Model(5) 2026

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Instructions for Answering

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Do not use checkmarks (✓) or crosses (✗).
If you need to change an answer, erase your previous mark completely.
Ensure that only one bubble is filled for each question.

Question 1

Element X, from the first transition series, is used in an acid-resistant alloy. Which of the following represents the electronic configuration of the X³⁺ ion?

a) [₁₈Ar] 3d³
b) [₁₈Ar] 3d⁶
c) [₁₈Ar] 3d⁷
d) [₁₈Ar] 3d⁴

✓ Correct Answer: (c) [₁₈Ar] 3d⁷

📚 Detailed Explanation

Nickel (Ni, Z=28) is widely used with steel to form acid-resistant alloys. Its neutral electronic configuration is [Ar] 4s² 3d⁸. Removing 3 electrons (two from 4s and one from 3d) to form Ni³⁺ leaves the configuration [Ar] 3d⁷.

Question 2

Element X, from the first transition series, has a completely filled (d) sublevel and no unpaired electrons in any oxidation state. Which of the following represents the element type and the electronic configuration of its ion in the compound XO?

a) Transition, [Ar] 3d¹⁰
b) Transition, [Ar] 3d⁹
c) Non-transition, [Ar] 3d¹⁰
d) Non-transition, [Ar] 3d⁹

✓ Correct Answer: (c) Non-transition, [Ar] 3d¹⁰

Zinc (Zn) has completely filled d-orbitals in both its atomic state (4s² 3d¹⁰) and its only oxidation state +2 (3d¹⁰). Because the d-sublevel is never partially filled, it is considered a non-transition element. In XO (ZnO), the Zn²⁺ ion has the configuration [Ar] 3d¹⁰.

Question 3

Which of the following expresses one of the reasons for using nickel as a catalyst in Oil hydrogenation?

a) Increases the surface area of the reactants
b) Decreases the chance of collisions between the reactant molecules.
c) Electrons of 4s, 3d share in concentrating reactants on its surface.
d) Increases the activation energy of the reactant molecules.

✓ Correct Answer: (c) Electrons of 4s, 3d share in concentrating reactants on its surface.

Transition metals act as ideal catalysts because the electrons of the 4s and 3d orbitals form weak bonds with reactant molecules, concentrating them on the metal's surface. This weakens the bonds in the reactant molecules and lowers the activation energy.

Question 4

Two non-consecutive transition elements, X and Y, are from the first transition series.
Element X: One of its compounds is used as a catalyst in the contact method.
Element Y: It has the same number of unpaired electrons as element X.
- Which of the following is correct?

(a) The radius of element (X) is larger than that preceding it in the series
(b) The effective nuclear charge of element (Y) is less than that of element (X)
(c) The density of element (Y) is less than the density of element X
(d) The atomic mass of element (Y) is greater than that follow it in the series.

✓ Correct Answer: (d) The atomic mass of element (Y) is greater than that follow it in the series.

Element X is Vanadium (V), as V₂O₅ is the catalyst in the contact process. V ([Ar] 4s² 3d³) has 3 unpaired electrons. Element Y, with 3 unpaired electrons and non-consecutive, is Cobalt (Co, [Ar] 4s² 3d⁷). Cobalt's atomic mass (58.9 u) is anomalously greater than the element following it, Nickel (58.7 u).

Question 5

(A, B, and C) three consecutive elements from the first transition series.
Element (A) has an anomalous electronic configuration than that of other elements in the series.
-Which of the following represents the correct order of magnetic moments for ions of these elements in the oxidation state (3+)?

a) A³⁺ > B³⁺ > C³⁺
b) B³⁺ > C³⁺ > A³⁺
c) B³⁺ > A³⁺ > C³⁺
d) C³⁺ > B³⁺ > A³⁺

✓ Correct Answer: (d) C³⁺ > B³⁺ > A³⁺

Element A with anomalous configuration is Chromium (Cr, Z=24). Since they are consecutive, B is Manganese (Mn, Z=25), and C is Iron (Fe, Z=26).
A³⁺ (Cr³⁺) is 3d³ (3 unpaired).
B³⁺ (Mn³⁺) is 3d⁴ (4 unpaired).
C³⁺ (Fe³⁺) is 3d⁵ (5 unpaired).
More unpaired electrons mean a higher magnetic moment, so Fe³⁺ > Mn³⁺ > Cr³⁺ (C > B > A).

Question 6

When a confirmatory reagent (X) is added to a salt solution (Y), the color of reagent (X) disappears and the salt (Y) oxidizes.
- Which of the following represents reagent (X) and salt (Y)?

Option	The reagent X	The salt Y
a)	Acidified potassium permanganate	sodium nitrite
b)	Magnesium sulphate	sodium carbonate
c)	Lead II acetate	sodium sulphide
d)	brown iodine solution	Sodium sulphate

✓ Correct Answer: (a)

Acidified potassium permanganate (KMnO₄) is a strong oxidizing agent with a distinct purple color. When it oxidizes sodium nitrite (NaNO₂) to sodium nitrate (NaNO₃), it is reduced to colorless Mn²⁺, causing its purple color to disappear.

Question 7

From the following diagram:

black ppt. (Y) ← (X) —— CuCl_2(aq) —— K₂CO_3(aq) → (Z) ppt.

-Which of the following represents substance X, Y, and Z?

