Maharashtra HSC Chemistry common mistakes
Maharashtra HSC Chemistry · Understand the traps and how to avoid them.
When asked to calculate the molar mass or density of a unit cell, the candidate plugs the edge length a directly into the formula in metres instead of converting picometres or Ångström to cm or m, producing answers off by 10^8 to 10^10.
Solid-state numericals are usually taught with a in pm and density in g/cm³, and candidates memorise d = ZM/(N_A·a³) without tracking that a³ must be in cm³ for the units to cancel. They also forget to cube after converting, or convert only one factor.
What to do instead
Before substituting, write a one-line unit plan: a (pm) → a (cm) = a × 10^-10, then a³ (cm³). Practise three cells (simple cubic, bcc, fcc) with a in pm and check the answer against the expected order of magnitude (metals are roughly 2–20 g/cm³). Also fix Z per lattice: 1 for simple cubic, 2 for bcc, 4 for fcc.
The candidate computes molality and molarity interchangeably in colligative-property problems, or uses the mass of solvent where the mass of solution belongs (and vice versa), so ΔTf = Kf·m or π = MRT comes out wrong even when the arithmetic is clean.
Both concentration terms are introduced together in Solutions and both are symbolised with 'm' or 'M', and textbook problems often give volume of solution plus mass of solute, tempting a direct division by litres for a formula that needs kg of solvent.
What to do instead
Adopt a fixed two-column habit: write 'solute mass → moles' on one side, then decide explicitly whether the denominator is kg solvent (molality, for ΔTf, ΔTb) or L solution (molarity, for osmotic pressure). For osmotic pressure remember to convert temperature to kelvin and to multiply by the van't Hoff factor i for electrolytes.
In ionic-equilibria numericals the candidate treats a weak acid as if it fully dissociated, or applies the Henderson–Hasselbalch/buffer expression to a solution that is not a buffer, and reports a pH that ignores the common-ion suppression of dissociation.
The chapter is full of short formulas (pH = –log[H⁺], pH = pKa + log([salt]/[acid])) and candidates pattern-match on the presence of an acid and a salt rather than checking whether the acid is weak and whether both components are present in comparable amounts.
What to do instead
Start every problem by classifying the system out loud: strong acid, weak acid, salt of weak acid/strong base, buffer, or mixture. Only then choose the relation. For weak acids use Ka = x²/(C – x) and state the approximation x ≪ C explicitly; for buffers confirm both conjugate pair members are present before using the ratio.
The candidate calculates ΔH or ΔG for a reaction and then reports spontaneity from the sign of ΔH alone, or forgets to convert ΔS from J to kJ before combining with ΔH in ΔG = ΔH – TΔS.
Enthalpy is emphasised first and 'exothermic means spontaneous' is a comfortable shortcut, while the entropy term is small in magnitude and its joule units are easy to overlook, so the T ΔS term is silently dropped or mis-scaled.
What to do instead
Always write ΔG = ΔH – TΔS with units attached to each term (kJ, K, kJ/K) and convert ΔS by dividing by 1000 before subtracting. Then reason about the four sign combinations of ΔH and ΔS, and note that ΔG < 0 is the criterion - not ΔH < 0.
While doing electrochemistry numericals the candidate mixes up anode and cathode in a galvanic cell, or uses the wrong number of electrons n in the Nernst equation, giving an EMF with the wrong sign or magnitude.
The same electrode can be anode in one cell and cathode in another, and the Indian convention of writing the anode on the left is easy to invert under exam pressure. The Nernst expression also requires n from the balanced half-reaction, which candidates often take from the molecular equation instead.
What to do instead
For every cell, write both half-reactions, balance electrons, and state 'oxidation at anode (left), reduction at cathode (right)' before computing E°cell = E°cathode – E°anode. For Nernst, take n from the balanced electron count and substitute the reaction quotient Q with the correct powers. Check that a spontaneous cell gives a positive EMF.
In chemical-kinetics problems the candidate confuses order with molecularity, or treats a first-order half-life as dependent on initial concentration when using t½ = 0.693/k.
Rate law, order, and molecularity are defined close together, and the first-order and zero-order half-life formulas look similar. Candidates also read the stoichiometric coefficients of the balanced equation as the exponents in the rate law.
What to do instead
Memorise the rule that order comes only from experiment, never from coefficients, and keep a small table: zero order t½ = [A]₀/2k (depends on concentration), first order t½ = 0.693/k (independent of concentration). Practise identifying order from a units-of-k question, since the units of k reveal the order directly.
The candidate writes a formula for an oxoacid or interhalogen of Groups 16/17/18 from memory with the wrong oxidation state or wrong number of oxygen atoms, e.g. giving chlorine the wrong oxoacid series or misassigning the oxidation state of S in peroxodisulphuric acid.
These chapters are memorisation-heavy with many similar species (HClO, HClO₂, HClO₃, HClO₄), and candidates learn names without learning the systematic oxidation-state logic that would let them derive the formula.
What to do instead
Build the oxoacid series from oxidation state: for chlorine, +1, +3, +5, +7 give HClO, HClO₂, HClO₃, HClO₄. Compute oxidation states by the rule that O is –2 and H is +1, and apply it to peroxo species where an O–O bond changes the count. Do a short drill writing formulas from oxidation state rather than the reverse.
When naming or drawing a coordination compound the candidate lists the ligands in the wrong order, omits the charge on the complex ion, or writes the formula with the metal first even when the complex is an anion.
IUPAC rules for coordination entities are layered (ligands alphabetical, anionic before neutral, metal oxidation state in Roman numerals), and candidates memorise example names without internalising the ordering rules, so a new compound breaks the pattern.
What to do instead
Apply the rules in a fixed sequence: identify the complex ion and its charge, name ligands alphabetically (ignoring prefixes like di-/tri- for ordering), then the metal with its oxidation state in parentheses. In formulas, place the metal first for cations and last for anions, and enclose the whole coordination entity in square brackets. Verify charge by summing ligand and metal charges.
In organic conversions the candidate identifies a reagent by name only and cannot predict the product, e.g. using alcoholic KOH where aqueous KOH is needed for substitution versus elimination, or applying a reagent to the wrong functional group in a multi-step sequence.
Halogen derivatives, alcohols, carbonyls, and amines are taught as separate reagent lists, and candidates memorise 'KOH gives alcohol' without the solvent qualifier that flips the outcome, so similar-sounding reagents are swapped.
What to do instead
Learn each reagent with its condition attached: KOH(aq) → substitution, KOH(alc.) → elimination; LiAlH₄ reduces acids and esters, NaBH₄ reduces aldehydes and ketones only. For multi-step conversions, draw the starting structure and write the functional group after each arrow, checking that the named reagent actually acts on the group present.
The candidate writes structural formulas for biomolecules and polymers without the correct linkage, e.g. showing a glycosidic or peptide bond without the loss of water, or drawing a polymer repeat unit that includes the double bond that opened during addition polymerisation.
These chapters are descriptive and candidates reproduce textbook pictures from memory rather than reasoning from the condensation or addition mechanism, so the linking atom or the loss of a small molecule is dropped.
What to do instead
For every polymer, state whether it is addition or condensation, then draw the repeat unit from the monomer: addition keeps the backbone carbons and removes the double bond; condensation shows the amide or ester linkage and the small molecule eliminated. For peptides and disaccharides, explicitly mark the –CO–NH– or –O– bridge and the water removed.