JEE Chemistry Physical and Inorganic Chemistry MCQs Set 03

Answer: (b) 100 cc

Explanation:
1. \(\ce{CO2 + C -> 2CO}\).
2. One volume of \(\mathrm{CO_2}\) gives two volumes of CO at the same temperature and pressure, so 50 cc gives 100 cc.

Teacher's Note:
1. Gay-Lussac's law lets volumes of gases be compared directly as mole ratios.
2. Carbon is a solid and does not contribute to volume.

Answer: (a) 10

Explanation:
1. The kernel (core) is the atom without its valence electrons.
2. Phosphorus has the configuration \(1s^22s^22p^63s^23p^3\), with 5 valence electrons.
3. Kernel electrons \(= 15 - 5 = 10\).

Teacher's Note:
1. Kernel = nucleus plus inner shell electrons.
2. Equivalent to the charge \(+5\) core of P.

Answer: (c) \(Q_c < K_c\)

Explanation:
1. If \(Q_c < K_c\), products are fewer than at equilibrium, so the reaction moves forward to form more products.
2. \(Q_c = K_c\) means equilibrium and \(Q_c > K_c\) means a backward shift.

Teacher's Note:
1. Compare Q with K: smaller Q means forward, larger Q means reverse.
2. Q has the same form as K but uses non-equilibrium concentrations.

Answer: (c) 4050 cal

Explanation:
1. For an adiabatic process \(q = 0\), so \(W_{on} = \Delta U = nC_v\Delta T\).
2. \(W = 3\times27\times(250 - 200) = 4050\) cal.

Teacher's Note:
1. In an adiabatic compression all work done on the gas raises its internal energy.
2. \(\Delta U = nC_v\Delta T\) is valid for any process of an ideal gas.

Answer: (a) Sn, O

Explanation:
1. Sn changes from +2 in \(\mathrm{SnCl_2}\) to +4 in \(\mathrm{SnCl_4}\), so it is oxidised.
2. \(\mathrm{CrO_5}\) is a peroxide with Cr at +6 and four oxygen atoms at \(-1\). In \(\mathrm{CrO_4^{2-}}\) Cr is still +6, while the peroxide oxygen goes from \(-1\) to \(-2\), so oxygen is reduced.

Teacher's Note:
1. Chromium stays at +6 in both species, so it is neither oxidised nor reduced.
2. Always check for peroxide linkages when assigning oxidation numbers.

Answer: (b) 136.6 - 241.6 pm

Explanation:
1. The NaCl structure has octahedral coordination, for which \(0.414 \le r^+/r^- \le 0.732\).
2. \(r^-\) ranges from \(\dfrac{100}{0.732} = 136.6\) pm to \(\dfrac{100}{0.414} = 241.5\) pm, i.e. about 136.6 to 241.6 pm.

Teacher's Note:
1. Radius ratio limits determine the coordination number.
2. Octahedral holes: 0.414 to 0.732.

Answer: (d) They are used as anion exchange resins.

Explanation:
1. Zeolites are aluminosilicates with a 3-D framework in which some Si atoms are replaced by Al.
2. The Al substitution gives a negative framework charge balanced by cations such as \(\mathrm{Na^+}\), so zeolites are cation exchangers, not anion exchange resins.

Teacher's Note:
1. Zeolites are used as molecular sieves, in catalysis (ZSM-5) and for softening hard water.
2. Their ion exchange is of cations.

Answer: (a) exothermic

Explanation:
1. \(\Delta H = E_{a,f} - E_{a,b} = 20.5 - 55.5 = -35\) kJ mol\(^{-1}\).
2. The negative value means the reaction is exothermic.

Teacher's Note:
1. Products lie lower in energy than the reactants when the backward barrier is larger than the forward barrier.
2. This difference gives the enthalpy change.

Answer: (d) \(\mathrm{[Cr(NH_3)_6]Cl_3}\)

Explanation:
1. Conductivity increases with the number of ions produced.
2. (a) gives 0 ions, (b) 2 ions, (c) 3 ions and (d) 4 ions: \(\ce{[Cr(NH3)6]^{3+} + 3Cl^-}\).

Teacher's Note:
1. Only the species outside the square bracket ionise.
2. More ions per formula unit means greater conductivity.

Answer: (b) 87%

Explanation:
1. \(\ce{Ba(NO3)2 -> Ba^{2+} + 2NO3^-}\), so \(n = 3\).
2. \(i = 1 + (n-1)\alpha = 1 + 2\alpha = 2.74\).
3. \(\alpha = 0.87\), i.e. 87%.