Option	X	Y	Z
a)	AgNO_3(aq)	AgCl_(s)	KCl_(s)
b)	HCl_(aq) + H₂S_(g)	CuCl_2(s)	CuCO_3(s)
c)	HCl_(aq) + H₂S_(g)	CuS_(s)	CuCO_3(s)
d)	Na₂S_(aq)	NaCl_(s)	KCl_(s)

✓ Correct Answer: (c) X = HCl_(aq) + H₂S_(g), Y = CuS_(s), Z = CuCO_3(s)

Reacting CuCl_2(aq) with K₂CO₃ causes a double substitution yielding Copper Carbonate (CuCO₃), a precipitate (Z). Passing H₂S gas in acidic medium (HCl) into CuCl₂ precipitates Copper (II) Sulfide (CuS), which is the characteristic black precipitate (Y) for Analytical Group II.

Question 8

100 mL of 0.5 M calcium hydroxide is added to 200 mL of 0.6 M hydrochloric acid according to the following equation: 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
- Which of the following represents the concentration of excess reactant?

a) 0.2 M
b) 0.0667 M
c) 0.633 M
d) 0.316 M

✓ Correct Answer: (b) 0.0667 M

Moles of Ca(OH)₂ = 0.1 L × 0.5 M = 0.05 moles.
Moles of HCl = 0.2 L × 0.6 M = 0.12 moles.
Reaction ratio is 1:2. 0.05 moles Ca(OH)₂ require exactly 0.10 moles of HCl. Thus, HCl is in excess by (0.12 - 0.10) = 0.02 moles.
Total final volume = 100 mL + 200 mL = 300 mL (0.3 L).
Concentration of excess = 0.02 moles / 0.3 L ≈ 0.0667 M.

Question 9

2.52 g of hydrated crystals of acid (A) [A·XH₂O]. It was dissolved in distilled water to make a 250 mL of 0.08 M solution. 25 mL of this solution was titrated with a sodium hydroxide solution containing 0.16 g of the solute.
(Molar mass of the acid = 90 g/mol, NaOH = 40 g/mol)
-Which of the following expresses the number of water molecules (X) and the type Of acid?

(a) X = 2 and the acid is diprotic
(b) X = 1 and the acid is monoprotic
(c) X = 2 and the acid is triprotic
(d) X = 3 and the acid is triprotic

✓ Correct Answer: (a) X = 2 and the acid is diprotic

Moles of hydrated acid in 250 mL = 0.25 L × 0.08 M = 0.02 moles.
Molar mass of hydrated acid = 2.52 g / 0.02 moles = 126 g/mol.
Mass of water of hydration = 126 - 90 = 36 g/mol. Since H₂O is 18 g/mol, X = 36 / 18 = 2.
Titration: 25 mL acid contains 0.002 moles. Moles NaOH = 0.16 g / 40 g/mol = 0.004 moles. Ratio of Acid:Base is 0.002:0.004 = 1:2. The acid is diprotic.

Question 10

All of the following reactions produce substances sparingly soluble in water except:

a) Magnesium sulphate solution with potassium bicarbonate solution
b) Dilute hydrochloric acid with potassium thiosulphate salt
c) Silver nitrate solution with potassium sulphite solution
d) Carbon dioxide with calcium hydroxide solution for a short time

✓ Correct Answer: (a) Magnesium sulphate solution with potassium bicarbonate solution

Reaction (a) yields Magnesium bicarbonate, Mg(HCO₃)₂, which is fully soluble in water (all bicarbonates are soluble). Options (b) produces elemental Sulphur ppt, (c) produces Silver sulphite ppt, and (d) produces Calcium carbonate ppt.

Question 11

Which of the following is true for reactions (A) and (B)? (When each is performed in a sealed container)?

Reaction (A)	Reaction (B)
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)	Mg_(s) + 2HCl_(aq) → MgCl_2(aq) + H_2(g)

a) Both reactions reach equilibrium at the same time.
b) Neither reaction reaches equilibrium.
c) Reaction (A) reaches equilibrium, but reaction (B) does not.
d) Reaction (B) reaches equilibrium, but reaction (A) does not.

✓ Correct Answer: (c) Reaction (A) reaches equilibrium, but reaction (B) does not.

Reaction (A) is a reversible gaseous reaction; in a sealed container, it achieves equilibrium. Reaction (B) is effectively a complete/irreversible reaction under standard conditions (metals reacting with strong acids go to completion), thus it does not establish a dynamic chemical equilibrium even in a closed system.

Question 12

Two solutions of two weak acids (HA) and (HB) have a pH = 3 for each
- Acid HA has a concentration of 0.1 M.
- Acid HB has a concentration of 0.01 M.
Which of the following is correct for acids HA and HB?

a) Acids HA and HB have a different [OH^-]
b) Acids HA and HB have the same value of K_a.
c) The value of K_a for acid HB is greater.
d) The value of K_a for acid HA is greater.

✓ Correct Answer: (c) The value of K_a for acid HB is greater.

Since pH = 3 for both, [H⁺] = 10^-3 M for both.
Using K_a ≈ [H⁺]² / C_a:
For HA: K_a = (10^-3)² / 0.1 = 10^-5
For HB: K_a = (10^-3)² / 0.01 = 10^-4
10^-4 > 10^-5, so acid HB has the greater K_a value.

Question 13

Which of the following equilibrium reactions shifts to the forward reaction by increasing the volume of the container?

a) N₂O_4(g) ⇌ 2NO_2(g)
b) H_2(g) + I_2(g) ⇌ 2HI_(g)
c) 2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
d) N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

✓ Correct Answer: (a) N₂O_4(g) ⇌ 2NO_2(g)

Increasing the volume decreases the pressure. According to Le Chatelier's Principle, the equilibrium shifts toward the side with the greater number of gaseous moles. In reaction (a), 1 mole of gas forms 2 moles of gas. Thus, it will shift forward to the right.

