d block Elements Class 12 Chemistry Notes

CBSE Class 12, Unit-4: d-Block Elements Notes

d-Block Elements:

The d-block elements are those elements in which the last electron enters the d–subshell of penultimate shell. The general electronic configuration of these elements is (n – 1) d^1-10ns^1-2 , where n is outermost shell. The d-block consisting of groups 3–12 occupies the large middle section of the periodic table.

Transition Elements:

The elements of d-block are known as transition elements as they possess properties that are transitional between the s and p block elements. A transition element is defined as an element which has incompletely filled d-orbitals in its ground state or any one of its oxidation states. There are four series of transition elements spread between group 3 and 12.

First transition series or 3d-series: Scandium (₂₁Sc) to Zinc (₃₀Zn)
Second transition series or 4d-series: Yttrium (₃₉Y) to Cadmium (₄₈Cd)
Third transition series or 5d-series: Lanthanum (₅₇La) and Hafnium (₇₂Hf) to Mercury (₈₀Hg) (Omitting ₅₈Ce to ₇₁Lu)
Fourth transition series or 6d-series: Begins with Actinium (₈₉Ac) is still incomplete.

Zinc, cadmium and mercury of group 12 have full d¹⁰ configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals. However, being the end elements of the three transition series, their chemistry is studied along with the chemistry of the transition elements.

General characteristics of Transition Elements:

Physical Properties:

1. All transition elements are metals.

NCERT Example Question:4.1

On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

Solution:
On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d¹), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d ¹⁰) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.

NCERT Intext Question:4.1

Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?

Solution:
Silver is considered as a transition element because, while it has a completely filled 4d orbital (4d¹⁰) in its ground state, it can exhibit a +2 oxidation state (4d⁹), resulting in an incompletely filled d-orbital.

2. All are malleable and ductile except mercury (liquid).

3. High thermal and electrical conductivity.

4. Metallic lustre and sonorous.

5. Atomic radii: Smaller than atomic size of s-block elements, larger than atomic size of p-block elements in a period. In a transition series, as the atomic number increases, the atomic radii first decreases till the middle, becomes constant and then increases towards end of the period. It usually increase down the group. The size of 4d elements is almost of the same size as of the 5d series elements. The filling of 4d before 5d orbitals results in regular decrease in atomic radii which is called as lanthanoid contraction.

6. Ionic radii: The ionic radii decrease with increase in oxidation state.

7. Density: From left to right in a period, density increases.

8. Ionisation enthalpy: Along the series from left to right, there is an increase in ionisation enthalpy. Irregular trend in the first ionisation enthalpy of 3d metals is due to irregularity in electronic configuration of 4s and 3d orbitals. In a group, IE decreases from 3d to 4d-series but increases from 4d to 5d series due to lanthanoid contraction.

NCERT Intext Question:4.5

How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Solution:
The irregular variation in ionisation enthalpies is due to the extra stability of the configuration like d⁰, d⁵, d¹⁰ because these states are extremely stable and have high ionisation enthalpies.

Chromium has low first ionization enthalpy because after losing an electron it attains stable configuration (d⁵). But in the case of Zinc, the first IE is very high, because we remove an electron from a stable configuration(3d¹⁰,4s²).

The second IE is extremely high in case of Cr⁺and Cu⁺ because they already in a stable state.

9. Metallic bonding: In metallic bonding, regular lattice of positive ions are held together by a cloud of free electrons, which can move freely through the lattice. Transition metal atoms are held together by strong metallic bonds.

10. Enthalpy of atomisation: Enthalpy of atomisation is the heat required to convert 1 mole of crystal lattice into free atoms. Transition elements have high enthalpy of atomisation. It first increases, becomes maximum in the middle of the series and then decreases regularly.

NCERT Example Question:4.2

Why do the transition elements exhibit higher enthalpies of atomisation?

Solution:
Transition elements exhibit higher enthalpies of atomisation because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

NCERT Intext Question:4.2

In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why?

Solution:
Zinc has the lowest enthalpy of atomization (126 kJ mol⁻¹) in the series Sc to Zn because it has a fully filled d-subshell (3d¹⁰4s²) and no unpaired electrons, leading to weaker metallic bonding compared to other transition metals in the series.

