JEE Chemistry Physical and Inorganic Chemistry MCQs Set 01

Answer: (c) 0

Explanation:
1. A hydrogen atom absorbs a photon only if its energy equals the difference between two energy levels.
2. From \( n = 1 \): \( n=2 \) needs 10.2 eV, \( n=3 \) needs 12.09 eV, ionisation needs 13.6 eV.
3. 11 eV matches none of these, so the photon is not absorbed and no spectral line is obtained.

Teacher's Note:
1. Unlike photoelectric effect, atomic excitation by a photon is a resonant process.
2. Only energies equal to exact level differences, or above ionisation energy, can be absorbed.

Answer: (c) \( 7 \times 10^{-4} \)

Explanation:
1. For a conjugate pair \( pK_a + pK_b = pK_w = 14 \).
2. \( pK_a = 14 - 10.84 = 3.16 \).
3. \( K_a = 10^{-3.16} = 10^{0.84} \times 10^{-4} \approx 6.9 \times 10^{-4} \approx 7 \times 10^{-4} \).

Teacher's Note:
1. Conjugate acid and base always satisfy \( pK_a + pK_b = 14 \) at 25 \( ^\circ \)C.
2. \( 10^{0.84} \approx 6.9 \).

Answer: (d) \( \ce{Xe(g) + 2F2(g) -> XeF4(s)} \)

Explanation:
1. \( \Delta H_f^\circ \) refers to the formation of exactly one mole of a compound from its elements in their standard states.
2. (a) starts with a compound (CO) and makes 2 mol. (b) uses ozone, which is not the standard state of oxygen. (c) has compounds as reactants and two products.
3. (d) forms 1 mol of \( \ce{XeF4} \) from the elements Xe and \( \ce{F2} \) in standard states, so \( \Delta H^\circ_{rxn} = \Delta H^\circ_f \).

Teacher's Note:
1. Check three things: elements only on the left, one product, one mole of it.
2. The standard state of oxygen is \( \ce{O2} \), not \( \ce{O3} \).

Answer: (a) \( \ce{HCOOH} \) (\( pK_a = 3.77 \))

Explanation:
1. A lower \( pK_a \) means a larger \( K_a \), i.e. a stronger acid.
2. The smallest \( pK_a \) is 3.77, that of formic acid.

Teacher's Note:
1. Electron-donating alkyl groups reduce acid strength; formic acid has none.
2. Compare \( pK_a \) values directly: smaller means stronger.

Answer: (c) 1.4 times that of He

Explanation:
1. Graham's law: \( \frac{r_{H_2}}{r_{He}} = \sqrt{\frac{M_{He}}{M_{H_2}}} \).
2. \( = \sqrt{\frac{4}{2}} = \sqrt{2} \approx 1.4 \).

Teacher's Note:
1. Lighter gases diffuse faster; rate is inversely proportional to the square root of molar mass.
2. \( \sqrt{2} = 1.414 \).

Answer: (a) catalyst

Explanation:
1. The rate constant k depends on temperature and on the catalyst (through the activation energy, \( k = Ae^{-E_a/RT} \)).
2. It does not depend on the concentrations of the reactants.
3. Pressure affects only the rate through concentrations, not k itself, so only the catalyst is correct.

Teacher's Note:
1. k is constant at a fixed temperature unless a catalyst changes the path.
2. Rate depends on concentration; rate constant does not.

Answer: (d) 4.8 L

Explanation:
1. Water is electrolysed: anode \( \ce{2H2O -> O2 + 4H+ + 4e-} \); cathode \( \ce{2H2O + 2e- -> H2 + 2OH-} \).
2. Per 4 electrons: 1 mol \( \ce{O2} \) and 2 mol \( \ce{H2} \).
3. So \( V_{H_2} = 2 \times 2.4 = 4.8 \) L.

Teacher's Note:
1. In aqueous \( \ce{Na2SO4} \), the ions are not discharged; water is.
2. \( \ce{H2} : \ce{O2} = 2 : 1 \) by volume.

Answer: (c) Mass %

Explanation:
1. Molality is mol/kg, molarity is mol/L and density is g/mL (or kg/m\(^3\)).
2. Mass percent is a ratio of two masses, so it has no unit.

Teacher's Note:
1. Ratios of like quantities are dimensionless.
2. Mole fraction and mass fraction are also unitless.

Answer: (c) 4.5%

Explanation:
1. Isotonic solutions have equal molar concentration (same osmotic pressure at the same temperature).
2. Urea: 1.5 g per 100 mL \( = \frac{1.5}{60} = 0.025 \) mol per 100 mL.
3. Glucose needed \( = 0.025 \times 180 = 4.5 \) g per 100 mL, i.e. 4.5% (w/v).

Teacher's Note:
1. Both solutes are non-electrolytes, so no van't Hoff factor is needed.
2. Equal moles in the same volume means isotonic.

Answer: (c) \( \ce{SF4} \)

Explanation:
1. \( \ce{XeF4} \) is square planar, \( \ce{BF3} \) is trigonal planar and \( \ce{SiF4} \) is tetrahedral: all symmetrical, so the dipoles cancel.
2. \( \ce{SF4} \) has one lone pair (see-saw shape), which gives an uneven dipole distribution, so it is polar.