Question 14

Two tubes (A, B) each contain equal masses of an active metal and the same acid.
If the reaction in tube (B) is ended in a shorter time than in tube (A).
Which of the following is the reason of the different in reaction time of the two reactions?

a) Division of the metal in tube (A)
b) Adding a catalyst to tube (A)
c) Increasing the temperature of tube (B)
d) Adding water to tube (B)

✓ Correct Answer: (c) Increasing the temperature of tube (B)

A shorter reaction time means a higher reaction rate. Increasing the temperature provides more kinetic energy to particles, increasing collision frequency and successful collisions, thereby speeding up the reaction in tube (B). (Adding water would slow it down, while options a and b would make tube A faster, contradicting the premise).

Question 15

Four solutions have the same concentration:
1- HCl solution 2- NaOH solution
3- CH₃COOH acid solution 4- NH₄OH solution
-The order of these solutions according to the hydronium ion concentration is:

a) 1 > 3 > 4 > 2
b) 2 > 4 > 3 > 1
c) 1 > 2 > 3 > 4
d) 1 > 3 > 2 > 4

✓ Correct Answer: (a) 1 > 3 > 4 > 2

Hydronium ion concentration [H₃O⁺] dictates acidity.
- HCl (1) is a strong acid (Highest [H⁺]).
- CH₃COOH (3) is a weak acid.
- NH₄OH (4) is a weak base (low [H⁺], high [OH^-]).
- NaOH (2) is a strong base (Lowest [H⁺], approaching ~10^-14).

Question 16

A saturated lead (II) chloride (PbCl₂) solution is in equilibrium according to the Following equation: PbCl_2(s) ⇌ Pb²⁺_(aq) + 2Cl^-_(aq)
-Which of the following is correct regarding the solubility of PbCl₂ when a small Amount of each of the following salts is added separately?

a) Solubility decreases with addition of sodium sulphate solution.
b) Solubility increases with addition of sodium nitrate solution.
c) Solubility decreases with addition of lead (II) acetate solution.
d) Solubility increases with addition of potassium nitrate solution.

✓ Correct Answer: (c) Solubility decreases with addition of lead (II) acetate solution.

Adding Lead (II) acetate provides additional Pb²⁺ ions to the solution. According to Le Chatelier's principle (Common Ion Effect), the equilibrium shifts to the left to consume the extra Pb²⁺, forcing more PbCl₂ to precipitate out and thus decreasing its solubility.

Question 17

A galvanic cell consists of:
An electrode (X) in XSO₄ solution
An electrode (Y) in YSO₄ solution
-If metal (X) is a stronger reducing agent than metal (Y), which of the following describes what happens in this cell?

a) Electrons move from (Y) to (X) through the external wire
b) The mass of electrode (X) increases by time
c) Negative ions move from the bridge to the half cell (X)
d) Reduction of (X²⁺) ions occurs at the anode

✓ Correct Answer: (c) Negative ions move from the bridge to the half cell (X)

If X is a stronger reducing agent, X is the Anode (undergoes oxidation: X → X²⁺ + 2e^-). As positive X²⁺ ions accumulate in the anode half-cell, negative ions (anions) from the salt bridge migrate towards it to neutralize the excess positive charge.

Question 18

An electrochemical cell undergoes the following reaction:
2Y_(s) + 3X⁺_(aq) → 2Y⁺_(aq) + 3X_(s) ; emf = +1.6 V
- Which of the following is correct for both electrode type (Y) and cell type?

Option	the Y-electrode type	type of cell
A	cathode	Galvanic
B	anode	Electrolytic
C	anode	Galvanic
d	cathode	Electrolytic

✓ Correct Answer: (C)

The positive EMF (+1.6 V) indicates a spontaneous reaction, making it a Galvanic cell. In the reaction, Y goes from oxidation state 0 to +1. Oxidation occurs at the anode, so the Y-electrode is the anode.

Question 19

Four electrodes [X], [Y], [Z], [L], where:

(X²⁺/X)	(Y²⁺/Y)	(Z/Z²⁺)	(L/L²⁺)
-0.44 V	+1.5 V	-0.34 V	+1.18 V

Which of the following reactions represents a cell that produces the greatest electromotive force (emf)?

a) Y + L²⁺ → Y²⁺ + L
b) X + L²⁺ → X²⁺ + L
c) L + Y²⁺ → L²⁺ + Y
d) X + Y²⁺ → X²⁺ + Y

✓ Correct Answer: (c) L + Y²⁺ → L²⁺ + Y

First, harmonize standard potentials to Reduction Potentials (E°red):
X: -0.44V, Y: +1.5V, Z: +0.34V (flipped from ox), L: -1.18V (flipped from ox).
Greatest EMF = E°red(Cathode) - E°red(Anode). Maximum gap is between highest (+1.5V for Y) and lowest (-1.18V for L).
L must oxidize (anode), Y must reduce (cathode): L + Y²⁺ → L²⁺ + Y.
EMF = 1.5 - (-1.18) = +2.68 V.

Question 20

Which of the following examples illustrates cathodic protection of metals against corrosion?

(a) Coating iron with an inorganic substance such as red primer
(b) Galvanizing iron by dipping it in zinc
(c) Coating iron with a metal that has a lower reduction potential than iron
(d) Coating iron with a metal that has a lower oxidation potential than iron

✓ Correct Answer: (d) Coating iron with a metal that has a lower oxidation potential than iron

Cathodic protection involves coating iron with a metal that is LESS active (acts as a cathode), meaning it has a lower oxidation potential (or higher reduction potential) than iron. An example is coating iron with tin (Sn). (Note: Galvanizing is *anodic* protection).

