Class 11 Chemistry — Important Board Questions with Answers

Law of conservation of mass: matter can neither be created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products. Law of definite proportions: a given chemical compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source or method of preparation.

Molar mass of CO2 = 12 + 2(16) = 44 g/mol. Number of moles = given mass / molar mass = 11 / 44 = 0.25 mol.

n = m / M

Number of molecules = 0.25 x 6.022 x 10^23 = 1.506 x 10^23 molecules. Each CO2 molecule has 3 atoms (1 C + 2 O), so number of atoms = 3 x 1.506 x 10^23 = 4.52 x 10^23 atoms.

N = n x N_A

One mole is the amount of a substance that contains as many elementary entities (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. This number of entities is Avogadro's number, N_A = 6.022 x 10^23 mol^-1.

1 mol = 6.022 x 10^23 particles

Divide each percentage by atomic mass: C = 24.27/12 = 2.02, H = 4.07/1 = 4.07, Cl = 71.65/35.5 = 2.02. Divide by the smallest (2.02): C = 1, H = 2, Cl = 1. Empirical formula = CH2Cl (empirical mass = 12 + 2 + 35.5 = 49.5). n = molar mass / empirical mass = 98.96/49.5 = 2. Molecular formula = C2H4Cl2.

n = molar mass / empirical formula mass

Molarity (M) is the number of moles of solute dissolved per litre of solution; unit mol/L (or M). Molality (m) is the number of moles of solute dissolved per kilogram of solvent; unit mol/kg (or m). Molality is independent of temperature since it uses mass, whereas molarity changes with temperature because volume changes.

M = n_solute / V(L); m = n_solute / mass_solvent(kg)

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = 5/40 = 0.125 mol. Volume = 450 mL = 0.450 L. Molarity = 0.125 / 0.450 = 0.278 M (approximately 0.28 M).

M = n / V(L)

Moles of H2 = 3/2 = 1.5 mol; moles of O2 = 29/32 = 0.906 mol. From the equation, 2 mol H2 need 1 mol O2. For 1.5 mol H2, O2 required = 0.75 mol, but 0.906 mol O2 is available, so O2 is in excess and H2 is the limiting reagent. Water formed = moles of H2 (since 2H2 -> 2H2O) = 1.5 mol = 1.5 x 18 = 27 g.

mass = moles x molar mass

Mole fraction of a component = number of moles of that component / total number of moles of all components in the solution; it is dimensionless and the sum of mole fractions equals 1. Mass percent of a component = (mass of the component / total mass of solution) x 100.

x_A = n_A / n_total; mass% = (mass component / total mass) x 100

1) The electron revolves around the nucleus only in certain fixed circular orbits of definite energy called stationary states, without radiating energy. 2) Angular momentum of the electron is quantised: mvr = nh/2pi, where n = 1, 2, 3... 3) Energy is emitted or absorbed only when an electron jumps from one orbit to another, and the energy difference equals hv (delta E = E2 - E1 = h v).

mvr = nh/2pi ; delta E = h v

Wavelength = 4000 Angstrom = 4000 x 10^-10 m = 4 x 10^-7 m. Energy E = hc/lambda = (6.626 x 10^-34 x 3 x 10^8) / (4 x 10^-7) = 1.988 x 10^-25 / 4 x 10^-7 = 4.97 x 10^-19 J.

E = hc / lambda

For a hydrogen atom, r_n = n^2 x r_1. For n = 3: r_3 = 3^2 x 0.529 = 9 x 0.529 = 4.761 Angstrom.

r_n = n^2 x r_1

It is impossible to determine simultaneously and with absolute accuracy both the position and the momentum (or velocity) of a microscopic particle such as an electron. The product of the uncertainties in position (delta x) and momentum (delta p) is at least of the order of h/4pi.

delta x . delta p >= h / 4pi

de Broglie wavelength lambda = h / (m v) = (6.626 x 10^-34) / (9.1 x 10^-31 x 2.05 x 10^7). Denominator = 1.866 x 10^-23. lambda = 3.55 x 10^-11 m = 0.355 Angstrom.

lambda = h / (m v)

1) Principal quantum number (n): gives the main energy level/shell and size of the orbital (n = 1, 2, 3...). 2) Azimuthal (angular momentum) quantum number (l): gives the subshell and shape of the orbital (l = 0 to n-1, i.e. s, p, d, f). 3) Magnetic quantum number (m_l): gives the orientation of the orbital in space (m_l = -l to +l). 4) Spin quantum number (m_s): gives the direction of electron spin (+1/2 or -1/2).

l = 0 to (n-1); m_l = -l ... +l

Cr (Z = 24): [Ar] 3d5 4s1 (not 3d4 4s2). Cu (Z = 29): [Ar] 3d10 4s1 (not 3d9 4s2). One 4s electron shifts to 3d because exactly half-filled (3d5) and completely filled (3d10) subshells have extra stability due to symmetrical distribution of electrons and greater exchange energy.

