CBSE Class 12 Chemistry D And F Block Elements Worksheet

Question. Transition metals are less reactive, high melting point and enthalpy of atomization.
Answer : Strong metallic bonding and d-d overlapping.

Question. Transition metals have high enthalpy of hydration.
Answer : Small size

Question. Transition metals show several oxidation states.
Answer : Comparable energies of (n-1)d and ns electrons.

Question. Transition metals form coloured complexes.
Answer : d-d transition

Question. Transition metals take part in catalytic reactions.
Answer : Can pass from one oxidation state to another very easily.

Question. Why does vanadium pentaoxide act as a catalyst?
Answer : Can pass from one oxidation state to another very easily.

Question. Transition metals are paramagnetic in nature.
Answer : Unparied electrons

Question. Transition metals form complexes.
Answer : Have vacant d orbitals to accept electrons.

Question. Transition metals have irregular E⁰ values.
Answer : Irregualr trend in IE, sublimation energy and hydration energy

Question. The E⁰M²⁺/M for copper is positive (0.34v) .Copper is the only metal in first series of transition elements showing this behavior, why?
Answer : Sum of enthalpies of sublimation and ionization ( enthalpy of atomization ) > hydration energy.

Question. Transition metals form alloys.
Answer : Almost same size.

Question. Transition metals form interstitial compounds.
Answer : Can accommodate small atoms in void position.

Question. Cu⁺ is unstable in aqueous solution.
Answer : Low hydration enthalpy and property to show disproportionation reaction.

Question. Cu²⁺ is stable in aqueous solution.
Answer : High hydration enthalpy.

Question. Zr and Hf exhibit almost same radii and properties.
Answer : Lanthanoid contraction

Question. The d¹ configuration is generally unstable in ions.
Answer : Hydration energy > Ionization energy.

Question. There is a in general increase in density of element from titanium(Z=22) to copper ( Z=29).
Answer : It is due to increase in mass per unit volume with increase in atomic no.

Question. The lowest oxides of transition metals is basic, the highest is amphoteric or acidic.
Answer : Lower oxides are ionic while higher oxides are covalent.

Question. Anhydrous CuSO₄ is white while hydrated Copper sulphate is blue.
Answer : No d-d transition in the absence of ligand.

Question. Co²⁺ is easily oxidized to Co³⁺ in presence of strong ligand.
Answer : Crystal field stabilisation energy of Co+3 ion is higher than Co²⁺ ion.

Two marks questions

Question. (a) Why HCl cannot be used in place of sulphuric acid to acidify KMnO₄ solution in volumetric analysis ?
(b) Potassium dichromate is a good oxidising agent in acidic medium, why?
Answer : (a) HCl can be oxidized into Cl₂.
(b) It evolve nascent oxygen

Question. Write the balanced ionic equations for reacting ions to represent the acidified potassium dichromate solution with :
(i) Potassium iodide solution
(ii) Acidified ferrous sulphate solution.
Answer : (a) Cr₂O₇^2- + 6I^- + 14 H ⁺ → 2 Cr³⁺ + 3I₂ + 7H₂O
(b) Cr₂O₇^2- + 6Fe²⁺ + 14 H ⁺ → 6Fe³⁺ + 2 Cr³⁺ + 7H₂O

Question. List some applications of d block elements.
Answer : Ni , Fe, V₂O₅ are used as catalyst in various industrial process.
AgBr is used in photography.
Pt compound are used in anticancer drug.
MnO₂ is used as OA in dry cells.

Question. Describe giving reasons which one of the following pairs has the properties indicated?
(a) Fe or Cu has higher melting point.
(b) Co²⁺ Or Nⁱ²⁺ has lower magnetic moment.
Answer : Fe
Ni²⁺

Question. Calculate the magnetic moment of a trivalent ion in aqueous solution whose atomic no. is 25.
Answer : 4 unpaired electrons M.M = [4(4+2)]^1/2

Question. Define transition elements. Explain why is Zn not considered as transition element while Cu does?
Answer : Transition elements are those whose neutral atom or stable ion has partly filled dorbitals.
Cu²⁺ has partly filled d-orbitals which are absent in Zn or Zn²⁺.

Question. What happens when Cu²⁺ is added to I-? Write the balanced chemical equation .
Answer : 2CU²⁺ + 4I^- → Cu₂I₂ + I₂

Question. Write the electronic configuration of₂₄ Cr and ₂₆Fe²⁺ .
Answer : Cr= [Ar] 3d⁵ 4s¹    Cu[Ar]3d¹⁰4s¹

Question. Compare non transition and transition elements on the basis of their Variability of oxidation states (ii) stability of oxidation states.
Answer : Oxidation states of transition elements differ from each other by unity. In non transitionelements oxidation states normally differ by a unit of two.

