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Case Study 1
The lanthanide series is a unique class of 15 elements with relatively similar chemical properties. They have atomic number ranging from 57 to 71, which corresponds to the filling of the 4f orbitals with 14 electrons. This configuration leads to phenomenon known as lanthanide contraction. The lanthanides are sometimes referred to as the ‘rare earth elements’, leading to misconception that they are rare. In fact many of the rare earth elements are more common than gold, silver and in some cases, lead. The lanthanides are commonly found in nature as a mixture in a number of monazite (LnPO4) and bastnaesite (LnCO3F) in the +3 oxidation state.
The chemical and physical properties of lanthanides provide the unique features that set them apart from other elements. Lanthanides are most stable in the +3 oxidation state. Yb and Sm though stable in the +3 state, also have accessible +2 oxidation states. The ease of accessibility of both oxidation states is quite important in chemical synthesis and these elements act as Lewis acid in the +3 oxidation state and single electron reductant in the +2 oxidation state.
Assertion (A): The elements scandium and yttrium are called “rare earths”.
Reason (R): Scandium and yttrium are rare on earth’s crust.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. The elements scandium and yttrium are called “rare earths” because they were originally discovered together with lanthanides in rare minerals and isolated as oxides or “earths”. Collectively, these metals are also called rare earth elements.
Teacher's Note:
1. The name “rare earths” is historical; as the passage notes, many of these elements are actually more common than gold or silver.
Assertion (A): Separation of lanthanide elements is difficult.
Reason (R): They have similar chemical properties.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Lanthanides have very similar ionic radii and chemical properties, so they occur together and are difficult to separate; ion-exchange methods are used.
Teacher's Note:
1. Similar size (because of lanthanoid contraction) is the root cause of their similar chemistry.
Assertion (A): There is continuous increase in size among lanthanides.
Reason (R): Lanthanides show lanthanide contraction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. In lanthanide series, with increasing atomic number, there is a progressive decrease in the atomic as well as on radii of trivalent ions from La3+ to Lu3+.
Teacher's Note:
1. Lanthanoid contraction means a steady decrease, not increase, in size across the series.
Assertion (A): Yb3+ is more stable than Yb2+.
Reason (R): Electronic configuration of Yb2+ is [Xe]4f7.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. As stated in the passage, lanthanides (including Yb) are most stable in the +3 oxidation state; Yb2+ is only an accessible lower state.
2. The electronic configuration of Yb2+ is [Xe]4f14 (Yb: [Xe]4f14 6s2), not [Xe]4f7. Hence the reason is wrong.
Teacher's Note:
1. [Xe]4f7 is the configuration of Eu2+ and Gd3+; Yb2+ has a completely filled 4f14 subshell.
Assertion (A): All lanthanides have similar chemical properties.
Reason (R): Because the lanthanoids differ only in the number of 4f-electrons.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. The differentiating electron enters the inner 4f subshell, which has little effect on chemical behaviour; the outer configuration and the common +3 state are the same, so the chemical properties are similar.
Teacher's Note:
1. 4f electrons are buried deep inside the atom and take little part in bonding, which is why the series behaves almost as one element.
Case Study 2
The transition elements have incompletely filled d-subshells in their ground state or in any of their oxidation states. The transition elements occupy position in between s- and p-blocks in groups 3-12 of the Periodic table. Starting from fourth period, transition elements consists of four complete series: Sc to Zn, Y to Cd and La, Hf to Hg and Ac, Rf to Cn. In general, the electronic configuration of outer orbitals of these elements is (n − 1)d1–10 ns0–2. The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n − 1)d10 ns2. All the transition elements have typical metallic properties such as high tensile strength, ductility, malleability. Except mercury, which is liquid at room temperature, other transition elements have typical metallic structures. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. Transition metal also forms alloys. An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
(a) High enthalpy of atomisation
(b) Variable oxidation states
(c) Paramagnetic behaviour
(d) Colour of hydrated ions
Show Answer & Explanation
Answer: (b) Variable oxidation states
Explanation:
1. The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states to form complexes.
Teacher's Note:
1. A catalyst must change oxidation state during the reaction and then return to it, which variable oxidation states make possible.
(a) same atomic number
(b) same electronic configuration
(c) nearly same atomic size
(d) same oxidation states.
Show Answer & Explanation
Answer: (c) nearly same atomic size
Explanation:
1. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals.
