FBISE Class 9th (SSC-l)
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FBISE Class 9th (SSC-l)
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.
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.
Answer: IR spectroscopy measures the absorption of infrared radiation by molecular vibrations (stretching and bending of bonds), revealing functional groups and molecular framework.
Answer: Chemical shift reflects the shielding/deshielding of a proton by surrounding electrons. Deshielded protons (near electronegative atoms) appear at higher delta values.
Answer: E1 is unimolecular: only the substrate is involved in the rate-determining step (carbocation formation), so Rate = k[substrate], making it first order.
Answer: As T increases, Ea/RT decreases, making e^(-Ea/RT) larger, so k increases exponentially with temperature.
Answer: Standard enthalpy of combustion is the enthalpy change when 1 mole of a substance undergoes complete combustion in excess O2 under standard conditions (298 K, 1 atm).
Answer: Tetrafluoroethylene (CF2=CF2) undergoes addition polymerization to form PTFE (Teflon), where the double bond opens to form long chains.
Answer: In high-spin d5, each of the five d-orbitals gets one electron (t2g3 eg2) with no pairing, so no pairing energy is required.
Answer: The slowest step in SN1 is ionization: the departure of the leaving group to form a carbocation intermediate. This step has the highest activation energy.
Answer: For first-order reactions, t1/2 = ln(2)/k = 0.693/k, which is independent of the initial concentration.
Answer: Numbering gives the double bond the lowest locant: C1=C2-C3(Br)-C4. Bromine is on C3, so the name is 3-bromobut-1-ene.
Answer: Sigma bonds are the strongest, so sigma -> sigma* transitions require the highest energy (shortest wavelength), typically in the far UV region.
Answer: The salt formed (e.g., NaCl) is neutral, so the pH at equivalence is exactly 7 for strong acid-strong base titrations.
Answer: A bidentate ligand has two donor atoms that simultaneously coordinate to the metal center, forming a chelate ring (e.g., ethylenediamine, oxalate).
Answer: The De Broglie equation states lambda = h/p = h/(mv), where h is Planck\'s constant, m is mass, and v is velocity, predicting wave-particle duality.
Answer: HBr adds across the C=C double bond in propene. The electrophilic H+ attacks the pi bond first, followed by nucleophilic Br- attack.
Answer: Faraday\'s first law: mass deposited (m) = (M x I x t)/(n x F), where M is molar mass, I is current, t is time, n is electrons transferred, and F = 96485 C/mol.
Answer: 1H NMR gives different splitting patterns: 1-propanol shows a triplet for the CH3, while 2-propanol shows a distinctive septet for the CH and a doublet for two equivalent CH3 groups.
Answer: Molecularity is the number of molecules that come together in an elementary step. Three simultaneous collisions give termolecular (trimolecular) molecularity.