NCERT Solutions for Class 12 Chemistry Chapter 4 – The D and F Block Elements

Intext Questions with Solutions of Class 12 Chemistry Chapter 4 – The D and F Block Elements

4.1

Ans – Ag (Z=47) has an external electronic structure of 4d¹⁰5s¹. In AgO & AgF₂, it displays +1 & +2 O.S. Additionally, the electrical arrangement in +2 O.S. is d⁹, meaning that the d-subshell is not full. As a result, it serves as a shift component.

4.2

Ans – The durability of the crystal lattice and the resilience of the bond between metals are closely related to the enthalpy of atomization. Because the orbital components are occupied in the instance of zinc (3d¹⁰4s²), none of the electrons from the 3d-orbitals play a role in the creation of metallic bonds. Nevertheless, a few d-electrons contribute to the metallic bonds in each additional component in the 3d group. This indicates that zinc has the smallest enthalpy of atomization in the 3d family and more fragile metallic bonds.

4.3

Ans – Manganese has the highest number of O.S. (Z = 25). Its external EC is 3d⁵4s², which explains this. It has the most e^-1 s to shed or exchange since 3d and 4s have similar energies. As a result, it displays O.S. between +2 and +7, which is the highest value.

4.4

Ans – Each metal’s E^o(M²⁺/M) is correlated with the total of the enthalpy shifts that occur in the subsequent stages:

Cu possesses a minimal enthalpy of hydration (Δ_hydH) and an elevated enthalpy of atomization (Δ_aH). The hydration enthalpy of Cu(s) is insufficient to offset the considerable energy needed to convert it to Cu²⁺(aq). E^o(Cu²⁺/Cu) is consequently affirmative.

4.5

Ans – Whenever an e^-1 is eliminated, the energy levels of the 4s and 3d orbitals alternate, causing a shift in the IEs of the 3d series. Ionization is therefore accompanied by a reorganization activity. In the dn arrangement, this leads to the discharge of exchange energy, which rises as the total amount of e^-1 s grows. Since the loss of 1 e- provides steady EC (3d⁶), Cr has a modest first IE. Since e~ must be eliminated from the steady configuration’s 4s orbital (3d¹⁰ 4s²), zinc has an exceptionally great IE. It gets challenging to remove the second e- after losing the first. Second IEs are therefore greater and often rise from left to right since the stable arrangement of Cr⁺ (3d⁵) & Cu⁺ (3d¹⁰) requires the removal of the second e–, Cr, and Cu exhibit substantially larger ratios.

4.6

Ans – The electronegativity ratings of oxygen & fluorine are fairly significant. The metallic substances can be oxidized to their maximum oxidation state by them. Consequently, the fluorides and oxides derived from metallic substances, especially transition alloys, exhibit the greatest oxidation states.

4.7

Ans – Compared to Fe²⁺, Cr²⁺ is a more potent reducing agent. In contrast to E°(Fe³⁺/Fe²⁺), which is positive (+ 0.77 V), E°(Cr³⁺/Cr²⁺) is negative (- 0.41 V). Therefore, whereas Fe²⁺ cannot readily oxidize to Fe³⁺, Cr²⁺ may readily oxidize to Fe³⁺.

4.8

Ans – Electronic configuration of M (Z = 27) is [Ar] 3d⁷4s²
⇒ Electronic configuration of M²⁺(aq) is [Ar] 3d⁷4s⁰

4.9

Ans – Whereas Cu²⁺ (aq) remains steady, Cu⁺ (aq) is unstable. This happens because Cu²⁺(aq) has a substantially greater Δ_hydH compared to Cu⁺(aq), which makes up for the second IE of Cu. Certain Cu(I) groups are therefore volatile in aqueous solution and have experienced the following uneven distribution: 2 Cu⁺ —–> Cu²⁺ + Cu.

4.10

Ans – Since 5/ electron particles have a worse shielding impact than 4f electrons, actinoid contraction—the reduction or contraction of atomic and ionic radii in actinoid elements—is greater than lanthanoid shrinkage. As a result, actinoid atoms are more susceptible to the physical contraction caused by a rise in the charge of their nuclei.

Exercise Questions with Solutions of Class 12 Chemistry Chapter 4 – The D and F Block Elements

4.1

Ans –

4.2

Ans – Mn²⁺ has an electronic structure of 3d⁵. This setup is reliable since it is partially filled. As a result, the third ionization enthalpy is extremely high, meaning that the 3rd electron is difficult to lose. Fe²⁺ has the electronic structure 3d⁶. It may readily lose a single electron to reach an intact 3d⁵ structure.