11. Variable oxidation state: Since the energies of ns and (n–1) d electrons are almost equal, therefore the electrons of both these orbitals take part in the reactions, due to which transition elements show variable oxidation states. Transition metal ions show variable oxidation states except the first and last member of the series.

NCERT Example Question:4.3

Name a transition element which does not exhibit variable oxidation states.

Solution:
Scandium (Z = 21) does not exhibit variable oxidation states. It has only one unpaired electron in d-orbital and due to this, it doesn't exhibit variable oxidation states.

NCERT Intext Question:4.3

Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Solution:
Manganese (Mn) exhibits the largest number of oxidation states in the 3d series of transition metals because of its electronic configuration, which allows it to have the most unpaired electrons in its d-orbitals. Manganese shows oxidation states from +2 to +7.

NCERT Intext Question:4.6

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Solution:
Highest oxidation state of a metal exhibited in its oxide or fluoride because oxygen and fluorine are very electronegative. This means they have a strong tendency to attract electrons.

12. Electrode potential: The electrode potential develops on a metal electrode when it is in equilibrium with a solution of its ions, leaving electrons from the electrode. Transition metals have lower value of reduction potential. Reduction potential is a measure of how easily a chemical species accept electrons and undergo reduction. The more positive the reduction potential, the more easily the chemical species accept electrons. Variation in E° value is irregular due to the regular variation in ionisation enthalpies (I.E₁ + I.E₂), sublimation and hydration enthalpies.

NCERT Intext Question:4.4

The E^o(M²⁺/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high ΔH^o and low Δ_hydH^o)

Solution:
The positive E^o_Cu⁺²/Cu value is due to the high enthalpy of atomization and ionization, and the low enthalpy of hydration of copper.

NCERT Intext Question:4.7

Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why ?

Solution:
Cr²⁺ is a stronger reducing agent than Fe²⁺ because, Cr²⁺ is more easily oxidized to Cr³⁺ than Fe²⁺ to Fe³⁺.

13. Catalytic properties: Many of the transition metals and their compounds, particularly oxides act as catalysts for a number of chemical reactions. Iron, cobalt, nickel, platinum, chromium, manganese and their compounds are the commonly used catalysts. All transitional metals show multiple oxidation states and have large surface area so, all metals work as a catalyst.

14. Magnetic properties: On the basis of the behaviour of substances in magnetic field, they are of two types:
Diamagnetic and Paramagnetic.
Diamagnetic substances have paired electrons only. e.g., Zn has no (zero) paired electrons.
In paramagnetic substances, it is necessary to have at least one unpaired electron. Paramagnetism increases with the increase in number of unpaired electrons.
Paramagnetism may be measured by magnetic moment.
Magnetic moment, (µ) = √n (n + 2) B.M.
where n = number of unpaired electrons in atom or ion and B.M. = Bohr Magneton (unit of magnetic moment). Diamagnetic and paramagnetic substances are repelled and attracted in the magnetic field respectively (Magnetic properties of transition elements).

NCERT Intext Question:4.8

Calculate the spin only magnetic moment of M²⁺(aq) ion (Z = 27).

Solution:
Co²⁺ has three unpaired electrons.
∴ Spin only magnetic moment (μ) = [n (n + 2)]^1/2 B.M
μ = [3 (3 + 2)]^1/2 B.M
or, μ = √15 B.M = 3.87 B.M.

15. Melting and boiling points: Except zinc, cadmium and mercury, all other transition elements have high melting and boiling points. This is due to strong metallic bonds and presence of partially filled d-orbitals in the shell of the atom of element.

16. Complex formation: They have tendency to form complex ions due to high charge on the transition metal ions and the availability of d-orbitals for accommodating electrons donated by the ligand atoms.

17. Formation of coloured compounds: Transition metals form coloured ions due to the presence of unpaired d-electrons. As a result, light is absorbed in the visible region to cause excitation of unpaired d-electrons (d – d transition) and colour observed corresponds to the complementary colour of the light absorbed. Cu⁺, Zn²⁺ and Cd²⁺ are colourless due to the absence of unpaired d-electron (d¹⁰).