Question. (a) Name a transition element which does not exhibit variable oxidation state.
(b) Name three elements of d block which are not regarded as transition element.
Answer : (a) Sc
(b) Zn, Cd, Hg

Three Marks questions

Question. Give chemical reactions for the following observations:
(i) Potassium permanganate is a good oxidising agent in basic medium.
(ii) Inter convertibility of chromate ion and dichromate ion in aqueous solution depends Upon pH of the solution.
(iii) Potassium permanganate is thermally unstable at 513K.
Answer : (i) 2MnO₄^- + 2H₂O + 3 e^- → 2MnO₂ + 4OH^-
(ii) CrO₄^2- ======= Cr₂O₇^2-
( if PH > 4 the CrO₄^2- ion will exist and if PH < 4 then Cr₂O₇^2- will exist)
ie. 2 CrO₄^2- + 2H+ → Cr2O7^2- + H₂O
Cr₂O₇^2- + 2OH- → 2 CrO4^2- + H₂O
(iii) 2 KMnO₄ + 513 K → K₂MnO₄ + MnO₂ + O₂

Question. Define lanthanoid Contraction. Ce⁴⁺ is a good oxidizing agent whereas Eu²⁺, Sm²⁺ is a good reducing agent, why ?
Answer : (a) Decrease in atomic or ionic radii with increase in atomic number.
(b) For lanthanoids common oxidation state is +3 . to acquire +3 oxidation state Ce⁴⁺ undergoes reduction and hence acts as oxidizing agent ,while Eu²⁺ undergoes oxidation and hence acts as reducing agent. 

Question. (a) From element to element the actinoid contraction is greater than lanthanoidcontraction,why?
(b) Name the lanthanoid element which forms tetra positive ions in the aqueous solution.
© The chemistry of actinoids is not as smooth as lanthanoids, why?
Answer : (a) It is due to poor shielding by 4f and 5f electrons.
(b) Ce ( Ce⁴⁺)
© Actinoids are radioactive in nature.

Question. Balance the following equations:
(i) MnO₄ ^- + S₂O₃^2- → ( Basic medium)
(ii) MnO₄ ^- + S₂O₃^2- + H₂O →
(iii) MnO₄ ^- + I^- → ( in neutral or alkaline medium)
Answer : i) & (ii) 8MnO₄^- + 3S₂O₃^2- + H₂O → 8MnO₂ + 2 OH^- + 6 SO₄^2-
(iii) 2MnO₄^- + I^- + H₂O → 2OH^- + 2MnO₂ + IO₃^-

Question. (a) The enthalpies of atomization of transition metals of 3d series do not follow a regular trend throughout the series.
(b) The enthalpy of atomization of zinc is lowest.
© Zn Cd Hg are soft and have low melting points.
Answer : (a) Bec. enthalpy of atomisation depends upon no. of unpaired electrons .
(b) No unpaired electrons .
© Absence of d-d overlapping and poor metallic bonding.

Question. Explain:
(a) The E° value for Ce⁴⁺/Ce³⁺ is 1.74 Volt.
(b) K₂Cr₂O₇ is used as Primary Standard in volumetric analysis.
© The third ionization energy of manganese (z=25) is exceptionally high.
Answer : (a) Ce⁴⁺ is strong oxidant, being Lanthanoid it reverts to Ce³⁺ as + 3 is most stable.
(b) K₂Cr₂O₇ is not much soluble in cold water. However, it is obtained in pure state and is not Hygroscopic in nature.
© Mn³⁺,extra stable due to half filled d⁵ configuration.

Question. Explain:
(a) Although Cu⁺ has configuration 3 d¹⁰ 4 s⁰ (stable) and Cu²⁺ has configuration 3 d⁹ (unstable configuration) still Cu²⁺ compounds are more stable than Cu⁺.
(b) Titanium (IV) is more stable than Ti (III) or Ti (II).
© The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
Answer : (a) It is due to much more negative Hydration enthalpy of Cu²⁺ (aq) than Cu⁺
(b) TiIV is more stable due to d⁰ configuration.
(c) Maximum no. of unpaired electrons are in middle.

Question. (a) Highest manganese flouride is MnF4 whereas the highest oxide is Mn₂O₇, why?
(b) Copper can notlibrate H₂ from dilacids,why?
(c) Which of the 3d- series of transition metals exhibits largest number of oxidation states and why?
Answer : (a) The ability of oxygen to form multiple bonds to metals, explain its superiority to show higher oxidation state with metal.
(b) Positive E° value (+ O . 34 Volt) accounts for its inability to liberate H₂ from acids.
(c) Mn, Maximum no. of unpaired electrons.