4.3

Ans – In this section, the j-orbitals lose two electrons, and as the number of atomic particles rises, the 3d-orbital eventually becomes saturated. The long-term reliability of the cations (M²⁺) rises from Sc²⁺ to Mn²⁺ as the amount of electrons with no pairings in the 3d orbital gets higher.

4.4

Ans – The oxidation states that result in precisely partially packed or entirely occupied d-orbitals are more durable in the initial set of transition materials. For instance, the electronic structure of Mn (Z = 25) is [Ar] 3d⁵ 4s². Mn (II) exhibits oxidation states ranging from + 2 to + 7. However, its partially filled structure [Ar] 3d⁵ makes it the most robust. Because its f-orbitals are partially occupied Sc³⁺ is less volatile than Sc⁺, while Fe³⁺ is a lot more durable than Fe²⁺.

4.5

Ans –

4.6

Ans –

4.7

Ans – Lanthanoid Contraction: The electrons that reside in the 4f-subshell of lanthanoids are filling up. The nuclear energy grows as one moves from left to right, and this rise should be offset by the four f-electron filtering impact expanding in strength. Nevertheless, the protective impact of the f-electrons is quite weak. Lanthanoid contraction is the result of the atomic & ionic radii decreasing from left to right.
Here are the Lanthanoid Contraction’s effects,
(a) Isolation Lanthanoids: It is challenging to distinguish among lanthanoids because of their comparable characteristics.
(b) Change in the basic strength of hydroxides: Lanthanoid contraction causes M³⁺ ion size to dip, which in turn causes the covalent nature of the M—OH bonding to rise. Hence, Lu(OH)₃ replaces La(OH)₃ as the basic character of oxides and hydroxides.
(c) The atomic dimensions of the substances belonging to the same class from the second and third transition periods are identical. The discrepancy between the values of Zr and Hf is much smaller than the distinction between the atomic radii of Y and La. The lanthanoid contraction is the reason behind this.
(d) Change in conventional reduction possibility: Because of lanthanoid contraction, the average reduction potential (E°) for the reduction operation increases slightly but steadily.
M³⁺ (aq) + 3e^– —–> 4 M(aq)
(e) Changes in physical characteristics, such as hardness, boiling & melting points.

4.8

Ans – Transitional components’ basic attributes.
(i) Electronic setup – (n -1) d^1-10 ns^1-2
(ii) Metallic traits: They generally exhibit conventional metallic shapes, with the notable neglect of Zn, Cd, and Hg.
(iii) Ionic and atomic dimensions: Ions with the same energy in a sequence exhibit a gradual reduction in diameter as the atomic number rises.
Oxidation condition: Variable, with a range of +2 to +7.
(v) Paramagnetism: Ions including electrons with no pairs exhibit paramagnetic behavior.
(vi) As the charge gets bigger, so does the ionization enthalpy.
The presence of electrons without pairings causes colored ions to form.
(viii) Complex compositions are formed as a result of the metal ions’ tiny dimensions and high energy saturation.
(ix) Because they can transform into various levels of oxidation, they have catalj¯c characteristics.
(x) Interstitial chemical creation.
(xi) Synthesis of alloys. They fall into the category of s & p-block components and are referred to as transitional substances because of their partially filled d-orbitals in the state of equilibrium or any acceptable oxidation condition. Since their ground states contain completely occupied d-orbitals, Zn, Cd, and Hg could not be considered transitional components.

4.9

Ans – Electronic equation of transition substances: (n – 1)d^1-10 ns^1-2. Non-transition components’ electronic arrangement: ns^1-2 or ns²np^1-6. Comparatively speaking, it is clear that non-transition ions lack d-orbitals in their atoms’ valence shells, whereas transitional substances have partial d-orbitals (or s-orbitals in certain situations). This explains why the components that correspond to various groups of structures have different features.

4.10

Ans – Lanthanides display oxidation states of +2, +3, and +4. The predominant oxidation state of lanthanoids is +3.

4.11

Ans – (i) Magnetic characteristics: Transitional substances and many of their analogs are paramagnetic, meaning that a magnetic field only faintly attracts them. This is because atoms, ions, or molecules contain electrons with no pairing as the number of electrons without pairings rises, the paramagnetic property will also get increased. The area of magnetic attraction is used to quantify the paramagnetic nature, and the result will be as follows,

(ii) They are capable of greater interatomic interactions which leads to deeper metallic bonds between atoms due to the huge amount of electrons without pairings in their d-orbitals, which raises the degree of atomization.