Teacher's Note:
1. Alloy formation needs atomic radii within about 15% of each other so that one metal can replace another in the lattice.
(a) [Xe]4f0 5d1 6s2
(b) [Xe]4f14 5d2 6s2
(c) [Xe]4f14 5d3 6s2
(d) [Xe]4f14 5d4 6s2
Show Answer & Explanation
Answer: (c) [Xe]4f14 5d3 6s2
Explanation:
1. Ta (Z = 73) is in group 5 of the 5d series: [Xe]4f14 5d3 6s2.
Teacher's Note:
1. For a 5d element after La, add the full 4f14 first; the 5d count then equals group number minus 2.
(a) 3d5 4s1
(b) 3d5 4s2
(c) 3d2 4s2
(d) 3d3 4s2
Show Answer & Explanation
Answer: (b) 3d5 4s2
Explanation:
1. Greater the number of valence electrons, more will be the number of oxidation states exhibited by the element.
Teacher's Note:
1. 3d5 4s2 is Mn, which shows oxidation states from +2 to +7, the widest range in the 3d series.
(i) all d- and f-block elements are metals
(ii) all d- and f-block elements form coloured ions
(iii) all d- and f-block elements are paramagnetic.
(a) (i) only
(b) (i) and (ii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Show Answer & Explanation
Answer: (a) (i) only
Explanation:
1. All the d-block elements are metals, they exhibit most properties of metals like lustre, malleability, ductility, high density, high melting and boiling point, hardness, conduction of heat and electricity, etc. All the f-block elements are also metals but they are not good conductors of heat and electricity.
Teacher's Note:
1. Watch for the word “all”: ions such as Zn2+, Sc3+ and La3+ are colourless and diamagnetic, which rules out (ii) and (iii).
Case Study 3
Transition metal oxides are compounds formed by the reaction of metals with oxygen at high temperature. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5. V2O4 dissolves in acids to give VO2+ salts. Transition metal oxides are commonly utilized for their catalytic activity and semiconductive properties. Transition metal oxides are also frequently used as pigments in paints and plastic. Most notably titanium dioxide. One of the earliest application of transition metal oxides to chemical industry involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid. Since then, many other applications have emerged, which include benzene oxidation to maleic anhydride on vanadium oxides; cyclohexane oxidation to adipic acid on cobalt oxides. An important property of the catalyst material used in these processes is the ability of transition metals to change their oxidation state under a given chemical potential of reductants and oxidants.
(a) VO
(b) V2O3
(c) VO2
(d) V2O5
Show Answer & Explanation
Answer: (a) VO
Explanation:
1. Oxide of V in lowest oxidation state, i.e., VO is basic and ionic in character.
Teacher's Note:
1. As the oxidation state of the metal rises, its oxides become more covalent and more acidic.
(a) VO2+
(b) VO2+
(c) V2O+
(d) VO43−
Show Answer & Explanation
Answer: (a) VO2+
Explanation:
1. Vanadyl ion is VO2+ where V is in +4 oxidation state.
Teacher's Note:
1. Do not mix up vanadyl VO2+ (V +4) with the pervanadyl or dioxovanadium(V) ion VO2+ (V +5).
(a) With fluorine vanadium can form VF5.
(b) With chlorine vanadium can form VCl5.
(c) Vanadium exhibits highest oxidation state in oxohalides VOCl3, VOBr3 and fluoride VF5.
(d) With iodine vanadium cannot form VI5 due to oxidising power of V5+ and reducing nature of I−.
Show Answer & Explanation
Answer: (b) With chlorine vanadium can form VCl5.
Explanation:
1. Only fluorine, the strongest oxidant, can stabilise V(+5) in a binary halide (VF5); the highest chloride of vanadium is VCl4, so VCl5 is not formed.
Teacher's Note:
1. The +5 state of V is reached with Cl and Br only in oxohalides (VOCl3, VOBr3), because oxygen helps stabilise high oxidation states.
(a) +5/2
(b) +7
(c) +5
(d) +6
Show Answer & Explanation
Answer: (c) +5
Explanation:
1. 2x + 5(−2) = 0 ⇒ x = +5
Teacher's Note:
1. +5 is also the group number of V, matching the passage: the highest oxidation number in oxides equals the group number.