Teacher's Note:
1. The van't Hoff factor for a salt giving n ions is \(1 + (n-1)\alpha\).
2. The factor can never exceed the number of ions n.

Answer: (d) metallic properties

Explanation:
1. A more electronegative element attracts electrons strongly and tends to gain them, which is non-metallic behaviour.
2. A less electronegative element tends to lose electrons easily, so metallic character varies inversely with electronegativity.

Teacher's Note:
1. Metallic character decreases across a period and increases down a group, the opposite of electronegativity.
2. Non-metallic character follows electronegativity.

Answer: (b) sp\(^2\), sp, sp\(^3\)

Explanation:
1. \(\mathrm{NO_3^-}\) has 3 bond pairs and no lone pair on N, which is sp\(^2\) (trigonal planar).
2. \(\mathrm{NO_2^+}\) has 2 double bonds and is linear, which is sp.
3. \(\mathrm{NH_4^+}\) has 4 bond pairs, which is sp\(^3\).

Teacher's Note:
1. Hybridisation = number of sigma bonds plus lone pairs on the central atom.
2. \(\mathrm{NO_2^+}\) is isoelectronic with \(\mathrm{CO_2}\).

Answer: (b) II < III < I

Explanation:
1. Thermal stability of carbonates increases as the polarising power of the cation decreases.
2. \(\mathrm{Mg^{2+}}\) is small and highly charged, so \(\mathrm{MgCO_3}\) decomposes easily.
3. Among alkali metals, stability increases down the group: \(\mathrm{Na_2CO_3} < \mathrm{K_2CO_3}\).
4. Order: II < III < I.

Teacher's Note:
1. Smaller, highly charged cations polarise the carbonate ion more and lower its stability.
2. Down group 1 and group 2 the stability of carbonates increases.

Answer: (c) \(\mathrm{Cu}\) + dil. \(\mathrm{H_2SO_4}\)

Explanation:
1. Zn reacts with NaOH to form sodium zincate and \(\mathrm{H_2}\).
2. \(\ce{NaH + H2O -> NaOH + H2}\) and \(\ce{2Na + 2H2O -> 2NaOH + H2}\).
3. Cu lies below hydrogen in the reactivity series and does not displace hydrogen from dilute acids.

Teacher's Note:
1. Only metals above hydrogen in the electrochemical series liberate \(\mathrm{H_2}\) from dilute non-oxidising acids.
2. Copper reacts only with oxidising acids and gives \(\mathrm{SO_2}\) with hot concentrated \(\mathrm{H_2SO_4}\).

Answer: (c) Zr and Hf have almost the same radius

Explanation:
1. In the lanthanoids the 4f electrons shield the nuclear charge poorly, so atomic radii decrease steadily from La to Lu.
2. This offsets the normal increase in size from the 4d to the 5d series, so Hf (5d) has almost the same radius as Zr (4d).
3. Their chemistry is therefore very similar and they occur together in nature.

Teacher's Note:
1. Poor shielding by 4f electrons causes the contraction.
2. Look for pairs of 4d/5d elements of the same group with equal radii: Zr-Hf, Nb-Ta, Mo-W.

Answer: (a) chlorides and sulphates of Mg and Ca

Explanation:
1. Permanent hardness is due to dissolved chlorides and sulphates of calcium and magnesium.
2. It cannot be removed by boiling, unlike temporary hardness caused by bicarbonates of Ca and Mg.
3. Sodium and potassium salts do not cause hardness.

Teacher's Note:
1. Boiling removes only temporary hardness.
2. Permanent hardness is treated with washing soda, calgon or ion exchange resins.

Answer: (c) the \(\mathrm{NH_4^+}\) ion does not have any lone pair

Explanation:
1. A ligand must be able to donate a lone pair of electrons to the metal ion.
2. \(\mathrm{NH_3}\) has a lone pair on N.
3. In \(\mathrm{NH_4^+}\) that pair is already used to bond with \(\mathrm{H^+}\), so it has no lone pair left.

Teacher's Note:
1. Ligands are Lewis bases.
2. NH\(_4^+\) is a Lewis acid-type species with all electrons in bonds.