Question. (a) O.S. of first transition series initially increase up to Mn and then decrease to Zn , why?
(b) Why is Cr²⁺ reducing and Mn³⁺ oxidizingwhile both have d⁴ configuration.
© Ti achieves tetrahalides while chromium forms heaxhalide, why?
Answer : (a) Number of unpaired electrons increases up to Mn and then decreases up to Zn.
(b) To acquire +3 O.S. Cr²⁺ has a tendency to lose the electron while Mn³⁺ has a tendency to accept an electron .
(c) To acquire d0 configuration.

Question. (a) Which form of Cu is paramagnetic and why?
(b) What is the oxidation no. of Cr in Cr₂O₇^2-?
Answer : (a) Cu²⁺, One unpaired electron.
(b) +6

Question. Complete the following reaction equations:
(a) MnO₂ + KOH (aq) + O₂ →
(b) Fe²⁺ + MnO₄^- + H⁺ →
(c) MnO₄ - + C₂O₄^2- (aq) + H⁺ →
(d) Cr₂O₇^2- + H₂S + H⁺ → 
(e) Cr₂O₇^2- + I^- + H ⁺ →
Answer : MnO₂ + 2 KOH (aq) + O₂ → K₂MnO₄ + H₂O
5 Fe²⁺ + MnO₄⁺ 8 H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
2MnO₄⁺ 5C₂O₄^2- (aq) +16 H⁺ → 2 Mn²⁺ + 8H₂O + 10 CO₂
Cr₂O₇^2- + H₂S + H⁺ → Do yourself
Cr₂O₇^2- + I^- + H ⁺ → Do yourself

Question. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing PH on a solution of potassium dichromate or Explain how the colour of K₂Cr₂O₇ solution depends on PH of the solution?
Answer : (a) 4FeCr₂O₄ + 8Na₂CO₃ + 7 O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2Na₂CrO₄ + 2H+ → Na₂Cr₂O₇ + 2Na+ + H₂O
Na₂Cr₂O₇ + 2 KCl  → K₂Cr₂O₇ + 2 NaCl
(b) CrO₄^2- ======= Cr₂O₇^2-)
(if PH > 4 the CrO₄^2- ion (yellow) will exist and if PH < 4 then Cr₂O₇^2- ion(orange) will exist

Question. (a) Describe the preparation of potassium permanganate.
(b) How does the acidifiedpermanganate solution react with (i) iron(II) ions (ii) SO₂ and (iii) oxalic acid?Write the ionic equations for the reactions.
Answer : (a) 2MnO₂ + 4KOH (aq) + O₂ → 2K₂MnO₄ + 2H₂O
MnO4^2- ---- electrolysis- →  MnO₄^-
5 Fe₂+ + MnO₄^- + 8 H+ → Mn²⁺ + 4H₂O + 5Fe³⁺
5 SO₂ + 2MnO₄^- + 2H₂O 2Mn²⁺ + 4H+ + 5SO₄^2-
2MnO₄^- + 5C₂O₄^2- (aq) +16 H⁺ → 2 Mn²⁺ + 8H₂O + 10 CO₂

Question. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: iodide (ii) iron(II) solution and (iii) H₂S
Answer : (a) Ch₃CH₂OH-----K₂Cr₂O₇/H⁺ → CH₃COOH
(b) (b)Cr₂O₇^2- + 6I- + 14H ⁺ → 2Cr³⁺ +3I₂ + 7 H₂O
Cr₂O₇^2- + 6 Fe²⁺+ 14 H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7 H₂O
Cr₂O₇^2- + 3S^2- + 14 H⁺ → 2Cr³⁺ + 3S+ 7 H₂O

Question. When a chrromite ore A is fused with sodium carbonate in free excess of air and the product is dissolved in water , a yellow solution of compound B is obtained .After treatment of this yellow solution with sulphuric acid compound C can be crystallize from the solution .When compound C is treated with KCl orange crystals of compound D is crystallizes out. Identify A to D and wtite the reaction from A to B.
Answer : A= FeCr₂O₄
B= Na₂CrO₄
C= Na₂Cr₂O₇
D = K₂Cr₂O₇4
FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

Question. Zn, Cd & Hg are not treated as true transition elements. Why?
Answer : Because they have completely filled d-orbitals in their atomic as well as stable ionic state.

Question. Cu & Ag are transition metals although they have completely filled d-orbitals. Why?
Answer : Cu²⁺ & Ag²⁺ have (n-1)d⁹ 4s⁰ configuration.

Question. Why some d-block elements have irregular (exceptional) electronic configuration?
Answer : Due to very small energy difference between (n-1) d & ns sub-shell

Question. Atomic size does not change appreciably in a row of transition metals. Why?
Answer : Along the rows nuclear charge increases but the penultimate d-sub shell has poor shielding effect so atomic and ionic size remain almost same.