(iii) An extremely frequent property of transition alloys is the development of colored substances, both in solidified form and in aqueous solution. This is because some ultraviolet radiation is absorbed, which causes the electrons in conversion metal atoms to undergo a d-d transition. Electrons can move from a single collection of d-orbitals to another group inside the identical sub-shell. The d-orbitals lack an identical amount of energy and can divide into two distinct groups of orbitals that have various energies when ligands are present. We refer to these transitions as d-d transitions. For those d-d adjustments, the energy disparity is in the optical range. When elements made from transition metals are exposed to white light, they collect a certain frequency and release the extra colors, giving the complicated substance its distinctive color. Since they don’t absorb any light in the evident spectrum, Zn²⁺ and Ti⁴⁺ salts appear white.

(iv) Catalytic assets: Many transition metals and their mixtures function as catalysts in a range of processes. For example, Haber’s process uses finely split iron to produce NH₃, while the Contact method uses Pt or V₂O₅ to produce H₂S0₄. The subsequent 2 variables are responsible for the catalytic process.
(a) A novel pathway to a reaction with less activation energy is made possible by the transition metal ion’s capacity to move “easily” between oxidation states.
(b) The transition metal’s surface functions as an excellent adsorption increasing the percentage of reacting substances on it and triggering the chemical process.

4.12

Ans – There are several interstitial molecules that transition metals may create. Microscopic atoms from substances such as H, G, N, B, and so on can be trapped in their crystal lattice and may even form weak connections with them. They become less malleable and ductile and more compressive due to the production of interstitial complexes. Compared to wrought iron, steel, and cast iron are harder because of confined carbon atoms in the interstitial gaps.

4.13

Ans – Owing to the involvement of (n – 1) d electrons alongside the ns electrons in the bond growth, transition alloys exhibit a variety of different states of oxidization. As a result, they show a wide range of different oxidation levels. However, the non-transition metals that are often found in the s-block do not exhibit varied oxidation states because they obtain the arrangement of the closest precious gas components through the elimination of valence s-electrons.
The inert pair impact in the p-block favors the more powerful constituents in the simpler oxidation states, but in the d-block, the converse is true. Mo(VI) and W(VI), for instance, have been demonstrated to be more resistant to corrosion than Cr(VI) in category 6. Therefore, while MoO₃ & WO₃ are weak oxidizers, Cr(VI) is in its current state of dichromate in an acidic media.

4.14

Ans – Chromite ore (FeCr₂O₃) is fused with sodium carbonate in an unrestrained surplus of air to produce chromate, which is then used to make potassium dichromate. The following is the outcome of what happens with sodium carbonate:
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Orange sodium dichromate (Na₂Cr,0₇.2H₂0) can be crystallized from the yellow sodium chromate mixture after it has been filtered & acidified with sulfuric acid.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Compared to potassium dichromate, sodium dichromate remains highly soluble. Consequently, the final product is made by adding potassium chloride to the sodium dichromate approach.
Na₂Cr₂O₇+ 2KCl → K₂Cr₂O₇ + 2NaCl
Potassium dichromate crystallizes as orange flakes. The pH of the fluid affects the chromates & dichromates. Potassium dichromate turns into yellow potassium chromate when its pH is raised.
2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
Cr₂O₇^2- + 2OH^– → 2CrO₄^2- + H₂O

4.15

Ans – K₂Cr₂O₇ is a powerful oxidising agent. In a dilute sulphuric acid solution, the oxidation state of chromium transitions from +6 to +3.

4.16

Ans – Pyrolusite (MnO₂), potassium hydroxide & O₂ are combined to create potassium permanganate (KMn04); initially, green potassium manganate is produced. 2K₂MnO₄ + 2H₂0 —> 2MnO₂ + 4KOH + O₂ Water is used for separating potassium manganate, which is subsequently disproportionated in an acidic or neutral solution to produce potassium permanganate.

4.17

Ans – (i) The reduction potential of Cr³⁺/Cr²⁺ is negative. Therefore, it is not possible to convert Cr³⁺ to Cr²⁺. The positive redox capacity of Mn³⁺/Mn²⁺ is high. Therefore, it is easy to decrease Mn³⁺ to Mn²⁺. The positive reduction capacity of Fe³⁺/Fe²⁺ is limited. Fe³⁺ is more volatile than Cr³⁺ but highly durable than Mn³⁺.

