Check Here the Most Important Questions of F-Block!!! (Download PDF)

By Neetesh Tiwari|Updated : August 22nd, 2021

Hello Gradians,

We hope you all are safe and healthy!

Are you looking for some short and reliable notes during your CSIR-NET preparations? Then, you have come to a perfect place!

While preparing for exams, revision is the key to success. Today, finding the study material for the exam preparation online is straightforward, but finding the best and authentic study material is very difficult. BYJU'S Exam Prep has always helped students with different phases of their preparation journey by providing time to time free series, free quizzes, etc. This time, we have come up with 'Important Questions on F-Block' for 'Chemical Sciences' to help you score good marks. So for the aspirants preparing for the CSIR NET Exam, these questions can be fruitful in the last-minute revision. These questions will help students to fetch more marks in the exams.

Our experienced subject-matter experts have meticulously designed this set of Important Questions on F-Block to give you the most standard set of study materials to be focused upon. After finishing every topic or component of the syllabus, it is recommended to solve questions either asked in the previous year papers or given in a mock. This article, including only questions, is aimed to make you understand and attempt a particular syllabus domain effectively. Read the full article and download the PDF.

Hello Gradians,

We hope you all are safe and healthy!

Are you looking for some short and reliable notes during your CSIR-NET preparations? Then, you have come to a perfect place!

While preparing for exams, revision is the key to success. Today, finding the study material for the exam preparation online is straightforward, but finding the best and authentic study material is very difficult. BYJU'S Exam Prep has always helped students with different phases of their preparation journey by providing time to time free series, free quizzes, etc. This time, we have come up with 'Important Questions on F-Block' for 'Chemical Sciences' to help you score good marks. So for the aspirants preparing for the CSIR NET Exam, these questions can be fruitful in the last-minute revision. These questions will help students to fetch more marks in the exams.

Our experienced subject-matter experts have meticulously designed this set of Important Questions on F-Block to give you the most standard set of study materials to be focused upon. After finishing every topic or component of the syllabus, it is recommended to solve questions either asked in the previous year papers or given in a mock. This article, including only questions, is aimed to make you understand and attempt a particular syllabus domain effectively. Read the full article and download the PDF.

Important Questions on F-Block

1. Due to lanthanide contraction- 

  1. All f-block ions have equal size. 
  2. All isoelectronic ions have equal size.
  3. Zr and Hf have equal sizes. 
  4. Fe, Co and Ni have equal size.

2. Which of the following statements concerning lanthanides elements is false? 

  1. Ionic radii of trivalent lanthanides steadily increase with the increase in atomic number. 
  2. Lanthanides are separated from one another by the ion-exchange method. 
  3. All lanthanides are highly dense metals. 
  4. The characteristic oxidation state of lanthanides elements is +3. 

3. Which of the following statements is not correct? 

  1. La(OH)3 is less basic than Lu(OH)3.
  2. The atomic radius of Zr and Hf are the same because of lanthanide contraction. 
  3. In the lanthanide series ionic radius of Ln3+ ion decreases. 
  4. La is actually an element of the transition series rather than lanthanide.

4. The lanthanide contraction is responsible for the fact that-

  1. Zr and Zn have the same oxidation state
  2. Zn and Y have about the same radii. 
  3. Zr and Nb have similar oxidation states. 
  4. Zr and Hf have about the same radii. 

5. The electronic configuration of Gd is? 

  1. [Xe]4f8 5d9 6s2 
  2. [Xe]4f7 5d1 6s2 
  3. [Xe]4f6 5d2 6s2 
  4. [Xe]4f3 5d3 6s2

6. Which one of the following elements shows the maximum number of different oxidation states in its compound?

  1. La 
  2. Eu 
  3. Lu 
  4. Gd 

7. Cerium (Z=58) is an important member of the lanthanide series. Which of the following statements about Cerium is incorrect? 

  1. Cerium IV acts as an oxidizing agent. 
  2. The +4 oxidation state of cerium is known in solution. 
  3. The +3 oxidation state is more stable than the +4 state. 
  4. The common oxidation states of cerium are +3 and +4.

8. Which of the following factors may be regarded as the main cause of lanthanide contraction: 

  1. Poor Shielding of one of the 4f electrons by another in the subshell. 
  2. Greater shielding of 5f electrons by 4f electrons. 
  3. Poor shielding of 5d electrons by 4f electrons. 
  4. Effective shielding of one 4f electron by another in the subshell. 

9. The lanthanide contraction is due to: 

  1. Filling of 5d before 4f 
  2. Filling of 4d before 4f 
  3. Filling of 4f before 5d
  4. Filling of 4f before 4d 

10. Ln3+ (trivalent lanthanide ions) have an electronic configuration. 

  1. [Xe]4f1 to [Xe]4f14 
  2. [Xe]4f1 4f1 to [Xe]4d1 4f14 
  3. [Xe]4d2 4f0 to [Xe]4d1 4f14 
  4. [Xe]4f0 to [Xe]4f14

Answer Key

 

  1. C
  2. A
  3. A
  4. D
  5. B
  6. B
  7. B
  8. C
  9. A
  10. A

Solutions

Solution 1:

Zirconium is [Kr] 4d² 5s² 

Hafnium is [Xe] 4f¹⁴ 5d² 6s² 

Due to Lanthanide contraction; there’s poor shielding of f-subshell, due to which the attraction of nucleus on outer shell electrons increases and thus the size of Hf decreases and size of Zr and Hf becomes almost the same. This is the reason why both of them have similar properties.

