JEE Chemistry Physical and Inorganic Chemistry MCQs Set 02

Answer: (b) \(\ce{C8H18}\)

Explanation:
1. For \(\ce{C_xH_y}\), 1 volume gives \(x\) volumes of \(\ce{CO2}\) and \(y/2\) volumes of water vapour.
2. \(x = \dfrac{2400}{300} = 8\) and \(\dfrac y2 = \dfrac{2700}{300} = 9\), so \(y = 18\).
3. The formula is \(\ce{C8H18}\).

Teacher's Note:
1. Gay-Lussac's law lets you use gas volumes directly as mole ratios at the same temperature and pressure.
2. Remember that water vapour contains two hydrogen atoms per molecule.

Answer: (a) \(4\times10^{-6}\) M

Explanation:
1. Precipitation starts when the ionic product equals \(K_{sp}\): \([\ce{Pb^{2+}}][\ce{SO4^{2-}}] = K_{sp}\).
2. \([\ce{Pb^{2+}}] = \dfrac{1.8\times10^{-8}}{0.0045} = 4\times10^{-6}\) M.

Teacher's Note:
1. A salt precipitates only when the ionic product exceeds its solubility product.
2. Use the given sulphate concentration, not the molar solubility of the salt.

Answer: (b) the heat absorbed by the system is equal to the increase in internal energy

Explanation:
1. In an isochoric process the volume is constant, so \(w = -P\Delta V = 0\).
2. From the first law, \(\Delta U = q + w = q_V\). The heat absorbed equals the increase in internal energy.

Teacher's Note:
1. At constant volume, heat becomes a state function because it equals \(\Delta U\).
2. Work is a path function and never a state function.

Answer: (b) \(-2226\) kJ

Explanation:
1. Formation reaction: \(\ce{12C + 11H2 + 11/2 O2 -> C12H22O11}\).
2. \(\Delta_fH = 12(-393.5) + 11(-286.2) - (-5644.2) = -4722 - 3148.2 + 5644.2\).
3. \(\Delta_fH = -2226\) kJ/mol.

Teacher's Note:
1. Use Hess's law: \(\Delta_fH\) equals the sum of combustion enthalpies of reactants minus that of the product.
2. Multiply each combustion enthalpy by the number of moles in the formation equation.

Answer: (a) \(r = 0.414R\)

Explanation:
1. An octahedral void is surrounded by six atoms. Along the edge of the square formed by four of them, \(\sqrt2\cdot 2R = 2(R + r)\).
2. \(R + r = \sqrt2R\), so \(r = (\sqrt2 - 1)R = 0.414R\).

Teacher's Note:
1. The radius ratio \(r/R = 0.414\) for an octahedral void and 0.225 for a tetrahedral void.
2. The void is smaller than the atoms that surround it.

Answer: (a) I only

Explanation:
1. The rate is decided by the slow step. In mechanism I the slow step is \(\ce{Cl2 + H2S}\), so rate \(= k[\ce{Cl2}][\ce{H2S}]\), which matches.
2. In mechanism II the slow step gives rate \(= k'[\ce{Cl2}][\ce{HS-}]\). The fast equilibrium gives \([\ce{HS-}] = K[\ce{H2S}]/[\ce{H+}]\), so rate \(= k''[\ce{Cl2}][\ce{H2S}]/[\ce{H+}]\), which does not match.
3. Only mechanism I is consistent.

Teacher's Note:
1. Always replace intermediates in the slow-step rate law using earlier fast equilibria.
2. A mechanism is acceptable only if it reproduces the observed rate law exactly.

Answer: (c) \(-0.272^\circ\)C

Explanation:
1. \(\Delta T_b = 0.15 = K_b m\), so \(m = \dfrac{0.15}{0.512} = 0.293\) mol/kg.
2. Diluting with an equal volume of water halves the molality: \(m' = 0.1465\).
3. \(\Delta T_f = K_f m' = 1.86 \times 0.1465 = 0.272\), so the freezing point is \(-0.272^\circ\)C.