V2O5 + 5Ca → 2V + 5CaO
(a) V2O5
(b) Ca
(c) V
(d) None of these
Show Answer & Explanation
Answer: (a) V2O5
Explanation:
1. V changes from +5 in V2O5 to 0 in V, so V2O5 is reduced and acts as the oxidising agent.
2. Ca changes from 0 to +2 (CaO), so Ca is the reducing agent.
Teacher's Note:
1. The oxidising agent is the species that is itself reduced, i.e. the one whose oxidation number falls.
Case Study 4
The unique behaviour of Cu, having a positive E° accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric acid) react with Cu, the acids being reduced. The stability of the half-filled (d5) subshell in Mn2+ and the completely filled (d10) configuration in Zn2+ are related to their E°(M3+/M2+) values. The low value for Sc reflects the stability of Sc3+ which has a noble gas configuration. The comparatively high value for Mn shows that Mn2+ (d5) is particularly stable, whereas a comparatively low value for Fe shows the extra stability of Fe3+ (d5). The comparatively low value for V is related to the stability of V2+ (half-filled t2g level).
(a) ZnO cannot be reduced to Zn by H2 under standard conditions
(b) Zn cannot liberate H2 with concentrated acids
(c) Zn2+ is a stronger oxidising agent than H+
(d) Zn is generally the cathode in an electrochemical cell.
Show Answer & Explanation
Answer: (a) ZnO cannot be reduced to Zn by H2 under standard conditions
Explanation:
1. The negative E° shows that Zn2+ is a weaker oxidising agent than H+ (E° = 0.00 V), so H2 cannot reduce Zn2+ (or ZnO) to Zn under standard conditions.
2. On the other hand, Zn reduces H+ and liberates H2 from acids, and it acts as the anode in an electrochemical cell.
Teacher's Note:
1. A negative reduction potential means the metal is a better reducing agent than H2; such metals displace hydrogen from dilute acids.
(a) Cr2+ acts as a reducing agent whereas Mn3+ acts as an oxidizing agent
(b) Cr2+ is more stable than Cr3+ state
(c) Mn3+ is more stable than Mn2+
(d) Cr2+ acts as an oxidizing agent whereas Mn3+ acts as a reducing agent.
Show Answer & Explanation
Answer: (a) Cr2+ acts as a reducing agent whereas Mn3+ acts as an oxidizing agent
Explanation:
1. Lesser and negative reduction potential indicates that Cr2+ is a reducing agent. Higher and positive reduction potential indicates that Mn3+ is a stronger oxidizing agent.
Teacher's Note:
1. Cr2+ (d4) goes to the stable d3 (half-filled t2g) Cr3+, while Mn3+ (d4) goes to the stable d5 Mn2+.
(a) O > N > M
(b) O > M > N
(c) M > N > O
(d) M > O > N
Show Answer & Explanation
Answer: (a) O > N > M
Explanation:
1. The electrode which has more reduction potential is a good oxidizing agent and has least reducing power.
2. Hence, reducing property: O (−3.13 V) > N (−1.13 V) > M (+2.46 V).
Teacher's Note:
1. Simply arrange the reduction potentials from most negative to most positive to get the order of decreasing reducing power.
(I) Mn2+ compounds are more stable than Fe2+ towards oxidation to +3 state.
(II) Titanium and copper both in the first series of transition metals exhibits +1 oxidation state most frequently.
(III) Cu+ ion is stable in aqueous solutions.
(IV) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+.
(a) (II) and (III)
(b) (I) and (IV)
(c) (I) and (III)
(d) (II) and (IV)
Show Answer & Explanation
Answer: (b) (I) and (IV)
Explanation:
1. (I) It is because Mn2+ has 3d5 electronic configuration which has extra stability.
2. (II) Not titanium but copper, because with +1 oxidation state an extra stable configuration, 3d10 results.
3. (III) It is not stable as it undergoes disproportionation; 2Cu+(aq) → Cu2+(aq) + Cu(s). The E° value for this is favourable.
4. (IV) Much larger third ionisation energy of Mn (where the required change is d5 to d4) is mainly responsible for this.
Teacher's Note:
1. For multi-statement questions, mark each statement true or false first and only then match with the options.
(a) more negative ΔhydH of Cu2+(aq)
(b) less negative ΔhydH of Cu2+(aq)
(c) more positive ΔhydH of Cu2+(aq)
(d) less positive ΔhydH of Cu2+(aq).