(ii) Based on the E° standards, the metal oxidizes to the divalent cation in the following sequence: Mn > Cr > Fe.

4.18

Ans – Particularly ions with empty d-orbitals will be colored. Colorless ions have either full or vacant d-orbitals. In light of this, the colored ions in the above table are:
Ti³⁺(3d¹), V³⁺(3d²), Mn²⁺(3d⁵), Fe³⁺(3d⁵), Co²⁺ (3d⁷)
Sc³⁺ (3d°) and Cu⁺ (3d¹⁰) ions remain colorless in nature.

4.19

Ans – Because the total of the initial and secondary ionization energy increases, the degree of stability of the +2 oxidation stage in the initial transition sequence often falls from left to right. Nevertheless, partially filled d-orbitals (3d⁵) make Mn²⁺ more durable, whereas filled d-orbitals (3d¹⁰) make Zn²⁺ have greater stability.

4.20

Ans – (i) Electronic structure: Lanthanoids’ typical electronic setup can be depicted as [Xe]⁵⁴ 4f^1-14 5d^0-1 6s² compared to actinoids is [Rn]⁸⁶ 5f^0-14 6d^0-1 7s². Actinoids are members of the 5f series, while lanthanoids are members of the 4f series.

(ii) Atomic & ionic dimensions: When we move from left to right in the +3 oxidation state, each of the atoms or ions within both actinoids & lanthanoids exhibit a drop in size. The reduction is known as lanthanoid constriction in lanthanoids and actinoid relaxation in actinoids. Because 5f electrons provide less protection, the contractility increases from element to element in actinides.

(iii) Oxidation condition: Actinoids exhibit +3, +4, +5, +6, and +7 oxidation stages, while lanthanoids exhibit a restricted number of oxidation types (+2, +3, +4), with +3 being the most prevalent. The significant energy difference that exists among the 4f, 5d, and 6s orbitals is the source of this. Nevertheless, due to the tiny energy gap between the 5f 6d and Is orbitals, actinoids exhibit a vast range of different levels of oxidation.

(iv) Reactivity of chemicals: The early lanthanoids perform progressively more like aluminium as their atomic no. Increases, yet they’re still highly reactive like calcium. Whenever metallic substances are gradually warmed in the gas, they mix with the H₂ as well. Warming the metals with carbon results in the formation of carbides, Ln₃C, Ln₂C₃, and LnC₂. They ignite in halogens to generate halides and release hydrogen from diluted acid. They produce the hydroxides M(OH)₃ & oxides M₂O₃.
Particularly whenever precisely split, actinoids are very reactive metals. They react with water that is boiling to form an equilibrium of hydride and oxide, and at relatively low temperatures, they combine with the majority of non-metals.
Each of the metals is attacked by HCl, although most are only marginally impacted by nitric acid because defensive oxide coatings develop; alkalis are ineffective. Because actinoids have larger atoms and smaller ionization energies than lanthanoids, they become increasingly reactive.

4.21

Ans – (i) While the E° measurement for Mn³⁺/Mn²⁺ is favorable (+1.57 V), the E° value for Cr³⁺/Cr²⁺ is adverse (-0-41 V). Thus, Mn³⁺ may readily experience a reduction to yield Mn²⁺ and so operate as an oxidizing substance, while Cr²⁺ ions can readily encounter oxidation to provide Cr³⁺ ions by serving to become powerful reducing agents.

(ii) Compared to Co (II), Co (III) is more likely to form coordinating groups. Therefore, Co (II) transforms into Co (III) to be readily oxidized with the inclusion of ligands.

(iii) To obtain a stable d° structure, ions having a dx arrangement are prone to lose the lone electron in the d-subshell. As a result, they become fragile and oxidize or disproportionate.

4.22

Ans – When a material experiences both oxidation and reduction or when a component’s no. of oxidation processes rises and falls to produce 2 distinct goods, this is known as a disproportionation event.

4.23

Ans – Since the cation or positive ion gains an upright state structure of d-orbitals (3d¹⁰) by shedding 1 electron, Cu with arrangement [Ar] 4s¹³d¹⁰ displays a +1 oxidation condition and produces the Cu⁺ ion.

