NEET Chapter: The d and f Block Elements

The elements belonging to groups 3 to 12 in the modern periodic table, are called d-block or transition elements. In d-block elements, the last differentiating electron enters into the d-orbitals of the penultimate shell i.e. (n–1) d where n is the last shell. These element properties are intermediate between the properties of s-block and p-block elements. These elements represent a change or transition in properties from more electropositive elements (i.e. s-block) to less electropositive elements (i.e. p-block) that is why these elements are called transition elements. 

The d- and f- Block Elements is an important topic of the NEET Exam. Physical properties, chemical properties, oxidation states, electronic configuration of d and f block elements  and some important compounds of the d-block element are all significant topics. We will explore everything about the characteristics of transition metals, in this article. This article provides an example of the types of questions that might be asked about this subject.

Important Topics of the p-Block Elements

• Electronic configuration of d-block element

• General properties of d-block element 

• Preparation and properties of K2Cr2O7

• Preparation and properties of KMnO4

• Electronic configuration of f-block element

• Oxidation states of f-block element

• Chemical reactivity of f-block element

• Lanthanoids contraction

• Comparison of Actinoids with lanthanoids

Electronic Configuration of the d- Block Elements

• In d-block elements, as the atomic number increase, the d-orbitals of the penultimate shell i.e. (n-1) d are gradually filled by electrons. The general electronic configuration of d- block element is, (n-1) d1-10, ns0-2

• Depending upon the d-orbitals of which the penultimate shell i.e. n = 4, 5, 6, 7 are filled, four rows (called series) as 10 elements in each are obtained. They consist of 3d, 4d, 5d and 6d subshells.

• Energy of the (n–1)d subshell is slightly greater than the ‘ns’ subshell therefore the ns orbital is filled first then (n – 1)d orbitals.

• First (3d) Transition Series (Sc - Zn)

• Second (4d) Transition Series (Y - Cd)

• Third (5d) Transition Series (La - Hg)

Characteristics of Transition Metals

• The elements belonging to a given transition series do not differ so much from one another because of representative elements of the same period. That is why in a transition series, there is no change in the number of electrons in the outermost shell but only change occurs in the (n - 1)d electrons from member to another member in a period. These elements show horizontal and vertical relationships or show similarities in a period as well as in group.   

General Trends in Properties of the First row Transition Metals

1. Metallic Character

• Except for mercury which is a liquid, All the transition elements are metals and exhibit most of the properties of metals i.e. metallic lustre, high density, high melting and boiling points, malleability, ductility, high tensile strength, hardness, brittleness, etc. These are good conductors of heat and electricity. They exhibit all 3 types of structures: face centred cubic (fcc), hexagonal closed packed (hcp) and body centred cubic (bcc). 

2. Ionization Enthalpy

• The first ionization enthalpies of d-block elements lie between s-block and p-block elements because they are higher than those of s-block elements and lesser than those of p-block elements. The ionization enthalpy gradually increases with increase in atomic number along a given transition series though some irregularities are observed. For example - the ionization enthalpy values of Sc, Ti, V and Cr are fairly close to one another. Similarly, Fe, Co, Ni, and Cu have essentially identical values.

3. Oxidation States

• With the exception of a few elements, most of the d-block elements show more than one oxidation state as they show variable oxidation states.

• All transition elements possess variable oxidation states except the last element in each series.

• Minimum oxidation state = Total number of electrons in 4s lost. Maximum oxidation state = (Total number of electrons in 4s + number of unpaired electrons in 3d lost).

• Across the period in the periodic table, oxidation state increases and it is maximum at the center then decreases even if atomic number increases. The element which shows the highest oxidation state occurs in the middle or near the middle of the series and then decreases.

• The relative stabilities of some oxidation states can be explained on the basis of the rule of extra stability, according to which d0, d5 and d10 are stable configurations.

• Transition metals also show zero oxidation states in metal carbonyl complexes.

4. Ionic Radii

• The trend followed by the ionic radii is the same as that followed by atomic radii. Ionic radii of transition metals are different in different oxidation states. Transition metals have smaller ionic radii than typical elements from the same period.

5. Colour

• These form coloured ions due to presence of incompletely filled d-orbitals and unpaired electrons, they can undergo d-d transition by absorbing colour from visible region and radiating complementary colour, e.g. Cu2+(blue), V2+(violet), Cr3+(green). And Cu+(3d10), Zn2+(3d10) are white due to the presence of no unpaired electrons and cannot undergo d-d transition.

6. Catalytic Property

• Most transition elements and their compounds have good catalytic properties because they possess variable oxidation states and they provide a large surface area for the reactant to be absorbed.