Show Answer & Explanation
Answer: (a) more negative ΔhydH of Cu2+(aq)
Explanation:
1. The stability of Cu2+(aq) rather than Cu+(aq) is due to the much more negative ΔhydH° of Cu2+(aq) than Cu+, which more than compensates for the second ionisation enthalpy of Cu.
Teacher's Note:
1. Hydration enthalpy is always negative (energy released); the smaller, more highly charged Cu2+ releases much more.
Case Study 5
The f-block elements are those in which the differentiating electron enters the (n − 2)f orbital. There are two series of f-block elements corresponding to filling of 4f and 5f-orbitals. The series of 4f-orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of f0, f7 and f14 configurations, though the most common oxidation states is +3. There is a regular decrease in size of lanthanides ions with increase in atomic number which is known as lanthanide contraction.
(a) 4f8, 4f11, 4f13
(b) 4f11, 4f8, 4f13
(c) 4f0, 4f2, 4f11
(d) 4f3, 4f7, 4f9
Show Answer & Explanation
Answer: (a) 4f8, 4f11, 4f13
Explanation:
1. Terbium (Tb3+): 4f8, Erbium (Er3+): 4f11, Ytterbium (Yb3+): 4f13.
Teacher's Note:
1. Shortcut: number of 4f electrons in Ln3+ = Z − 57 (so 65 − 57 = 8, 68 − 57 = 11, 70 − 57 = 13).
(a) Gd
(b) At
(c) Xe
(d) Te
Show Answer & Explanation
Answer: (a) Gd
Explanation:
1. Gd (Z = 64) is a lanthanide; At, Xe and Te are p-block elements.
Teacher's Note:
1. Lanthanide contraction is a property of the 4f series (Ce to Lu), so pick the only lanthanide in the list.
(a) [Xe]4f10 6s2
(b) [Xe]4f1 5d1 6s2
(c) [Xe]4d14 5d10 6s2
(d) [Xe]4f7 5d1 6s2
Show Answer & Explanation
Answer: (c) [Xe]4d14 5d10 6s2
Explanation:
1. (a) is Dy, (b) is Ce and (d) is Gd.
2. [Xe]4d14 5d10 6s2 is impossible: a d subshell holds at most 10 electrons and 4d is already filled in [Xe].
Teacher's Note:
1. A quick check: any configuration showing more than 10 electrons in a d subshell or more than 14 in an f subshell is invalid.
(a) Cerium (Z = 58)
(b) Europium (Z = 63)
(c) Lanthanum (Z = 57)
(d) Gadolinium (Z = 64)
Show Answer & Explanation
Answer: (a) Cerium (Z = 58)
Explanation:
1. Ce4+ has the stable 4f0 (noble gas) configuration, so cerium is well known to exhibit the +4 oxidation state.
Teacher's Note:
1. Europium is the classic example of the +2 state (4f7), while cerium is the classic +4 example.
(a) Lanthanoid contraction is the accumulation of successive shrinkages.
(b) The different radii of Zr and Hf is due to consequence of the lanthanoid contraction.
(c) Shielding power of 4f electrons is quite weak.
(d) There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.
Show Answer & Explanation
Answer: (b) The different radii of Zr and Hf is due to consequence of the lanthanoid contraction.
Explanation:
1. The almost identical radii of Zr (160 pm) and Hf (159 pm), a consequence of lanthanoid contraction.
Teacher's Note:
1. Lanthanoid contraction makes Zr and Hf almost the same size, not different; this is why they are hard to separate.
Case Study 6
Within the 3d series, manganese exhibits oxidation states in aqueous solution from +2 to +7, ranging from Mn2+(aq) to MnO4−(aq). Likewise, iron forms both Fe2+(aq) and Fe3+(aq) as well as the FeO42− ion. Cr and Mn form oxoanions CrO42−, MnO4−, owing to their willingness to form multiple bonds. The pattern with the early transition metals—in the 3d series up to Mn, and for the 4d, 5d metals upto Ru and Os—is that the maximum oxidation state corresponds to the number of “outer shell” electrons. The highest oxidation states of the 3d metals may depend upon complex formation (e.g., the stabilisation of Co3+ by ammonia) or upon the pH (thus MnO42−(aq) is prone to disproportionation in acidic solution). Within the 3d series, there is considerable variation in relative stability of oxidation states, sometimes on moving from one metal to a neighbour; thus, for iron, Fe3+ is more stable than Fe2+, especially in alkaline conditions, while the reverse is true for cobalt. The ability of transition metals to exhibit a wide range of oxidation states is marked with metals such as vanadium, where the standard potentials can be rather small, making a switch between states relatively easy. [CBSE QB 2021]
Assertion (A): Highest oxidation state is exhibited by transition metals lying in the middle of the series.
