Multiple Choice Questions (MCQs) for JEE Chemistry: The D and F Block Elements
Review structured MCQ sets for JEE Chemistry The D and F Block Elements. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
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Multiple Choice Questions
(a) \( H_3PO_4 \)
(b) \( H_3PO_2 \)
(c) \( N_2H_4 \)
(d) \( H_3PO_3 \)
Show Answer & Explanation
Answer: (a) \( H_3PO_4 \)
Explanation:
1. Phosphorus in H3PO4 is already in its highest oxidation state (+5), so it cannot reduce Ag+. H3PO2, H3PO3 and hydrazine are reducing agents.
Teacher's Note:
1. Hypophosphorous acid is the strongest reductant.
(a) \( CH_3COONa \)
(b) \( AgNO_3 \)
(c) \( Na_2S_2O_3 \)
(d) \( SrCl_2 \)
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Answer: (d) \( SrCl_2 \)
Explanation:
1. M is \( FeCl_3 \). It reacts with acetate (hydrolysis), silver nitrate (AgCl) and thiosulphate (reduction), but not with strontium chloride.
Teacher's Note:
1. No common ion or redox change with SrCl2.
(a) it disproportionates in neutral solution to give \( KMnO_4 \)
(b) it is green
(c) it is stable in acidic medium
(d) it gives brown MnO2 when it disproportionates
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Answer: (c) it is stable in acidic medium
Explanation:
1. \( 3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O \): in acid it disproportionates.
Teacher's Note:
1. Manganate is stable only at high pH.
(a) 5 and 4
(b) 4 and 8
(c) 4 and 5
(d) 8 and 9
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Answer: (b) 4 and 8
Explanation:
1. Chromate changes to dichromate in acid (about pH 4) and dichromate changes to chromate in base (about pH 8).
Teacher's Note:
1. Acidic solutions are orange, basic yellow.
(a) \( Mn^{2+}(aq) < Mn^{3+}(aq) \)
(b) \( Co^{2+}(aq) < Co^{3+}(aq) \)
(c) \( Fe^{3+}(aq) < Fe^{2+}(aq) \) in all conditions
(d) \( Cr^{2+}(aq) < Cr^{3+}(aq) \)
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Answer: (d) \( Cr^{2+}(aq) < Cr^{3+}(aq) \)
Explanation:
1. \( Cr^{3+} \) with \( t_{2g}^3 \) is very stable and \( Cr^{2+} \) is a strong reductant.
2. \( Mn^{3+} \) and \( Co^{3+} \) are strong oxidants, so \( Mn^{2+} \) and \( Co^{2+} \) are more stable in water.
Teacher's Note:
1. Compare the E0 values of the M3+/M2+ couples (Cr: -0.41 V, Mn: +1.57 V, Co: +1.97 V).
(a) with NaOH and with water
(b) with water and with \( NH_3 \)
(c) with NaOH and with HCl
(d) with HCl and with \( NH_3 \)
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Answer: (c) with NaOH and with HCl
Explanation:
1. \( Zn(OH)_2 + 2NaOH \to Na_2[Zn(OH)_4] \) (acidic behaviour).
2. \( Zn(OH)_2 + 2HCl \to ZnCl_2 + 2H_2O \) (basic behaviour).
Teacher's Note:
1. An amphoteric hydroxide reacts with both an acid and a base.
(a) \( ZnCO_3 \)
(b) \( Zn(NO_3)_2 \)
(c) \( Zn(OH)_2 \)
(d) Zn metal
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Answer: (b) \( Zn(NO_3)_2 \)
Explanation:
1. Zinc carbonate and hydroxide react with acid, zinc metal liberates H2, but zinc nitrate has no reaction with dilute HCl.
