer: Sc³⁺

Explanation: Diamagnetic d-block elements possess paired electrons exclusively in the d-orbitals. They exhibit repulsion when subjected to a magnet. This occurs because the paired electrons within them possess opposite magnetic spins, resulting in the cancellation of their magnetic properties.

21^Sc = [18^Ar]4s²3d¹

21^Sc3+ = [18^Ar]4s⁰3d⁰

24^Cr = [18^Ar]4s¹3d⁵

24^Cr3+ = [18^Ar]4s⁰3d³

25^Mn = [18^Ar]4s²3d⁵

25^Mn2+ = [18^Ar]4s⁰3d⁵

29^Cu = [18^Ar]4s¹3d¹⁰

29^Cu2+ = [18^Ar]4s⁰3d⁹

The absence of electrons in the d-orbital of the Scandium 3+ ion leads to its diamagnetic properties.

78. In actinides, the differentiating electrons enter successively

1. 3d
2. 4d
3. 4f
4. 5f

Answer: 5f

Explanation: Actinides, the elements of the 5f-block series in the periodic table, possess incomplete 5f and 6d orbitals alongside complete 7s orbitals. Their general electronic configuration is represented by "5f^1-14 6d^0-1 7s²". The actinide series includes elements with atomic numbers ranging from 90 to 103, encompassing Thorium to Lawrencium.Therefore, in actinides, the last differentiating electron occupies the 5f subshell.

79. The red colour of blood in red blood cells (erythrocytes) is due to a globular protein haemoglobin consisting of four polypeptide chains and a non-protein constituent 'heme'which in turn contains four subunits. Each subunit consists of a porphyrin complex of a bivalent metal ion (coordination number 6) bound to 4N atoms. The metal ion is____________________

1. Fe²⁺
2. Mg²⁺
3. Co²⁺
4. Cu²⁺

Answer: Fe²⁺

Explanation: Hemoglobin, a conjugated protein composed of heme and globin, imparts red blood cells their distinct color. It can be seen as an approximate tetramer of myoglobin. Hemoglobin comprises four heme groups attached to four protein chains. Upon oxygenation, two heme groups dissociate while the remaining two draw closer together. Heme, a complex organic pigment containing ferrous iron, is a constituent of hemoglobin.

80. The function of haemoglobin in red blood cells is ____________________

1. To transport dioxygen from lungs to tissues
2. To convert CO₂ of the tissues (absorbed from environment) to HCO^-₃ ions
3. The O₂ transforms haemoglobin to oxyhaemoglobin which is faciliated by Fe²⁺ ions for storage of energy in lungs, but not in muscles
4. The affinity for haemoglobin towards O₂ decreases with p^H and ligands like CO, CN etc. reduce it to deoxyhaemoglobin, which is not good O² carrier

Answer: To transport dioxygen from lungs to tissues

Explanation: Blood transports oxygen from the respiratory organs to tissue cells for oxidation and also carries carbon dioxide from tissue cells to the respiratory surface for elimination. Oxygen is transported in two forms in the blood: firstly, as a small amount dissolved in plasma, and secondly, as a larger quantity diffusing into red blood cells, where it combines with the respiratory pigment hemoglobin to form oxyhemoglobin, a bright red compound.

Hemoglobin + Oxygen ⇄ Oxyhemoglobin

4Hb + 4O₂ ⇄ Hb₄O₈

This reaction is reversible and regulated by oxygen concentration. In the respiratory organs, with low oxygen concentration and high carbon dioxide concentration, the reaction reverses, leading oxyhemoglobin to break down into hemoglobin and oxygen. Oxygen then diffuses from the blood into tissue fluid and cells, while hemoglobin returns to the lungs for reuse.

81. The central metal ion in (cyanocobalamin) Vitamin B₁₂ is____________

1. Fe³⁺
2. Cr³⁺
3. Co³⁺
4. V³⁺

Answer: Co³⁺

Explanation: Cyanocobalamin, commonly referred to as Vitamin B12, contains a cobalt metal ion within its structure. This essential vitamin is primarily found in animal-derived foods like meat, eggs, fish, and dairy products. Regular intake of these Vitamin B12-rich foods is crucial for preventing anemia and neurological disorders associated with its deficiency. This is because Vitamin B12 plays a vital role in brain function and the formation of red blood cells.