Why are most transition metal compounds colored?

Solution:
Most transition metal compounds are colored because their partially filled d-orbitals allow electrons to absorb specific wavelengths of visible light, leading to d-d transitions and the emission of complementary colors.

18. Formation of alloys: Alloy formation is due to almost similar size of the metal ions, their high ionic charges and the availability of d-orbitals for bond formation. Therefore, these metals can mutually substitute their position in their crystal lattice to form alloys. e.g., steel, brass.

Why do transition metals form alloys readily?

Solution:
Transition metals readily form alloys because their similar atomic sizes allow atoms of one metal to easily substitute those of another in the crystal lattice and form solid solutions.

19. Formation of interstitial compounds: Interstitial compounds are known for transition metals as small-sized atoms of H, B, C, N, etc. can easily occupy positions in the voids present in the crystal lattices of transition metals. Characteristics of interstitial compounds:
A. High melting points
B. Hard
C. Chemically inert
D. Retain metallic conductivity
E. Non-stoichiometric

What are interstitial compounds? Why are such compounds well known for transition metals?

Solution:
Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice. Interstitial compounds are well known for transition metals because small-sized atoms of H, B, C, N, etc., can easily occupy positions in the voids present in the crystal lattices of transition metals.

NCERT Intext Question:4.9

Explain why Cu⁺ ion is not stable in aqueous solutions?

Solution:
The Cu⁺ ion is unstable in aqueous solutions because it disproportionates into Cu²⁺ and Cu. This reaction occurs because Cu²⁺ has a higher hydration energy than Cu⁺, making Cu²⁺ more stable in solution.
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s).

Preparation and Properties of Potassium Dichromate (K₂Cr₂O₇)

Preparation of Potassium Dichromate:

It is prepared from chromate ore in the following steps:
1. Chromite ore is fused with sodium carbonate in the presence of air to give sodium chromate.
4FeCr₂O₄ + 8Na₂CO₃ +7O₂ → 2Fe₂O₃ + 8Na₂CrO₄ + 8CO₂
2. Na₂CrO₄ is filtered and acidified with conc. H₂SO₄ to give Na₂Cr₂O₇.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O.
3. Sodium dichromate solution is treated with KCl to give
K₂Cr₂O₇. Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Properties of Potassium Dichromate:

1. It is an orange, crystalline solid.
2. Reaction with alkali: Cr₂O₇^2– + 2OH^– → 2CrO₄^2– + H₂O
Chromate ion(Yellow)
3. Reaction with acid: 2CrO₄^2– + 2H⁺ → Cr₂O₇^2– + H₂O
Dichromate ion (orange red)
In acidic solutions, oxidising action is
Cr₂O₇^2– + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O
4. It is a powerful oxidising agent

What is the effect of increasing pH on a solution of potassium dichromate?

Solution:
On increasing the pH, i.e., on making the solution alkaline, dichromate ions (orange coloured) are converted into chromate ions and thus, the solution turns yellow.

Uses of Potassium Dichromate:

1. In leather industry for chrome tanning.
2. Preparation of azo compounds.
3. As a primary standard in volumetric analysis for the estimation of reducing agent.

Structures of Chromate and Dichromate Ions:

Preparation and Properties of Potassium Permanganate (KMnO₄)

Preparation of Potassium Permanganate:

Potassium permanganate is commercially prepared from MnO₂ (Pyrolusite). The preparation involves two steps:

1. Conversion of MnO₂ to potassium manganate (K₂MnO₄) by fusing with KOH in presence of air(or an oxidizing agent like KNO₃).
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

2. Potassium manganate is oxidised to potassium permanganate either by electrolysis or by acidification.
3MnO₄^2- + 4H⁺ → 2MnO₄^- + MnO₂ + 2H₂O

Properties of Potassium Permanganate:

Potassium permanganate forms dark purple crystals which are iso-structural with those of potassium perchlorate (KClO₄).
The colour of permanganate is due to ligand to metal charge transfer. i.e. an electron is transferred from oxygen atom to the vacant d-orbital of Mn.
When heated it decomposes and liberate O₄.₂
2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

Structure of Potassium permanganate:

The manganate and permanganate ions are tetrahedral. The green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.