Teacher's Note:
1. Dilution to double the volume halves the molality when the solution is dilute.
2. Freezing point depression is subtracted from \(0^\circ\)C.

Answer: (c) Fe is oxidised to \(\ce{Fe^{3+}}\) and oxygen is reduced to \(\ce{O^{2-}}\)

Explanation:
1. In rusting, iron acts as the anode and is oxidised, \(\ce{Fe -> Fe^{2+} + 2e-}\); in the presence of air and water it is oxidised further to \(\ce{Fe^{3+}}\), which forms hydrated ferric oxide (rust).
2. The cathode reaction is the reduction of dissolved oxygen: \(\ce{O2 + 4H+ + 4e- -> 2H2O}\), in which oxygen goes to the \(\ce{O^{2-}}\) state.
3. Water is not reduced to \(\ce{O2^{-}}\) or \(\ce{O2^{2-}}\), so options (a) and (d) are wrong. Option (c) is the only one in which oxygen is the species reduced.

Teacher's Note:
1. Rust is hydrated iron(III) oxide, \(\ce{Fe2O3.xH2O}\).
2. Oxygen, not water, is the oxidising agent in atmospheric corrosion.

Answer: (a) \(\ce{KClO4}\)

Explanation:
1. Isomorphous compounds have the same crystal structure and similar formula type.
2. \(\ce{KMnO4}\) and \(\ce{KClO4}\) both have the form \(\ce{KXO4}\), with a tetrahedral anion, and both crystallise in the orthorhombic system.
3. So \(\ce{KMnO4}\) is isomorphous with \(\ce{KClO4}\).

Teacher's Note:
1. Look for the same ratio of ions and the same type of anion geometry.
2. Isomorphous salts can grow mixed crystals.

Answer: (d) \(\ce{N + O+ -> N+ + O}\)

Explanation:
1. Compare the ionisation energies for the cation processes. In (a), the ionisation energy of Cl (1251 kJ/mol) is lower than that of F (1681 kJ/mol), so \(\ce{Cl+}\) forms from F+ easily and the reaction is spontaneous.
2. In (b) and (c) the electron goes to the atom with the more negative electron gain enthalpy (S more than O, B more than Be), so they are spontaneous.
3. In (d), \(\ce{O+}\) would transfer its charge to N. This needs \(\text{IE(N)} = 1402\) kJ/mol, which is higher than \(\text{IE(O)} = 1314\) kJ/mol. This requires energy, so it is non-spontaneous.

Teacher's Note:
1. The atom with the lower ionisation energy is the one that keeps the positive charge.
2. Nitrogen has a half-filled p subshell, so its first ionisation energy is higher than that of oxygen.

Answer: (a) \(\ce{XeO3}\) : trigonal planar

Explanation:
1. \(\ce{XeO3}\) has three bond pairs and one lone pair on Xe (\(sp^3\)), so its shape is trigonal pyramidal and not trigonal planar.
2. \(\ce{ClF3}\) (three bond pairs, two lone pairs) is T-shaped, \(\ce{XeOF4}\) (five bond pairs, one lone pair) is square pyramidal, and \(\ce{XeF2}\) (two bond pairs, three lone pairs) is linear.

Teacher's Note:
1. Count bond pairs and lone pairs to apply VSEPR theory.
2. Lone pairs take equatorial positions in a trigonal bipyramid.

Answer: (a) \(\theta = 90^\circ\)

Explanation:
1. The resultant of two equal bond dipoles \(\mu_0\) is \(\mu = 2\mu_0\cos\dfrac\theta2\).
2. \(\cos(\theta/2)\) decreases as \(\theta\) increases, so \(\mu\) is greatest for the smallest angle listed, \(90^\circ\). At \(180^\circ\) it is zero.

Teacher's Note:
1. A linear molecule has zero resultant dipole as the bond dipoles cancel.
2. The smaller the bond angle, the larger the resultant of two equal vectors.