82. The most stable oxidation state of lanthanides is __________

1. +4
2. +2
3. +3
4. +1

Answer: +3

Explanation: All lanthanides demonstrate their most stable oxidation state of +3 in both aqueous solutions and solid compounds. The reasons behind the formation of this tri-positive state are not fully elucidated based on the electronic configuration of lanthanides. However, the ionization and hydration energies of these elements contribute to the stability of the +3 state over the +2 or +4 states in aqueous solutions. In solid compounds, the collective effect of ionization and lattice energies results in a more negative value for the +3 state compared to the +2 or +4 states. Consequently, the +3 state is more prevalent than the +2 or +4 states.

83. The systematic name of [Cr(en)₃]³⁺

1. Triethylene diamine chromium ion
2. Tris-ethylene diamine chromium (III) ion
3. Ethylene diamine chromium (III) ion
4. Chromium ethylene diamine ion

Answer: Tris-ethylene diamine chromium (III) ion

Explanatin: When naming a complex ion, the ligands' names and numbers precede the central metal ion's name. If polydentate ligands involve numerical prefixes, terms like "bis," "tris," and "tetrakis" are employed. For example, three molecules of ethylene diamine are denoted as "tris-ethylene diamine." The oxidation state of the central metal atom is indicated by Roman numerals in parentheses immediately following the metal atom's name.

84. The indicator employed in the complexometric titration of magnesium using EDTA is_______________

1. Phenolpthalein
2. Methyl Orange
3. Erichrome Black-T
4. Catechol Violet

Answer: Erichrome Black-T

Explanation: Complexometric titration is a type of volumetric analysis wherein the endpoint of the titration is determined by the formation of a colored complex. It is particularly valuable for determining a mixture of various metal ions in a solution. During this titration, an indicator is employed to produce a distinct color change, signaling the endpoint. Indicators like calcein and Eriochrome Black T are commonly utilized in complexometric titrations.

For instance, Eriochrome Black T transitions from blue to wine red when it forms a colored chelate with metal ions, such as Mg²⁺. Conversely, it remains blue when free from metal ions, indicating the release of Mg²⁺ to form a chelate with EDTA.

85. An example of a semi-metal is ___________

1. Fe
2. Mg
3. Ge
4. Nb

Answer: Ge

Explanation: Germanium (Ge) exhibits properties that are intermediate between those of metals and non-metals, categorizing it as a semimetal.

86. Which of the following conditions should atoms taking part in covalent bonding obey?

1. The atomic orbitals should have unpaired electrons
2. They should have empty orbitals
3. They should have only paired electrons
4. They should have electrons of same valency

Answer: The atomic orbitals should have unpaired electrons

Explanation: Covalent bonds form under specific conditions: (1) The combining atoms must possess half-filled orbitals in their valence shell with electrons having opposite spins. (2) Each of the two atoms should have 5, 6, or 7 valence electrons, allowing them to achieve the nearest noble gas configuration by sharing 3, 2, or 1 electron pair.

87. According to Fajan's rules, the bond will acquire covalent character, if the cation is

1. Large size
2. Small size
3. Largest size
4. Independent of its size

Answer: Small size

Explanation: Fajan's rule illustrates that when oppositely charged ions come close, the positively charged cation draws the outermost electrons of the anion while repelling its positively charged nucleus. This leads to the distortion or polarization of the anion, followed by partial electron sharing between the ions, making the bond somewhat covalent. According to Fajan's rule, greater polarization of the anion is favored by: larger ion charges, smaller cations, and larger anions.

88. At very low temperature silicon and germanium can act as

1. Superconductors
2. Insulators
3. Conductors
4. Semiconductors

Answer: Insulators

Explanation: At low temperatures, silicon and germanium behave as insulators because their electrons lack sufficient energy to bridge the band gap of 3 eV, preventing them from transitioning from the valence band to the conduction band. Consequently, they act as insulators.

89. The outershell configuration of the element with atomic number 45 is ___________

1. 3d⁶4s²
2. 3d⁸4s²
3. 4d⁸5s¹
4. 4d¹⁰5s⁰

Answer: 4d⁸5s¹

Explanation: The Aufbau principle applies to nearly all elements, particularly those with lower atomic numbers. Exceptions occur when half-full or full shells or subshells are more stable than partially filled ones.
In cases where there's a small energy level difference between two subshells, an electron might move to the higher level shell to either fill or half-fill it. This electron occupies the higher energy level shell, contradicting the Aufbau principle, but it enhances the atom's stability.
For example, Rhodium, with an atomic number of 45, has an electronic configuration of [18^Ar]4d⁸5s¹. Despite the higher energy level of the 4d orbital compared to the 5s orbital, the transfer of electrons from 5s to 4d enhances its stability.