Uses of Potassium permanganate:

1. It is used as an oxidising agent in acidic, basic and neutral medium.
2. It is used as a primary standard in volumetric analysis.
3. It is used for the bleaching of wool, cotton, silk and other textile fibres and also for the decolourisation of oils.

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Important Questions Answer

Zinc, cadmium and mercury of group 12 are not regarded as transition metals, Why ?

Zinc, cadmium and mercury of group 12 have full d¹⁰ configuration ( d orbitals are completely filled ) in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals.

The outer electronic configuration of Cr is 3d⁵ 4s¹ instead of 3d⁴ 4s², why?

Half filled (3d⁵) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital.

The outer electronic configuration of Cu is 3d¹⁰ 4s¹ instead of 3d⁹ 4s², why?

Completely filled (3d¹⁰) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital.

Account for high melting point and boiling points of transition metals.

The melting and boiling points of transition metals are high because of the involvement of greater number of electrons from (n-1)d orbitals in addition to the ns electrons in the inter atomic metallic bonding.

What is the trend in melting points of transition metals in a series?

The melting points of the transition metals in a series rise to a maximum at the middle of the series (i.e. Cr or Mo or W - element with d⁵ configuration ) and fall regularly as the atomic number increases.

Why do transition metals have higher enthalpies of atomization?

Involvement of a large number of unpaired electrons of d orbitals favour stronger inter atomic interactions resulting in stronger bonds between the atoms of a metal and higher enthalpies of atomization.

Transition metals exhibit variable oxidation states in its compounds, why?

Transition metals exhibit variable oxidation states in its compounds due to the availability of both ns & (n – 1 ) d electrons for bond formation.

Name 3d series metal which shows highest oxidation state.

The highest oxidation state shown by 3d series transition metals is +7 by Mn.

Name a metal in the 3d series of transition metals which exhibit +1 oxidation state most frequently.

Copper exhibit +1 oxidation state in 3d seties of transition metals.

Why transition metals and their compounds shows paramagnetic behavior ?

The transition metal ions are generally containing one or more unpaired electrons in them & hence their compounds are generally paramagnetic.

The transition metals generally form coloured compounds, why?

The compounds of transition elements shows colour due to presence of unpaired electron & ability to undergo d-d transition. When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region.

Give reason transition metals and their many compounds acts as good catalysts.

Transition metals and their many compounds acts as good catalysts,it is due to
i. partially filled (n-1) d orbital
ii. variable oxidation state and provide a suitable surface for the reaction to take place.

What is the action of neutral or faintly alkaline permanganate solution on iodide ?

Alkaline permanganate solution oxidize iodide to iodate.

What happens when potassium permanganate is heated to 513 K ?

Potassium permanganate decomposes at 513K to potassium manganate, manganese dioxide and oxygen.

What is Actinoid contraction?

There is a gradual decrease in the size of atoms or M3+ ions across the series. This is known as the actinoid contraction.

Actinoid contraction is more than lanthanoid contraction. Give reason.

The actinoid contraction is, more than lanthanoid contraction due to poor shielding by 5f electrons from nuclear charge.

Actionoids show larger number of oxidation states than lanthanoids. Why?

In actinoids 5f, 6d and 7s levels are of comparable energies, hence electrons from these orbitals are available to lose or share.

Give one use of Mischmetall.

Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Transition metals form a large number of complex compounds.Give reason.

Transition metals for complex compounds due to
i. small sizes of metal cations
ii. their ionic charges and
iii. availability of d orbitals for bond formation.

Name the metal of the 1st row transition series that
i. has highest value for magnetic moment
ii. has zero spin only magnetic moment in its +2 oxidation state.
iii. exhibit maximum number of oxidation states.

i. Chromium
ii. Zinc
iii. Manganese

What is the composition of mischmetall? Give its one use.