Question. Transition elements have variable oxidation states. Why?
Answer : Due to very small energy difference between (n-1)d & ns sub-shell electrons from both the sub-shell take part in bonding

Question. Transition metals have high melting and boiling points. Why?
Answer : A large number of unpaired electrons take part in bonding so they have very strong metallic bonds and hence high m.pt & b.pt

Question. Transition metals have high enthalpy of atomization. Why?
Answer : A large number of unpaired electrons take part in bonding so they have very strong metallic bonds and hence high enthalpy of atomization

Question. Transition metals show catalytic properties .Why?
Answer : Because they have variable oxidation states and hence can form different intermediates. They also provide large surface area

Question. Transition metals and their salts are generally colored .Why?
Answer : Because they have partially filled d-sub shell and hence d-d electron transition takes place when they absorb radiations from visible region and transmit complementary colors.

Question. Why transition metals form coordination compounds?
Answer : Because they have large number of vacant orbitals in (n-1)d, ns, np & ns sub shells so they can accept electron pairs from ligands

Question. Why transition metals form alloys?
Answer : They have comparable atomic size and hence can be mixed uniformly.

Question. Transition metals form interstial compounds. Why?
Answer : Because small atoms like H, C, N etc can be entrapped in their metallic crystals

Question. Zn, Cd & Hg have low boiling points and Hg is liquid. Why?
Answer : They have full filled 3d-orbitals and no electrons from d-orbitals are taking part in metallic bonding so they have weak metallic bonding. Due to larger atomic size Hg is liquid.

Question. Transition metals and many of their compounds show paramagnetic behavior .Why?
Answer : Because they have unpaired electrons

Question. d¹ configuration is very unstable in ions .Why?
Answer : Because by losing one electron they get extra stability

Question. Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidizing. Why?
Answer : E0 value for Cr³⁺/Cr²⁺ is negative but that of Mn³⁺/Mn²⁺ is positive so Cr²⁺ can lose electron to form Cr³⁺ while Mn³⁺ accepts electron to form Mn²⁺. In case of Cr d4 to d³ occurs for Cr²⁺ to Cr³⁺. d³ is stable.

Question. Cobalt(II) is stable in aqueous solution but in presence of complexing agents it gets oxidized. Why?
Answer : Oxidation state changes from +2 to +3 because in presence of ligands d-orbitals split up into t2g and eg having the stable configuration t²g⁶ eg⁰.

Question. Mn²⁺ compounds are more stable than Fe²⁺ .Why?
Answer : Mn²⁺ has half-filled d-orbitals i.e 3d⁵ 4s⁰ configuration

Question. Fe³⁺ is stable compared to Fe²⁺.Why?
Answer : Due to half-filled configuration i.e 3d⁵ 4s⁰ configuration.

Question. Transition metals exhibit highest oxidation states in oxides and fluorides. Why?
Answer : Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.

2 Mark Questions

Question. The highest oxidation state of transition metal is exhibited in oxoanions For the first row transition metals the Eo values are:
E^o V Cr Mn Fe Co Ni Cu
(M²⁺/M) –1.18 – 0.91 –1.18 – 0.44 – 0.28 – 0.25 +0.34
Explain the irregularity in the above values.?
Answer : The E⁰ (M²⁺/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (Δ i H 1 + Δ i H 2) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Question. Why is the E⁰ value for the Mn³⁺/Mn²⁺ couple much more positive than that for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺? Explain.?
Answer : Much larger third ionisation energy of Mn (where the required change is d⁵ to d⁴) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance

Question. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest.Why?
Answer : In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds.

Question. Explain why Cu⁺ ion is not stable in aqueous solutions?
Answer : Cu⁺ in aqueous solution undergoes disproportionation, i.e.,
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s) The E⁰ value for this is favorable

Question. Actinoid contractions are greater from element to element than lanthanoid contraction.Why?
Answer : The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor shielding from element to element in the series

Question. K₂PtCl₆ is well known compound and corresponding Ni⁴⁺ Salt isunknown . Whereas Ni⁺² is more stable than Pt⁺².
Answer : The stability of the compounds depends upon sum of ionization enthalpies:
IE₁ + IE₂ < IE₁ + IE₂
in Ni in Pt
Ni²⁺ is stable than Pt^+2.
IE₁ + IE₂ + IE₃ + IE₄ < IE₁ + IE₂ + IE₃ + IE₄
in Pt⁴⁺ in Ni⁴⁺
Pt⁴⁺ is stable, K₂PtCl₆ is well known compound.

