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JEE Main Chemistry Mock Test- 5 Free Online Test 2026


Full Mock Test & Solutions: JEE Main Chemistry Mock Test- 5 (25 Questions)

You can boost your JEE 2026 exam preparation with this JEE Main Chemistry Mock Test- 5 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Total Questions: 25
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Section I, Section II

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JEE Main Chemistry Mock Test- 5 - Question 1
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1-Phenylethanol can be prepared by the reaction of benzaldehyde with

Detailed Solution: Question 1

1-Phenylethanol can be synthesised through the Grignard reaction, which involves the reaction of benzaldehyde with a suitable organomagnesium reagent. The most effective reagent for this purpose is methyl iodide combined with magnesium.

  • Benzaldehyde reacts with methyl iodide in the presence of magnesium.
  • This reaction forms a new carbon-carbon bond, resulting in the production of 1-phenylethanol.
  • The use of Grignard reagents is crucial as they are highly reactive and facilitate this transformation.

Other reagents such as ethyl iodide or methyl bromide are less suitable for this specific reaction. In summary, the combination of methyl iodide and magnesium is the correct choice for effectively preparing 1-phenylethanol from benzaldehyde.

JEE Main Chemistry Mock Test- 5 - Question 2
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Lucas reagent is

Detailed Solution: Question 2

Lucas reagent is a chemical reagent used to classify and identify alcohols based on their reactivity with zinc chloride and hydrochloric acid.

  • Concentration: The Lucas reagent must be concentrated for effectiveness.
  • Components: It typically consists of hydrochloric acid (HCl) and zinc chloride (ZnCl₂).
  • Formulation: The reagent can be prepared with either anhydrous or hydrous zinc chloride.

In this context, the reagent options include:

  • Concentrated HCl and anhydrous ZnCl₂
  • Concentrated HNO₃ and hydrous ZnCl₂
  • Concentrated HCl and hydrous ZnCl₂
  • Concentrated HNO₃ and anhydrous ZnCl₂

The correct formulation for the Lucas reagent is the combination of conc. HCl and anhydrous ZnCl₂.

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JEE Main Chemistry Mock Test- 5 - Question 3
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Which of the following sequence is correct as per Aufbau principle?

Detailed Solution: Question 3

According to the Aufbau principle, the correct order of electron filling in atomic orbitals is important for understanding electron configuration.

  • The Aufbau principle states that electrons fill orbitals starting from the lowest energy level to the highest.
  • In the provided options, we need to identify the correct sequence of filling orbitals.
  • The correct order of energy levels is:
    • 1s (lowest energy level)
    • 2s
    • 2p
    • 3s
    • 3p
    • 4s
    • 3d
    • 4p
    • 5s
    • 4d
    • 5p
    • 5d
  • From the options provided, the sequence 1s < 2p="">< 4s=""><> is accurate.

This sequence reflects the correct filling order based on energy levels, making it essential for predicting the arrangement of electrons in an atom.

JEE Main Chemistry Mock Test- 5 - Question 4
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Which of the following is the least polar molecule?

Detailed Solution: Question 4

Polarity in molecules is determined by the distribution of electrical charge. In simple terms, a molecule is polar if it has a positive end and a negative end, leading to an uneven distribution of charge.

The least polar molecule among the given options is Cl₂. Here’s why:

  • Cl₂ (Chlorine gas) is a diatomic molecule made of two identical atoms, which means the charge is evenly distributed, resulting in no dipole moment.
  • HCl (Hydrochloric acid) and HBr (Hydrobromic acid) both have significant differences in electronegativity between hydrogen and chlorine or bromine, making them polar.
  • BrCl (Bromine monochloride) is also polar due to the difference in electronegativity between bromine and chlorine.

In summary, since Cl₂ has no polar character due to its identical atoms, it is the least polar molecule compared to the others listed.

JEE Main Chemistry Mock Test- 5 - Question 5
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Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively :

Detailed Solution: Question 5

The ground state electron configuration of a Chromium (Cr) atom, which has an atomic number of 24, can be understood by examining the distribution of its electrons among different energy levels and subshells.

In this configuration:

  • Electrons are arranged according to the principle quantum number (n), starting from the lowest energy level.
  • The azimuthal quantum number (l) defines the shape of the orbital. For this case:
    • l = 1 corresponds to the p orbitals.
    • l = 2 corresponds to the d orbitals.