Wave number (nu bar) = R (1/n1^2 - 1/n2^2) = 1.097 x 10^7 (1/2^2 - 1/3^2) = 1.097 x 10^7 (1/4 - 1/9) = 1.097 x 10^7 x (5/36) = 1.523 x 10^6 m^-1. This is the H-alpha line of the Balmer series.

nu bar = R (1/n1^2 - 1/n2^2)

Pauli's exclusion principle: no two electrons in an atom can have the same set of all four quantum numbers; an orbital can hold at most two electrons with opposite spins. Hund's rule of maximum multiplicity: electron pairing in orbitals of the same subshell (degenerate orbitals) does not occur until each orbital is singly occupied, and all singly filled orbitals have parallel spin.

Aufbau principle: in the ground state of an atom, electrons are filled into orbitals in order of increasing energy, i.e. the lowest energy orbital is filled first. (n + l) rule: the orbital with the lower (n + l) value has lower energy and is filled first; if two orbitals have the same (n + l) value, the one with the lower n is filled first (e.g. 4s (n+l=4) is filled before 3d (n+l=5)).

energy order by increasing (n + l)

Modern periodic law: the physical and chemical properties of elements are a periodic function of their atomic numbers. Mendeleev's law was based on atomic mass, whereas the modern law is based on atomic number (number of protons), which removed anomalies such as the position of argon and potassium.

Across a period (left to right), atomic radius decreases because nuclear charge increases while electrons are added to the same shell, so the increased effective nuclear charge pulls the electron cloud closer. Down a group, atomic radius increases because a new shell is added at each step and the number of inner shielding electrons increases, outweighing the rise in nuclear charge.

Nitrogen has the configuration 1s2 2s2 2p3 with a half-filled 2p subshell, which is extra stable due to symmetry and exchange energy, so removing an electron requires more energy. Oxygen (1s2 2s2 2p4) has one paired electron in 2p; removing it relieves electron-electron repulsion and gives a stable half-filled configuration, so less energy is needed. Hence IE1 of nitrogen > oxygen.

Ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Across a period it increases because nuclear charge increases and atomic size decreases, so the electron is held more tightly. Down a group it decreases because atomic size increases and shielding by inner electrons increases, so the outer electron is more easily removed.

M(g) -> M+(g) + e-

Although fluorine is smaller, its 2p subshell is very compact, so the incoming electron experiences strong inter-electronic repulsion from the already crowded 2p electrons. In chlorine, the larger 3p subshell accommodates the incoming electron with less repulsion, so more energy is released. Hence chlorine has a more negative (more exothermic) electron gain enthalpy than fluorine.

Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons in a covalent bond towards itself. It increases across a period (as size decreases and nuclear charge increases) and decreases down a group (as size increases). Fluorine is the most electronegative element. The Pauling scale is commonly used.

All four species have 10 electrons (isoelectronic). For isoelectronic species, the greater the nuclear charge (number of protons), the smaller the radius. Nuclear charges: O2- (8), F- (9), Na+ (11), Mg2+ (12). Increasing order of radius: Mg2+ < Na+ < F- < O2-.

Elements are classified by the subshell into which the last (differentiating) electron enters: s-block (last electron in s), p-block (in p), d-block (in d) and f-block (in f). d-block elements are called transition elements because they lie between the s-block (metals) and p-block (non-metals) and show a gradual transition in properties; their atoms or common ions have partially filled d orbitals.

An ionic (electrovalent) bond is formed by the complete transfer of one or more electrons from one atom to another, producing oppositely charged ions held by electrostatic attraction (e.g. NaCl). A covalent bond is formed by the mutual sharing of electron pairs between atoms (e.g. H2 or Cl2). Ionic bonds form between metals and non-metals; covalent bonds form between non-metals.

