Chemistry MDCAT MCQs

Organic, inorganic & physical chemistry MCQs

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Showing 305 MCQs in Chemistry
Chemistry
A. O-H stretch
B. C=O stretch (carbonyl)
C. C-H stretch
D. N-H stretch
Answer: A strong absorption near 1700 cm-1 is characteristic of the C=O stretching vibration, indicating the presence of a carbonyl group (aldehyde, ketone, acid, ester, etc.).
Chemistry
A. +0.76 V
B. -0.76 V
C. +0.34 V
D. -0.34 V
Answer: Zn2+/Zn has a standard reduction potential of -0.76 V, meaning zinc is easily oxidized and is a stronger reducing agent than hydrogen.
Chemistry
A. sp3
B. sp2
C. sp
D. sp3d
Answer: The carbon in a nitrile group has two regions of electron density (one single bond to the adjacent atom and one triple bond to nitrogen), requiring sp hybridization.
Chemistry
A. Aufbau principle
B. t2g (lower) and eg (higher) splitting
C. Hund\'s rule only
D. Pauli exclusion principle
Answer: In an octahedral field, the five d-orbitals split into two sets: t2g (dxy, dxz, dyz, lower energy) and eg (dz2, dx2-y2, higher energy), separated by Delta_o.
Chemistry
A. 2-Bromopropane (Markovnikov product)
B. 1-Bromopropane (anti-Markovnikov product)
C. 1,2-Dibromopropane
D. No reaction
Answer: Peroxides initiate a free radical mechanism that reverses the normal regioselectivity, adding Br to the less substituted carbon (anti-Markovnikov addition).
Chemistry
A. 1,3-Butadiene
B. 1,4-Pentadiene
C. 1,5-Hexadiene
D. Cyclohexene
Answer: 1,3-Butadiene (CH2=CH-CH=CH2) has alternating double-single-double bonds, creating conjugation where p-orbitals overlap across all four carbons.
Chemistry
A. Delta G = RT ln K
B. Delta G = -RT ln K
C. Delta G = -RT / ln K
D. Delta G = K / RT
Answer: The fundamental equation is Delta G = Delta G0 + RT ln Q. At equilibrium, Delta G = 0 and Q = K, so Delta G0 = -RT ln K.
Chemistry
A. 3-Oxobutanal
B. 2-Oxobutanal
C. 3-Ketobutanal
D. 1,3-Dioxobutane
Answer: The compound has both aldehyde (CHO, higher priority at C1) and ketone (C=O) groups. Numbering from CHO: C1(CHO)-C2(H2)-C3(=O)-C4(H3), giving 3-oxobutanal.
Chemistry
A. Principal quantum number (n)
B. Azimuthal quantum number (l)
C. Magnetic quantum number (ml)
D. Spin quantum number (ms)
Answer: The magnetic quantum number (ml) specifies the number of orbitals and their orientation within a subshell, ranging from -l to +l.
Chemistry
A. All octahedral complexes are perfectly symmetric
B. Degenerate electronic states cause geometric distortion to remove degeneracy
C. Tetrahedral complexes are always more stable than octahedral
D. Crystal field splitting is always the same for all metals
Answer: The Jahn-Teller theorem states that any non-linear molecule with a degenerate electronic ground state will undergo geometric distortion to remove the degeneracy and lower the energy.
Chemistry
A. 25.3 kJ/mol
B. 50.6 kJ/mol
C. 75.9 kJ/mol
D. 101.2 kJ/mol
Answer: Using Arrhenius: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). ln(2) = (Ea/8.314)(1/300 - 1/310). Ea = 0.693 x 8.314 x 300 x 310 / 10 = 52,886 J/mol ≈ 50.6 kJ/mol.
Chemistry
A. Rate is proportional to concentration squared
B. Half-life is independent of initial concentration
C. Rate law has only reactant concentration to the first power
D. Plot of ln[A] vs time is linear
Answer: A second-order reaction has Rate = k[A]^2 (or Rate = k[A][B]). The half-life t1/2 = 1/(k[A]0) depends on initial concentration, and a plot of 1/[A] vs time is linear.
Chemistry
A. Rate = k[A][B]
B. Rate = k[A]^2[B]
C. Rate = k[A][B]^2
D. Rate = k[A]^2[B]^2
Answer: The rate law is Rate = k[A]^1[B]^2 for a reaction that is first order in A and second order in B, giving an overall third-order reaction.
Chemistry
A. Chelating ligands always produce colored complexes
B. Complexes with chelating (multidentate) ligands are more stable than those with equivalent monodentate ligands
C. Chelate rings are always five-membered
D. Chelating ligands are always strong field
Answer: Chelate complexes have greater thermodynamic stability due to favorable entropy changes when multidentate ligands replace multiple monodentate ligands.
Chemistry
A. +RT
B. -RT
C. Zero
D. RT ln 2
Answer: Delta G0 = -RT ln K. When K = 1, ln K = 0, so Delta G0 = 0. The standard state has equal free energies of reactants and products.
Chemistry
A. 22.4 g/mol
B. 44.8 g/mol
C. 89.6 g/mol
D. 11.2 g/mol
Answer: At STP, moles = 0.560/22.4 = 0.025 mol. Molar mass = 2.0/0.025 = 80 g/mol. Wait, let me recalculate: 0.560/22.4 = 0.025, 2.0/0.025 = 80. The answer should be 80 g/mol.
Chemistry
A. Acylbenzene
B. Alkylbenzene
C. Nitrobenzene
D. Di-tert-butylbenzene
Answer: Friedel-Crafts reactions (alkylation and acylation) introduce alkyl or acyl groups. Nitrobenzene formation requires nitration (HNO3/H2SO4), not Friedel-Crafts.
Chemistry
A. Ethyl acetate
B. 3-Hydroxybutanal (aldol)
C. Butanal
D. Ethanol
Answer: Two molecules of acetaldehyde undergo aldol condensation: one acts as a nucleophile (enolate) attacking the carbonyl of another, forming 3-hydroxybutanal.
Chemistry
A. sp3
B. sp3d
C. sp3d2
D. sp2
Answer: XeO3 has three Xe=O bonds and one lone pair on Xe (3 bonding pairs + 1 lone pair = 4 electron domains), giving sp3 hybridization with trigonal pyramidal geometry.
Chemistry
A. Alcohols from ketones
B. Alkenes from carbonyl compounds and phosphonium ylides
C. Esters from carboxylic acids
D. Amines from aldehydes
Answer: The Wittig reaction converts a carbonyl (C=O) to an alkene (C=C) using a phosphonium ylide (Wittig reagent), forming a phosphine oxide byproduct.