Reason (R): The highest oxidation state exhibited corresponds to number of (n−1)d electrons.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Higher oxidation states are usually exhibited by the transition metals lying in the middle of the series because of higher number of unpaired electrons in (n − 1)d and ns orbitals at the middle of the series.
Teacher's Note:
1. The maximum oxidation state equals the total of (n−1)d and ns electrons (up to Mn), e.g. Mn: 3d54s2 gives +7; counting only d electrons gives the wrong value.
Assertion (A): Fe3+ is more stable than Fe2+.
Reason (R): Fe3+ has 3d5 configuration while Fe2+ has 3d6 configuration.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Fe3+ is more stable than Fe2+ because of the presence of half-filled d-orbital (3d5) in Fe3+ ion.
Teacher's Note:
1. This is why Fe2+ salts slowly get oxidised to Fe3+ in air; the reverse is true for Mn, where Mn2+ is the 3d5 ion.
Assertion (A): Vanadium had the ability to exhibit a wide range of oxidation states.
Reason (R): The standard potentials of vanadium are rather small, making a switch between oxidation states relatively easy.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Vanadium shows oxidation states from +2 to +5 (V2+, V3+, VO2+, VO2+).
2. As stated in the passage, the standard potentials between these states are small, so vanadium can switch easily from one oxidation state to another.
Teacher's Note:
1. Small E° differences between neighbouring states mean no single state is strongly favoured, so several states are accessible.
Assertion (A): Transition metals like Fe, Cr and Mn form oxoanions.
Reason (R): Oxygen is highly electronegative and has a tendency to form multiple bonds.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Transition metals like Fe, Cr and Mn form oxoions because oxygen is strong oxidising agent due to its high electronegativity and small size.
2. Due to these properties, and its ability to form multiple bonds with the metal (as stated in the passage), oxygen is able to oxidise the metal to its highest oxidation state.
Teacher's Note:
1. In oxoanions such as MnO4− and CrO42− the metal–oxygen bonds are multiple (M=O) bonds; this is why oxygen stabilises higher oxidation states better than fluorine.
Assertion (A): The highest oxidation states of the 3d metals depends only on electronic configuration of the metal.
Reason (R): The number of electrons in the (n−1)d and ns subshells determine the oxidation states exhibited by the metal.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. The passage states that the highest oxidation states of the 3d metals also depend upon complex formation (e.g., Co3+ stabilised by ammonia) and upon the pH, so they do not depend only on the electronic configuration. Hence the assertion is false.
2. The reason is true because the oxidation states are decided by the number of (n−1)d and ns electrons.
Teacher's Note:
1. Watch for words like “only” or “always” in an assertion; one counter-example from the passage makes it false.
Case Study 7
The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of upto some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d-electron bonding levels are involved, the cations exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group. [CBSE QB 2021]
Assertion (A): Cations of transition elements occur in various valence states.
Reason (R): Large number of oxides of transition elements are possible.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Transition element can occur in variable oxidation states due to maximum number of electrons are present in (n − 1)d and ns subshell that determines oxidation state.
2. Oxides of transition metals stabilises highest oxidation states, from Sc2O3 to Mn2O7 and no above Fe2O3, higher oxides are known.
Teacher's Note:
1. Cause and effect are reversed here: variable valency leads to many oxides, not the other way round, so R cannot explain A.
Assertion (A): Crystal structure of oxides of transition metals often show defects.
Reason (R): Ligand field effect cause distortions in crystal structures.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. The crystal field theory explains the model for interaction between transition metals and ligands. It causes attraction between charges of metal ion with ligands by degeneracy of orbital states, producing electric field by charge distribution.
2. It affects different properties like colour, distorted structures, magnetic properties etc.
Teacher's Note:
1. The passage names two causes of departure from ideal arrays: packing of ions of different sizes and ligand field effects.
Assertion (A): Transition metals form protective oxide films.