BF3: central B has 3 bond pairs and no lone pair -> trigonal planar, bond angle 120 deg. NH3: central N has 3 bond pairs and 1 lone pair -> pyramidal (trigonal pyramidal), bond angle about 107 deg. H2O: central O has 2 bond pairs and 2 lone pairs -> bent/angular, bond angle about 104.5 deg. Lone pairs repel more than bond pairs, reducing the angle from the ideal 109.5 deg in NH3 and H2O.

Both molecules are based on sp3 hybridisation with an ideal angle of 109.5 deg. NH3 has one lone pair, while H2O has two lone pairs. Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion, and water has an extra lone pair, so its bond pairs are pushed closer together. Hence the bond angle of H2O (104.5 deg) is smaller than that of NH3 (107 deg).

In ethyne each carbon is sp hybridised. The two sp hybrid orbitals on each carbon form sigma bonds: one C-C sigma bond and one C-H sigma bond, giving a linear molecule with a bond angle of 180 deg. The remaining two unhybridised p orbitals on each carbon overlap sideways to form two pi bonds, so there is a triple bond (1 sigma + 2 pi) between the carbon atoms.

H-C(triple bond)C-H

O2 has 16 electrons. Configuration: sigma(1s)2 sigma*(1s)2 sigma(2s)2 sigma*(2s)2 sigma(2pz)2 pi(2px)2 pi(2py)2 pi*(2px)1 pi*(2py)1. Bonding electrons Nb = 10, antibonding Na = 6. Bond order = (Nb - Na)/2 = (10 - 6)/2 = 2. The two unpaired electrons in the pi* antibonding orbitals make O2 paramagnetic.

Bond order = (Nb - Na)/2

N2 has 14 electrons. Number of bonding electrons Nb = 10 and antibonding electrons Na = 4. Bond order = (Nb - Na)/2 = (10 - 4)/2 = 3. This corresponds to a nitrogen-nitrogen triple bond, which explains the very high stability and bond dissociation energy of N2.

Bond order = (Nb - Na)/2

A coordinate bond is a covalent bond in which the shared pair of electrons is contributed by only one of the two bonded atoms (the donor). In the ammonium ion (NH4+), nitrogen in NH3 has a lone pair which it donates to a proton (H+) that has no electrons, forming the fourth N-H bond as a coordinate bond. Once formed, all four N-H bonds are identical.

NH3 + H+ -> NH4+

A hydrogen bond is a weak electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O or N) and the lone pair of another electronegative atom. Conditions: hydrogen must be attached to a small, highly electronegative atom. Types: intermolecular hydrogen bonding (between different molecules, e.g. in water and HF) and intramolecular hydrogen bonding (within the same molecule, e.g. o-nitrophenol).

Oxygen is much more electronegative and smaller than sulphur, so water molecules form strong intermolecular hydrogen bonds, whereas H2S molecules are held only by weak van der Waals (dipole) forces. Extra energy is needed to break the hydrogen bonds in water, so water has a much higher boiling point than H2S even though H2S has a higher molar mass.

Fajans' rules: covalent character in an ionic compound increases when (1) the cation is small, (2) the anion is large, and (3) the cation has a high charge (all three increase the polarising power/polarisability). Also, cations with a pseudo-noble gas (18-electron) configuration cause greater polarisation than those with a noble gas configuration. Greater polarisation of the anion by the cation increases covalent character.

Dipole moment (mu) is the product of the magnitude of charge and the distance between the centres of positive and negative charge; it is a vector quantity (unit: debye). CO2 is linear (O=C=O) and its two C=O bond dipoles are equal and opposite, so they cancel and the net dipole moment is zero. H2O is bent/angular, so its two O-H bond dipoles do not cancel and add up to give a net dipole moment (1.85 D). Hence CO2 is non-polar but H2O is polar.

mu = q x d

A sigma bond is formed by the head-on (axial) overlap of orbitals along the internuclear axis, while a pi bond is formed by the sidewise (lateral) overlap of parallel p orbitals. A sigma bond is stronger because axial overlap is more effective and greater, giving a larger region of electron density between the nuclei; pi bonds have smaller lateral overlap and are weaker and more reactive.

Boyle's law: at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. V is proportional to 1/P, so PV = constant. Charles's law: at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature, i.e. V is proportional to T, so V/T = constant.

