VITEEE Practice Test 2026

About the VITEEE exam

The VIT Engineering Entrance Examination (VITEEE) is a engineering examination administered in India. It is designed to assess candidates' knowledge and proficiency across 5 key subject areas including Physics, Chemistry, Mathematics, English. This exam is widely recognized as an essential benchmark for aspiring engineers.

Use short VITEEE practice sets to identify topics to revisit. Choose a subject and difficulty, answer questions, then review the explanations. New accounts get 30 question credits; further new questions use paid credits. Retaking previously seen questions is free.

Whether you're a first-time candidate or retaking the VITEEE to improve your score, InfinityMock helps you identify your strengths and weaknesses through detailed performance analytics. Our platform covers Physics, Chemistry, Mathematics, English, Aptitude, with questions available at Easy, Medium, and Hard difficulty levels. Track your progress, review detailed explanations, and build the confidence you need to pass on exam day.

VITEEE — quick answers

VITEEE at a glance

8 free VITEEE practice questions (2026)

Answer these first — no signup, no card. Every answer is explained. For more practice, sign up for 30 free question credits and choose a subject above.

1. Which of the following coordination complexes is expected to exhibit the most pronounced Jahn-Teller distortion in its ground state?

   • [Cr(H2O)6]3+
   • [Mn(H2O)6]3+
   • [Fe(H2O)6]3+
   • [Ni(H2O)6]2+
   Show answer & explanation

   Answer: B. [Mn(H2O)6]3+

   [Mn(H2O)6]3+ is a high-spin d4 system (t2g3 eg1). The single electron in the eg orbital (which points directly towards the ligands) causes strong asymmetric electronic distribution, leading to significant Jahn-Teller distortion. d3 (Cr3+), high-spin d5 (Fe3+), and d8 (Ni2+) systems have symmetrically filled t2g or eg subshells, resulting in negligible or no distortion.

   Topic: Chemistry · hard

2. Consider a concentration cell: Pt | H2(g, 1 bar) | HCl (a1) || HCl (a2) | H2(g, 1 bar) | Pt. The electromotive force (EMF) of this cell at 298 K is found to be 0.118 V. If the mean ionic activity coefficient of HCl in the first compartment is 0.90 at a concentration of 0.01 M, what is the activity (a2) of the second compartment?

   • 0.10
   • 0.90
   • 0.09
   • 0.009
   Show answer & explanation

   Answer: B. 0.90

   The cell potential for a concentration cell with transference is given by E = (2 * R * T / F) * ln(a2 / a1) or for a simple concentration cell without transference with hydrogen electrodes: E = (RT/F) * ln(a2/a1) at 298 K. Here, E = 0.0591 * log10(a2/a1). 0.118 = 0.0591 * log10(a2/a1) => log10(a2/a1) = 2 => a2/a1 = 100. Given a1 = m * gamma = 0.01 * 0.90 = 0.009. Therefore, a2 = 100 * 0.009 = 0.90.

   Topic: Chemistry · hard

3. When (R)-2-phenylpropanamide is treated with Br2 and KOH, the resulting amine obtained is:

   • (R)-1-phenylethanamine with retention of configuration
   • (S)-1-phenylethanamine with inversion of configuration
   • A racemic mixture of 1-phenylethanamine
   • (R)-2-phenylethanamine
   Show answer & explanation

   Answer: A. (R)-1-phenylethanamine with retention of configuration

   The reaction described is the Hoffmann bromamide degradation. The mechanism involves the migration of the alkyl/aryl group from the carbonyl carbon to the nitrogen atom. This intramolecular migration occurs with complete retention of configuration at the migrating chiral center. Thus, (R)-2-phenylpropanamide yields (R)-1-phenylethanamine.

   Topic: Chemistry · hard

4. For a consecutive first-order reaction A -> B -> C with rate constants k1 and k2 respectively, under what condition will the concentration of the intermediate B reach its maximum value at the shortest possible time?

   • k1 >> k2
   • k1 = k2
   • k2 >> k1
   • The time to reach maximum concentration of B is independent of k1 and k2
   Show answer & explanation

   Answer: A. k1 >> k2

   The time at which the concentration of B is maximum is given by t_max = ln(k2/k1) / (k2 - k1). If k1 >> k2, the value of (k2 - k1) is negative and the ratio k2/k1 is very small, leading to a small positive t_max. Conceptually, if k1 is extremely large, B is formed almost instantly, reaching its maximum concentration rapidly before slowly decaying via the slower k2 path.