For Chromium:

  • The total number of electrons with l = 1 (p orbitals) is 12.
  • The total number of electrons with l = 2 (d orbitals) is 5.

Thus, the correct counts for the azimuthal quantum numbers are:

  • 12 electrons in p orbitals (l = 1).
  • 5 electrons in d orbitals (l = 2).

JEE Main Chemistry Mock Test- 5 - Question 6
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Which of the following alkyl group has the maximum + I effect?

Detailed Solution: Question 6

The +I effect, or positive inductive effect, refers to the ability of alkyl groups to donate electron density through sigma bonds. This effect stabilises positive charges in a molecule. In evaluating which alkyl group has the maximum +I effect, we can consider the structure of each option:

  • CH₃ - (Methyl group): This group has a limited +I effect due to its small size and fewer carbon atoms.
  • CH₃CH₂ - (Ethyl group): This group has a stronger +I effect than the methyl group, as it has one more carbon atom to donate electron density.
  • (CH₃)₂CH - (Isopropyl group): This group has an even greater +I effect due to the presence of two methyl groups, enhancing its ability to donate electrons.
  • (CH₃)₃C - (Tert-butyl group): This group exhibits the strongest +I effect among the options. The three methyl groups significantly increase its electron-donating ability.

In summary, the tert-butyl group (CH₃)₃C - demonstrates the maximum +I effect due to its three methyl groups, which effectively stabilise positive charges and enhance electron donation.

JEE Main Chemistry Mock Test- 5 - Question 7
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Which of the following factors do not favour electrovalency?
1. Low charge on ions
2. High charge on ions
3. Large cation and small anion
4. Small cation and large anion 

Detailed Solution: Question 7

Electrovalency is favoured when polarization is minimum. According to Fajans’ rules, high charge on ions and a small cation with a large anion increase polarization, thereby reducing ionic (electrovalent) character. Hence, 2 and 4 do not favour electrovalency.

JEE Main Chemistry Mock Test- 5 - Question 8
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Alkyl groups are o- and p-directing because of

Detailed Solution: Question 8

Alkyl groups are considered o- and p-directing due to their electronic effects. These effects can be summarised as follows:

  • Inductive Effect: Alkyl groups are electron-donating through the sigma bond. This effect decreases with distance but helps to stabilise positive charges and increases electron density on the aromatic ring.
  • Mesomeric Effect: Alkyl groups can donate electron density through resonance. This stabilisation of the ring's electron system makes it more reactive towards electrophiles, particularly at the ortho (o) and para (p) positions.
  • Hyperconjugation Effect: The overlap of the sigma bond electrons from the alkyl group with the pi system of the aromatic ring enhances the stability of the o- and p- positions, making them more favourable for substitution reactions.

In summary, the directing effects of alkyl groups in electrophilic aromatic substitution reactions arise mainly from their ability to donate electron density, stabilising the intermediate formed during the reaction.

JEE Main Chemistry Mock Test- 5 - Question 9
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Glucose reacts with acetic anhydride to form

Detailed Solution: Question 9

Glucose reacts with acetic anhydride to produce different types of acetates, which are derivatives formed by the addition of acetic acid. The specific product depends on the number of acetate groups attached to the glucose molecule.

  • The reaction can yield a monoacetate, where one acetate group is added.
  • A tetra-acetate results from the addition of four acetate groups.
  • A penta-acetate forms when five acetate groups are attached.
  • The final product, a hexa-acetate, occurs when all six hydroxyl groups of glucose are replaced by acetate groups.

In this case, the correct answer is the formation of the penta-acetate, which indicates that five acetate groups have reacted with glucose.

JEE Main Chemistry Mock Test- 5 - Question 10
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Which of the following shows geometrical isomerism ?

Detailed Solution: Question 10

Geometrical isomerism occurs when compounds have the same molecular formula but differ in the spatial arrangement of their atoms. This phenomenon is particularly common in alkenes where double bonds restrict rotation.

  • But-1-ene: Does not exhibit geometrical isomerism as it has no substituents on the double bond.
  • But-2-ene: Exhibits geometrical isomerism. It can exist in two forms:
    • cis form: Both substituents are on the same side of the double bond.
    • trans form: Substituents are on opposite sides.
  • Propene: Lacks geometrical isomerism due to its simple structure with no different substituents on the double bond.
  • 1,1-Dichlorobutane: Does not show geometrical isomerism as both chlorine atoms are attached to the same carbon, limiting spatial variation.