PV = constant (Boyle); V/T = constant (Charles)

Convert temperatures to Kelvin: T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K. By Charles's law V1/T1 = V2/T2, so V2 = V1 x T2/T1 = 300 x 400/300 = 400 mL.

V1/T1 = V2/T2

The ideal gas equation is PV = nRT, where P = pressure, V = volume, n = number of moles, T = absolute temperature and R = universal gas constant. R = 0.0821 L atm K^-1 mol^-1 = 8.314 J K^-1 mol^-1.

PV = nRT

Using PV = nRT, V = nRT/P = (2 x 0.0821 x 300) / 2 = 49.26/2 = 24.63 L.

V = nRT / P

1) A gas consists of a large number of tiny particles (molecules) whose actual volume is negligible compared with the volume of the container. 2) There are no forces of attraction or repulsion between the molecules. 3) The molecules are in constant, rapid, random motion and collide with one another and with the walls of the container. 4) The collisions are perfectly elastic (no loss of kinetic energy). 5) The average kinetic energy of the molecules is directly proportional to the absolute temperature.

KE(avg) proportional to T

The van der Waals equation is (P + a n^2/V^2)(V - nb) = nRT. The constant 'a' corrects for the intermolecular forces of attraction (it accounts for the pressure being lower than ideal), and the constant 'b' corrects for the finite volume actually occupied by the gas molecules (excluded volume). Real gases deviate from ideal behaviour at high pressure and low temperature.

(P + a n^2/V^2)(V - nb) = nRT

A system is the specified part of the universe under study; the surroundings are the rest of the universe outside the system that can interact with it. The three types are: open system (exchanges both matter and energy with surroundings), closed system (exchanges only energy, not matter) and isolated system (exchanges neither matter nor energy).

The first law of thermodynamics states that energy can neither be created nor destroyed, only transformed from one form to another; the total energy of an isolated system remains constant. Mathematically, delta U = q + w, where delta U is the change in internal energy, q is the heat added to the system (positive when absorbed) and w is the work done on the system (positive when done on the system).

delta U = q + w

Heat supplied to the system q = +1 kJ = +1000 J. Work is done BY the gas, so work done on the system w = -200 J. By the first law, delta U = q + w = 1000 + (-200) = 800 J. The internal energy increases by 800 J.

delta U = q + w

Enthalpy (H) is the total heat content of a system at constant pressure, defined as H = U + PV. For a reaction at constant pressure and temperature, delta H = delta U + P delta V. For ideal gases, P delta V = delta ng RT, where delta ng is the change in the number of moles of gaseous species. Hence delta H = delta U + delta ng RT.

delta H = delta U + delta ng RT

We want C(s) + 1/2 O2(g) -> CO(g). By Hess's law, subtract the second equation from the first: delta H(reqd) = delta H1 - delta H2 = (-393.5) - (-283.0) = -110.5 kJ. So the enthalpy of formation of CO is -110.5 kJ/mol.

delta H(reqd) = delta H1 - delta H2

Hess's law states that the total enthalpy change of a reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and final conditions are the same. It follows from the fact that enthalpy is a state function. Applications: it is used to calculate enthalpies of formation, bond enthalpies and reaction enthalpies that cannot be measured directly.

Entropy (S) is a thermodynamic state function that measures the degree of randomness or disorder of a system. When ice (a highly ordered solid) melts into water (a more disordered liquid), disorder increases, so the entropy increases and delta S is positive.

delta S = q(rev) / T

The Gibbs energy change is given by delta G = delta H - T delta S. Criteria of spontaneity: if delta G < 0 (negative), the process is spontaneous; if delta G = 0, the system is at equilibrium; if delta G > 0 (positive), the process is non-spontaneous (the reverse is spontaneous). A reaction is always spontaneous when delta H is negative and delta S is positive.

delta G = delta H - T delta S

Convert delta S to kJ: -198 J/K = -0.198 kJ/K. delta G = delta H - T delta S = -92.4 - (298 x -0.198) = -92.4 - (-59.0) = -92.4 + 59.0 = -33.4 kJ. Since delta G is negative, the reaction is spontaneous at 298 K.

delta G = delta H - T delta S

Standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states (at 298 K and 1 bar); e.g. for CO2 it is -393.5 kJ/mol. Standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions; it is always negative (exothermic).

