JEE Chemistry The D And F Block Elements MCQs Set 21

Answer: (d) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Explanation:
1. 4f orbitals have a diffuse shape and shield the outer electrons poorly.
2. The increasing nuclear charge is therefore not fully balanced, and the size decreases steadily from Ce to Lu.

Teacher's Note:
1. Poor shielding by f electrons is the cause.

Answer: (d) \( \ce{HgNH2Cl} \)

Explanation:
1. \( \ce{Hg2Cl2 + 2NH4OH -> Hg(NH2)Cl + Hg + NH4Cl + 2H2O} \).
2. The products are the white mercury(II) amidochloride and black finely divided mercury, which gives a black residue.

Teacher's Note:
1. Black mercury deposit is the basis of the test for mercurous ions.

Answer: (b) Zr and Hf have about the same radius

Explanation:
1. The steady decrease in size across the lanthanoids makes the radius of the 5d element Hf almost equal to that of the 4d element Zr (160 pm and 159 pm).
2. This is why Zr and Hf are so similar and hard to separate.

Teacher's Note:
1. Elements of the 4d and 5d series in the same group have similar radii.

Answer: (c) The +4 oxidation state of cerium is not known in solutions

Explanation:
1. Ce\(^{4+}\) is well known in aqueous solution (for example in ceric ammonium sulphate and ceric sulphate titrations).
2. So the statement that +4 is not known in solutions is incorrect.

Teacher's Note:
1. Ce(IV) is used in volumetric analysis.

Answer: (d) \( (n-1)d^{5}ns^{2} \)

Explanation:
1. Transition elements use their ns and (n-1)d electrons for bonding.
2. The configuration (n-1)d\(^5\)ns\(^2\) (Mn) has 7 such electrons, which gives a maximum of +7, while the others give +6, +5 or +3 or +4 at most.

Teacher's Note:
1. The highest oxidation state of the early elements equals the total of ns and (n-1)d electrons.

Answer: (a) Cu, Ag, Au

Explanation:
1. Copper, silver and gold were traditionally used for making coins.

Teacher's Note:
1. The group 11 metals are the coinage metals.

Answer: (d) Silver nitrate

Explanation:
1. Ferric, copper and manganese nitrates leave oxides (\( \ce{Fe2O3} \), CuO, \( \ce{MnO2} \)).
2. Silver oxide is unstable at the temperature, so \( \ce{2AgNO3 -> 2Ag + 2NO2 + O2} \) leaves silver metal.

Teacher's Note:
1. Oxides of noble metals decompose on heating.

Answer: (b) From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl

Explanation:
1. AgCl dissolves in ammonia as \( \ce{[Ag(NH3)2]^+} \), but AgI does not because its solubility product is too low.
2. Manganese gives an amethyst bead in the oxidising flame and a colourless one in the reducing flame, ferric ions give Prussian blue, and boiling gives CaCO\(_3\), so the other statements are wrong.

Teacher's Note:
1. Solubility of silver halides in ammonia decreases from AgCl to AgI.

Answer: (b) \( \ce{HgI2} \)

Explanation:
1. Red HgI\(_2\) dissolves in KI forming \( \ce{K2[HgI4]} \).
2. On heating it decomposes into iodine (violet fumes) and mercury droplets.

Teacher's Note:
1. HgI\(_2\) is red and dissolves in excess iodide.

Answer: (b) In acidic solutions, protons coordinate with ammonia molecules forming \( \ce{NH4+} \) ions and \( \ce{NH3} \) molecules are not available

Explanation:
1. Ammonia is a base and combines with H\(^+\) in acid to form \( \ce{NH4+} \).
2. Free NH\(_3\) molecules are then not available to act as ligands for the copper ion.

Teacher's Note:
1. A ligand with a lone pair is removed by protonation in acid.