   Topic: Chemistry · hard

5. In a normal spinel structure, the divalent metal ions occupy:

   • One-eighth of the tetrahedral voids, and trivalent ions occupy half of the octahedral voids
   • Half of the tetrahedral voids, and trivalent ions occupy one-eighth of the octahedral voids
   • One-fourth of the tetrahedral voids, and trivalent ions occupy one-fourth of the octahedral voids
   • All octahedral voids, and trivalent ions occupy tetrahedral voids
   Show answer & explanation

   Answer: A. One-eighth of the tetrahedral voids, and trivalent ions occupy half of the octahedral voids

   In a normal spinel (AB2O4), the oxide ions form an FCC lattice. Divalent cations (A2+) occupy 1/8th of the tetrahedral voids, and trivalent cations (B3+) occupy 1/2 of the octahedral voids.

   Topic: Chemistry · hard

6. One mole of an ideal gas undergoes a reversible adiabatic expansion from volume V1 to V2. The entropy change of the universe (delta S_univ) for this process is:

   • R ln(V2/V1)
   • -R ln(V2/V1)
   • Zero
   • Cv ln(T2/T1)
   Show answer & explanation

   Answer: C. Zero

   For any reversible process, the total entropy change of the universe (system + surroundings) is exactly zero. Since it is also an adiabatic process, the entropy change of the system is zero (dq_rev = 0) and the entropy change of the surroundings is also zero.

   Topic: Chemistry · hard

7. Silicon tetrachloride (SiCl4) undergoes rapid hydrolysis to give silica gel, whereas carbon tetrachloride (CCl4) is inert towards water under normal conditions. This difference is primary because:

   • The Si-Cl bond is weaker than the C-Cl bond
   • Silicon has vacant 3d orbitals that can accept a lone pair from the oxygen atom of water
   • SiCl4 is non-polar, whereas CCl4 is polar
   • Silicon is more electronegative than carbon
   Show answer & explanation

   Answer: B. Silicon has vacant 3d orbitals that can accept a lone pair from the oxygen atom of water

   Silicon is a third-period element and possesses vacant 3d orbitals in its valence shell. During hydrolysis, water molecules act as nucleophiles, donating lone pairs of electrons to these vacant d-orbitals to form a coordinate bond, facilitating the displacement of Cl- ions. Carbon (second period) lacks vacant d-orbitals and cannot expand its coordination number to coordinate with water.

   Topic: Chemistry · hard

8. Which of the following symmetry elements is present in a chiral molecule of (2R,3R)-tartaric acid?

   • Plane of symmetry (sigma)
   • Center of inversion (i)
   • Alternating axis of symmetry (Sn)
   • Simple axis of symmetry (Cn)
   Show answer & explanation

   Answer: D. Simple axis of symmetry (Cn)

   A chiral molecule cannot possess any improper symmetry element (such as a plane of symmetry, center of inversion, or alternating axis of symmetry). However, (2R,3R)-tartaric acid possesses a C2 simple axis of symmetry. Rotating the molecule by 180 degrees around this axis yields an identical spatial arrangement.

   Topic: Chemistry · hard

Why Practice with InfinityMock's VITEEE Mock Test?

• Fresh practice questions — Use credits for new questions; retake seen questions free
• Instant Results & Explanations — Learn from every question with detailed answer explanations
• Performance Analytics — Track your progress with grade reports, difficulty breakdowns, and time analysis
• Independent practice — Use alongside current official VITEEE materials
• Multiple Difficulty Levels — Practice Easy, Medium, or Hard VITEEE questions

VITEEE Exam Subjects

Available practice subjects: Physics, Chemistry, Mathematics, English, Aptitude. Select a subject for focused preparation. Coverage is not a guarantee of the complete official syllabus.

📖 VITEEE Study Guide

Read our comprehensive How to Pass the VITEEE in 2026 — Complete Study Guide for expert preparation tips, study plans, and strategies.

🔗 Related Exams

Students preparing for VITEEE also practice for:

Frequently Asked Questions

📚 Browse All 275+ Exams →