Question 21

Which of the following represents the anode half-reaction during the electrolysis of molten sodium hydride (NaH) between graphite electrodes?

a) 2Na⁺ + 2e^- → 2Na
b) 2Na → 2Na⁺ + 2e^-
c) 2H₂ → 2H⁺ + 2e^-
d) 2H^- → H₂ + 2e^-

✓ Correct Answer: (d) 2H^- → H₂ + 2e^-

In a metal hydride like NaH, hydrogen exists as the hydride ion (H^-). During electrolysis, anions migrate to the anode where they are oxidized (lose electrons). 2H^- lose 2 electrons to form H₂ gas.

Question 22

Which of the following metals that deposits 18 g when 1.5 F passes through one of its molten salts? (Na = 23, Mg = 24, Ca = 40, K = 39)

a) Na
b) Mg
c) Ca
d) K

✓ Correct Answer: (b) Mg

According to Faraday's law, Equivalent Mass = Mass / Faradays = 18 g / 1.5 F = 12 g/eq.
We check the equivalent mass for each metal:
Na (23/1 = 23), Mg (24/2 = 12), Ca (40/2 = 20), K (39/1 = 39).
Magnesium (Mg) matches.

Question 23

Two organic compounds, (X) and (Y):
X: An aliphatic compound that, upon oxidation, produces a compound used in Polyester production.
Y: An aromatic compound whose sodium salt is used to prevent growth of fungi.
-Which of the following is correct?

a) Compound (X) obey to the formula C_nH_2n+2O₂ and does not contain a carboxyl group.
b) Compound (Y) does not obey the formula C_nH_2nO₂ and contains a carboxyl group.
c) Compound (X) obey to the formula C_nH_2n+2O₂ and contains a carboxyl group.
d) Compound(Y) obey to the formula C_nH_2nO₂ and contains a carboxyl group.

✓ Correct Answer: (B) Compound (Y) does not obey the formula C_nH_2nO₂ and contains a carboxyl group.

X is Ethylene or ethene (C₂H₆O₂), an aliphatic Has (no carboxyl group),and does not matches the general formula C_nH_2n+2O₂. It oxidizes to form acids used in polyesters (like Dacron). Y is Benzoic acid (C₇H₆O₂), an aromatic acid (its sodium salt, Sodium Benzoate, is a food preservative).

Question 24

Which of the following hydroxyl compounds cannot be obtained by ammonolysis of an ethyl benzoate isomer?

a) C₆H₅OH
b) C₆H₅CH₂CH₂OH
c) C₆H₅CH₂OH
d) C₂H₅OH

✓ Correct Answer:(d) C₂H₅OH

The correct answer is d) C₂H₅OH.

🧪 Chemical Explanation & Equations

The Isomer Criteria: Ethyl benzoate has the molecular formula C₉H₁₀O₂. Ammonolysis of an ester breaks it down into an amide and an alcohol/hydroxyl compound. The question strictly requires the starting ester to be an isomer of ethyl benzoate, not ethyl benzoate itself.
Why Ethanol (d) is the Exception: Ethanol (C₂H₅OH) is obtained directly from ethyl benzoate, which cannot be an isomer of itself:
\(\text{C}_6\text{H}_5\text{COOC}_2\text{H}_5 \text{ (Ethyl benzoate)} + \text{NH}_3 \rightarrow \text{C}_6\text{H}_5\text{CONH}_2 \text{ (Benzamide)} + \{\text{C}_2\text{H}_5\text{OH}}\)

💡 Formation from True Isomers (Why Options a, b, c Fail)

Option a (Phenol): Obtained by the ammonolysis of phenyl propanoate (an isomer with formula C₉H₁₀O₂):
\(\text{C}_2\text{H}_5\text{COOC}_6\text{H}_5 + \text{NH}_3 \rightarrow \text{C}_2\text{H}_5\text{CONH}_2 \text{ (Propanamide)} + \{\text{C}_6\text{H}_5\text{OH}}\)
Option b (2-Phenylethanol): Obtained by the ammonolysis of phenethyl formate (an isomer with formula C₉H₁₀O₂):
\(\text{HCOOCH}_2\text{CH}_2\text{C}_6\text{H}_5 + \text{NH}_3 \rightarrow \text{HCONH}_2 \text{ (Formamide)} + \{\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{OH}}\)
Option c (Benzyl alcohol): Obtained by the ammonolysis of benzyl acetate (an isomer with formula C₉H₁₀O₂):
\(\text{CH}_3\text{COOCH}_2\text{C}_6\text{H}_5 + \text{NH}_3 \rightarrow \text{CH}_3\text{CONH}_2 \text{ (Acetamide)} + \{\text{C}_6\text{H}_5\text{CH}_2\text{OH}}\)

Question 25

The following equation illustrates one of the thermal catalytic cracking processes:

C₂₀H_42(s) → X_(l) + Y_(g) + Z_(l)

X: An alkene containing 30 mol of atoms per mole.
Y: The heaviest gaseous alkene.
- Which of the following does not represent compound (Z)?

a) It is produced from the hydrogenation of hexyne.
b) It reacts with hydrogen halides (HBr) according to Markownikoff’s rule.
c) Its catalytic reforming yields a compound used as organic solvent.
d) It has the general formula C_nH_2n+2.

✓ Correct Answer: (b) It reacts with hydrogen halides (HBr) according to Markownikoff’s rule.

X is C₁₀H₂₀ (10 C + 20 H = 30 atoms). Y is Butene (C₄H₈, heavily gaseous alkene). By balancing carbon: 20 - 10 - 4 = 6 carbons. Z is Hexane (C₆H₁₄). Hexane is an alkane (saturated); it does NOT undergo addition reactions with HBr, making option (b) the incorrect statement regarding Z.