7. Magnetic Properties

• In the first transition elements series the orbital angular magnetic moment is insignificant; the orbital contribution is quenched by the electric fields of the surrounding atoms so the magnetic moment is equal to the spin magnetic moment only.

μeff = √n (n + 2) BM (where n = number of unpaired electrons)

Maximum transition elements and compounds are paramagnetic due to presence of unpaired electrons.

8. Interstitial Compounds

• The transition metals form a large number of interstitial compounds in which small atoms such as hydrogen, carbon, boron and nitrogen occupy the empty spaces (interstitial sites) in their lattices. They are represented by formulae like TiC, TiH2, Mn4N, Fe3H, Fe3C etc. They are very hard and rigid and have high melting points which are higher than those of the pure metals. They show conductivity like that of pure metal and acquire chemical inertness.

9. Alloy Formation

• Transition metals have similar atomic radii and other characteristics, hence they form alloys very readily. Alloys are generally harder, have higher melting points and are more resistant to corrosion than the individual metals. Commonly used are ferrous alloys, the metals chromium, vanadium, molybdenum, tungsten and manganese are used in the formation of alloy steels and stainless steels.

Preparation and Properties of K2Cr2O7 and KMnO4 

Lanthanoids - Electronic configuration, oxidation states, chemical reactivity, lanthanoids contraction and its consequences 

Electronic Configuration: The atom  these elements have a common electronic structure of 6s2 but a varied occupancy of the 4f level. All tripositive ions (the most stable oxidation state of all lanthanoids) have electronic configurations of the form 4fn (n=1 to 14 with increasing atomic number).

Oxidation States 

• The sum of the first three ionization energies for each element are low. Thus the oxidation state (+3) is ionic and Ln3+ dominates the chemistry of these elements. The Ln2+ and Ln4+ ions that do occur are always less stable than Ln3+. Oxidation numbers (+II) and (+IV) do occur, particularly when they lead to :

1. a noble gas configuration e.g. Ce4+ (f 0)

2. a half filled f - shell, e.g. Eu2+ and Tb4+ (f7)

3. a completely filled f - shell, e.g. Yb2+ (f14).

Chemical Reactivity 

• Hydroxides of Lanthanoids:  The hydroxides are both ionic and basic in nature. They are less basic than Ca(OH)2 but more basic than the amphoteric Al(OH)3 . By adding NH4OH to aqueous solutions, the hydroxides Ln(OH)3 form gelatinous precipitate and. In dilute acids, the metals, oxides, and hydroxides all dissolve and form salts. Ln(OH)3 are sufficiently basic to absorb CO2 and create carbonates from the air. From Ce to Lu, the basicity diminishes as the ionic radius lowers. As a result, Ce(OH)3 is the most basic, while Lu(OH)3 is the least basic, falling somewhere between scandium and yttrium in terms of basic strength. The hydroxides of the subsequent elements dissolve in hot concentrated NaOH, creating complexes, demonstrating the decline in basic characteristics.

Lanthanoids Contraction and its Consequences

• Due to increased nuclear charge and electrons entering the inner (n-2) f orbital, the atomic size or ionic radii of tri-positive lanthanide ions drop continuously from La to Lu. Lanthanide contraction is the progressive decrease in size as the atomic number increases.

Actinoids - Electronic configuration, oxidation states and comparison with lanthanoids

Electronic configuration 

• Actinides are the second series of f-block elements with the electrical configuration [Rn]5f1-14 6d0-1 7s2. Because the energies of 5f and 6d electrons are similar, electrons enter the 5f orbital.

Oxidation States 

• Actinides possess a large number of oxidation states because of the very small energy gap between 5f, 6d and 7s subshells. So, all their electrons can take part in bond formation. The dominant oxidation state of these elements is +3 (similar to Lanthanides). Besides +3 state, Actinides also exhibit an oxidation state of +4. Some Actinides show higher oxidation states. The maximum oxidation state first increases upto the middle of the series and then decreases, e.g., it increases from + 4 from Th to +5, +6, and +7 for Pa, U and Np but decreases in the succeeding elements.

Comparison with Lanthanoids:

Difference Between d and f Block Elements: 

Solved Examples From the Chapter

Example 1: Lanthanide contraction is caused due to:

(a) appreciable shielding on outer electrons by 4f electrons from the nuclear charge

(b) the appreciable shielding on outer 'electrons by 5d electrons from the nuclear charge

(c) the same effective nuclear charge from Ce to Lu.

