NCERT Solutions Class 9 Science Chapter 4

NCERT Solutions Class 9 Science Chapter 4: Structure of the Atom

Science plays a vital role in our life. For students who are desirous of building a career in the science and engineering fields, Class 9 will be a transitioning phase. It will prepare them for the upcoming board examinations and entrance tests. Science opens many opportunities for students, and it demands a fair understanding of the concepts.

Class 9 Science Chapter 4: Structure of the Atom starts with an introduction to the infinite world of atoms. Here, students will learn new concepts and theories designed by many researchers. It will be a fascinating experience to learn the features of subatomic particles and their behaviour in an atom. This Chapter plays a vital role as it lays the foundation for the upcoming Chapters in this Class. 

Students can visit the Extramarks website and access NCERT Solutions for Class 9 Science Chapter 4. Extramarks offers theoretical notes, illustrations, Examples and previous years’ papers. With the help of their NCERT Solutions, students can revise the complete syllabus and score high marks in board examinations. The language used while preparing the NCERT Solutions Class 9 Science Chapter 4 is proficient and aligns with the conceptual understanding of the Class 9 students. 

Extramarks is one of the leading  online learning platforms trusted   by students who  refer to their study materials. NCERT Solutions Class 9 Chapter 4 are prepared by subject matter experts keeping in mind the latest CBSE updates regarding the examination pattern. These solutions are useful for both the teachers and the students and they can be accessed anywhere without much hassle.

Students can visit the Extramarks website to get the latest updates  related to the CBSE syllabus. They can also refer to the Solutions of other  classes, including NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 11 and NCERT Solutions Class 12. 

Key Topics  covered In NCERT Solutions Class 9 Science Chapter 4

Extramarks NCERT Solutions Class 9 Science Chapter 4 covers all  the topics in  detail and answers the questions mentioned in the Class 9 Science Chapter 4. It will help students clear their basic concepts. Academic experts with many years of experience have prepared  NCERT Solutions. With the help of NCERT Solutions Class 9 Science Chapter 4, students can understand the concepts related to the structure of the atom, its discovery, and orbital shells. 

Some of the key sub-topics in NCERT Solutions Class 9 Science Chapter 4 are as follows: 

Sections Sub-topics
4.1 Charged Particles in Matter 
4.2 The Structure of an Atom 
4.3 How are Electrons Distributed in Different Orbits?
4.4 Valency
4.5  Atomic Number and Mass Number
4.6  Isotopes 

Charged Particles in Matter 

The matter is defined as any substance which has mass and occupies space. Besides, all matter comprises the smallest unit, the “Atom”. An atom is composed of protons and electrons, mutually balancing their charges. In addition, the protons were in the atom’s interior, whereas electrons could easily be removed but not protons. 

The Structure of an Atom

As per Dalton’s atomic theory, the atom was indivisible and indestructible. However, the discovery of two fundamental particles inside the atom usually leads to the failure of this aspect of Dalton’s atomic theory. Later, it was considered necessary to know how electrons and protons are arranged within an atom. Therefore, the scientists proposed various atomic models. At the same time, J.J. Thomson was the first to propose a model for the structure of an atom. 

  1. i) Thomson’s model of an atom 

  • Based on Thomson,

(i) An atom is composed of a positively charged sphere, and electrons are contained within. 

(ii) The magnitudes of positive and negative charges are the same. Therefore, the atom as a whole is electrically neutral.

  • The first theory of the atom is proposed and considered.
  • The edible red-coloured portion of watermelon is compared to the atom’s positive charge.
  • The black seeds of the watermelon are contrasted with the electrons embedded in it.
  1. ii) Rutherford’s model of an atom

In an experiment carried out by Rutherford, the alpha particles were made to fall on a thin gold foil. Based on it Rutherford concluded that:

  • A significant portion of α-particles hurled towards the gold sheet went through without deflection. Therefore, the majority of the space inside an atom is empty.
  • α-particles, in some cases, were dispersed due to the gold sheets’ small angles, and consequently, the positive charge inside an atom was not evenly dispersed.
  • Very few particles were deflected from their path, indicating that the positive charge of the atoms occupies very little space.
  • The majority of the α-particles were deflected, except a few particles with nearly 180-degree angles of deflection. The volume of positively charged particles within an atom is small compared to the atom’s overall amount.