Question 26

X, Y, and Z are three hydrocarbon derivatives:
X: has the general formula C_nH_2nO₂ and reacts with active metals.
Y: has the general formula C_nH_2nO₂ and does not react with active metals.
Z: has the general formula C_nH_2n+2O and reacts with active metals.
If the molar masses of (X, Y, and Z) are equal, which of the following is correct regarding their boiling points?

a) (Y) has a higher boiling point than (Z).
b) (Z) has a higher boiling point than (X).
c) (Y) has a higher boiling point than (X).
d) (X) has a higher boiling point than (Z).

✓ Correct Answer: (d) (X) has a higher boiling point than (Z).

X is a Carboxylic Acid (forms 2 hydrogen bonds per pair). Y is an Ester (no hydrogen bonding). Z is an Alcohol (forms 1 hydrogen bond). Thus, for similar molar masses, Boiling Point order is: Acid (X) > Alcohol (Z) > Ester (Y).

Question 27

An organic compound with the molecular formula C₅H₁₀O₂ has a complex Functional group of two functional groups.
- Which of the following could be the IUPAC name for this compound?

a) 2,2-Dimethyl butanoic acid
b) Methyl butanoate
c) Ethyl propanoate
d) 2,2-Dimethyl propanoic acid

✓ Correct Answer: (d) 2,2-Dimethyl propanoic acid

"A complex functional group of two functional groups" refers to the Carboxyl group (-COOH), which combines a Carbonyl (C=O) and Hydroxyl (-OH). The compound must be a carboxylic acid with 5 carbons. 2,2-Dimethyl propanoic acid has 3 carbons in the main chain + 2 methyl branches = 5 carbons total (C₅H₁₀O₂).

Question 28

Which of the following represents the correct order of some organic compounds according to the amount of oxygen required for the complete combustion of one mole?

a) Methane > Ethane > Ethyne
b) Ethane > Ethene > Ethyne
c) Ethyne > Ethene > Ethane
d) Ethyne > Ethene > Methane

✓ Correct Answer: (b) Ethane > Ethene > Ethyne

Combustion stoichiometry:
- Ethane (C₂H₆): requires 3.5 moles O₂.
- Ethene (C₂H₄): requires 3 moles O₂.
- Ethyne (C₂H₂): requires 2.5 moles O₂.
- Methane (CH₄): requires 2 moles O₂.

Question 29

Compound A, produced by the fractional distillation of coal tar, the following reactions, in order, under the standard conditions for each occurs for compound A:
(Chlorination by substitution - Reaction with sodium hydroxide - Nitration)
-Which of the following represents a use of the produced compound?

a) Insecticide
b) Widen arteries
c) Explosive substance
d) Bakelite polymer

✓ Correct Answer: (c) Explosive substance

Fractional distillation of coal tar yields Benzene (A). Chlorination -> Chlorobenzene. Reaction with NaOH at 300°C/300atm -> Phenol. Nitration of Phenol yields 2,4,6-trinitrophenol (Picric Acid), which is highly explosive and used to treat burns.

Question 30

Which of the following is the IUPAC name of the compound shown?

a) 3,6-dinitro benzoic acid
b) 2,5-dinitro benzoic acid
c) Meta, para-dinitro benzoic acid
d) 3,5-dinitro benzoic acid

✓ Correct Answer: (b) 2,5-dinitro benzoic acid

Numbering starts at the principal functional group (COOH) as carbon #1. To give the lowest locants to the substituents, we number around the ring towards the closest NO₂ group. It lands on positions 2 and 5.

Question 31

Which of the following describes the properties of each of the following types of alcohols?

a) Primary and secondary alcohols react with sodium metal, while tertiary do not.
b) Primary alcohols are oxidized only to form a mixture of aldehyde and ketone at the end of the reaction.
c) Primary alcohols have a continuous carbon chain, while secondary and tertiary alcohols have branched carbon chains.
d) Primary and secondary alcohols react with acidified potassium permanganate, while tertiary do not.

✓ Correct Answer: (d) Primary and secondary alcohols react with acidified potassium permanganate, while tertiary do not.

Tertiary alcohols lack hydrogen atoms on the carbinol carbon (C-OH), making them highly resistant to oxidation under normal conditions with reagents like KMnO₄ or K₂Cr₂O₇. Primary and secondary alcohols are easily oxidized.

Question 32

Three organic compounds are used in the manufacture of polymers:
- (Reaction of compound (1) with compound (2)) gives polymer used to treat heart disease.
- (Reaction compound 3 with an aldehyde) gives polymer used in the manufacture Of cigarette ash trays.
- Which of the following represents each of the three compounds?

Option	Compound 1	Compound 2	Compound 3
a)	Ethyne	Hydrogen fluoride	Teflon
b)	Ethylene glycol	Phthalic Acid	Carbolic acid
c)	Ethylene Glycol	Terphthalic Acid	Carbolic acid
d)	Phthalic Acid	Carbolic acid	Ethylene Glycol

✓ Correct Answer: (c) Ethylene Glycol / Terphthalic Acid / Carbolic acid

Dacron, used for artificial heart valves/vessels, is a polyester formed by the condensation of Ethylene Glycol and Terephthalic Acid. Bakelite, used for heat-resistant ash trays, is a thermosetting polymer formed from Carbolic acid (Phenol) and an aldehyde (Formaldehyde).