90. Which of the following ion has a magnetic moment of 3.87 BM?

1. Mn²⁺
2. Fe³⁺
3. Cr³⁺
4. Sc³⁺

Answer: Cr³⁺

Explanation: The paramagnetic strength of transition element or its ion is expressed in terms of magnetic moment (μ).

μ = [n(n+2)]^1⁄₂ BM

Where, n = number of unpaired electrons
BM stands for Bohr magneton

25^Mn = [18^Ar]4s²3d⁵

25^Mn2+ = [18^Ar]4s⁰3d⁵

μ = [5(5+2)]^1⁄₂ = 5.92 BM

24^Cr = [18^Ar]4s¹3d⁵

24^Cr3+ = [18^Ar]4s⁰3d³

μ = [3(3+2)]^1⁄₂ = 3.87 BM

26^Fe = [18^Ar]4s²3d⁶

26^Fe2+ = [18^Ar]4s⁰3d⁶

μ = [4(4+2)]^1⁄₂ = 4.89 BM

21^Sc = [18^Ar]4s²3d¹

21^Sc3+ = [18^Ar]4s⁰3d⁰

μ = [0(0+2)]^1⁄₂ = 0 BM

91. The effective atomic number of Co in [Co(NH₃)₆]³⁺ is____________

1. 33
2. 36
3. 38
4. 52

Answer: 36

Explanation: Sidgwick proposed that the metal ion will accept electron pairs until the combined number of electrons in the metal ion and those donated by ligands equals that of the next higher noble gas. This combined count of electrons is referred to as the effective atomic number of the metal.

EAN = (Atomic number of metal ion - Oxidation state of metal ion) + (Coordination number X 2)

For example- In [Co(NH₃)₆]³⁺, the effective atomic number of Co can be calculated as

Atomic number of Cobalt = 27

Number of electrons in Co³⁺ = 24

Electrons contributed by 6NH₃ ligands = 6 X 2 = 12

The EAN of Co³⁺ in the complex =24 + 12 = 36

92. The bond order in O^-₂ ion is __________________

1. 2
2. 1
3. 0
4. 1.5

Answer: 1.5

Explanation: O^-₂ ion is formed by the linear combination of O^- and O atomic orbitals.

Ground state of O^-₂ is as follows:

(σ^b1s)² (σ^*1s)²(σ^b2s)² (σ^*2s)² (σ^b2p_z)² (π^b2p_x)² = (π^b2p_y)² (π^*2p_x)² = (π^*2p_y)¹

Bond order  =  (^{N_b - N_a})⁄₂

Where,N_b = The number of electrons in bonding molecular orbits.

N_a = The number of electrons in antibonding molecular orbits

Bond order  =  ^{(10 - 7)}⁄₂  =  ³⁄₂  =  1.5

93. The hybridization in IF₇ molecule is __________________

1. dsp²
2. d²sp³
3. sp³d³
4. dsp³

Answer: sp³d³

Explanation: Iodine atomic number is 53. It's electronic configuration in the ground state can be;

53^I = [36^Kr]4d¹⁰5s²5p⁵5d⁰

In the first excited state, the electronic configuration of iodine is: [36^Kr]4d¹⁰5s²5p⁴5d¹

In the second excited state, the electronic configuration of iodine is: [36^Kr]4d¹⁰5s²5p³5d²

In the third excited state, the electronic configuration of iodine is: [36^Kr]4d¹⁰5s¹5p³5d³

Now, one 5s, three 5p, and three 5d orbitals each containing one unpaired electron undergo sp³d³ hybridization, resulting in the formation of seven sp³d³ hybrid orbitals.

Similarly, fluorine, with atomic number 9, shows the following electronic configuration in its ground state.

9^F = 1s²,2s²,2p⁵

Each of the sp³d³ hybrid orbitals of the central iodine atom overlaps with the half-filled 2p_z hybrid orbital of the fluorine atom, resulting in the formation of seven sp³d³-p sigma bonds.

With the central iodine atom possessing 7 bond pairs and zero lone pairs, the shape of Iodine heptafluoride is pentagonal bipyramidal.

Among the seven hybrid orbitals, 5 fluorine atoms are bonded to the central iodine atom through equatorial bonds within the pentagonal ring, while 2 fluorine atoms are bonded to the central iodine atom through axial bonds.

The bond angle between two equatorial fluorine atoms is 72 degrees, and the angle between an axial bond and an equatorial bond is 90 degrees. The axial bond length is 179 pm. And the equatorial bond length is 186 pm.