In summary, only But-2-ene displays geometrical isomerism due to its ability to form distinct cis and trans configurations.

JEE Main Chemistry Mock Test- 5 - Question 11
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The co-ordination no. of a metal in a co-ordination compounds is

Detailed Solution: Question 11

The coordination number of a metal in coordination compounds refers to the number of ligands bonded to the central metal atom. This number is crucial for understanding the structure and properties of the compound.

  • The coordination number is primarily determined by the valency of the metal.
  • It is also influenced by factors such as the size of the metal ion and the nature of the ligands.
  • In many cases, the coordination number corresponds to the secondary valency of the metal.
  • The most common coordination numbers are 4 and 6, but other numbers can occur depending on the metal and ligands involved.

Understanding coordination numbers helps predict the geometry of the compound, which can be tetrahedral, octahedral, or other shapes. This knowledge is essential in fields such as coordination chemistry, as it affects the reactivity and stability of the compounds.

JEE Main Chemistry Mock Test- 5 - Question 12
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Four successive members of the first row transition elements are listed below with their atomic no. Which of them is expected to have the highest third ionisation enthalpy?

Detailed Solution: Question 12

The third ionisation enthalpy is the energy required to remove the third electron from an atom. This property is influenced by several factors, including the electronic configuration and the stability of the resulting ion.

  • Iron (Z=26): Iron has a stable electronic configuration, leading to a relatively moderate ionisation enthalpy.
  • Chromium (Z=24): Chromium's unique configuration, which includes a half-filled d-subshell, makes it somewhat stable, resulting in a higher ionisation energy than iron.
  • Vanadium (Z=23): Vanadium has a less stable configuration compared to chromium, hence its third ionisation enthalpy is lower.
  • Manganese (Z=25): Manganese also has a half-filled d-subshell, contributing to a higher stability, which increases its third ionisation enthalpy.

Among these elements, manganese is expected to have the highest third ionisation enthalpy due to its stable electron configuration, which provides significant energy requirements to remove an additional electron.

JEE Main Chemistry Mock Test- 5 - Question 13
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Which is a planar molecule?

Detailed Solution: Question 13

Planar molecules are those that lie in a single plane. Understanding the geometry of these molecules is essential in chemistry. Here are some key points regarding the characteristics of planar molecules:

  • Geometry: Planar molecules typically have a geometry that allows for all atoms to be in the same flat plane.
  • Bond Angles: The bond angles in planar molecules are usually 120 degrees, which is common in trigonal planar structures.
  • Examples: Certain molecules, like XeF4, exhibit this planar characteristic due to their specific arrangements of atoms and lone pairs.
  • Lone Pairs: The presence of lone pairs can influence the molecular geometry, affecting whether a molecule is planar or not.

In conclusion, identifying whether a molecule is planar involves examining its molecular structure and the arrangement of its atoms and bonds.

JEE Main Chemistry Mock Test- 5 - Question 14
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In decinormal solution, CH₃COOH acid is ionised to the extent of 1.3%. If log 1.3=0.11. What is the pH of the solution?

Detailed Solution: Question 14

To calculate the pH of a decinormal solution of acetic acid (CH₃COOH) that is ionised to 1.3%, follow these steps:

  • Determine the concentration of acetic acid: In this case, it is 0.1 M (decinormal).
  • Calculate the concentration of ionised acetic acid:
    • Ionisation percentage is 1.3%.
    • Ionised concentration: 0.1 M × 0.013 = 0.0013 M.
  • Use the formula for pH:
    • pH = -log[H⁺]
    • Substituting the ionised concentration: pH = -log(0.0013).
  • Given that log(1.3) = 0.11, we can find:
    • log(0.0013) = log(1.3) - 3 (since 0.0013 = 1.3 × 10-3).
    • This gives us: log(0.0013) = 0.11 - 3 = -2.89.
  • Finally, calculate the pH:
    • pH = -(-2.89) = 2.89.

The pH of the solution is therefore 2.89.