Rutherford’s Model of an Atom:

Rutherford concluded the following results from his model of the atom that he developed from the a-particle scattering test; he defined the nuclear model of an atom:

  • It is possible to find a positively charged centre inside an atom, referred to as the nucleus. The majority of the mass in an atom is contained within the nucleus.
  • The electrons move around the nucleus in precise orbits.
  • The nucleus size is very small compared to the size of the atom.

iii) Bohr’s model of an atom

Bohr developed these theories to counter the criticisms of Rutherford’s model.

  • Electrons orbit around the nucleus in stable orbits and do not emit radiant energy. Each orbit is characterised by a specific energy, known as an energy shell or energy level.
  • An energy level or orbit is identified as K L or N shells. When an electron is at the lowest energy level, it is believed to be in the ground state.
  • An electron releases or absorbs energy when it moves from one orbital or energy level to the next.
  • It releases energy if it jumps from a more energy-rich level to a lower one. It absorbs energy as it moves from a low energy level to a higher level.

  1. iv) Neutrons

Later, J. Chadwick discovered another atomic particle with no charge, and its mass was nearly equal to that of a proton. Eventually, it was named Neutron. Neutrons are usually present in the nucleus of all atoms. However, they are not present in hydrogen. It is represented as “n”. The mass of an atom is therefore given by the sum of the masses of protons and neutrons present in the nucleus. 

How are Electrons Distributed in Different Orbits?

Bohr and Bury suggested the distribution of electrons into different orbits of an atom. Below are the rules for writing the number of electrons in different energy levels: 

  1. i) The maximum number of electrons which appear in a shell is given by the formula square of 2n, where “n” represents the number of energy level index. 
  2. ii) The maximum number of electrons accommodated in the outermost orbit is 8. 

iii) Electrons cannot be accommodated in a given shell, only on the condition that the inner shells are filled. 

Valency

The electrons in the outermost shells of atoms are known as electrons that are valence.

The electrons in the outermost shell of an atom are known as valence electrons. The key features of valency elaborated in NCERT Solutions Class 9 Science Chapter 4 are: 

  • The ability to combine atoms of the same elements and create molecules or elements is called the valency of the atom.
  • Atoms of elements with the outermost shell filled exhibit only a small amount of chemical activity.
  • Their capacity or valency is 0.
  • For instance, we know that the number of electrons in the outermost hydrogen shell is one, and in magnesium, it’s 2.
  • Thus, the valency of hydrogen is one since it can quickly lose one electron before becoming stable.
  • However, magnesium has a number 2 since it can shed two electrons in minutes and achieve stability.

Atomic Number and Mass Number 

  1. i) Atomic Number 

 Protons are present in the nucleus of an atom; it is the number of protons in an atom that helps to determine. The atomic number is denoted by “Z”. Each atom of an element has the same atomic number, and the number of protons defines the element. 

  1. ii) Mass Number 

Protons and neutrons reside in the nucleus. Therefore, the mass number represents the sum of these neutrons and protons.

Isotopes 

Isobars are the atoms (nuclides) of various chemical elements that differ in chemical properties; however, they share similar physical properties. Therefore, we can define isobars as elements with different chemical numbers but the same mass.

Isotopes are those elements where the number of neutrons is different and the number of protons is equal. Based on the above definition of atomic mass and the numbers in the atomic structure, we can conclude that isotopes refer to elements with the same atomic numbers and different mass numbers.

