Chemistry MDCAT MCQs

Organic, inorganic & physical chemistry MCQs

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Showing 305 MCQs in Chemistry
Chemistry
A. Reducing agent
B. Oxidizing agent
C. Catalyst
D. Dehydrating agent
Answer: MnO2 oxidizes HCl to produce Cl2: MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O. Here MnO2 is the oxidizing agent.
Chemistry
A. +2
B. +3
C. +4
D. +6
Answer: In Fe2O3: 2(Fe) + 3(-2) = 0, so 2Fe = 6, Fe = +3.
Chemistry
A. NMR spectroscopy
B. UV-Vis spectroscopy
C. IR spectroscopy
D. Mass spectrometry
Answer: IR spectroscopy measures the absorption of infrared radiation by molecular vibrations (stretching and bending of bonds), revealing functional groups and molecular framework.
Chemistry
A. SN1
B. SN2
C. E1
D. E2
Answer: Primary alkyl halides favor SN2 because they have minimal steric hindrance, and strong nucleophiles with polar aprotic solvents further promote bimolecular substitution.
Chemistry
A. 0
B. 1
C. 2
D. 3
Answer: Rate = k[A]^n. If 2^n = 4, then n = 2, meaning the reaction is second order with respect to A.
Chemistry
A. Maximum
B. Equal to E0
C. Zero
D. Negative
Answer: At equilibrium, Q = K and Delta G = 0, so from Delta G = -nFE, we get E = 0. The cell has no driving force at equilibrium.
Chemistry
A. Linkage isomerism
B. Geometric (cis-trans) isomerism
C. Ionization isomerism
D. Coordination isomerism
Answer: The complex can exist as cis (Cl ligands adjacent) and trans (Cl ligands opposite) geometric isomers due to its octahedral geometry.
Chemistry
A. First order
B. Second order
C. Third order
D. Zero order
Answer: E1 is unimolecular: only the substrate is involved in the rate-determining step (carbocation formation), so Rate = k[substrate], making it first order.
Chemistry
A. Energy is conserved in all chemical reactions
B. The total enthalpy change is independent of the path taken between initial and final states
C. All exothermic reactions have negative activation energy
D. Equilibrium constant depends on temperature only
Answer: Hess\'s law states that the total enthalpy change for a reaction is the same regardless of whether it occurs in one step or multiple steps, because enthalpy is a state function.
Chemistry
A. Complete retention of configuration
B. Complete inversion of configuration
C. Racemization with some inversion
D. Complete retention only
Answer: SN1 proceeds through a planar carbocation intermediate. Nucleophilic attack from both faces gives racemization, but ion pairing often causes slight excess inversion.
Chemistry
A. NaCl
B. MgO
C. KBr
D. CsI
Answer: MgO has the highest lattice energy due to the small ionic radii and high charges (+2 and -2) of Mg2+ and O2-, resulting in very strong electrostatic attractions.
Chemistry
A. I-
B. Br-
C. CN-
D. F-
Answer: CN- is a strong field ligand at the top of the spectrochemical series, causing the largest crystal field splitting (Delta_o).
Chemistry
A. Rate = k[CH3Br]
B. Rate = k[OH-]
C. Rate = k[CH3Br][OH-]
D. Rate = k[CH3Br]^2[OH-]
Answer: SN2 is bimolecular: both the substrate and nucleophile are involved in the rate-determining step, so Rate = k[CH3Br][OH-].
Chemistry
A. Propanal
B. Propan-2-one
C. Propanoic acid
D. Propan-1-ol
Answer: Propanone has the structure CH3COCH3. The carbonyl group is at carbon 2, so its IUPAC name is propan-2-one (commonly called acetone).
Chemistry
A. Ethane
B. Ethene
C. Ethyne
D. Butane
Answer: Ethyne (acetylene, C2H2) contains a carbon-carbon triple bond (C≡C), making it an alkyne.
Chemistry
A. Number of equivalent protons
B. The electronic environment of the proton
C. The number of neighboring protons
D. The molecular weight
Answer: Chemical shift reflects the shielding/deshielding of a proton by surrounding electrons. Deshielded protons (near electronegative atoms) appear at higher delta values.
Chemistry
A. +200 J
B. -200 J
C. 0 J
D. +100 J
Answer: At constant pressure, q = Delta H. Since heat is released (exothermic), Delta H = -200 J by sign convention.
Chemistry
A. 6, +2
B. 6, +3
C. 3, +3
D. 6, +6
Answer: Six NH3 ligands give coordination number 6. Each NH3 is neutral and 3 Cl- balance the charge, so Co is +3.
Chemistry
A. They must be gauche
B. They must be anti-periplanar (180 degrees dihedral angle)
C. They must be syn-periplanar (0 degrees dihedral angle)
D. There is no geometric requirement
Answer: E2 requires an anti-periplanar arrangement where the C-H and C-LG bonds are at 180 degrees, allowing maximum orbital overlap for the developing pi bond.
Chemistry
A. Cell potential increases
B. Cell potential decreases
C. Cell potential remains unchanged
D. Cell potential becomes zero
Answer: As Q increases, the logarithmic term (RT/nF)lnQ becomes more positive, which reduces E from E0, decreasing the cell potential.
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