A reversible reaction is one that proceeds in both forward and backward directions under the same conditions. Characteristics of chemical equilibrium: (1) it is dynamic in nature, i.e. the forward and backward reactions continue at equal rates; (2) the observable properties (concentration, pressure, colour) remain constant with time; (3) it can be attained from either direction and is disturbed by changing conditions.

rate(forward) = rate(backward)

Kc = [NH3]^2 / ([N2][H2]^3). The relation between Kp and Kc is Kp = Kc (RT)^(delta ng), where delta ng = (moles of gaseous products) - (moles of gaseous reactants). Here delta ng = 2 - (1 + 3) = -2, so Kp = Kc (RT)^-2.

Kp = Kc (RT)^(delta ng)

Kc = [NO2]^2 / [N2O4] = (0.04)^2 / 0.02 = 0.0016 / 0.02 = 0.08 mol/L. The units are mol/L because delta ng = 1 for this reaction.

Kc = [products]^coeff / [reactants]^coeff

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to counteract (reduce) the effect of that change. Increasing the pressure shifts this equilibrium in the forward direction (towards NH3), because the forward reaction reduces the number of gas moles from 4 (1 + 3) to 2, thereby lowering the pressure.

Temperature: increasing temperature favours the endothermic direction and decreasing temperature favours the exothermic direction (as per Le Chatelier's principle); temperature also changes the value of the equilibrium constant K. Catalyst: a catalyst speeds up both the forward and backward reactions equally, so it helps the system reach equilibrium faster but does not shift the position of equilibrium or change the value of K.

According to the Bronsted-Lowry concept, an acid is a substance that donates a proton (H+) and a base is a substance that accepts a proton. A conjugate acid-base pair is a pair of species that differ by a single proton; for example, in HCl + H2O -> H3O+ + Cl-, HCl/Cl- and H2O/H3O+ are conjugate acid-base pairs.

acid <=> base + H+

HCl is a strong acid and dissociates completely, so [H+] = 0.001 M = 1 x 10^-3 M. pH = -log[H+] = -log(10^-3) = 3. The solution is acidic (pH < 7).

pH = -log[H+]

NaOH is a strong base and dissociates completely, so [OH-] = 0.01 M = 10^-2 M. pOH = -log[OH-] = -log(10^-2) = 2. Since pH + pOH = 14, pH = 14 - 2 = 12. The solution is basic (pH > 7).

pH + pOH = 14

The ionic product of water Kw is the product of the molar concentrations of hydrogen and hydroxide ions in water: Kw = [H+][OH-]. At 298 K, Kw = 1.0 x 10^-14 mol^2 L^-2. Taking negative logarithm gives pKw = pH + pOH = 14 at 298 K. For pure (neutral) water, [H+] = [OH-] = 10^-7 M, so pH = 7.

Kw = [H+][OH-] = 1.0 x 10^-14

A buffer solution is one that resists a change in its pH on the addition of a small amount of acid or base. An acidic buffer is made from a weak acid and its salt with a strong base, e.g. acetic acid + sodium acetate (CH3COOH + CH3COONa). The Henderson-Hasselbalch equation is pH = pKa + log([salt]/[acid]).

pH = pKa + log([salt]/[acid])

AgCl dissociates as AgCl <=> Ag+ + Cl-. If solubility = s mol/L, then [Ag+] = [Cl-] = s. Ksp = [Ag+][Cl-] = s x s = s^2. So s = sqrt(Ksp) = sqrt(1.8 x 10^-10) = 1.34 x 10^-5 mol/L.

Ksp = s^2 (for AB type salt)

The common ion effect is the suppression of the degree of dissociation (ionisation) of a weak electrolyte by the addition of a strong electrolyte that provides an ion common to the weak electrolyte. For example, adding NH4Cl (which provides NH4+) to a solution of NH4OH suppresses the ionisation of NH4OH, decreasing the OH- concentration. This is used in qualitative analysis and to control pH.

Oxidation is the loss of electrons or an increase in oxidation number of an element; reduction is the gain of electrons or a decrease in oxidation number. Both occur simultaneously in a redox reaction. For example, in Zn -> Zn2+ + 2e-, zinc is oxidised (oxidation number 0 -> +2).

