NCERT solutions for class 12 chemistry chapter 8 The d and f block elements - The general and physical properties of d-block and f-block elements have been discussed here in this chapter and along with that NCERT solutions for class 12 chemistry chapter 8 The d and f block elements has all the answers pertaining to questions based on such properties. The d-block elements are elements of groups 3 to 12 of the periodic table . The elements of the f -block are those in which the 4f and 5f orbitals are filled. Also, you will study that these elements are formal members of group 3 from which they have been taken out to form a separate f-block of the periodic table. In NCERT solutions for class 12 chemistry chapter 8 The d and f block elements, you will find all the topic wise questions as well as the exercise questions which will make your learning easier. The d-block elements are often referred to as transition metals and f-block elements as inner transition metals. There are three series of the d block elements, 3d series (Sc to Zn), 4d se ries (Y to Cd) and 5d series (La to Hg, except Ce to Lu). The fourth 6d series which begins with Ac is still incomplete. The two series of the f-block elements, (4f and 5f) are known as lanthanoids and actinoids respectively. You will find all the NCERT solutions for class 12 chemistry chapter 8 The d and f block elements here in this article once you scroll down.
This chapter is not lengthy like chapter 7 the p-block elements but it is an important chapter because it holds 5 marks out of 70 in the CBSE Board exam. The NCERT solutions are available for other classes and other subjects as well which are going to help you in competitive exams like JEE, NEET, BITS, VITEE and KVPY etc. In NCERT solutions for class 12 chemistry chapter 8 The d a nd f block elements, there are eight sub-topics which cover essential concepts of d and f-block elements and to check your knowledge, 38 questions are given in the exercise at the end of the chapter. The NCERT solutions for class 12 chemistry chapter 8 The d and f block elements are created by subject experts to give a clear understanding of the concept used to solve the question. By referring to the NCERT solutions for class 12 , students can understand all the important concepts and practice questions well enough before their examination.
In chapter 8 The d and f block elements you will first read about the occurrence, electronic configuration and general characteristics of transition elements and discuss important trends in the properties of the first row (3d) transition metals, preparation and properties of some important compounds. After this, the chapter will talk about oxidation states, electronic configurations and chemical reactivity of the inner transition metals.
Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 8 The d and f block elements-
8.1 Position in the Periodic Table
8.2 Electronic Configurations of the d-Block Elements
8.3 General Properties of the Transition Elements (d-Block)
8.4 Some Important Compounds of Transition Elements
8.5 The Lanthanoids
8.6 The Actinoids
8.7 Some Applications of d- and f-Block Elements
Find Class 12 Chemistry Chapter 8 The d and f block elements below:
Solutions to In-Text Questions Ex 8.1 to 8.10
Question 8.7 Which is a stronger reducing agent or and why ?
Cr +2 is a better reducing agent as compared to Fe +2 , as this can be explained on the basis of standard electrode potential of Cr +2 (-0.41) and Fe +2 (+0.77).
It can also be explained on the basis of their electronic configuration achieved. Cr +2 obtained d 3 configuration whereas Fe +2 gets d 5 configuration upon reduction. It is known that d 3 is more stable than d 5 . So Cr +2 is a better reducing agent as compared to Fe +2 .
NCERT Solutions for Class 12 Chemistry Chapter 8 The d and f block elements- Exercise Questions
Question 8.1(i) Write down the electronic configuration of:
Chromium has atomic number 24. So, nearest noble gas element is Argon ( )
So electronic configuration of =
Question 8.3 Explain briefly how state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
According to our observation, except scandium, all other elements of the first row shows +2 oxidation state. On moving from Sc to Mn the atomic number increases from 21 to 25 and also the increasing number of electrons in 3d orbitals from . when metals lose two electrons from its 4s orbital then they achieve +2 oxidation state. Since the number of d electrons in (+2) state increases from , the stability of the +2 oxidation state increases as d-orbitals is becoming more and more half filled.
Mn(+2) has configuration, which is half filled (it makes it highly stable)
Question 8.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms :
Vanadium (atomic number- 23)
E.C = ,
So the stable oxidation states are (+2, +3, +4, +5)
Manganese (atomic number = 25)
E.C = ,
So the stable oxidation state are (+2, +4, +6, +7)
chromium (atomic number = 24)
E.C = ,
So the stable oxidation state are (+3, +4, +6)
No elements has electronic configuration in their ground state.