The difference between isotopes and isobars is elaborated in NCERT Solutions Class 9 Science Chapter 4: 

Isobars Isotopes
  • Isobars are chemical elements which have the same mass. 
  • The atomic masses are equal 
  • Often the physical properties are similar 
  • Atomic numbers are different 
  • Chemical elements are different.
  • Isotopes are the different atomic structures of the same element.
  • The atomic masses are different
  • Usually, physical properties are different
  • The atomic number is the same
  • Same chemical elements have different mass numbers.

NCERT Solutions Class 9 Science Chapter 4: Exercise &  Solutions

Subject matter experts curate Extramarks NCERT Solutions by taking a concept-based approach. Students can access  solutions by registering on the Extramarks website. NCERT Solutions Class 9 Science Chapter 4 has explanations in an easy-to-understand  language to provide the best solutions to clear their doubts and strengthen their base. 

Students can utilise the NCERT Solutions for Class 9 Science Chapter 4 as a quick reference to understand complex topics with ease. 

Click on the following links to access the Class 9 Science Chapter 4 Question Answer for different exercise questions:

  • Chapter 4: Exercise 4.1 – Charged particles in matter 2 Question ( 2 short)
  • Chapter 4: Exercise 4.2 – The structure of an atom 4 Question ( 4 short)
  • Chapter 4: Exercise 4.2.4 – Neutrons 2 Question ( 2 short)
  • Chapter 4: Exercise 4.3 – How are electrons distributed
  • Chapter 4: Exercise 4.4 – Valency 1 Question ( 1 long)
  • Chapter 4: Exercise 4.5 – Atomic number and mass number 2 Question ( 2 long)
  • Chapter 4: Exercise 4.6 – Isotopes 2 Question ( 1 long, one short)
  • Chapter 4: Exercise Solutions – 19 Questions ( 6 long, nine short, 4 MCQ)

Along with this, students can also check additional study materials provided by Extramarks for Class 9: 

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11
  • NCERT Solutions Class 12

NCERT Exemplar Class 9 Science

Science is both theory and practice. Students are required to understand the theories and concepts well enough to grasp the topics covered in each chapter. Regular revision of concepts and practising questions and answers is crucial for students to retain the knowledge they have studied.

NCERT exemplars’ questions require high-order thinking for conceptual questions, multiple-choice questions, fill inthe blanks, and matching the following and true or false types of questions. 

The exemplar also contains:

  • Objective questions.
  • Short-answer type questions.
  • Long type questions.

For students who want to take part in competitive Examinations,  NCERT Solutions for Class 9 Science Chapter 4 has questions to help them in  their preparation. Students can explore various topics and questions as well as their answers.

Key Features of NCERT Solutions Class 9 Science Chapter 4

Extramarks has a dedicated team of subject matter experts who provide appropriate Solutions for every topic. Students who have difficulty solving the questions can refer to the answers offered in the NCERT Solutions. 

The characteristic features of Extramarks NCERT Solutions Class 9 Science Chapter 4 are: 

  • Questions in NCERT Solutions Class 9 Science Chapter 4 have been answered in  a concise manner and  written in a simple language by the subject matter experts..
  • Key concepts and fundamentals of Chapter 4 have been provided in an easy to learn manner with the help of illustrations and tables wherever necessary. 
  • Class 9 Science Chapter 4 question and answers are elaborated simply with the pictorial presentation. Students will be able to understand every concept and answer any question easily. This encourages the students to master the topic and it increases their confidence level in achieving a higher grade. 

Q.1 Compare the properties of electrons, protons and neutrons.

Answer

Electron Proton Neutron
They are present outside the nucleus. They are present in the nucleus. They are present in the nucleus.
They are negatively charged. They are positively charged. They are neutral.
They are represented as e–. They are represented as p+. They are represented as n0.

Q.2 What are the limitations of J.J. Thomson’s model of the atom?

Answer

According to J.J. Thomson’s model of an atom, an atom consists of a positively charged sphere and the electrons are embedded in it. However, later it was found that the positively charged particles are present at the centre of the atom (nucleus) and electrons revolve around the nucleus.