In KMnO4: K = +1, O = -2 (four O = -8). Let Mn = x. Then +1 + x + (-8) = 0, so x = +7. In K2Cr2O7: 2 K = +2, 7 O = -14. Let each Cr = y. Then +2 + 2y - 14 = 0, so 2y = 12, y = +6. Thus Mn is +7 and Cr is +6.

sum of oxidation numbers = charge on species

Reduction half: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O. Oxidation half: Fe2+ -> Fe3+ + e-. To balance electrons, multiply the oxidation half by 5: 5Fe2+ -> 5Fe3+ + 5e-. Add the two halves: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+. This is the balanced equation.

An oxidising agent is a substance that oxidises another substance by accepting electrons and is itself reduced; e.g. KMnO4, O2. A reducing agent is a substance that reduces another substance by donating electrons and is itself oxidised; e.g. H2, C (carbon). In a redox reaction the oxidising agent gains electrons and the reducing agent loses electrons.

A disproportionation reaction is a redox reaction in which the same element in a single oxidation state is simultaneously oxidised and reduced (to a higher and a lower oxidation state). Example: 2H2O2 -> 2H2O + O2, where oxygen in the -1 state is both oxidised (to 0 in O2) and reduced (to -2 in H2O). Another example is Cl2 + 2NaOH -> NaCl + NaOCl + H2O.

Zinc goes from 0 to +2 (loses electrons), so Zn is oxidised and acts as the reducing agent. Copper goes from +2 (in CuSO4) to 0 (in Cu) by gaining electrons, so Cu2+ is reduced and CuSO4 acts as the oxidising agent. Thus Zn is the reducing agent and CuSO4 is the oxidising agent.

Zn + Cu2+ -> Zn2+ + Cu

Hydrogen resembles both alkali metals (group 1) and halogens (group 17), so its position is anomalous. Like alkali metals, it has one valence electron (1s1), forms H+ and shows +1 oxidation state. Like halogens, it is one electron short of a noble gas configuration, is diatomic (H2), and can gain an electron to form the hydride ion (H-). Because it fits neither group perfectly, its placement is debated.

Hard water is water that does not give lather easily with soap because it contains dissolved calcium and magnesium salts. Temporary hardness is due to bicarbonates of Ca and Mg (Ca(HCO3)2, Mg(HCO3)2) and can be removed by boiling or by Clark's method (adding calculated slaked lime). Permanent hardness is due to chlorides and sulphates of Ca and Mg and is removed by adding washing soda (Na2CO3) or by the ion-exchange (permutit/resin) method.

In H2O2 the oxidation number of oxygen is -1, which is intermediate between 0 (in O2) and -2 (in H2O). Therefore it can be reduced to -2 (acting as an oxidising agent) or oxidised to 0 (acting as a reducing agent), depending on the other reactant. For example, it oxidises PbS to PbSO4 (oxidising agent) and reduces acidified KMnO4 (reducing agent).

Alkali metals have low ionization enthalpies because of their large size and a single loosely held valence electron. They readily lose this electron to form unipositive ions, i.e. they are easily oxidised. Since a substance that is easily oxidised is a good reducing agent, alkali metals are strong reducing agents. Their reducing power generally increases down the group.

M -> M+ + e-

A diagonal relationship is the similarity in properties between an element and the element placed diagonally to its lower right in the periodic table (e.g. Li and Mg, Be and Al, B and Si). Lithium and magnesium resemble each other because they have similar atomic and ionic sizes and nearly the same charge-to-size (polarising power) ratio. For example, both form nitrides directly with nitrogen and both form covalent, water-soluble compounds unlike the rest of their groups.

Lithium differs from the other alkali metals because of its very small atomic and ionic size, high charge density (high polarising power) and the absence of d-orbitals in its valence shell. As a result, its compounds have appreciable covalent character; for example, LiCl is soluble in organic solvents, Li forms a nitride (Li3N) and its carbonate and hydroxide decompose on heating, unlike those of Na and K.

Solubility of hydroxides increases down the group (Mg(OH)2 is sparingly soluble while Ba(OH)2 is fairly soluble) because lattice energy decreases faster than hydration energy. Solubility of sulphates decreases down the group (MgSO4 is soluble but BaSO4 is almost insoluble) because hydration energy decreases faster for the larger ions. Flame colours: calcium gives brick-red, strontium gives crimson-red and barium gives apple-green.