Question 33

Fe₂O₃ can be obtained by all of the following methods except:

a) Addition of dil H₂SO₄ on Fe₃O₄ reduction product / alkali / heat >200°C"
b) Passing hot air on red hot iron / Oxidation
c) Leaving iron (II) sulphate in air for a period of time / Addition of alkali / Heating above 200°C
d) Addition of iron to diluted oxalic acid / Heating in the absence of air / Oxidation

✓ Correct Answer: (a) Addition of (dil.) H₂SO₄ to the product of reduction Fe₃O₄ / Addition of alkali / Heating above 200°C

Why Option (a) fails to produce Fe₂O₃: Reducing \(\text{Fe}_3\text{O}_4\) at high temperatures (\(>700^\circ\text{C}\)) yields pure Iron (\(\text{Fe}\)). Adding diluted \(\text{H}_2\text{SO}_4\) to Iron produces Iron(II) sulfate, which reacts with an alkali to form Iron(II) hydroxide. Heating \(\text{Fe(OH)}_2\) above \(200^\circ\text{C}\) yields Iron(II) oxide (\(\text{FeO}\)), not \(\text{Fe}_2\text{O}_3\):
\(\text{Fe} + \text{H}_2\text{SO}_4(dil) \rightarrow \text{FeSO}_4 + \text{H}_2\)
\(\text{FeSO}_4 + 2\text{NaOH} \rightarrow \text{Fe(OH)}_2 \downarrow + \text{Na}_2\text{SO}_4\)
\(\text{Fe(OH)}_2 \xrightarrow{\Delta >200^\circ\text{C}} \{\text{FeO}} + \text{H}_2\text{O}\)

💡 Why Other Options Successfully Produce Fe₂O₃

Option b: Passing hot air over red-hot iron produces magnetic iron oxide (\(\text{Fe}_3\text{O}_4\)), which can then be completely oxidized to form Iron(III) oxide:
\(3\text{Fe} + 2\text{O}_2 \xrightarrow{\Delta} \text{Fe}_3\text{O}_4\)
\(2\text{Fe}_3\text{O}_4 + \frac{1}{2}\text{O}_2 \xrightarrow{\Delta} \{3\text{Fe}_2\text{O}_3}\)
Option c: Leaving \(\text{FeSO}_4\) in air oxidizes it to an Iron(III) salt. Adding an alkali precipitates Iron(III) hydroxide, which easily thermally decomposes above \(200^\circ\text{C}\) to give Iron(III) oxide:
\(\text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe(OH)}_3 \downarrow\)
\(2\text{Fe(OH)}_3 \xrightarrow{\Delta >200^\circ\text{C}} \{\text{Fe}_2\text{O}_3} + 3\text{H}_2\text{O}\)
Option d: Reacting iron with oxalic acid yields Iron(II) oxalate. Heating it in the absence of air yields \(\text{FeO}\), which is then oxidized in the final step to form Iron(III) oxide:
\(\text{Fe(COO)}_2 \xrightarrow{\Delta \text{ (no air)}} \text{FeO} + \text{CO} + \text{CO}_2\)
\(2\text{FeO} + \frac{1}{2}\text{O}_2 \xrightarrow{\Delta} \{\text{Fe}_2\text{O}_3}\)

Question 34

Which of the following processes is necessary to produce steel iron from a sample of siderite?

a) Roasting - Oxidation - Removing impurities - Addition of carbon.
b) Roasting - Reduction - Removing impurities - Addition of carbon.
c) Removing impurities - Roasting - Addition of carbon - Reduction.
d) Removing impurities - Reduction - Addition of carbon - Roasting.

✓ Correct Answer: (b) Roasting - Reduction - Removing impurities - Addition of carbon.

Siderite (FeCO₃) is first roasted to yield Fe₂O₃. It is then reduced in a blast furnace to produce molten iron. The iron is moved to a steel converter (oxygen furnace) where impurities are removed by oxidation, and finally, precise amounts of carbon are added to form the steel alloy.

Question 35

5 g of impure sample of aluminum sulphate Al₂(SO₄)₃ was dissolved in water. An excess amount of ammonia solution was added until aluminum hydroxide was completely precipitated. The precipitate was separated and dried, yielding a mass of 2.03 g. Which of the following represents the percentage of aluminum in the sample?
[Al=27, S=32, N=14, O=16, H=1]

a) 85.95%
b) 14.05%
c) 52.5%
d) 8.75%

✓ Correct Answer: (b) 14.05%

Molar mass of Al(OH)₃ = 27 + (3 × 17) = 78 g/mol.
Moles of precipitate = 2.03 g / 78 g/mol = 0.0260 mol of Al.
Mass of Al = 0.0260 mol × 27 g/mol = 0.702 g.
Percentage in 5 g sample = (0.702 / 5) × 100 ≈ 14.05%.

Question 36

A solution containing (Ag⁺, K⁺, Cu²⁺) cations.
-Which of the following expresses the possible color of the precipitate formed when excess of HCl solution is added and then H₂S gas is passed through this solution?

(a) Black precipitate
(b) White precipitate
(c) A mixture of white and black precipitates
(d) No precipitate is formed

✓ Correct Answer: (c) A mixture of white and black precipitates

Adding HCl precipitates Group I cations: Ag⁺ forms White AgCl. Then, adding H₂S in the acidic medium precipitates Group II cations: Cu²⁺ forms Black CuS. Thus, the resulting mixture contains both white and black precipitates.

Question 37

Lead (II) chloride is sparingly soluble in water, its solubility product [Ksp=1.2x 10^-5] at 25°C
-Which of the following represents the volume of a saturated solution containing 0.1 g of salt at the same temperature? (PbCl₂ = 278 g/mol)

(a) 25 mL
(b) 50 mL
(c) 100 mL
(d) 150 mL

✓ Correct Answer: (a) 25 mL

PbCl₂ dissociates into 3 ions, Ksp = 4S³ = 1.2 × 10^-5.
S = &root3(1.2 × 10^-5 / 4) ≈ 0.0144 M.
Moles of 0.1 g = 0.1 / 278 ≈ 3.6 × 10^-4 moles.
Volume = Moles / Molarity = 3.6 × 10^-4 / 0.0144 ≈ 0.025 L = 25 mL.