Q.3 What are the limitations of Rutherford’s model of the atom?

Answer

The limitations of Rutherford’s model of the atom are:

  • According to the electromagnetic theory, if a charged particle moves around another charged particle, it accelerates and continuously loses energy in the form of the radiant energy. Loss of energy would slow down the speed of the electron and eventually, the electron would fall into the nucleus. However, such a collapse did not occur and Rutherford’s model was unable to explain the reason behind it.
  • It does not say anything about the distribution of electrons around the nucleus and the energy of electrons.

Q.4 Describe Bohr’s model of the atom.

Answer

  1. Electrons revolve around the nucleus in certain discrete orbits (energy levels) without losing any energy.
  2. An electron present in a particular orbit possesses a definite amount of energy.
  3. An electron does not radiate energy (lose energy) even though it accelerates the motion around the nucleus.

Q.5 Compare all the proposed models of an atom given in this chapter.

Answer

Thomson’s Model

Rutherford’s Model

Bohr’s Model
  1. An atom consists of a positively charged sphere and the electrons are embedded in it.
  1. There is a positively charged centre in the atom, which is called the nucleus. Almost the whole mass of an atom resides in the nucleus.
  1. The electrons revolve around the nucleus in well-defined orbits.
  1. The size of a nucleus is very small as compared to the size of an atom.
  1. Electrons revolve around the nucleus in certain discrete orbits (energy levels) without losing any energy.
  2. An electron present in a particular orbit possesses a definite amount of energy.
  3. An electron does not radiate energy (lose energy) even though it accelerates the motion around the nucleus.

Q.6 Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.

Answer

The rules for distributing the electrons in various shells are given below.

1.The maximum number of electrons present in a shell (n) is given by the formula 2n2, where ‘n’ is an integer.

For example,

Maximum number of electrons in K shell is 2(1)2 = 2 electrons

Maximum number of electrons in L shell is 2(2)2 = 8 electrons

2.The maximum number of electrons in the outermost orbit of an atom is 8.

For example,

Total number of electrons 13

Expected distribution is 2,11 but actual distribution is 2,8,3

3.Electrons are not accommodated in a given shell, unless the inner shells are filled, i.e., shells are filled in a step-wise manner.

For example,

In order to accommodate 13 electrons in K, L and M shell, the correct distribution is 2,8,3 and not 2,7,4.

Q.7 Define valency by taking examples of silicon and oxygen.

Answer

Valency is the number of electrons gained, lost or shared by an atom so as to make the octet of the electrons in the outermost shell.

For example,

  1. Atomic number of Silicon is 14.

Electronic configuration of Silicon is 2,8,4.

Since the number of valence electrons is 4 therefore, valency of Silicon is 4.

  1. Atomic number of Oxygen is 8.

Electronic configuration of Oxygen is 2,6.

Since the number of valence electrons is more than 4 therefore, valency of Oxygen is 8 – 6 i.e., 2.

Q.8 Explain with examples
(i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.

Answer

(i) Atomic number:

  • It represents the total number of protons present in the nucleus of an atom.
  • It is denoted by Z.
  • All the atoms of an element have the same atomic number.
  • The number of electrons and protons in an atom are equal.

For example,

Atomic number of Boron is 5

Atomic number of Aluminium is 13

Atomic number of Argon is 18

(ii) Mass number:

  • It represents the total number of protons and neutrons present in the nucleus of an atom.
  • It is denoted by A.
  • The mass of an atom is due to the protons and neutrons present in the nucleus of an atom.
  • Protons and neutrons are collectively called the nucleons.

For example,

Mass number of Boron is 11

Mass number of Aluminium is 27

Mass number of Argon is 40

(iii) Isotopes:

  • These are the atoms of the same element that have the same atomic number but different mass number.
  • Isotopes of a particular element have the same chemical properties but different physical properties.