Question. Why KMnO₄ is bright in colour ?
Answer : It is due to charge transfer. In MnO₄^– an electron is momentarily transferred from O to the metal, thus momentarily O^2– is changed to O^– and reducing the oxidation state of the metal from Mn (VII) to Mn (VI).

Question. CrO is basic but Cr₂O₃ is amphoteric?
It is due to charge transfer. In MnO₄^– an electron is momentarily transferred from O to the metal, thus momentarily O^2– is changed to O^– and reducing the oxidation state of the metal from Mn (VII) to Mn (VI).
Answer : CrO Cr₂O₃
O. N. + 2 + 3
Higher the oxidation states higher the acidity. In lower oxidation state some of valence e^– of the metal atom are not involved in bonding, can donate e^– and behave as base. In higher oxidation state e^– are involved in bonding and are not available, rather it can accept e^– and behave as an acid.

Question. In the titration of Fe²⁺ ions with KMnO4 in acidic medium, why dil. H₂SO₄ is used and not dilHCl.
Answer : KMnO₄ produce Cl₂ KMnO₄ in presence of dil. HCl acts as oxidising agent, Oxygen produced is used up partly for oxidation of HCl :
2KMnO₄ + 3 H₂SO₄ ——— K₂SO₄ + 2 MnSO₄ + 3 H₂O + 5 (O)
KMnO₄ + 4 HCl ——— 2 KCl + 2 MnCl₂ + 2 H₂O + 6 (O)
2 HCl + (O) ——— H₂O + Cl₂

Question. K₂Cr₂O₇ is used as Primary Standard in volumetric analysis.Why?
Answer : K₂Cr₂O₇ is not much soluble in cold water. However, it is obtained in pure state and is not Hygroscopic in nature.

3 MARKS QUESTIONS(31-40)

Question. (a) Although Cu⁺ has configuration 3 d10 4 s0 (stable) and Cu²⁺ has configuration 3 d9 (unstable configuration) still Cu²⁺ compounds are more stable than Cu+.
(b) Titanium (IV) is more stable than Ti (III) or Ti (II).
Answer : (a) It is due to much more (–) Hydration H– of Cu²⁺ (aq) than Cu⁺, which is more than compensates for the II ionization enthalpy of Cu.
(b) 22Ti = 3 d² 4 s²
TiIII = 3 d¹
TiII = 3 d²
TiIV = 3 d°
most stable configuration.
TiIV is more stable than TiIII and TiII.

Question. The actinoids exhibit more number of oxidation states and give their common oxidation states.
Answer : As the distance between the nucleus and 5 f orbitals (actinoides) is more than the distance between the nucleus and 4 f (lanthanoids) hence the hold of the nucleus on valence electrons decrease in actinoids. For this reason the actinoids exhibit more number of oxidation states in general.
Common O. N. exhibited are + 3 (similar to Canthanoids) besides + 3 state, also show + 4, maximum oxidation state in middle of series i. e. Pu and Np. have anoidation state upto + 7.

Question. (a) Give reason CrO₃ is an acid anhydride.
(b) Give the structure of CrO₅.
Answer : (a) CrO₃ + H₂O ——— H₂CrO₄ i. e. CrO₃ is formed by less of one H₂O molecule from chromic acid : – H₂O
H₂CrO₄ ——— CrO₃
(b)

Question. Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have d4 configuration ?
Answer : Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the d³ has half-filled t_2g level. n the other hand, the change from Mn²⁺ to Mn³⁺ results in the half filled (dS) configuration which has extra stability.

Question. (a) In MnO₄^– ion all the bonds formed between Mn and Oxygen are covalent. Give reason.
(b) Beside + 3 oxidation state Terbium Tb also shows + 4 oxidation state. (Atomic no. = 65)
Answer : (a) In MnO₄^–, O. N. is + 7, but it is not possible to lose 7 electrons because very high energy is required to remove 7 electrons. Therefore it forms covalent bonds.
(b) Tb = 65 E. C. is 4 f⁹ 6 s²
Tb⁴⁺ = 4 f⁷ 6 s⁰
half-filled f-orbital stable. after losing 4 e^– it attains half-filled orbital

Question. (a) Highest manganese flouride is MnF₄ whereas the highest oxide is Mn₂O₇.
(b) Copper can not librate H₂ from dil acids :
Note : Although only oxidising acids (HNO₃ and hot conc. H₂SO₄) react with Cu light.
Answer : (a) The ability of oxygen to form multiple bonds to metals, explain its superiority to show higher oxidation state with metal.
(b) Positive E° value (+ O – 34 Volt) accounts for its inability to liberate H₂ from acids. The high energy to transform Cu (s) to Cu²⁺ (aq) is not balanced by its Hydration enthalpy.
Note : For (b) Consult Fig. 8.4 in NCERT