Solution 2:

  • The ionic radii of trivalent lanthanides decrease with the increasing atomic number due to the poor shielding effect with an increase in the number of electrons.
  • A process known as ion exchange is then used to separate the lanthanides from each other. In this process, a solution of the lanthanides in ionic, soluble form is passed down a long column containing a resin. The lanthanide ions "stick" to the resin with various strengths based on their ion size. The lanthanum ion, being smallest, binds most tightly to the resin, whereas the largest ion, lutetium, binds the weakest. The lanthanides are then washed out of the ion exchange column with various solutions, emerging one at a time, and so are separated. Each is then mixed with acid, precipitated as the oxalate compound, and then heated to form the oxide.
  • All lanthanides are highly dense metals. The lanthanides have a high atomic number as well as atomic mass. Lutetium, the last member of the lanthanide series, is the densest lanthanide with the highest melting point.
  • Lanthanides show variable oxidation states. They also show +2, +3, and +4 oxidation states. But the most stable oxidation state of Lanthanides is +3.

Solution 3:

 

  • Due to Lanthanide contraction, as the size of lanthanide ions decreases from La+3 to Lu+3, the covalent character of the hydroxides increases and hence the basic strength decreases. So, La(OH)3 is the most basic and Lu(OH)3 is the least basic.

Solution 4:

Zirconium is [Kr] 4d² 5s² 

Hafnium is [Xe] 4f¹⁴ 5d² 6s² 

Due to Lanthanide contraction; there’s poor shielding of f-subshell, due to which the attraction of nucleus on outer shell electrons increases and thus the size of Hf decreases and size of Zr and Hf becomes almost the same. This is the reason why both of them have similar properties.

Solution 5:

The electronic configuration of Gd (Z = 64) is [Xe] 4f75d16s2

Solution 6:

Element La shows only one oxidation state (+3)

Element Eu exhibit two oxidation states (+2 and +3)

Element Lu exhibit only one oxidation state (+3)

Element Gd exhibit only one oxidation state (+3) 

 

Solution 7:

  • Cerium (IV) is a powerful oxidizing agent which finds immense applications in the analysis of several pharmaceuticals. 
  • The +4 oxidation state of Cerium is not known in solutions.
  • [Xe] 4f¹ 5d¹ 6s²

The +3 oxidation state of Cerium is more stable than the +4 oxidation state due to the greater stabilization of 4f orbital than 5d and 6s orbitals i.e., 4f>5d>6s.

  • We know that Cerium is a member of Lanthanide. [Xe] 4f¹ 5d¹ 6s²

So to attain the stability of noble gas Xenon (Z=54), Cerium has to lose 4 electrons to attain noble gas stability.
Lanthanides exhibit different oxidation states i.e. +2,+3,and+4
When Cerium shows a +2 oxidation state then the electronic configuration is 4f1 5d1 6s0
When Cerium shows a +3 oxidation state the electronic configuration is 4f1 5d0 6s0
When Cerium shows a +4 oxidation state the electronic configuration is 4f0 5d0 6s0
The +3 oxidation state of Cerium is more stable than the +4 oxidation state due to the greater stabilization of 4f orbital than 5d and 6s orbitals i.e. 4f > 5d > 6s.

As 4f orbital is closest to the nucleus, the attraction of electrons is more in 4f orbital and thus penetration of electrons from 4f is difficult. As it requires more Ionization energy for penetration of electrons from 4f orbital.
That’s why Cerium(IV)acts as an oxidising agent.
While penetrating electrons from 5d and 6s orbitals require less ionization energy. And because of the high penetration energy of 4f orbital, the +3 oxidation state is the most stable one among all other oxidation states.
So from the above explanation, we can say that the most common oxidation state of Cerium is +3 and +4.

Solution 8:

The cause of the lanthanide contraction is the poor shielding effect of 4f orbital due to which the electrons of s-orbitals attract more towards the nucleus due to which the size of the atoms decreases.

Solution 9:

The lanthanide contraction is due to the filling of 5d before 4f.

Solution 10:

 Ln3+ (trivalent lanthanide ions) have electronic configuration [Xe]4f1 to [Xe]4f14 

Since lanthanides have configuration of [Xe]4f1-14 5d1 6s2

So, their trivalent ion has a configuration of [Xe]4f1 to [Xe]4f14 

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Neetesh TiwariNeetesh TiwariMember since Aug 2020
Working Experience of more than 2 years in Education Industry. Currently associated to CSIR-NET at Gradeup. Guiding students to crack the exam and fulfil their dreams.
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