94. In diborane boron atoms undergo _________________ hybridization

1. sp
2. sp³
3. sp²
4. both sp² and sp³

Answer: sp³

Explanation: Boron, with an atomic number of 5, exhibits a ground state electronic configuration of 1s² 2s² 2p_x¹ and transitions to an excited state with a configuration of 1s² 2s¹ 2p_x¹2p_y¹.

In this excited state, one 2s orbital and three 2p orbitals of boron, having similar energy levels, hybridize to form four sp³ hybrid orbitals. Among these orbitals, three contain one unpaired electron each, while the fourth remains empty.

The formation of bonds in diborane involves unique configurations. The bridging hydrogen atoms' 1s orbitals overlap with one half-filled and one empty sp³ hybrid orbital from each boron atom to form the H—B---H three-center bonds.

Simultaneously, the terminal hydrogen atoms bond with the sp³ hybrid orbitals of the boron atoms through s-sp³ sigma bonds.

In the diborane molecule, two boron atoms are surrounded by four terminal hydrogen atoms, forming a flat plane, while two bridging hydrogen atoms, situated above and below this plane, link the boron atoms.

The bridging hydrogen atoms acquire a positive charge, resulting in repulsion between them. This repulsion causes the three-center bonds to bend away from each other in the middle, resulting in a banana-shaped configuration. These unique hydrogen bridges are commonly referred to as banana bonds or tau bonds.

Overall, each boron atom in diborane adopts sp³ hybridization, leading to a tetrahedral geometry in the molecule.

95. The compound formed on partial hydrolysis of XeF₄ is

1. XeOF₂
2. XeO₂
3. XeO₃
4. XeOF₄

Answer: XeOF₂

Explanation: At a temperature of 193K, xenon tetrafluoride undergoes a slow and incomplete hydrolysis, resulting in the formation of the unstable compound known as xenon oxydifluoride.

Xenon tetrafluoride + Water -----→Xenon oxydifluoride + Hydrogen fluoride

                    193K

XeF₄ + H₂O-------→XeOF₂ + 2HF

96. The molecular shape of XeF₄ is __________________________

1. Linear
2. Square
3. Square planar
4. Pentagonal bipyramid

Answer: Square planar

Explanation: The atomic number of xenon is 54.The ground state electronic configuration of xenon is;

54^Xe = [36^Kr]4d¹⁰5s²5p⁶5d⁰

In the first excited state, the electronic configuration of xenon is: [36^Kr]4d¹⁰5s²5p⁵5d¹

In the second excited state, the electronic configuration of xenon is: [36^Kr]4d¹⁰5s²5p⁴5d²

Similarly, fluorine, with atomic number 9, shows the following electronic configuration in its ground state.

9^F = 1s²,2s²,2p⁵

In its excited state, the central Xenon atom possesses four unpaired electrons in its valence shell, while each Fluorine atom has one unpaired electron in its valence shell. Through a process of hybridization, involving one 5s, three 5p, and two 5d orbitals, Xenon forms six sp³d² hybrid orbitals.

Among these, two orbitals contain lone pairs of electrons, while the remaining four are half-filled. When the half-filled 2p orbitals of the four Fluorine atoms overlap with the half-filled sp³d² hybrid orbitals of Xenon, four sp³d²-p sigma bonds are formed.

The expected geometry for XeF₄ is octahedral. However, due to the repulsion between lone pairs of electrons being greater than that between lone pair-bond pair and bond pair-bond pair, the molecule's shape is distorted.

Consequently, XeF₄ adopts a distorted octahedral geometry with two lone pairs of electrons, essentially resembling a square planar geometry.

97. In an acidic medium, using H₂S as a reagent, which can be identified separately from the other three ions from the following?

1. Cu²⁺
2. Cd²⁺
3. Ni²⁺
4. Hg²⁺

Answer: Ni²⁺

Explanation: In Group II, sparingly soluble sulphides ((Hg²⁺, Pb²⁺, Cu²⁺, Cd²⁺, Bi³⁺) ions precipitate when hydrogen sulphide is passed through an acidic solution with low hydrogen ion concentration.

The radicals found in Group IV include CO²⁺, Ni²⁺, Mn²⁺, and Zn²⁺. When H₂S gas is passed through the ammonical solution of these salts, they precipitate as sulphides. Therefore, the reagent for this group is H₂S gas in the presence of NH₄Cl and NH₄OH.

The K_sp value of Group II sulphides is lower than that of Group IV sulphides. Consequently, while hydrogen sulphide is passed through an acidic solution to precipitate Group II cation sulphides, it is done in an alkaline medium to precipitate Group IV cation sulphides when the S^2- ion concentration is low.