Question. A metal which is strongly attracted by a magnet is attacked slowly by the HCl liberating a gas and producing a blue solution. The addition of water to this solution causes it to turn pink, the metal is
Answer : The metal is CO
CO + 2 HCl ——— COCl₂ + H₂
blue solution
COCl₂ in solution is [CO (H₂O)₆]²⁺
blue                                  pink

Question. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition metal ?
Answer : The following points justify that the given statement is true:-
(i) Ionization enthalpies of heavier transition elements are higher than the elements of 3d series.
Consequently, heavier transition elements are less reactive in comparison to 3d-elements.
(ii) Melting points of heavier transition elements are higher than 3d-elements.
(iii) Higher oxidation states of heavier transition elements are stable whereas lower oxidation states are stable in 3d-elements.

Question. The paramagnetic character in 3d-transition series elements increases upto Mn and then decreases.
Explain why?
Answer : In the 3d-transition series as we move from Sc (21) to Mn (25) the number of unpaired electrons increases and hence paramagnetic character increases. After Mn, the pairing of electrons in the d-orbital starts and the number of unpaired electrons decreases and hence, paramagnetic character decreases.

Question. Why are Mn²⁺ compounds more stable than Fe²⁺ compounds towards oxidation to +3 state?
Answer : The electronic configuration of Mn²⁺ is [Ar] 3d⁵, i.e. all five d-orbitals are singly occupied. Thus this is stable electronic configuration and further loss of electron requires high energy .on other hand side the electronic configuration of Fe²⁺ is [Ar] 3d6, i.e. Loss of one electron requires low energy

5 MARK QUESTIONS(41-45)

Question. A wellknown orange crystalline compound (A) when burnt impart violet colour to flame. (A) on treating (B) and conc. H₂SO₄ gives red gas (C) which gives red yellow solution (D) with alkaline water. (D) on treating with acetic acid and lead acetate gives yellow p. pt. (E). (B) sublimes on heating. Also on heating (B) with NaOH gas (F) is formed which gives white fumes with HCl. What are (A) to (F) ?
Answer : (i) K₂Cr₂O₇ + 4 NH4Cl + 3 H2SO₄ ——— K₂SO₄ +2 Cr₂O₂Cl₂ + 2 (NH₄)₂SO₄ + 3 H₂O (A) (B) Sublime Chromyl Chloride red gas(C)
(ii) CrO₂Cl₂ + 4 NaOH ——— Na₂CrO₄ + 2 NaCl +2 H₂O
(D) Yellow Soln.
(iii) Na₂CrO₄ + (CH₃COO)₂ Pb ——— PbCrO₄ + 2 CH₃COONa
Yellow p. pt. (E)

Question. Give reasons for the following:
(a) Among the lanthanoids, Ce(III) is easily oxidised to Ce(IV).
(b) Fe³⁺ / Fe²⁺ redox couple has less positive electrode potential than Mn³⁺ / Mn²⁺couple.
(c) The second and third transition series elements have almost similar atomic radii.
(d) Transition metals and many of their compounds show paramagnetic behaviour.
(e) KMnO4 titration is not carried out using HCl as acid medium.
(iii) The transition metals generally form coloured compounds.
Answer : (a) vacant (n-2) f subshell in Ce(IV).
(b) extra stability of Fe³⁺ than Mn³⁺ ion (c) Due to lanthanoid contraction
(d) they both will be reacting. (e) unpaired d –electrons and d-d transitions

Question. a) A Complex having scandium in +3 oxidation-state was found colorless why?
b) Show the splitting of d, orbitals of Ti in [Ti(H₂O)₆]⁺³
c) [Ti(H₂O)₆]⁺³ is coloured why?
d) Differentiate between. Lanthanides and actinides w.r t.
(i) Oxidation state (ii) electronic configuration
Answer : a) Due to absence of unpaired electron in d orbtal
b) Correct splitting of d orbitals in octahesral feild into t2g and eg c) Due to presence of single electron
d) lathanoid Actinoid
(i) Shows common It shows common oxidation state of +3+4+5 oxidation state +3
(ii) 4 f arefielled 5f are filled

Question. 1. Mixed oxide of iron and Chromium FeO.Cr₂O₃ is fused with Sodium Carbonate in the presence of air to form yellow compound (A). On acidification Compound (A) forms an orange coloured compound (B) which is an oxidizing agent
i) Identify A and B.
ii) Write balanced chemical equation for each
2. Among the lanthanoids, Ce(III) is easily oxidised to Ce(IV).
3. Fe³⁺ / Fe²⁺ redox couple has less positive electrode potential than Mn³⁺ / Mn²⁺couple.
Answer : (1) A= Na₂CrO₄ ; B= k₂Cr₂ O7
4FeCr₂O₄ + 8 NaCrO₄ +7O₂ -----→ 8 Na₂CrO₄ + 2 Fe₂O₃+8 CO₂
2Na₂CrO₄ + 2 H⁺ -----------→ Na₂Cr₂O₇ + 2Na⁺ +H₂O
(2) vacant (n-2) f subshell in Ce(IV).
(3) extra stability of Fe³⁺ than Mn³⁺ ion