JEE Main Chemistry Mock Test- 5 - Question 15
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Sodium nitroprusside reacts with sulphide ion to give a purple colour due to the formation of :

Detailed Solution: Question 15

Sodium nitroprusside interacts with the sulphide ion, resulting in a purple colour. This colour change is due to the formation of a specific complex. The key product formed is:

  • [ Fe(CN)5 NOS ]4-

This complex is characterised by the presence of:

  • Iron (Fe) at its core
  • A combination of cyanide (CN) ligands
  • The nitrosyl group (NOS) which contributes to the purple colour

The formation of this complex occurs when sodium nitroprusside reacts with the sulphide ions, leading to a notable visual change that can be observed in chemical analyses.

JEE Main Chemistry Mock Test- 5 - Question 16
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Which are isomers?

Detailed Solution: Question 16

Isomers are compounds that share the same molecular formula but have different structural arrangements. Here are some examples of isomers:

  • Ethyl alcohol (ethanol) and dimethyl ether have the same formula (C2H6O) but differ in structure and properties.
  • Acetone and acetaldehyde both contain three carbon atoms but differ in functional groups and reactivity.
  • Propanoic acid and propanone are both C3H6O compounds; however, their functional groups lead to different characteristics.
  • Methyl alcohol (methanol) and dimethyl ether share the same molecular formula but have different arrangements of atoms.

Understanding isomers is crucial in chemistry as they can exhibit vastly different behaviours despite having the same chemical composition.

JEE Main Chemistry Mock Test- 5 - Question 17
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Chlorination of toluene with excess of chlorine in the presence of light and heat followed by treatment with aqueous NaOH gives

Detailed Solution: Question 17

The process of chlorinating toluene with an excess of chlorine in the presence of light and heat produces several products. After this step, treatment with aqueous NaOH leads to specific outcomes. Here’s a simplified breakdown of the reaction products:

  • o-cresol: This compound is formed as one of the possible products.
  • p-cresol: Another isomer resulting from the chlorination process.
  • 2, 4-Dihydroxytoluene: This compound may also be produced during the reaction.
  • Benzoic acid: The final product of interest, particularly after the treatment with NaOH.

Among these, the key product resulting from the reaction sequence is benzoic acid, which is a significant organic compound used in various applications, including as a food preservative and in the synthesis of other chemicals.

JEE Main Chemistry Mock Test- 5 - Question 18
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In a polymer, 30% of molecules have a molecular weight of 20,000, 40% of molecules have a molecular weight of 30,000, and the rest, 30% of molecules, have a molecular weight of 60,000. The weight average molecular weight of the polymer is

Detailed Solution: Question 18

Given:

  • 30% of the molecules have a molecular weight of 20,000

  • 40% of the molecules have a molecular weight of 30,000

  • 30% of the molecules have a molecular weight of 60,000

Step 1: Convert percentages to decimals (fractions)

  • 30% = 0.30

  • 40% = 0.40

  • 30% = 0.30

Step 2: Use the formula for Weight Average Molecular Weight (Mw)

Formula:
Mw = (weight fraction × molecular weight) for all components, added together.

Now calculate:

  • 0.30 × 20,000 = 6,000

  • 0.40 × 30,000 = 12,000

  • 0.30 × 60,000 = 18,000

Now add them:

  • 6,000 + 12,000 + 18,000 = 36,000

Final Answer: 36,000

JEE Main Chemistry Mock Test- 5 - Question 19
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Electrolysis of molten sodium chloride leads to the formation of

Detailed Solution: Question 19

The process of electrolysis involves breaking down a compound using electricity. When sodium chloride (NaCl) is melted and subjected to electrolysis, the following key products are formed:

  • Sodium (Na): This element is produced at the negative electrode (cathode) due to the reduction of sodium ions.
  • Chlorine (Cl₂): This diatomic molecule is released at the positive electrode (anode) through the oxidation of chloride ions.

During this process:

  • The sodium ions gain electrons, transforming into neutral sodium metal.
  • The chloride ions lose electrons, forming chlorine gas.

It is important to note that:

  • Electrolysis of molten sodium chloride does not produce hydrogen (H₂) or oxygen (O₂).
  • Instead, the primary products are sodium and chlorine.

In summary, the correct products of the electrolysis of molten sodium chloride are:

  • Sodium (Na)
  • Chlorine (Cl₂)

JEE Main Chemistry Mock Test- 5 - Question 20
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In acidic medium dichromate ion oxidizes ferrous ion to ferric ion. If the gram molecular weight of potassium dichromate is 294 grams, its gram equivalent weight is....grams.