Answer: (b) \(\mathrm{[Mn(CN)_6]^{3-}}\)

Explanation:
1. \(\mathrm{CN^-}\) is a strong field ligand and pairs up the d electrons.
2. In \(\mathrm{[Mn(CN)_6]^{3-}}\), Mn\(^{3+}\) is \(d^4\), and the electrons pair to occupy the \(t_{2g}\) orbitals, leaving two d orbitals free for \(d^2sp^3\) hybridisation.
3. This is inner orbital and low spin. \(\mathrm{F^-}\) is a weak field ligand giving outer orbital (\(sp^3d^2\)) high spin complexes, and \(\mathrm{[NiCl_4]^{2-}}\) is tetrahedral \(sp^3\).

Teacher's Note:
1. Strong field ligands (CN\(^-\), CO) give low spin inner orbital complexes.
2. Weak field ligands (halides) give high spin outer orbital complexes.

Answer: (c) \(\mathrm{BiF_5}\)

Explanation:
1. \(\mathrm{BiF_5}\) exists since the small fluoride ion can stabilise Bi in the +5 state.
2. \(\mathrm{BiI_5}\) does not exist because iodide reduces Bi(V).
3. \(\mathrm{MnF_7}\) is unknown (Mn in +7 is only seen with oxygen), and \(\mathrm{SCl_6}\) does not exist because six large Cl atoms cannot fit around S.

Teacher's Note:
1. Fluorine stabilises the highest oxidation states because of its small size and high electronegativity.
2. The inert pair effect makes the +5 state of Bi strongly oxidising.

Answer: (b) dilute acid

Explanation:
1. \(\ce{CaOCl2 + H2SO4 -> CaSO4 + H2O + Cl2}\).
2. The chlorine liberated by a dilute acid is called the available chlorine.
3. NaOH, CO and \(\mathrm{SO_2}\) do not liberate chlorine in this way (\(\mathrm{SO_2}\) is oxidised by it).

Teacher's Note:
1. Bleaching action depends on the available chlorine content (about 35 percent).
2. Chlorine is also set free by atmospheric \(\mathrm{CO_2}\) acting slowly.

Answer: (c) Sn, Zn

Explanation:
1. Oxides of Sn and Zn are reduced by carbon: \(\ce{SnO2 + 2C -> Sn + 2CO}\), \(\ce{ZnO + C -> Zn + CO}\).
2. Cu is obtained from its sulphide by self reduction in a Bessemer converter, Al by electrolysis of alumina, and Ag by cyanide leaching and displacement with zinc.

Teacher's Note:
1. Metals less reactive than carbon at the temperature of reduction can be extracted by carbon.
2. Highly reactive metals like Al need electrolysis.

Answer: (a) 1.0

Explanation:
1. Freundlich equation: \(\log\dfrac xm = \log k + \dfrac1n\log p\).
2. Slope \(= \dfrac1n = \tan45^\circ = 1\) and \(\log k = 0.6990\), so \(k = 5\).
3. \(\dfrac xm = 5\times0.2 = 1.0\).

Teacher's Note:
1. The slope of the log-log plot gives 1/n and the intercept gives log k.
2. A slope of 45 degrees means n = 1.

Answer: (b) A, B, C

Explanation:
1. In homogeneous catalysis the catalyst and reactants are in the same phase.
2. In A all species are gaseous, in B the acid and sucrose are both in aqueous solution, and in C all are liquids.
3. In D the catalyst is a solid and the reactants are gases, so it is heterogeneous. The correct set is A, B and C.

Teacher's Note:
1. Count the phases of catalyst and reactants before deciding.
2. Lead chamber and acid hydrolysis are the classic homogeneous examples.

Answer: (a) gel (liquid dispersed in solid)

Explanation:
1. In opal, a liquid is the dispersed phase and a solid is the dispersion medium.
2. Such a colloid is a gel.

Teacher's Note:
1. Pumice stone and foam rubber are solid sols with gas in solid.
2. Pearl and cheese are other examples of gels.

Answer: (b) associated colloid

Explanation:
1. Soap molecules aggregate to form micelles above the critical micelle concentration and Kraft temperature.
2. Colloids formed by such aggregation are associated colloids.

Teacher's Note:
1. Associated colloids behave as normal electrolytes at low concentration.
2. The micelle formation is the basis of the cleaning action of soaps.

Answer: (c) 50

Explanation:
1. 10 ml of 1 N \(\ce{Na2SO4}\) contains 10 milli-equivalents, which is 5 millimoles (the salt is divalent).
2. This quantity coagulates 100 ml = 0.1 L of sol, so the coagulation value \(= \dfrac{5}{0.1} = 50\) millimoles per litre.

Teacher's Note:
1. Coagulation value is the minimum millimoles of electrolyte per litre of sol needed for coagulation.
2. A smaller coagulation value means a more effective coagulating ion.