4.24

Ans – Mn³⁺ = 3d¹ = 4 electrons without any pairs, Cr³⁺ = 3d³ = 3 electrons pairs, V³⁺ = 3d² = 2 set of electrons, Ti³⁺=3d¹ = l electron. From the above electron ranges, Cr³⁺ remains highly stable in an aqueous solution due to the half-filled t_2g stage.

4.25

Ans – (i) Owing to every atom of metal possessing a reduced oxidation range, the less reactive oxide of a transition metal is neutral, while the more reactive oxides are acidic by virtue of their elevated oxidation position. MnO, for instance, is fundamental. While Mn₂O₇ is acidic because they are ionic, oxides at less oxidation states are base. Because they are covalent, compounds at greater forms of oxidation are acidic.

(ii) Since oxygen & fluorine are tiny, relatively electronegative components with the highest oxidizing properties, a transition metal has greater oxidation modes in oxides & fluorides. Vanadium has an oxidation temperature of +5 in V₂O₅, while osmium exhibits an oxidation value of +6 in O₅F₆.

(iii) The greatest degree of oxidation is found in oxo metallic anions, such as Cr in Cr₂O₇^2-, which has an oxidation stage of +6, and Mn in MnO^4-, which has an oxidation position of +7. Once more, this results from the metal’s interaction with oxygen, a powerful oxidizer and strong electronegative.

4.26

Ans – (i)

(ii)

4.27

Ans – A uniform combination of several metals or non-metal substances is called a metal alloy. Misch metal is a cerium (Ce) alloy. Iron (Fe), neodymium (Nd), lanthanum (La), and minor quantities of aluminium, sulphur, and carbon, among other elements. Jet engine components are made with it.

4.28

Ans – Inner-transition components are f-block features where the final electron reaches the f-sub shell. They consist of actinoids (Z=90 to 103) & lanthanoids (Z=58 to 71). Therefore, the interior transition components are those that have atomic no. of 59, 95 & 102.

4.29

Ans – Lanthanoids exhibit only three oxidation states: +2, +3 & +4 (with +3 being probably the most prevalent). The significant energy difference between the 4f, 5d, and 6s subshells is the source of instability. Actinoids exhibit a variety of oxidation types. However, their predominant oxidation level is + 3. For instance, uranium (Z = 92) & plutonium (Z – 94) exhibit + 3, + 4, + 5 and + 6, neptunium (Z = 94) displays + 3, +4, + 5 and + 7 respectively. The following is due to the little power differential among the actinoids’ 5f, 6d & 7s orbital elements.

4.30

Ans – Lawrencium (Z = 103) is the last actinoid.
Electronic configuration will be [Rn]⁸⁶ 5f¹⁴ 6d¹ 7s² 
The possible state of oxidation is +3.

4.31

Ans –

4.32

Ans – +4 oxidation state: ₅₈Ce, ₅₉Pr, ₆₅Tb, +2 oxidation state: ₆₀Nd, ₆₂Sm, ₆₃Eu, ₆₉Tm, ₇₀Yb.
Materials possessing the structure 5d⁰6s² often display the +2 oxidation level, which makes it easy to lose 2 electrons. In a similar way elements that shed 4 electrons and gain an arrangement that is either near 4f⁰ or 4f⁷ exhibit the +4 oxidation condition.

4.33

Ans – (i) Electronic arrangement: Actinoids have continuously occupied 5f-orbitals, while lanthanoids have consistently occupied 4f-orbitals.

(ii) Oxidation reads as follows: The oxidation state of lanthanoids is +3. Additionally, multiple substances exhibit +2 & +4 oxidation states. The oxidation levels of actinoids are +3, +4, +5, +6, and +7. Even so, the more popular ones are +3 and +4.

(iii) Chemical responsiveness: Because actinoids have larger atoms and smaller ionization energies than lanthanoids, they have a greater reactive.

4.34

Ans –

4.35

Ans – (i) Electronic arrangement: Overall, the electronic configurations of each component in a single vertical row are identical. There are just 2 outliers to the initial transition sequence. Namely Cr = 3d⁵ 4s¹ & Cu = 3d¹⁰ 4s¹. However, there are more exemptions in the subsequent transition sequence, such as Y = 4d¹ 5s², Nb = 4d¹ 5s¹, Mo = 4d⁵ 5s¹, Ru = 4d¹ 5s¹, Rh = 4d⁸ 5s¹, Pd = 4d¹⁰ 5s°, and Ag = 4d¹⁰ 5s¹. Pt = 5d⁹ 6s¹ & Au = 5d¹⁰ 6s¹ are the 2 outliers to the 3rd phase. As a result, the electrical arrangement of each part of 3 groups is not always comparable in an identical vertical section.

