FBISE Class 9th (SSC-l)
Explore subjects and practice MCQs
3
Chapters
1,610
Total MCQs
FBISE Class 9th (SSC-l)
Answer: Kp = Kc(RT)^(Delta n). Delta n = 2 - (1+3) = -2, so Kp = Kc(RT)^(-2) = Kc/(RT)^2.
542
Answer: E2 reactivity increases with alkyl substitution because more substituted alkenes (Zaitsev product) are more stable, and tertiary substrates have more beta-hydrogens.
Answer: The spin-only formula gives magnetic moment mu = sqrt(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons.
Answer: Hot, concentrated KMnO4 is a strong oxidizing agent that cleaves the C=C double bond in cyclohexene, oxidizing it completely to adipic acid (HOOC-CH2-CH2-CH2-CH2-COOH).
Answer: The SN2 transition state has a pentacoordinate carbon with the nucleophile and leaving group on opposite sides, forming a trigonal bipyramidal arrangement.
Answer: The magnetic quantum number (ml) specifies the number of orbitals and their orientation within a subshell, ranging from -l to +l.
Answer: Two molecules of acetaldehyde undergo aldol condensation: one acts as a nucleophile (enolate) attacking the carbonyl of another, forming 3-hydroxybutanal.
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.
Answer: 1,3-Butadiene (CH2=CH-CH=CH2) has alternating double-single-double bonds, creating conjugation where p-orbitals overlap across all four carbons.
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.
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.).
Answer: Chelate complexes have greater thermodynamic stability due to favorable entropy changes when multidentate ligands replace multiple monodentate ligands.
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.
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.
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.
Answer: Friedel-Crafts reactions (alkylation and acylation) introduce alkyl or acyl groups. Nitrobenzene formation requires nitration (HNO3/H2SO4), not Friedel-Crafts.
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.
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.
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.
Answer: Peroxides initiate a free radical mechanism that reverses the normal regioselectivity, adding Br to the less substituted carbon (anti-Markovnikov addition).