Question 38

In the following equilibrium reaction: A_(g) + B_(g) ⇌ 2C_(g)
- The concentrations of reactants and products at 200°C in 1L container are:
[A] = 0.200 M, [B] = 3.00 M, [C] = 0.500 M
- Which of the following represents the number of moles of (A) that must be added To make [C] = 0.700 M at the same temperature?

a) 0.225 mol
b) 0.305 mol
c) 0.417 mol
d) 0.610 mol

✓ Correct Answer: (b) 0.305 mol

K_c = (0.500)² / (0.200 × 3.00) = 0.4167.
To increase [C] from 0.5 to 0.7, 0.2 moles of C must form, meaning 0.1 moles of A and B react. New [B] = 3.0 - 0.1 = 2.9 M.
At new equilibrium: (0.700)² / ([A]_eq × 2.9) = 0.4167 → [A]_eq ≈ 0.405 M.
The total A required before reaction shift is 0.405 + 0.1 = 0.505 M.
We started with 0.200 M, so we must add 0.505 - 0.200 = 0.305 mol.

Question 39

From the opposite figure:

All of the following are correct for cells (1) and (2) except:

a) Cells (2) is used to plate metal (X)
b) Cell (1) is used to purify metal (X)
c) The potential of battery is slightly greater than That of metal (X)
d) The oxidation potential of substance (A) is higher than that of (X)

✓ Correct Answer: (d) The oxidation potential of substance (A) is higher than that of (X)

(A) represents an external DC source forcing a non-spontaneous reaction (Electrolytic cell). By definition, for the electrolysis to proceed, the applied potential (EMF of the battery) must be greater than the cell's opposing galvanic potential. However, it doesn't directly mean A's chemical oxidation potential is higher, it merely acts as a power source overcoming the cell's EMF.

Question 40

In Daniell cell, if the mass of consumed zinc is 6.5 g, then the mass of deposited copper is: (Zn = 65, Cu = 63.5)

a) 6.5 g, deposited at the positive electrode
b) 6.35 g, deposited at the anode half-cell
c) 6.35 g, deposited at the positive electrode
d) 3.25 g , deposited at the cathode half-cell

✓ Correct Answer: (c) 6.35 g, deposited at the positive electrode

Moles of Zinc consumed (oxidized) = 6.5 g / 65 g/mol = 0.1 mol.
Since both Zn and Cu are divalent (exchange 2 e^-), 0.1 mol of Cu is deposited.
Mass of Cu = 0.1 mol × 63.5 g/mol = 6.35 g.
In a Galvanic (Daniell) cell, Copper acts as the Cathode, which is the POSITIVE electrode.

Question 41

From the following diagram:

-Which of the following is true?

a) (X): Used in the manufacture of insecticides, (Y): Dibasic acid
b) (X): Used in car radiators, (Y): Removes the color of acidified KMnO₄
c) (X): Used in the manufacture of polymers, (Y): Is C₂H₂O₄
d) (X): Used in the treatment of burns, (Y): Is C₂H₂O₄

✓ Correct Answer: (C) (X): Used in the manufacture of polymers, (Y): Is C₂H₂O₄

Reacting Ethene (C₂H₄) with alkaline KMnO₄ (Baeyer's Test) yields Ethylene Glycol (X), used as an antifreeze in car radiators. Reacting it with acidic KMnO₄ causes the double bond to cleave or fully oxidize, rapidly destroying the purple color of the permanganate.

Compound (X) is Ethylene Glycol (\(\text{C}_2\text{H}_6\text{O}_2\)): Produced via Baeyer's reaction by reacting ethene with cold alkaline \(\text{KMnO}_4\). It is an essential monomer used alongside terephthalic acid in the condensation polymerization industry to manufacture Dacron polyester fibers.
\(\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} + [\text{O}] \xrightarrow{\text{Alkaline } \text{KMnO}_4} \{\text{CH}_2\text{OH}-\text{CH}_2\text{OH}}\)
Compound (Y) is Oxalic Acid (\(\text{C}_2\text{H}_2\text{O}_4\)): Formed when both primary alcohol groups of ethylene glycol undergo full, rigorous oxidation with hot acidic \(\text{KMnO}_4\), yielding a dicarboxylic acid structure (\(\text{HOOC}-\text{COOH}\)) matching the formula C₂H₂O₄.
\(\text{CH}_2\text{OH}-\text{CH}_2\text{OH} + 4[\text{O}] \xrightarrow{\text{Acidic } \text{KMnO}_4} \{\text{COOH}-\text{COOH}} + 2\text{H}_2\text{O}\)

Option a: Ethylene glycol is not an insecticide agent; it is primarily utilized as a polymer precursor or automotive coolant.
Option b: While ethylene glycol is famously utilized in automotive car radiators, oxalic acid (\(\text{Y}\)) is already at its ultimate dicarboxylic oxidation state and cannot act as a reducing agent under typical introductory conditions to rapidly decolorize \(\text{KMnO}_4\) without high-temperature heat assistance.
Option d: Ethylene glycol is toxic and never utilized in direct clinical skin medical therapy for treatment of thermal burns.

Question 42

By performing the following processes (in order) on Organic acid that has a pungent odor and freezes at 16°C):
(Neutralization - Dry distillation - Heating then rapid cooling - Trimerization).
- Which of the following represents the final product of these processes?

a) Branched aromatic hydrocarbon.
b) Unsaturated aliphatic hydrocarbon.
c) Unbranched aromatic hydrocarbon.
d) Saturated aliphatic hydrocarbon.