Teacher's Note:
1. A molecule is polar if the bond dipoles do not cancel.
2. Lone pairs frequently destroy symmetry.

Answer: (c) ethyne, \( \ce{H-C#C-H} \)

Explanation:
1. Bond strength increases with the s-character of the carbon orbital in the bond.
2. Carbon is sp\(^3\) (25% s) in methane, sp\(^2\) (33% s) in ethene and sp (50% s) in ethyne.
3. Hence the C-H bond is strongest in ethyne.

Teacher's Note:
1. More s-character means the orbital is closer to the nucleus, giving a shorter and stronger bond.
2. Ethyne is also the most acidic of the three.

Answer: (d) \( \ce{H2} \)

Explanation:
1. At high temperature sodium reacts with ammonia to form sodamide: \( \ce{2Na + 2NH3 -> 2NaNH2 + H2} \).
2. The gas evolved is hydrogen.

Teacher's Note:
1. Dissolved in liquid ammonia sodium gives a blue solution; on heating it forms sodamide and hydrogen.
2. Metals displace hydrogen from weakly acidic N-H bonds.

Answer: (c) potassium chromate

Explanation:
1. Potassium chromate \( \ce{K2CrO4} \) is lemon yellow.
2. In acid it forms dichromate: \( \ce{2CrO4^2- + 2H+ -> Cr2O7^2- + H2O} \) (orange).
3. On adding alkali the equilibrium reverses to yellow chromate.

Teacher's Note:
1. Chromate and dichromate are interconverted by changing the pH.
2. Copper nitrate is blue and ferric chloride is brown or yellow-brown.

Answer: (b) a non-planar open-book structure with a dihedral angle of \( 94^\circ \) between the two O-H planes

Explanation:
1. Hydrogen peroxide has a non-planar, open-book structure with the two O-H bonds in different planes.
2. The dihedral angle is \( 94.8^\circ \) in the gas phase (about \( 111.5^\circ \) in the solid state), and the H-O-O angle is about \( 97^\circ \).

Teacher's Note:
1. Repulsion between lone pairs and O-H bonds twists the molecule out of planarity.
2. Only the skew shape agrees with the dipole moment of 2.1 D.

Answer: (d) potassium carbonylpentacyanoferrate(II)

Explanation:
1. Charge balance: \( 3(+1) + x + 5(-1) + 0 = 0 \), so the oxidation state of Fe is +2.
2. Ligands are named in alphabetical order: carbonyl before cyano (prefix penta- ignored). Anionic complex ends in -ate.
3. Name: potassium carbonylpentacyanoferrate(II).

Teacher's Note:
1. Alphabetical order ignores numerical prefixes such as penta-.
2. CO is neutral, so it does not contribute to the charge.

Answer: (d) (i) and (ii) only

Explanation:
1. An organometallic compound has at least one direct bond between a metal and a carbon atom.
2. (i) has Al-C bonds and (ii), Zeise's salt, has a Pt-C bond (pi bond with ethene).
3. (iii) is trimethylamine, with only C-N bonds (nitrogen is a non-metal). (iv) has Ti-O bonds, so it is a metal alkoxide, not organometallic.

Teacher's Note:
1. Look for a metal-carbon bond, not just carbon in the ligand.
2. Metal alkoxides and amines are not organometallic.

Answer: (b) \( \ce{HF2-} \) and \( \ce{XeF5+} \) ions

Explanation:
1. \( \ce{XeF6} \) acts as a fluoride ion donor towards the acidic solvent HF.
2. \( \ce{XeF6 + HF -> XeF5+ + HF2-} \).
3. The solution is a good conductor because it contains \( \ce{XeF5+} \) and \( \ce{HF2-} \) ions.

Teacher's Note:
1. \( \ce{XeF6} \) loses \( \ce{F-} \) to form \( \ce{XeF5+} \), which is square pyramidal.
2. The fluoride ion is taken up by HF, forming the bifluoride ion.

Answer: (d) \( \ce{BeCl2} \) (solid)

Explanation:
1. In the dimer of \( \ce{BeCl2} \) each Be makes two covalent bonds and one coordinate bond, giving only 6 electrons around Be.
2. In solid \( \ce{BeCl2} \) (polymeric chain) each Be is bonded to four Cl atoms: two normal covalent bonds and two coordinate bonds from bridging Cl atoms. This gives 8 electrons: a complete octet.
3. \( \ce{BeH2} \) is electron-deficient in both forms (three-centre two-electron bonds, no lone pairs to donate).

Teacher's Note:
1. Chlorine has lone pairs to donate; hydrogen has none, so \( \ce{BeH2} \) cannot complete the octet by coordinate bonds.
2. Count shared electrons around Be.

Answer: (d) greater solubility of the impurity in the molten state than in the solid

Explanation:
1. A heater moves slowly along a rod of impure metal, melting a narrow zone.
2. Impurities are more soluble in the melt than in the solid, so they move with the molten zone towards one end.
3. Repeating the process leaves very pure metal behind; the impure end is cut off.

Teacher's Note:
1. Zone refining gives ultra-pure metals such as Ge, Si, Ga and In for semiconductors.
2. The principle is fractional crystallisation: impurities stay in the liquid phase.