(ii) Oxidation condition: In broad terms, the oxidation states of each element in the identical transverse column are comparable. The substances in the center of each sequence display the most forms of oxidation, while the ones at the extremes display the fewest.

(iii) Enthalpies of ionization: Although there are occasional outliers in every region, the initial ionization enthalpies in every series typically rise progressively as we proceed from left to right. Several components of the second (4d) series have greater initial ionization energy values than the constituents of the third family in the identical vertical column, while others have less than values than average ones. The 3rd (5d) series’ initial ionization energy levels. Nevertheless, are greater in comparison to those of the 3d & Ad families. This results from the nucleus’s modest insulation by 4f-electrons in the series of 5d arrangement.

(iv) Atomic dimensions: Although the drop is relatively slight, ions with identical charge levels or atoms in a certain series often exhibit an ongoing reduction in circumference as the atomic number rises. However, due to lanthanoid shrinkage, the dimension of the constituent parts of the 5d family is almost equal to that of the Ad sequence, while the atoms of the Ad category are bigger than the equivalent components of the 3d species.

4.36

Ans –

4.37

Ans – The 4th (4d), fifth (5d) & sixth (6d) transition sequences are where the stronger transition components are found. For several causes, their characteristics ought to vary from those of the components in the initial (3d) collection:
(i) The substances in the Ad & 5d series have bigger atomic radii because they possess larger outer electron shells. Even so, due to lanthanoid contraction, the disparity between the Ad and 5d transitional components is relatively smaller.
(ii) The components of the Ad and 5d series have greater m.p. and b.p. due to tighter interatomic bonds.
(iii) Once we proceed from a single sequence to the next, ionization enthalpies should drop. Yet, because of lanthanoid contractions, the probabilities for each component in the 5d categories are greater than those of the remaining 2 categories.
As a result, the nominal nuclear charge rises while the dimension of an atomic particle shrinks. In the scenario of 3D components, this leads to a rise in ionization power.

4.38

Ans –

(i) K₄[Mn(CN)₆] In this compound, Mn exhibits a +2 oxidation state and a μ of 2.2 BM. This indicates that it possesses only one unpaired electron. As CN^– ligands approach the Mn²⁺ ion, the electrons in the 3d orbital undergo pairing.

Therefore, CN^– acts as a strong ligand. The hybridisation is d²sp³, resulting in an inner-orbital octahedral complex.

(ii) [Fe(H₂O)₆]²⁺:- In this compound, Fe exhibits a +2 oxidation state and a μ of 5.3 BM. This indicates that there are 4 unpaired electrons in 3d. Also 3d electrons do not pair up when the H₂O molecules approach. Thus H₂O is a weak ligand.

The hybridisation is sp³d², resulting in an outer-orbital octahedral complex.

(iii) K₂[MnCl₄] In this compound, Mn exhibits a +2 oxidation state and a μ of 5.92 BM. This indicates that it possesses 5 unpaired electrons.

The hybridisation is sp³, resulting in tetrahedral complex. Cl is a weak ligand

Related Study Resources of Chapter 4 – The D and F Block Elements

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NCERT Solutions for Class 12 The D and F Block Elements

Inorganic chemistry depends much on the d- and f-block elements. These elements show properties not like those of the s- and p-block elements. Covering all Intext and Exercise Questions, this page offers D and F Block Elements Class 12 NCERT Solutions, thereby enabling the students to acquire vital ideas efficiently to enable good test preparation.
Class 12 Chemistry Chapter 4 explains d and f block elements. These are transition and inner transition metals. Electronic configurations, oxidation states, and chemical properties are explained for transition elements, which show variable valency, catalytic behavior, and colored ion formation. Lanthanides and actinides have a different kind of trend: atomic size shows contraction. It is important for the understanding of metal chemistry and its industrial importance. Solutions at NCERT provided here are giving detailed answers to all the Intext and Exercise Questions, thus making exam preparation easier and structured.
Exceeding your Class 12 chemistry curriculum requires mastery of the ideas D and F block elements provide. The NCERT answers for the intext questions and exercises will help you to increase your knowledge and let you pass your tests. To properly base your knowledge in this part of chemistry, practise often and get assistance if needed.