✓ Correct Answer: (c) Unbranched aromatic hydrocarbon.

The acid is Acetic Acid (glacial acetic acid freezes at 16°C).
1. Neutralization → Sodium Acetate (CH₃COONa)
2. Dry distillation → Methane gas (CH₄)
3. Heating (1500°C) & rapid cooling → Ethyne / Acetylene (C₂H₂)
4. Trimerization in a red-hot Ni tube → Benzene (C₆H₆), which is an unbranched aromatic hydrocarbon.

Question 43

Which of the following represents the correct order of processes required to obtain the compound with the formula C₆H₅COOH from a compound with the formula C₇H₁₅COOH?

a) Catalytic reforming - Dry distillation - Oxidation - Neutralization
b) Neutralization - Dry distillation - Oxidation - Catalytic reforming
c) Thermal catalytic cracking - Catalytic reforming - Neutralization
d) Neutralization - Dry distillation - Catalytic reforming - Oxidation

✓ Correct Answer: (d) Neutralization - Dry distillation - Catalytic reforming - Oxidation

📚 Detailed Explanation

To convert Octanoic acid (C₇H₁₅COOH) to Benzoic acid (C₆H₅COOH):

1. Neutralization: Reacting octanoic acid with NaOH produces sodium octanoate (C₇H₁₅COONa).
2. Dry Distillation: Heating sodium octanoate with soda lime produces Heptane (C₇H₁₆).
3. Catalytic Reforming: Passing heptane over heated Pt reshapes it into an aromatic ring, yielding Toluene (C₆H₅CH₃).
4. Oxidation: Oxidizing toluene with oxygen (V₂O₅ cat.) yields Benzoic acid (C₆H₅COOH).

Question 44

Three organic compounds:
(A): An aliphatic organic compound with the general formula C_nH_2n+2O₂
(B): An aromatic acid containing two different functional groups
(C): A liquid organic compound used in the manufacture of artificial fats and as a Base for manufacture of compound with a general formula C_nH_2n+2O₃
- Which of the following is true for compounds A, B, and C?

a) Compound (A) effervesces when sodium bicarbonate is added to it.
b) Compound (B) turns potassium dichromate from orange to green.
c) Compound (C) removes the color of alkaline potassium permanganate solution.
d) Compound (A) gives a purple color with Iron III chloride solution.

✓ Correct Answer: (c) Compound (C) removes the color of alkaline potassium permanganate solution.

- Compound A (e.g. Ethylene Glycol) is an alcohol and does not react with NaHCO₃ or FeCl₃.
- Compound B (Salicylic Acid) is resistant to dichromate oxidation.
- Compound C is an unsaturated vegetable oil. It contains C=C double bonds which undergo addition with alkaline KMnO₄ (Baeyer's test), rapidly removing the purple color.

Question 45 (Essay)

You have four test tubes containing different salt solutions:

Write the chemical name of the reagent used to distinguish between the solid Salts of the four solutions by one experiment.
Which of the four solutions gives a precipitate with an excess of sodium Hydroxide solution?
Which of the above solutions gives a gas and a precipitate when dilute sulphuric Acid is added?
Which of the above salts cannot detects its cation by a normal chemical reagent?

✓ Model Answers

Concentrated Sulfuric Acid (H₂SO₄): Will cause effervescence (CO₂) with A, brown fumes (NO₂) with B, acidic fumes (HCl) with D, and no reaction with C.
FeSO₄ (Solution C): Forms a greenish-white precipitate of Fe(OH)₂ which does not dissolve in excess NaOH. (Al³⁺ in B forms a precipitate that dissolves in excess).
Ca(HCO₃)₂ (Solution A): Reacting with dil. H₂SO₄ yields CO₂ gas and sparingly soluble CaSO₄ (precipitate).
KCl (Solution D): Potassium (K⁺) salts are highly soluble; detection usually requires a physical flame test (violet flame) rather than standard chemical precipitation.

Question 46 (Essay)

From the following diagram:

If you know that compound (E) is used as a medicinal drug and is recommended to be taken with plenty of water to prevent stomach ulcers:
- Write the chemical formulas of compounds (Z) and (Y).
- Write the chemical formula of the acid produced by the acidic hydrolysis of Compound (F).
- Mention one use of compound (F).

✓ Model Answers

1. Chemical formulas of compounds (Z) and (Y):

C₇H₈O

₃

₆

₄

C₇H₇Cl

₃

₆

₄

2. Chemical formula of the acid produced by the acidic hydrolysis of Compound (F):
- Salicylic acid: C₇H₆O₃ (or HOC₆H₅COOH)

3. One use of compound (F) (Methyl salicylate):
- Used as a topical analgesic (pain reliever) in medical ointments, liniments, and counter-irritant rubs to soothe muscle pain and joint stiffness.

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44 Multiple Choice + 2 Essay Questions

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Total Score: ((1:32)x1M =32M +(33:42)x2M =24M MCQ ) = 56M + 2 Essay x2M = 4M total 56+4=60M

Filled circle = Correct Answer | Empty circle = Incorrect Option

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44 Multiple Choice + 2 Essay Questions

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Total Score: 44 MCQ × 2.5 pts = 110 points + 2 Essays = 40 points

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Chemistry 2026

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🧪 Chemical Explanation & Equations

💡 Formation from True Isomers (Why Options a, b, c Fail)

Question 25

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Question 33

💡 Why Other Options Successfully Produce Fe₂O₃

Question 34

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Question 45 (Essay)

Question 46 (Essay)

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