The inert pair effect is the reluctance of the two outermost s-electrons (ns2) to take part in bonding, which becomes more pronounced down a group. As a result the lower oxidation state (2 less than the group valency) becomes more stable for heavier elements. For example, in group 14 the +2 state becomes more stable than +4 down the group, so Pb2+ is more stable than Pb4+, while carbon and silicon prefer +4.

In BF3, boron has only six electrons in its valence shell after forming three B-F bonds, so it has an incomplete octet and an empty p-orbital. This makes BF3 electron-deficient, so it can accept a lone pair of electrons from a donor (Lewis base) such as NH3 to complete its octet (forming F3B<-NH3). A species that accepts an electron pair is a Lewis acid, hence BF3 is a Lewis acid.

BF3 + :NH3 -> F3B-NH3

Diborane (B2H6) has two boron atoms and six hydrogen atoms. Four hydrogen atoms (two on each boron) are terminal and are bonded by normal two-centre two-electron (2c-2e) B-H bonds. The remaining two hydrogen atoms are bridging: each forms a three-centre two-electron (3c-2e) B-H-B bond, often called a banana bond. Each boron is sp3 hybridised. The molecule is electron-deficient because it does not have enough valence electrons for normal two-electron bonds throughout.

B2H6 (2 bridging + 4 terminal H)

Catenation is the self-linking of atoms of the same element to form long chains or rings. Carbon shows catenation to the maximum extent because the C-C bond is very strong (high bond energy) due to the small size of the carbon atom, allowing effective overlap. Three crystalline allotropes of carbon are diamond, graphite and fullerene (C60).

In diamond each carbon is sp3 hybridised and bonded to four others in a rigid three-dimensional tetrahedral network, so it is extremely hard and a non-conductor. In graphite each carbon is sp2 hybridised and bonded to three others in flat hexagonal layers held together by weak van der Waals forces, so the layers slide easily (soft/lubricating). The fourth (unhybridised) electron of each carbon is delocalised over the layer, allowing graphite to conduct electricity.

(CH3)2CHCH2CH3 has a four-carbon main chain (butane) with a methyl group on the second carbon, so its IUPAC name is 2-methylbutane. CH3COOH is a two-carbon carboxylic acid, so its IUPAC name is ethanoic acid (common name acetic acid).

Structural (constitutional) isomerism occurs when compounds have the same molecular formula but different arrangements of atoms. Three types: (1) Chain isomerism - different carbon skeletons, e.g. n-butane and isobutane (C4H10). (2) Position isomerism - same skeleton but a substituent or functional group in different positions, e.g. 1-propanol and 2-propanol. (3) Functional isomerism - same molecular formula but different functional groups, e.g. ethanol (alcohol) and dimethyl ether (C2H6O).

The inductive effect is the permanent displacement of the shared sigma-bond electron pair towards the more electronegative atom, transmitted through a chain of carbon atoms and decreasing with distance. Groups that push electrons away from themselves (electron-releasing, e.g. alkyl groups -CH3) show the +I effect, while groups that pull electrons towards themselves (electron-withdrawing, e.g. -NO2, -Cl) show the -I effect.

Resonance is the representation of a molecule or ion by two or more structures (canonical/contributing structures) that differ only in the position of electrons, none of which alone describes it fully; the actual structure is a resonance hybrid. Resonance energy is the difference in energy between the resonance hybrid and the most stable contributing structure; the greater the resonance energy, the more stable the molecule. Example: benzene, which is a hybrid of two Kekule structures and is more stable than expected.

Hyperconjugation is the delocalisation of the sigma electrons of a C-H bond (adjacent to a positively charged carbon or a double bond) into the empty p-orbital or pi-system; it is also called no-bond resonance. In carbocations, the more alkyl groups attached to the positive carbon, the more C-H bonds are available for hyperconjugation, so more delocalisation of charge occurs. Hence stability order is tertiary > secondary > primary > methyl carbocation.

An electrophile (electron-loving) is an electron-deficient species that accepts a pair of electrons; examples: H+, NO2+ (also positively charged or neutral electron-deficient species like BF3). A nucleophile (nucleus-loving) is an electron-rich species that donates a pair of electrons; examples: OH-, CN- (also neutral species with lone pairs like NH3 and H2O).