(d) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Solution: The decrease in the atomic, as well as ionic radii with an increase in the atomic number, is referred to as lanthanoid contraction. While increasing from La+3 to Lu+3 in a lanthanoid series, the size of the ion reduces. This reduction in the size in the series is called the lanthanoid contraction. This phenomenon arises because of the imperfect shielding of a 4f electron by a different election from the same subshell.

Therefore, the correct answer is (d).

Key Point to Remember: The atomic radius depends on effective nuclear charge.

Example 2: The spin only magnetic moment (in units of Bohr magneton) of Ni2+ in aqueous solution would be (At. No. of Ni = 28) :

(a) 2.84                    (b) 4.90                 (c) 0                      (d) 1.73

Solution:  Ni2+ has 3d8 configuration, therefore 2 unpaired electrons are present. 

Thus, magnetic moment = √n(n + 2) =  √8 = 2.84 B.M. (where n = number of unpaired electrons). Thus, the correct answer is option (a) 2.84

Key point to remember: The spin magnetic moment is,

μeff = √n (n + 2) BM (where n = number of unpaired electrons)

Solved Questions From The Previous Year Question Papers

Question 1:  Which one of the following ions exhibits d-dtransition and paramagnetism as well?

(a) CrO42–                                                                    (b) MnO4–

(c) Cr2O72–                                                                   (d) MnO42–

Solution: 

CrO42–     =       Cr6+       = [Ar];            Unpaired electron (n) = 0;  Diamagnetic

Cr2O72–    =       Cr6+       = [Ar];            Unpaired electron (n) = 0;  Diamagnetic

MnO42–   =       Mn6+     = [Ar] 3d1 ;    Unpaired electron (n) = 1;   Paramagnetic

MnO4–     =       Mn7+    = [Ar] ;            Unpaired electron (n) = 0;   Diamagnetic

Hence, the correct answer is option (d) MnO42–.

Trick: Paramagnetic behavior depends on the unpaired electrons.

Question 2: The calculated spin only magnetic moment 2+Cr ion is :

(a) 5.92 BM

(b) 2.84 BM

(c) 3.87 BM

(d) 4.90 BM

Solution: Cr2+ ion has 3d6 configuration, therefore 4 unpaired electrons are present. 

Thus, magnetic moment = √n(n + 2) (where n = number of unpaired electrons).

=  √(4(4+2)

= 4.90 B.M. 

Thus, the correct answer is option (d) 4.90 B.M

Trick: The spin only magnetic moment is approximately equal to the number of unpaired electrons as the formula for magnetic moment = √n(n + 2). 

Question 3: Identify the incorrect statement.

(a) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.

(b) The oxidation state of chromium in CrO42- and Cr2O72- are not the same.

(c) Cr2+(d4) is a stronger reducing agent than Fe2+ (d6) in water.

(d) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.

Solution: The size of H, C, and N is smaller in comparison to metals used in crystal lattices of metals such as iron. Therefore, the statement “Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals” is correct.

Standard electrode potential for Cr2+(d4) is less than that of Fe2+ (d6). So, Fe2+ (d6) has more tendency to reduce than Cr2+(d4). Therefore, the statement “Cr2+(d4) is a stronger reducing agent than Fe2+ (d6) in water” is correct.

The statement “The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes” is also true due to the surface area provided by transition metals and their compounds for catalytic activity.

The oxidation state of chromium in CrO42- and Cr2O72- is 6. Therefore, the statement, “The oxidation state of chromium in CrO42- and Cr2O72- are not the same” is incorrect.

The correct option is (c).

Trick: The transition metals have a tendency to form interstitial compounds and act as catalysts. Reducing power depends on standard electrode potential.

Practice Questions

Question 1: The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are 23, 24, 25 and 26 respectively. Which one of these may be expected to have the highest second ionization enthalpy?

(a) V                     (b) Cr                (c) Mn                     (d) Fe

Answer: (b) Cr

Question 2: What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

(a) Cr3+ and Cr2O72— are formed 

(b) Cr2O72— and H2O are formed

(c) Cr2O72— is reduced to + 3 state of Cr 

(d) Cr2O72— is oxidized to + 7 state of Cr

Answer: (b) Cr2O72— and H2O are formed

Conclusion

In this article, we studied the d- and f- Block Elements. And we studied electronic configuration of d and f block elements, oxidation states, magnetic properties, catalytic property, colour, alloy formation and the preparation and properties of some compounds. And also we know the physical and chemical properties of the d- and f- Block elements as well as their some common compounds. This article explained difference between d and f block elements.

Hence, this chapter is important not only for competitive exams like JEE or NEET but also in better understanding the study of d- and f- Block elements and their compounds.