Teacher's Note:
1. There is no gas, precipitate or weak electrolyte formed.
(a) KF (aq)
(b) KCl (aq)
(c) KBr (aq)
(d) KI (aq)
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Answer: (a) KF (aq)
Explanation:
1. Dichromate oxidises I-, Br- and (on heating with conc. acid) Cl-, but never fluoride.
Teacher's Note:
1. Fluoride is the weakest reducing halide ion.
(a) \( La^{3+} < Ce^{3+} < Pm^{3+} < Yb^{3+} \)
(b) \( Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+} \)
(c) \( Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+} \)
(d) \( Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+} \)
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Answer: (d) \( Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+} \)
Explanation:
1. Ionic radii decrease steadily from La to Lu (lanthanoid contraction).
2. Order of atomic number: La (57) < Ce (58) < Pm (61) < Yb (70), so radii: La > Ce > Pm > Yb.
Teacher's Note:
1. Higher atomic number means smaller Ln3+ in the series.
(a) Mn
(b) Fe
(c) Cr
(d) V
Show Answer & Explanation
Answer: (c) Cr
Explanation:
1. Removing the second electron from Cr+ (\( 3d^5 \)) breaks the stable half filled \( d^5 \) configuration, so it needs the most energy.
Teacher's Note:
1. Cr and Cu have unusually high second ionisation enthalpies.
(a) \( Cr^{3+} \) and \( Cr_2O_7^{2-} \) are formed
(b) \( Cr_2O_7^{2-} \) and \( H_2O \) are formed
(c) \( Cr_2O_7^{2-} \) is reduced to Cr(III)
(d) \( Cr_2O_7^{2-} \) is oxidised to Cr(VII)
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Answer: (b) \( Cr_2O_7^{2-} \) and \( H_2O \) are formed
Explanation:
1. \( 2CrO_4^{2-} + 2H^+ \to Cr_2O_7^{2-} + H_2O \). Nitric acid is not a reductant, so the oxidation state of Cr stays +6.
Teacher's Note:
1. Acidification only changes chromate into dichromate; no redox takes place.
(a) 0.85 \( \AA \)
(b) 1.6 \( \AA \)
(c) 1.4 \( \AA \)
(d) 1.06 \( \AA \)
Show Answer & Explanation
Answer: (a) 0.85 \( \AA \)
Explanation:
1. Lanthanoid contraction reduces the ionic radius from about 1.06 Angstrom (La3+) to about 0.85 Angstrom (Lu3+).
Teacher's Note:
1. The contraction is caused by poor shielding by 4f electrons.
(a) Ce(IV) acts as an oxidising agent
(b) The +3 state is more stable than the +4 state
(c) The common oxidation states are +3 and +4
(d) The +4 oxidation state of cerium is not known in solutions
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Answer: (d) The +4 oxidation state of cerium is not known in solutions
Explanation:
1. Ce4+ is known in aqueous solution (e.g. ceric ammonium sulphate, a common oxidant), so the statement that it is not known is wrong.
Teacher's Note:
1. Ce(IV) has the stable noble gas configuration but is a strong oxidant.
(a) Evolved \( I_2 \) is reduced
(b) \( Na_2S_2O_3 \) is oxidised
(c) \( CuI_2 \) is formed
(d) \( Cu_2I_2 \) is formed
Show Answer & Explanation
Answer: (c) \( CuI_2 \) is formed
Explanation:
1. \( 2Cu^{2+} + 4I^- \to Cu_2I_2 + I_2 \): cuprous iodide forms, not \( CuI_2 \) (which does not exist).
2. \( I_2 + 2S_2O_3^{2-} \to 2I^- + S_4O_6^{2-} \).
Teacher's Note:
1. Cupric iodide cannot exist because Cu2+ oxidises iodide.
(a) \( NH_2\text{-}Hg\text{-}Hg\text{-}Cl \)
(b) \( HgNH_2Cl \) (with black Hg)
(c) \( Hg_2O \)
(d) HgO
Show Answer & Explanation
Answer: (b) \( HgNH_2Cl \) (with black Hg)
Explanation:
1. \( Hg_2Cl_2 + 2NH_3 \to Hg(NH_2)Cl + Hg + NH_4Cl \). The black deposit of mercury gives the name calomel (beautiful black).