Question. (I) Account for the following :
(a) Zirconium & Hafnium exhibit almost similar properties
(b) Zinc salts are white while Cu²⁺ salts are coloured
(c) The transition elements have high enthalpies of atomization.
(d) Among transition metals, the highest oxidation state is exhibited in oxoanins of a metal.
(e) Zn²⁺ salts are white while Cu²⁺ salts are blue
Answer : 1(a). Because of same size
(b). Because Zn⁺² ion does not have unpaired electrons while Cu⁺² have one unpaired electron
©.Because of many unpaired electrons they have many metallic binds
(d) In these oxoanions the oxygen atoms are directly bonded to the transition metal.
Since oxygen is highly electronegative, the oxoanions bring out the highest oxidation state of the metal.
(e) Zn²⁺ ion has all its orbitals completely filled whereas in Cu²⁺ ion there is one half-filled 3d-orbital. It therefore has a tendency to form coloured salts whereas Zn²⁺ has no such tendency. 1

1. For the complex [NiCL4]^2-, write (i) the IUPAC name.

(ii) thehybridization type. (iii) the shape of the complex. (3)

2. What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configurationof d⁴ in terms of t2g and eg in an octahedral field when Δ₀≥P (ii) Δ^o≤P (3)

3. Give reason for the following:

(i) Mn³⁺ is a good oxidising agent.

(ii) E⁰_M2+/M values are not regular for first row transition metals (3d series)

(iii) Although “F” is more electronegative than “O” the highest Mn fluoride is MnF₄, whereas the higest oxide is Mn₂O₇. (3)

4. Complete the following equations: (2)

(a) 2CrO₄^2- + 2H⁺ →

(b) KMnO₄^Δ →

5. (a) Why do transition elements show variable oxidation staes?

(b) name the element showing maximum number of oxidation state in first transition series.

(c) Name the element which shows only +3 oxidation state.

(d) What is lanthanoid contraction?

ONE MARK QUESTIONS

1.What is Misch metal? Give its use.

2.Why do most of the transition metal ions exhibit characteristic colour in aqueous solution?

TWO MARK QUESTIONS

1.Why do transition elements show variable oxidation states? How is the variability in oxidation states of d-block different from that of the p-block elements? 

2.What is Lanthanoid contraction? Give its consequences.

3.Explain with equations, how the colour of a solution of K₂Cr2O₇ depends on pH.

4.Complete and balance the following chemical equations:
a) Cr₂O₇^2- + I- + H⁺→
b) MnO₄^- + SO₃^2- + H⁺→

5.Answer the following questions:
a) Why do actonoids in general exhibit a greater range of oxidation states than the Lanthanoids?
b) Which element in the first series of transition elements does not exhibit variable oxidation states and why? 

6.Describe the preparation of
a) Potassium dichromate from sodium chromate and
b) KMnO₄ from K₂MnO₄

THREE MARK QUESTIONS

1. a) E0 value for the Mn³⁺/ Mn²⁺ couple is positive (+ 1.5 V) whereas that of Cr³⁺/ Cr²⁺ is negative (-0.4 V). Why?
b) The chemistry of actinoids is not sosmooth as that of lanthanoids.
c) Complete the following equation :
2MnO₄^- + 16 H⁺ + 5C₂O4^2- →

2. Explain the following observations:
a) Transition metals generally form coloured compounds.
b) Zinc is not regarded as a transition metal.
c) Transition elements and their compounds are generally found to be good catalysts in chemical reactions.

3. Account for the following:
a) The enthalpy of atomization of the transition metals is high.
b) The lowest oxide of a transition metal is basic; the highest is amphoteric/acidic.
c) Cobalt (II) is stable in aqueous solution but in the presence of complexing agents, it is easily oxidized.

FIVE MARKQUESTIONS

1. i) Complete and balance the following chemical equations:
a) Cr₂O₇^2- + I^- + H⁺→
b) MnO₄^- + SO₃^2- + H+→
ii) How would you account for the following:
a) The oxidizing power of oxoanions are in the order VO₂⁺ < Cr₂O₇^2- < MnO₄^-
b) The third ionization enthalpy of manganese (Z = 25) is exceptionally high.
c) Cr²⁺ is a stronger reducing agent than Fe²⁺.

