Science Group

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Science Group

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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.
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.
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.
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.
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.
A. Covalent bond energy
B. Lattice energy of ionic compounds
C. Ionization energy of metals
D. Electron affinity of nonmetals
Answer: The Born-Haber cycle applies Hess\'s law to calculate lattice energy by combining steps: sublimation, ionization, dissociation, electron affinity, and formation.
A. Addition of a diene to an alkene forming a six-membered ring
B. Elimination reaction forming a conjugated system
C. Nucleophilic substitution on a diene
D. Radical polymerization of dienes
Answer: The Diels-Alder reaction is a [4+2] cycloaddition between a conjugated diene (4 pi electrons) and a dienophile (2 pi electrons), forming a cyclohexene ring.
A. K = kf + kr
B. K = kf / kr
C. K = kr / kf
D. K = kf x kr
Answer: At equilibrium, forward rate equals reverse rate: kf[reactants] = kr[products], so K = [products]/[reactants] = kf/kr.
A. Formaldehyde
B. Acetaldehyde
C. Acetone
D. Acetic acid
Answer: CH3COCH3 (propan-2-one) is commonly known as acetone, the simplest ketone and an important industrial solvent.
A. IR spectroscopy
B. NMR spectroscopy
C. Mass spectrometry
D. UV-Vis spectroscopy
Answer: Mass spectrometry ionizes molecules and measures the mass-to-charge ratio (m/z), providing precise molecular mass and structural fragmentation patterns.
A. Ethane
B. Ethanal (acetaldehyde)
C. Ethanoic acid
D. Ethyl acetate
Answer: Mild oxidation of ethanol with acidified K2Cr2O7 produces ethanal (acetaldehyde). With excess oxidant and heating, it further oxidizes to ethanoic acid.
A. Tetrahedral
B. Square planar
C. Octahedral
D. Trigonal bipyramidal
Answer: d8 metals (like Ni2+, Pd2+, Pt2+) with strong field ligands form square planar complexes because the large crystal field splitting makes this geometry more stable than tetrahedral.
A. Temperature and pressure of a gas
B. Absorbance to concentration, path length, and molar absorptivity
C. Rate of reaction to concentration
D. Voltage to current and resistance
Answer: Beer-Lambert Law states absorbance (A) = molar absorptivity (epsilon) x path length (l) x concentration (c), forming the basis of quantitative UV-Vis analysis.
A. Bromination of benzene with Br2/FeBr3
B. Chlorobenzene reacting with NaOH at high temperature
C. Friedel-Crafts alkylation of benzene
D. Nitration of benzene
Answer: Chlorobenzene reacts with NaOH at high T/pressure (Dow process) via nucleophilic aromatic substitution, where OH- replaces Cl on the aromatic ring.
A. Kc = [SO3]^2 / ([SO2]^2 x [O2])
B. Kc = [SO2]^2 x [O2] / [SO3]^2
C. Kc = [SO3] / ([SO2] x [O2])
D. Kc = [SO3]^2 / [SO2]^2
Answer: Kc = [products]^coefficients / [reactants]^coefficients = [SO3]^2 / ([SO2]^2 x [O2]^1).
A. Polyethylene
B. Polyvinyl chloride
C. Polylactic acid (PLA)
D. Polystyrene
Answer: PLA is a biodegradable polyester derived from lactic acid (from corn starch). Microorganisms can break down its ester linkages in the environment.
A. An ester
B. An amide (N-methylacetamide)
C. A ketone
D. An ether
Answer: CH3NH2 (methylamine) reacts with CH3COOH (ethanoic acid) in a condensation reaction to form N-methylacetamide (CH3CONHCH3) and water.
A. Increases K
B. Decreases K
C. Does not change K
D. Makes K zero
Answer: A catalyst speeds up both forward and reverse reactions equally by lowering activation energy. It helps reach equilibrium faster but does not change the position of equilibrium (K).
A. 2-Methylbutanoic acid
B. 3-Methylbutanoic acid
C. 2-Ethylpropanoic acid
D. 2-Methylpentanoic acid
Answer: The longest chain containing COOH has 4 carbons (butanoic acid). Numbering from COOH, the methyl is on C3, giving 3-methylbutanoic acid.
A. s-s overlap
B. p-p overlap
C. sp-sp overlap
D. s-p overlap
Answer: In H2, each hydrogen has a 1s orbital. The sigma bond is formed by the head-on overlap of these two 1s orbitals (s-s overlap).