RBSE SS Chemistry common mistakes
RBSE SS Chemistry · Understand the traps and how to avoid them.
For a Daniell cell, writes the cell notation as Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s) and calculates E°cell by adding the two standard reduction potentials.
The candidate remembers that oxidation occurs at the anode but places the cathode on the left and treats both tabulated potentials as oxidation potentials.
What to do instead
Use the convention: anode half-cell on the left, cathode on the right, salt bridge in the middle. Write Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s), then calculate E°cell = E°cathode(red) - E°anode(red), keeping the sign that makes the cell potential positive.
For the reaction 2A → B, writes rate = k[A]^2 only because the balanced equation has coefficient 2 in front of A, and gives the first-order rate constant units as mol L^-1 s^-1.
The candidate confuses stoichiometric coefficients with experimentally determined reaction order and memorises units without linking them to the rate law.
What to do instead
Determine order from experimental rate data, not from the balanced equation. Use rate = k[A]^m and derive units from mol L^-1 s^-1 = k(mol L^-1)^m; first order has s^-1, second order has L mol^-1 s^-1. For reactant disappearance, include the negative sign: rate = -(1/2)d[A]/dt.
Calculates the freezing-point depression of 0.10 m NaCl as ΔTf = Kf × 0.10, omitting the van’t Hoff factor i.
The candidate treats an ionic solute as a single particle and applies the colligative formula exactly as memorised for glucose or urea.
What to do instead
Count the ions produced per formula unit: i = 2 for NaCl, 3 for CaCl2, 1 for glucose. Use ΔTf = iKf m and ΔTb = iKb m; for associating solutes in nonpolar solvents, use the experimentally relevant i less than 1.
States that a body-centred cubic unit cell contains 4 atoms and has coordination number 6, while an face-centred cubic unit cell contains 2 atoms and has coordination number 8.
The candidate mixes the atom-count and coordination-number values of simple cubic, bcc and fcc lattices because the three structures are studied together.
What to do instead
Memorise the triple: simple cubic = 1 atom, CN 6, packing 52.4%; bcc = 2 atoms, CN 8, packing 68%; fcc = 4 atoms, CN 12, packing 74%. Derive atom count from corner, face and body-centre contributions: corner 1/8, face 1/2, body centre 1.
Names [Co(NH3)5Cl]Cl2 as chloropentaamminecobalt(II) chloride and puts chloro before ammine in the name.
The candidate ignores the charge balance from the two outside chloride ions and does not apply alphabetical ligand ordering.
What to do instead
Let cobalt be x: x + 5(0) + (-1) + 2(-1) = 0, so x = +3. Name ligands alphabetically by their names, ignoring prefixes: ammine before chloro. The correct name is pentaamminechlorocobalt(III) chloride.
Writes the ground-state configuration of Cr as [Ar]3d4 4s2 and the configuration of Cu+ as [Ar]3d9 4s1.
The candidate applies the aufbau order literally and forgets that half-filled and fully filled d subshells are more stable, and that 4s electrons are removed before 3d electrons when transition-metal cations form.
What to do instead
Use Cr = [Ar]3d5 4s1 and Cu = [Ar]3d10 4s1. For ions, remove electrons from the highest n orbital first: Cu+ = [Ar]3d10, Cu2+ = [Ar]3d9, Cr3+ = [Ar]3d3. Check by counting total electrons.
Predicts that the H–O–H bond angle in water is larger than the H–N–H bond angle in ammonia because oxygen has more lone pairs.
The candidate knows lone-pair repulsion is important but reverses its effect, assuming more lone pairs open the angle instead of compressing it.
What to do instead
Apply VSEPR: CH4 109.5°, NH3 about 107°, H2O about 104.5°. Lone pairs occupy more space than bonding pairs, so each additional lone pair on the central atom reduces the bond angle.
Claims that 2-bromo-2-methylpropane reacts faster than 1-bromobutane in an SN2 reaction because tertiary carbocations are more stable.
The candidate memorises “tertiary is more reactive” from SN1 and applies it to SN2 without checking the mechanism.
What to do instead
For SN2, reactivity follows methyl > primary > secondary > tertiary because the nucleophile attacks from the back and bulky groups block approach. For SN1, reactivity follows tertiary > secondary > primary because carbocation stability controls the rate. Match the mechanism before ranking substrates.
Uses aqueous NaHCO3 to distinguish phenol from acetic acid and states that phenol dissolves with effervescence of CO2.
The candidate remembers that phenol is acidic but does not compare its acid strength with carbonic acid.
What to do instead
Phenol is a weaker acid than carbonic acid, so it does not liberate CO2 from NaHCO3. Carboxylic acids such as acetic acid do give CO2 with NaHCO3. Use NaHCO3 to distinguish carboxylic acids from phenols; phenol reacts with NaOH but not with NaHCO3.
States that acetone gives a positive Tollens and Fehling test, and that benzaldehyde gives a positive iodoform test.
The candidate memorises “carbonyl compounds react” without separating aldehyde-specific oxidation from the methyl-ketone requirement of the iodoform test.
What to do instead
Tollens and Fehling are aldehyde tests; ketones such as acetone are negative. Aromatic aldehydes reduce Tollens but generally not Fehling. Iodoform requires CH3CO– or CH3CH(OH)–; acetone is positive, benzaldehyde and other non-methyl ketones are negative.