Question 8.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
On moving along the lanthanoid series, the atomic number is gradually increased by one. It means the no. of electrons and protons of the atom is also increases by one. And because of it the effective nuclear charge increases (electrons are adding in the same shell, and the nuclear attraction overcomes the interelectronic repulsion due to adding of a proton). Also, with the increase in atomic number, the number of electrons in orbital also increases. Due to the poor shielding effect of the electrons, the effective nuclear charge experienced by an outer electron is increased, and also the attraction of the nucleus for the outermost electron is increased. As a result, there is a gradual decrease in the atomic size as an increase in atomic number. This is known as lanthanoid contraction.
Consequences of Lanthanoid contraction-
- Similarities in the properties of second and third transition series
- Separation of lanthanoid can be possible due to LC.
- Due to LC, there is variation in basic strength of hydroxide of lanthanoid. (basic strength decrease from ).
Question 8.11(iii) Explain giving reasons:
(iii) The transition metals generally form coloured compounds.
Most of the complex of transition elements are coloured. This is due to the absorption of radiation from visible light region to excite the electrons from its one position to another position in d-orbitals. In the presence of ligands, d-orbitals split into two sets of different orbital energies. Here transition of electron takes place and emit radiation which falls on the visible light region.
Question 8.17(ii) For and systems the values for some metals are as follows:
Use this data to comment upon:
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
From the values of , the order of oxidation of the given metal to the divalent cation is-
Question 8.19 Compare the stability of +2 oxidation state for the elements of the first transition series.
According to our observation, except scandium, all other elements of the first row shows +2 oxidation state. On moving from to the number of oxidation states increases but from to number of oxidation states decreases due to a decrease in unpaired electrons. The stability of +2 oxidation state increase on moving from to due to increase in difficulty level of removal of the third electron from d -orbital.
| +2 |
| +2 |
| +2 |
| +2 |
| +2 |
Question 8.20(iv) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(iv) chemical reactivity.
In lanthanoid series, an earlier member of the series is more reactive, and that is comparable to . with an increase in atomic number, lanthanoids starts behaving similar to aluminium.
Actinoids are highly reactive metals, especially when they are finally divided. When we add them into the water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperature. Alkalies have no action on these actinoids metals
Question 8.26(i) Indicate the steps in the preparation of:
(i) from chromite ore .
(i) P otassium dichromate is obtained from the fusion of chromite ore with sodium and potassium carbonate in the free supply of air.
(ii) Sodium chromate is filtered and acidified with sulphuric acid ( ) to form sodium dichromate, can be crystallised
(iii) Sodium dichromate is more soluble than potassium dichromate. So, treat the solution of dichromate with the potassium chloride( )
The chromate and dichromate are interconvertible in aqueous solution at pH 4
Structures of chromate and dichromate ion
Question 8.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
E lements of the first transition series possess many properties different from those of heavier transition elements in the following ways-
- The atomic size of the 1st transition series is smaller than those of 2nd and 3rd series elements. But due to lanthanoid contraction, atomic size of the 2nd series elements are nearly the same as 3rd series element of the corresponding same vertical group.
- In 1st transition series +2 and +3 oxidation states are more common but in the 2nd and 3rd series higher oxidation states are more common.
- The enthalpy of atomisation of first series elements is lower than 2nd and 3rd series elements.
- The melting and boiling point of the 1st transition series is less than that of heavier metals. This is because of strong metallic bonding in heavier metals.
Question 8.38 What can be inferred from the magnetic moment values of the following complex species ?
Example Magnetic Moment (BM)
Magnetic moment is given as -
Putting the value on n = 1, 2, 3, 4, 5 (number of unpaired electrons in d-orbital)
we get the value of are 1.732, 2.83, 3.87, 4.899, 5.92 respectively.
By comparing with our calculation we get the values n nearest to 1. It means, in above compound d-orbital has one unpaired electron( ), which means is astrong field ligand that cause force pairing of the electron.
After comparing with our calculation the nearest value of n = 4. Here iron is in +2 oxidation state ( configuration). So, we can say that is a weak field ligand, which not cause any force pairing.
By observing we get the nearest value of n is 5. So, in this complex Manganese has configuration. So, we conclude that ligand does not cause any force pairing and hence it is a weak ligand.
NCERT Solutions Class 12 Chemistry
NCERT Solutions for Class 12 Subject wise
Benefits of NCERT solutions for class 12 chemistry chapter 8 The d and f block elements
- The comprehensive answers given in the NCERT solutions for class 12 chemistry chapter 8 The d and f block elements will help you to understand chapter easily.
- Revision will be easy because with the help of the detailed solutions you will always remember the concepts and get very good marks in your class.
- Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 12 chemistry and you are stress-free.
If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.