Answer: (b) \( \ce{Mn^{2+}} \) and \( \ce{O2} \)

Explanation:
1. \( \ce{2MnO4^- + 5H2O2 + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O} \).
2. Permanganate is reduced to Mn\(^{2+}\) and H\(_2\)O\(_2\) acts as a reducing agent, giving oxygen.

Teacher's Note:
1. Hydrogen peroxide is oxidised to O\(_2\) by permanganate.

Answer: (d) They are chemically reactive

Explanation:
1. Interstitial compounds are hard, have high melting points and retain metallic conductivity.
2. They are chemically inert, so the statement that they are chemically reactive is incorrect.

Teacher's Note:
1. Small atoms (H, B, C, N) occupy the voids.

Answer: (b) \( \ce{Yb^{2+}} \)

Explanation:
1. Eu\(^{2+}\) is 4f\(^7\), Sm\(^{2+}\) is 4f\(^6\) and Ce\(^{2+}\) has a partly filled 4f or 5d subshell, so all three have unpaired electrons.
2. Yb\(^{2+}\) is 4f\(^{14}\), completely filled, so it is diamagnetic.

Teacher's Note:
1. Completely filled subshells give diamagnetism.

Answer: (b) Steel

Explanation:
1. Steel contains iron with carbon, a non-metal.
2. Invar (Fe, Ni), bell metal (Cu, Sn) and bronze (Cu, Sn) are alloys of metals only.

Teacher's Note:
1. Carbon is the non-metallic component of steel.

Answer: (b) \( \ce{Na2Cr2O7} \) is preferred over \( \ce{K2Cr2O7} \) in volumetric analysis

Explanation:
1. H\(_2\)S is oxidised to colloidal sulphur (milky). Dichromate is orange in acid and yellow (chromate) above pH 7.
2. Na\(_2\)Cr\(_2\)O\(_7\) is hygroscopic, so K\(_2\)Cr\(_2\)O\(_7\) (a primary standard) is preferred, and (b) is untrue.

Teacher's Note:
1. K\(_2\)Cr\(_2\)O\(_7\) is a primary standard.

Answer: (c) Ac

Explanation:
1. Actinium (5f\(^0\) 6d\(^1\) 7s\(^2\)) loses three electrons and shows only +3, similar to lanthanum.
2. U, Th and Pa show higher oxidation states such as +4, +5 and +6.

Teacher's Note:
1. Ac is the first actinoid and resembles La.

Answer: (c) Cu (Z = 29)

Explanation:
1. The \( E^\ominus_{M^{2+}/M} \) values are Co -0.28 V, Ni -0.25 V, Cu +0.34 V and Fe -0.44 V.
2. Only copper has a positive value, because of its high enthalpy of atomisation and low hydration enthalpy.

Teacher's Note:
1. Cu is the only metal in the first series with a positive \( E^\ominus_{M^{2+}/M} \).

Answer: (b) \( \ce{Ti^{3+}} \) (Z = 22)

Explanation:
1. Colour requires unpaired electrons for d-d or f-f transitions.
2. Ti\(^{3+}\) is d\(^1\), so it is coloured (purple). La\(^{3+}\) (f\(^0\)), Lu\(^{3+}\) (f\(^{14}\)) and Sc\(^{3+}\) (d\(^0\)) are colourless.

Teacher's Note:
1. Empty or completely filled subshells give colourless ions.

Answer: (d) A-(iv), B-(ii), C-(iii), D-(i)

Explanation:
1. Sulphuric acid is made by the Contact process, steel by the Bessemer process, sodium hydroxide (caustic soda) by the Leblanc process, and ammonia by Haber's process.
2. This gives A-(iv), B-(ii), C-(iii), D-(i).

Teacher's Note:
1. Match each industrial product to its classical process.

Answer: (d) Mn

Explanation:
1. The oxygen evolving complex of photosystem II contains a cluster of manganese ions.
2. It is this Mn cluster that oxidises water to O\(_2\).

Teacher's Note:
1. Manganese is essential for photosynthetic oxygen evolution.