Reason (R): Oxides of transition metals are always stoichiometric.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Transition metals like Zn, Cr are protected by formation of oxide film to present from corrosion.
2. These oxides of transition metals are likely to form non-stoichiometry due to variable oxidation states and their defects in the lattices.
Teacher's Note:
1. Non-stoichiometric oxides such as Fe0.94O are typical of transition metals because of their variable valency.
Assertion (A): CrO crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by chromium ions.
Reason (R): Transition metal oxide may be hexagonal close-packed lattice of oxide ions with metal ions filling the octahedral voids.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. In a structure with hcp oxide ions and two-thirds of the octahedral holes filled, the ratio Cr : O = 2 : 3, i.e., the formula is Cr2O3, not CrO. Hence the assertion is false.
2. The reason is true, as the passage states that oxide structures are often classified as close-packed lattices of one set of ions with the other filling octahedral or tetrahedral interstices.
Teacher's Note:
1. Number of octahedral voids = number of close-packed ions; filling 2/3 of them gives M2O3, filling all of them gives MO.
Case Study 8
The d block elements are the 40 elements contained in the four rows of ten columns (3-12) in the periodic table. As all the d block elements are metallic, the term d-block metals is synonymous. This set of d-block elements is also often identified as the transition metals, but sometimes the group 12 elements (zinc, cadmium, mercury) are excluded from the transition metals as the transition elements are defined as those with partly filled d or f shells in their compounds. Inclusion of the elements zinc, cadmium and mercury is necessary as some properties of the group 12 elements are appropriate logically to include with a discussion of transition metal chemistry. The term transition element or transition metal appeared to derive from early studies of periodicity such as the Mendeleev periodic table of the elements. His horizontal table of the elements was an attempt to group the elements together so that the chemistry of elements might be explained and predicted. In this table there are eight groups labeled I-VIII with each subdivided into A and B subgroups. Mendeleev recognized that certain properties of elements in Group VIII are related to those of some of the elements in Group VII and those at the start of the next row Group I. In that sense, these elements might be described as possessing properties transitional from one row of the table to the next. [CBSE QB 2021]
Assertion (A): Group 12 elements are not considered as transition metals.
Reason (R): Transition metals are those which have incompletely filled d sub-shell in their compounds.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Elements of Group 12 (Zn, Cd, Hg) are not considered as transition metals because of their fully filled d-orbitals [(n − 1)10ns2], whereas transition metals are those metals whose d-orbitals are partially filled.
Teacher's Note:
1. Zn, Cd and Hg have d10 configuration both as atoms and in their common ions (Zn2+, Cd2+, Hg2+), so they fail the definition.
Assertion (A): All d-block elements are metallic in nature.
Reason (R): The d-block elements belong to Group 3-12 of the periodic table.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Group 3 to Group 12 are the d-block elements, out of which all are considered as transition metals except Group 12 elements.
2. Transition metals show metallic properties due to low ionization energies and their vacant d-orbitals in the valenced shell.
Teacher's Note:
1. Position in groups 3–12 only locates the elements; their metallic nature comes from low ionisation enthalpies and few valence electrons.
Assertion (A): Group VII elements of Mendeleev periodic table are transition elements.
Reason (R): Group I–VIII in Mendeleev periodic table is divided into two subgroups, A and B.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Mendeleev periodic table is divided into two subgroups, A & B that possess same physical property and chemical properties. In subgroup B of Group III to VII defines for transition elements.
2. According to the passage, it is the Group VIII elements that Mendeleev described as transitional, so the assertion is false.
Teacher's Note:
1. Read the passage carefully: it is Group VIII (Fe, Co, Ni etc.) whose properties are “transitional” between Group VII and Group I of the next row.
Assertion (A): Nickel is a transition element that belongs to group 10 and period 4 of the modern periodic table.
Reason (R): Electronic configuration of Nickel is [Ar]3d84s2.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Ni (Z = 28) has the configuration [Ar]3d84s2: the 4s shell places it in period 4, and the 10 electrons in (n−1)d + ns place it in group 10. Its incomplete 3d subshell makes it a transition element.
Teacher's Note:
1. For d-block elements, group number = number of (n−1)d electrons + number of ns electrons.
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Useful Resources and Notes for Class 12 Chemistry Unit 4 The d- and f-Block Elements
Core Study Kit: Class 12 Chemistry Unit 4 The d- and f-Block Elements
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