The composition of mischmetall is lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Which gases liberated when
i. crystals of potassium permanganate is heated to 513K ?
ii. acidified potassium permanganate is treated with oxalate ion at 333K?

i. When crystals of potassium permanganate is heated to 513K Oxygen (O₂) gas is liberated.
ii. Acidified potassium permanganate when treated with oxalate ion at 333K liberates Carbon dioxide (CO₂) gas.

What happens when
i. A lanthonoid reacts with dilute acids ?
ii. A lanthonoid reacts with water?

i. When lanthonoid reacts with dilute acids , it liberates hydrogen gas.
ii. When lanthonoid reacts with water , it forms lanthanoid hydroxide and liberate hydrogen gas.

What is lanthanoid contraction? Write any one consequence of lanthanoid contraction.

Steady decrease in the size of lanthanides with increase in atomic number is known as lanthanoid contraction.
Due to lanthanoid contraction radii of members of 3rd transition series are very much similar to corresponding members of 2nd series.

Write any two consequences of lanthanoid contraction.

Two consequences of lanthanoid contrations are
i. The radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. Ex. The almost identical radii of Zr (160 pm) and Hf (159 pm) & Nb (146pm) & Ta (146pm)
ii. Difficulty in separation of lanthanoids due to similarity in chemical properties.

Zn⁺² salts are white while Cu⁺² salts are coloured. Why?

Zn⁺² salts are white because it does not have unpaired electron, whereas Cu⁺² salts are coloured because it has unpaired electron and undergoes d-d transition by absorbing light from visible region and radiate blue colour.

E° value for the Mn⁺³/Mn⁺² couple is positive (+1.5 V) whereas that of Cr⁺³/Cr⁺² is negative (-0.4 V). Why?

Mn⁺² is more stable than Mn⁺³ due to half filled d-orbitals (3d⁵), whereas Cr⁺³ is more stable than Cr⁺² due to half filled orbitals.

The Mn⁺² compounds are more stable than Fe⁺² towards oxidation to their +3 state.

Mn⁺² has 3d⁵ (stable electronic configuration), therefore, it does not get oxidised to Mn⁺³, whereas Fe⁺² has 3d⁶ which readily changes to Fe⁺³ (3d⁵), which has stable electronic configuration.

Following ions of 3d transition series are given:
V³⁺, Cr²⁺, Cu⁺, Fe²⁺, (Atomic number : V = 23, Cr = 24, Cu = 29, Fe = 26)
Identify the ion which is
i. unstable in aqueous solution.
ii. a strong reducing agent in aqueous solution.
iii. colourless in aqueous solution?
Give suitable reason in each. CBSE 2022, 3M

i. Cu⁺ is unstable in aqueous solution because, in aqueous solution, Cu⁺ tends to disproportionate into Cu²⁺ and Cu^o (solid copper).
2Cu⁺(aq) → Cu²⁺(aq) + Cu^o(s)

ii. Cr²⁺ is a strong reducing agent in aqueous solution because after losing an electron it becomes Cr³⁺ (3d³), which has more stable half-filled 3d subshell.

iii. Cu⁺ is colourless in aqueous solution because, it has a completely filled 3d subshell (3d¹⁰) and no unpaired electron is present. So, no d-d transitions are possible, and the ion appears colourless.

Define transition elements. Write two characteristics of transition elements. CBSE 2022, 3M

Transition elements are those elements that have partially filled d orbitals in their atomic or ionic states. They are placed in the d-block of the periodic table from group IB to VIIB and VIII.

Two Characteristics of Transition Elements
i. Variable Oxidation States
ii. Formation of Colored Compounds

Variable Oxidation States: Transition elements exhibit variable oxidation states in their compounds because, they can lose electrons from both their valence shell, leading to the formation of ions with different charges. For example, iron can exhibit +2, +3, and +6 oxidation states in various compounds.

Formation of Colored Compounds: Many compounds of transition elements are colored due to the presence of partially filled d orbitals, which allow for d-d transitions of electrons. When visible light falls on these compounds, some wavelengths are absorbed, while others are transmitted, resulting in the observed color. For example, copper sulfate (CuSO₂) is blue in color due to the Cu²⁺ ion.