Detailed Solution: Question 20

In an acidic medium, the dichromate ion plays a crucial role in the oxidation of ferrous ions to ferric ions. To understand the gram equivalent weight of potassium dichromate, we can break it down as follows:

  • The gram molecular weight of potassium dichromate is 294 grams.
  • To find the gram equivalent weight, we need to consider the number of electrons transferred in the reaction. In this case, the dichromate ion can accept 6 electrons when it reduces to chromium(III).
  • The formula for calculating gram equivalent weight is:
    • Gram Equivalent Weight = Gram Molecular Weight / n
  • Where n is the number of electrons transferred. Here, n = 6.
  • Thus, the calculation is:
    • Gram Equivalent Weight = 294 grams / 6 = 49 grams

Therefore, the gram equivalent weight of potassium dichromate is 49 grams.

*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 5 - Question 21
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The sum of number of unpaired electrons in
[CoCl6]–3, [Cr(NH3)6]+3, [Zn(NH3)4]+2


Detailed Solution: Question 21

The total number of unpaired electrons in the following complexes is calculated as follows:

  • [CoCl6]⁢3: Cobalt has an oxidation state of +3. Its electronic configuration is [Ar] 3d6. In this complex, all six chloride ions are weak field ligands, leading to high spin configuration. This results in 4 unpaired electrons.
  • [Cr(NH3)6]⁢3: Chromium in this complex has an oxidation state of +3, with the configuration [Ar] 3d3. The ammonia ligands are strong field ligands, causing pairing of electrons. This gives 3 unpaired electrons.
  • [Zn(NH3)4]⁢2: Zinc has an oxidation state of +2 and an electronic configuration of [Ar] 3d10. As all d-orbitals are fully filled, there are no unpaired electrons here.

Summing the unpaired electrons from each complex:

  • 4 (from [CoCl6]⁢3)
  • 3 (from [Cr(NH3)6]⁢3)
  • 0 (from [Zn(NH3)4]⁢2)

This results in a total of 7 unpaired electrons.

*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 5 - Question 22
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Sum of number of ions in aqueous solution of CrCl3.5NH3 and CrCl3.4NH3.


Detailed Solution: Question 22

The sum of ions in an aqueous solution of CrCl3.5NH3 and CrCl3.4NH3 can be calculated as follows:

  • Each CrCl3 compound dissociates into:
    • 1 Cr3+ ion
    • 3 Cl- ions
  • For CrCl3.5NH3:
    • It produces 4 ions: 1 Cr3+ + 3 Cl-
  • For CrCl3.4NH3:
    • It also produces 4 ions: 1 Cr3+ + 3 Cl-
  • Combining both solutions:
    • 4 ions from CrCl3.5NH3
    • 4 ions from CrCl3.4NH3
  • Total ions: 4 + 4 = 8 ions.

Therefore, the total number of ions present in the solution is 8.

*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 5 - Question 23
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Find the planner species out of the following species
SF2, SF4, SF6, SO2, SO3


Detailed Solution: Question 23

To identify planner species among the given compounds:

  • SF2 – This species is not a planner molecule due to its bent shape.
  • SF4 – This compound has a seesaw geometry, which is not planner.
  • SF6 – With an octahedral structure, this molecule is also not planner.
  • SO2 – This species is bent and thus is a planner molecule.
  • SO3 – Exhibiting trigonal planar geometry, this molecule is planner as well.

In summary, the planner species identified from the list are SO2 and SO3.

*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 5 - Question 24
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The total number of contributing structures showing hyperconjugation (involving C–H bonds) for the following carbocation is :


*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 5 - Question 25
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How many isomeric amines have the molecular formula C4H11N :


Detailed Solution: Question 25

Isomeric amines with the molecular formula C4H11N can be identified based on their structural variations. Here are the key points regarding the isomers:

  • Isomeric amines are compounds that share the same molecular formula but differ in their structure.
  • For the formula C4H11N, there are a total of 8 distinct isomeric amines.
  • These isomers can include:
    • Primary amines (where the nitrogen atom is attached to one carbon chain).
    • Secondary amines (where the nitrogen atom is bonded to two carbon chains).
    • Tertiary amines (where the nitrogen atom is connected to three carbon chains).
  • The variations arise from different arrangements of the carbon skeleton and the position of the amine group.

Understanding these isomers is important in organic chemistry, as they can exhibit different chemical properties despite having the same molecular formula.

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