98. If 4 g. of a solute (molecular weight = 40) is present in 200ml of the solution, the molarity of the solution is ___________

1. 0.5M
2. 0.25M
3. 0.75M
4. 1.0M

Answer: 0.5M

Explanation: The molarity of a solution is equal to the number of moles of solute divided by the litres of a solution. This is calculated using the following formula:

Molarity of a solution = ^{number of moles of solute}⁄_{volume of solution in litres}

Number of moles of solute = ^{Mass of solute in grams}⁄_{Molecular mass of solute}

Number of moles of the given solute = ⁴⁄₄₀  = 0.1

Volume of solution = 200ml = 0.2L

Molarity of the given solution = ^0.1⁄_0.2  = 0.5M

99. A radioactive nuclide emits one alpha and two beta particles in succession. The daughter nuclide is _________

1. Isobar
2. Isotone
3. Isotope
4. Isomer

Answer: Isotope

Explanation: It should be noted that the emission of an alpha particle, followed by the emission of two beta particles, leads to the formation of an isotope of the parent element.

           -α particle            -β particle               -β particle

92^U238-----------→90^Th234-----------→91^Pa234-----------→92^U234

92^U238  and   92^U234  are isotopes.

100. The IUPAC name for   CH₃-CH-CH-CH₃  is________________

                                            ⃒    ⃒

                                            CH₃OH

1. 2-methyl butyl alcohol
2. 2-methyl butanol-1
3. 3-methyl butanol-2
4. 2-methyl butanol and 2-ol

Answer: 3-methyl butanol-2

Explanation: In accordance with IUPAC nomenclature, the longest continuous chain of carbon atoms serves as the parent chain, with various alkyl groups considered as substituents.

When the parent carbon chain contains two or more substituents, it is numbered to minimize the sum of the locants, with each number indicating the position of a substituent, known as a locant.

The hydroxyl group takes priority over alkyl groups when numbering the parent chain.

The parent structure is denoted by a "suffix," while the substituent is indicated by a "prefix." Alcohols are named by replacing the suffix "-ane" of the corresponding alkane with "-anol".

In the given compound, butane serves as the parent chain, while the methyl group is treated as a substituent. The parent chain is numbered from right to left to ensure that the hydroxyl group receives the lowest possible number during numbering.

Answers to the questions in APPGCET Chemical Sciences Sample Paper-1

Table-1: APPGCET Chemical Sciences Sample Paper-1 with Answers
Question number	Correct answer	Question number	Correct answer
Q-1	2	Q-51	4
Q-2	4	Q-52	3
Q-3	3	Q-53	3
Q-4	3	Q-54	2
Q-5	3	Q-55	2
Q-6	1	Q-56	2
Q-7	4	Q-57	4
Q-8	4	Q-58	3
Q-9	2	Q-59	1
Q-10	1	Q-60	4
Q-11	2	Q-61	4
Q-12	3	Q-62	2
Q-13	1	Q-63	2
Q-14	4	Q-64	3
Q-15	1	Q-65	4
Q-16	2	Q-66	2
Q-17	3	Q-67	1
Q-18	4	Q-68	4
Q-19	4	Q-69	1
Q-20	3	Q-70	2
Q-21	3	Q-71	2
Q-22	4	Q-72	4
Q-23	1	Q-73	1
Q-24	1	Q-74	3
Q-25	3	Q-75	3
Q-26	1	Q-76	2
Q-27	2	Q-77	4
Q-28	1	Q-78	4
Q-29	3	Q-79	1
Q-30	4	Q-80	1
Q-31	1	Q-81	3
Q-32	4	Q-82	3
Q-33	1	Q-83	2
Q-34	3	Q-84	3
Q-35	2	Q-85	3
Q-36	1	Q-86	1
Q-37	3	Q-87	2
Q-38	3	Q-88	2
Q-39	1	Q-89	3
Q-40	3	Q-90	3
Q-41	2	Q-91	2
Q-42	1	Q-92	4
Q-43	2	Q-93	3
Q-44	4	Q-94	2
Q-45	3	Q-95	1
Q-46	3	Q-96	3
Q-47	3	Q-97	3
Q-48	2	Q-98	1
Q-49	3	Q-99	3
Q-50	1	Q-100	3

Conclusion:

Our blog topic discusses the APPGCET Chemical Sciences Sample Paper-1 With Answers. We believe that our approach to providing answers helps students gain a deep understanding of the concepts.

Additional references:

For further details, please refer to the APPGCET instruction booklet. If you have any additional queries, feel free to explore our article titled "Top 50 Questions You Should Ask about APPGCET."