VALUE BASED QUESTION

1. Six adults and twenty children were shot and killed at a Connecticut elementary school in US by a man who opened fire inside the two class rooms.
a) Which alloy is used to produce bullets and shells?
b) What do you think may be lacking in man who opened fire in terms of values?
c) What are the uses of Lanthanoids?

Important Questions for NCERT Class 12 Chemistry The d and f Block Elements

Question. More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is
(a) more active nature of the actinoids
(b) more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
(c) lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
(d) greater metallic character of the lanthanoids than that of the corresponding actinoids.

Answer    C

Question. Which one of the following elements shows maximum number of different oxidation states in its compounds?
(a) Gd
(b) La
(c) Eu
(d) Am

Answer    D

Question. HgCl₂ and I₂ both when dissolved in water containing I– ions, the pair of species formed is
(a) HgI₂, I^–
(b) HgI₄^2–, I₃^–
(c) Hg₂I₂, I^–
(d) HgI₂, I^3–
Answer    B

Question. Which of the following elements is responsible for oxidation of water to O₂ in biological processes?
(a) Cu
(b) Mo
(c) Fe
(d) Mn

Answer    C
 
Question. When calomel reacts with NH₄OH, we get
(a) Hg₂O
(b) HgO
(c) HgNH2Cl
(d) NH2–Hg–Hg–Cl

Answer    C

Question. Photographic films and plates have an essential ingredient of
(a) silver nitrate
(b) silver bromide
(c) sodium chloride
(d) oleic acid.

Answer    B

Question. Which of the following radioisotopes is used as anticancerous? 
(a) Na-24
(b) C-14
(c) U-235
(d) Co-60

Question. Which of the following compound is coloured?
(a) TiCl₃
(b) FeCl₃ 
(c) CoCl₂
(d) All of these

Question. The purple colour of KMnO₄ is due to the transition 
(a) C.T. (L → M)
(b) C.T. (M → L)
(c) d – d
(d) p – d

Question. The compounds containing complex oxyanion CrO4^2- are intensely yellow coloured because
(a) Chromium ion in CrO4² is a transition metal ion 
(b) Cr and O are Π-bonded and the allowed Π → Π* transition occurs in the visible region
(c) Cr → O charge transfer is responsible for yellow colour
(d) Compounds have strong absorption at ~ 640 nm

Question. The only cations present in a slightly acidic solutions are Fe3+, Zn²⁺ and Cu²⁺. The reagent that when added in excess to this solution would identify and separate Fe³⁺ in one step is
(a) 2 M HCI
(b) 6 M NH₃ 
(c) 6 M NaOH
(d) H₂S gas

Question. The colour of copper sulphide is 
(a) Blue
(b) Black
(c) Red
(d) Green

Question. Which one of the following statements concerning lanthanide elements is false? 
(a) Lanthanides are separated from one another by ion exchange method
(b) The ionic radii of trivalent lanthanides steadily increase with increase in atomic number
(c) All lanthanides are highly dense metals
(d) Most characteristic oxidation state of lanthanides is +3

Question. White silver surface turns black when O₃ is passed over it. This is due to the formation of
(a) silver hydroxide 
(b) freshly reduced silver which is black in colour
(c) silver oxide
(d) a complex compound of silver and ozone

Question. In which of the following cases, the stability of two oxidation states is correctly represented
(a) Ti³⁺ > Ti⁴⁺
(b) Mn²⁺ > Mn³⁺
(c) Fe²⁺ > Fe³⁺
(d) Cu⁺ > Cu²⁺

Question. Cuprous ion is colourless while cupric ion is coloured because 
(a) Both have half filled p-and d-orbitals
(b) Cuprous ion has incomplete d-orbitaland cupric ion has a complete d-orbital
(c) Both have unpaired electrons in the d-orbitals
(d) Cuprous ion has complete d-orbital and cupric ion has an incomplete d-orbital.

Question. A compound of a metal ion Mx+ (Z = 24) has a spin only magnetic moment of 15 Bohr Magnetons. The number of unpaired electrons in the compound are 
(a) 2
(b) 4
(c) 5
(d) 3

Question. Identify the incorrect statement among the following : 
(a) Lanthanoid contraction is the accumulation of successive shrinkages.
(b) As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements.
(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.

Question. Consider the following statements 
(I) La(OH)₃ is the least basic among hydroxides of lanthanides.
(II) Zr⁴⁺ and Hf⁴⁺ posses almost the same ionic radii.
(III) Ce⁴⁺ can as an oxidizing agent.
Which of the above is/are true ?
(a) (I) and (III)
(b) (II) and (III)
(c) (II) only
(d) (I) and (II)