In homolytic fission, a covalent bond breaks so that each bonded atom takes one of the shared electrons, producing neutral free radicals (species with an unpaired electron); it usually occurs in the presence of heat or light in non-polar bonds. In heterolytic fission, the bond breaks so that one atom takes both shared electrons, producing oppositely charged ions (a cation and an anion, e.g. a carbocation and an anion).

Crystallisation: used to purify solids; the impure solid is dissolved in a suitable hot solvent, filtered, and cooled so that the pure compound crystallises out while soluble impurities remain in solution. Simple distillation: used to separate a volatile liquid from a non-volatile impurity or two liquids with a large difference in boiling points; the liquid is boiled and the vapour is condensed and collected. Steam distillation: used to purify liquids that are steam-volatile and immiscible with water; steam is passed through the mixture so the compound distils over below its normal boiling point (e.g. aniline).

Wurtz reaction: two molecules of an alkyl halide react with sodium metal in dry ether to give a symmetrical alkane with double the number of carbon atoms, e.g. 2CH3Cl + 2Na -> CH3-CH3 + 2NaCl (ethane). Decarboxylation: the sodium salt of a carboxylic acid is heated with soda lime (NaOH + CaO) to give an alkane with one carbon less, e.g. CH3COONa + NaOH -> CH4 + Na2CO3 (methane).

2CH3Cl + 2Na -> C2H6 + 2NaCl

Markovnikov's rule states that when an unsymmetrical reagent (HX) adds to an unsymmetrical alkene, the negative part of the reagent (X) attaches to the carbon bearing fewer hydrogen atoms, while hydrogen adds to the carbon bearing more hydrogen atoms. So HBr adds to propene (CH3-CH=CH2) to give mainly 2-bromopropane (CH3-CHBr-CH3), because the more stable secondary carbocation is formed. In the presence of peroxide, addition is anti-Markovnikov (peroxide/Kharasch effect), giving 1-bromopropane.

CH3-CH=CH2 + HBr -> CH3-CHBr-CH3

Ethene is prepared by heating ethanol with concentrated sulphuric acid at about 443 K (170 deg C), which removes a molecule of water (dehydration): CH3CH2OH -> CH2=CH2 + H2O. Test for unsaturation: (1) alkenes decolourise reddish-brown bromine water (Br2/H2O). (2) alkenes decolourise cold dilute alkaline KMnO4 (Baeyer's reagent), turning the purple colour colourless and forming a diol. These tests confirm a carbon-carbon double bond.

CH3CH2OH -> CH2=CH2 + H2O

Ozonolysis is the reaction of an alkene with ozone (O3) to form an unstable ozonide, which on reductive cleavage (with Zn and water) breaks the carbon-carbon double bond and gives carbonyl compounds (aldehydes and/or ketones). It is used to locate the position of the double bond. Propene (CH3-CH=CH2) on ozonolysis gives ethanal (CH3CHO) and methanal (HCHO).

CH3-CH=CH2 -> CH3CHO + HCHO

Huckel's rule states that a planar, cyclic, fully conjugated ring is aromatic if it contains (4n + 2) pi electrons, where n = 0, 1, 2, 3... Benzene is aromatic because it is planar and cyclic, every carbon is sp2 hybridised with continuous conjugation (a delocalised pi system), and it has 6 pi electrons, which fits (4n + 2) with n = 1. This delocalisation gives benzene extra stability (resonance/aromatic stabilisation).

(4n + 2) pi electrons

Benzene has a stable, delocalised aromatic pi-electron cloud (6 pi electrons). Addition reactions would destroy this aromatic stability, so benzene prefers substitution, which preserves the aromatic ring. The pi cloud attracts electrophiles, so it undergoes electrophilic substitution. Examples: nitration (with conc. HNO3 + conc. H2SO4, electrophile NO2+) and halogenation (with Cl2/FeCl3); others are sulphonation and Friedel-Crafts alkylation/acylation.

Ethane (alkane) does not react with bromine water or Baeyer's reagent (no decolourisation) as it is saturated. Ethene (alkene) decolourises both bromine water and cold dilute alkaline KMnO4 (Baeyer's reagent) but gives no precipitate with ammoniacal silver nitrate. Ethyne (terminal alkyne) also decolourises bromine water and Baeyer's reagent, and in addition forms a white precipitate of silver acetylide with ammoniacal silver nitrate (and a red precipitate with ammoniacal cuprous chloride), because of its acidic terminal hydrogen.