Teacher's Note:
1. Calomel turns black with ammonia.
(a) Zr and Hf have about the same radius
(b) Zr and Y have about the same radius
(c) Zr and Nb have similar oxidation states
(d) Zr and Zn have similar oxidation states
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Answer: (a) Zr and Hf have about the same radius
Explanation:
1. The 4f electrons shield poorly, so Hf (after the lanthanoids) has nearly the same size as Zr.
2. This is why Zr and Hf are chemically so similar and hard to separate.
Teacher's Note:
1. Similar radii are also seen for Nb and Ta, Mo and W.
(a) the same effective nuclear charge from Ce to Lu
(b) appreciable shielding by 4f electrons
(c) appreciable shielding by 5d electrons
(d) imperfect shielding of outer electrons by 4f electrons from the nuclear charge
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Answer: (d) imperfect shielding of outer electrons by 4f electrons from the nuclear charge
Explanation:
1. The 4f subshell has diffuse shape and shields poorly, so effective nuclear charge rises along the series, shrinking the size.
Teacher's Note:
1. A nuclear charge increase of 1 per element is not cancelled by an added 4f electron.
(a) d-block elements show irregular and erratic chemical properties among themselves
(b) The chemistry of the various lanthanoids is very similar
(c) 4f and 5f orbitals are equally shielded
(d) La and Lu have partially filled d orbitals and no other lanthanoid has partially filled d orbitals
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Answer: (c) 4f and 5f orbitals are equally shielded
Explanation:
1. 5f orbitals are less buried and are shielded less effectively than 4f orbitals, so the statement is wrong.
Teacher's Note:
1. This is why actinoids show more oxidation states than lanthanoids.
(a) 5f orbitals are more buried than 4f orbitals
(b) 5f orbitals extend further from the nucleus than 4f orbitals
(c) actinoids are more reactive
(d) 4f and 5f have similar angular wave functions
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Answer: (b) 5f orbitals extend further from the nucleus than 4f orbitals
Explanation:
1. 5f electrons are less tightly held, and the 5f, 6d and 7s energies are close, so more electrons take part in bonding.
Teacher's Note:
1. This also explains why actinoid ions are more complexing.
(a) In the highest oxidation states the metal shows basic character and forms cationic complexes
(b) In the highest oxidation states of Sc to Mn all 4s and 3d electrons are used for bonding
(c) Once d5 is exceeded, the tendency to involve all 3d electrons decreases
(d) Zero oxidation state is also shown in complexes
Show Answer & Explanation
Answer: (a) In the highest oxidation states the metal shows basic character and forms cationic complexes
Explanation:
1. In the highest oxidation states transition metals form oxides and oxoanions that are acidic, and they form anionic complexes, for example MnO4- and CrO42-.
Teacher's Note:
1. Acidic character rises with oxidation state.
(a) Ionic sizes of Ln(III) decrease with increasing atomic number
(b) Ln(III) hydroxides are mainly basic
(c) Bonding in Ln(III) compounds is predominantly ionic
(d) Ln(III) compounds are generally colourless
Show Answer & Explanation
Answer: (d) Ln(III) compounds are generally colourless
Explanation:
1. Most Ln3+ ions with partly filled 4f orbitals are coloured because of f-f transitions, so the statement that they are generally colourless is wrong.
Teacher's Note:
1. La3+ (f0), Lu3+ (f14) and Ce3+ (f1) are colourless, but most others are coloured.
(a) Radii decrease gradually with atomic number
(b) All members exhibit the +3 state
(c) Availability of 4f electrons results in +4 compounds for all members
(d) Separation is difficult because of similar properties
Show Answer & Explanation
Answer: (c) Availability of 4f electrons results in +4 compounds for all members
Explanation:
1. Only a few members (Ce, Pr, Tb) show +4 and a few show +2 (Eu, Yb).
Teacher's Note:
1. +3 is the characteristic state of all of them.
(a) Ferrous compounds are less volatile than ferric compounds
(b) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds
(c) Ferrous oxide is more basic than ferric oxide
(d) Ferric salts are more covalent than ferrous salts
Show Answer & Explanation
Answer: (b) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds
Explanation:
1. Fe3+ has a higher charge and smaller size, so it is hydrolysed more readily than Fe2+. The given statement is therefore incorrect.
Teacher's Note:
1. Higher oxidation state means more covalent, more volatile and more easily hydrolysed.
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The D and F Block Elements Objective Questions & Solutions for JEE Chemistry
About The D and F Block Elements MCQs for JEE Chemistry
Review structured objective questions for JEE Chemistry The D and F Block Elements. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.
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FAQs
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