Answer: (d) Both (b) and (c)

Explanation:
1. \(\ce{MgCO3}\) decomposes on heating to \(\ce{MgO + CO2}\).
2. The carbonates of sodium and potassium are very stable thermally and melt without decomposing, so they do not give oxides.
3. So both \(\ce{Na2CO3}\) and \(\ce{K2CO3}\) do not give oxides.

Teacher's Note:
1. Thermal stability of carbonates increases down group 1 and group 2.
2. Lithium carbonate is the only group 1 carbonate that decomposes easily.

Answer: (b) 2 mol \(\ce{NH3}\)

Explanation:
1. \(\ce{Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3}\).
2. One mole of magnesium nitride gives 2 moles of ammonia.

Teacher's Note:
1. The nitride ion is strongly basic and takes protons from water.
2. Balance magnesium, hydrogen and oxygen in turn.

Answer: (b) Pb(IV) is a strong oxidising agent and \(\ce{I-}\) is a strong reducing agent

Explanation:
1. Because of the inert pair effect, Pb(IV) is a strong oxidising agent.
2. Iodide is a good reducing agent, so \(\ce{Pb^{4+}}\) oxidises \(\ce{I-}\) to \(\ce{I2}\) and is itself reduced to \(\ce{Pb^{2+}}\). So \(\ce{PbI4}\) cannot exist.
3. Option (c) is true, but it does not explain the specific absence of \(\ce{PbI4}\), while \(\ce{PbCl4}\) exists.

Teacher's Note:
1. The inert pair effect makes the +2 state more stable than +4 for heavier group 14 elements.
2. Fluoride and chloride resist oxidation, so PbF4 and PbCl4 can exist.

Answer: (a) \(\ce{BF6^{3-}}\)

Explanation:
1. Boron is a second-period element with no d orbitals, and its maximum covalency is 4.
2. So \(\ce{BF6^{3-}}\) cannot form. Aluminium, silicon and germanium use d orbitals and can have a covalency of 6.

Teacher's Note:
1. The maximum covalency of the second-period elements is 4.
2. \(\ce{BF4-}\) exists, but \(\ce{BF6^{3-}}\) does not.

Answer: (c) \(\ce{[Ma5b]}\)

Explanation:
1. \(\ce{[Ma3b3]}\) shows facial and meridional isomers. \(\ce{[Ma4b2]}\) and \(\ce{[Ma2b4]}\) show cis and trans isomers.
2. In \(\ce{[Ma5b]}\) all six positions are equivalent relative to the single ligand b, so there is only one arrangement and no geometrical isomerism.

Teacher's Note:
1. Geometrical isomerism needs at least two ligands of each kind in an octahedral complex.
2. \(\ce{[M(abcdef)]}\) has 15 geometrical isomers.

Answer: (d) \(\ce{MgCO3}\)

Explanation:
1. Hardness is caused by the soluble bicarbonates (temporary hardness), and the chlorides and sulphates (permanent hardness) of calcium and magnesium.
2. \(\ce{Mg(HCO3)2}\), \(\ce{CaCl2}\) and \(\ce{MgSO4}\) are all soluble and cause hardness.
3. \(\ce{MgCO3}\) is sparingly soluble, so it does not put ions in the water to cause hardness.

Teacher's Note:
1. Only dissolved \(\ce{Ca^{2+}}\) and \(\ce{Mg^{2+}}\) ions make water hard.
2. Boiling removes temporary hardness by precipitating the carbonates.

Answer: (d) \(\ce{HgS}\)

Explanation:
1. Roasting \(\ce{HgS}\) in air first gives \(\ce{HgO}\): \(\ce{2HgS + 3O2 -> 2HgO + 2SO2}\).
2. \(\ce{HgO}\) is unstable at that temperature and decomposes: \(\ce{2HgO -> 2Hg + O2}\). So the metal is obtained directly.
3. \(\ce{PbCO3}\) is calcined to \(\ce{PbO}\), and \(\ce{FeS2}\) and \(\ce{ZnS}\) give oxides on roasting.

Teacher's Note:
1. Mercury is below hydrogen in the electrochemical series, so its oxide decomposes on heating.
2. Cinnabar is the main ore of mercury.