For example,

Carbon has three isotopes 12C, 13C, 14C

Boron has two isotopes 10B, 11B

(iv) Isobars:

  • These are the atoms that have the same mass numbers but different atomic numbers.
  • Isobars have different chemical as well as physical properties.

For example,

Ar40, 19K40, 20Ca40

C14, 7N14

Uses of Isotopes

  1. An isotope of uranium is used as fuel in the nuclear reactor.
  2. An isotope of cobalt is used in the treatment of cancer.

Q.9 Na+ has completely filled K and L shells. Explain.

Answer

Atomic number of Sodium (Na) is 11.

Electronic configuration of Sodium (Na) is 2,8,1.

In order to complete its octet it will lose its valence electron and form Sodium ion (Na+).

Electronic configuration of sodium ion (Na+) is 2,8.

In the electronic configuration of sodium ion both K and L shell are completely filled.

Q.10 If bromine atom is available in the form of, say, two isotopes 35Br79 (49.7%) and 35Br81 (50.3%), calculate the average atomic mass of bromine atom.

Answer

Atomic mass of bromine = 79u x 49.7 + 81u x 50.3 49.7+50.3 = 8000.6 100 u = 80.0 u MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiaabgeacaqG0bGaae4Baiaab2gacaqGPbGaae4yaiaabccacaqGTbGaaeyyaiaabohacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabkgacaqGYbGaae4Baiaab2gacaqGPbGaaeOBaiaabwgacaqGGaGaaeypaiaabccadaWcaaqaaiaabEdacaqG5aGaaeyDaiaabccacaqG4bGaaeiiaiaabsdacaqG5aGaaeOlaiaabEdacaqGGaGaae4kaiaabccacaqG4aGaaeymaiaabwhacaqGGaGaaeiEaiaabccacaqG1aGaaeimaiaab6cacaqGZaaabaGaaeinaiaabMdacaqGUaGaae4naiaabUcacaqG1aGaaeimaiaab6cacaqGZaaaaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeypaiaabccadaWcaaqaaiaabIdacaqGWaGaaeimaiaabcdacaqGUaGaaeOnaaqaaiaabgdacaqGWaGaaeimaaaacaqG1baabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG9aGaaeiiaiaabIdacaqGWaGaaeOlaiaabcdacaqGGaGaaeyDaaaaaa@A0AE@

Q.11 The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16/8X and 18/8 X in the sample?

Answer

Let the percentage of the isotope X 8 16 be y 1 Hence, the percentage of the isotope X 8 18 is = 100 – y 1 Average atomic mass of X = (Percentage of X 8 16 x 16u) + (100 – % of X 8 16 ) x 18u 100 or 16.2 u = (y 1 x 16 u) + (100 – y 1 ) x 18 u 100 or 1620 u = 16 y 1 u + 1800 u –18 y 1 u or 18 y 1 u – 16 y 1 u = 1800u – 1620 u or 2y 1 u = 180 u or y 1 = 180 2 = 90 Now, Percentage of X 8 16 = 90 % Percentage of X 8 18 = 100 – 90 = 10 % MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiaabYeacaqGLbGaaeiDaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiCaiaabwgacaqGYbGaae4yaiaabwgacaqGUbGaaeiDaiaabggacaqGNbGaaeyzaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyAaiaabohacaqGVbGaaeiDaiaab+gacaqGWbGaaeyzaiaabccadaqhbaWcbaGaaeioaaqaaiaabgdacaqG2aaaaOGaaeiwaiaabccacaqGIbGaaeyzaiaabccacaqG5bWaaSraaSqaaiaabgdaaeqaaaGcbaGaaeisaiaabwgacaqGUbGaae4yaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaaeyzaiaabkhacaqGJbGaaeyzaiaab6gacaqG0bGaaeyyaiaabEgacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGPbGaae4Caiaab+gacaqG0bGaae4BaiaabchacaqGLbGaaeiiamaaDeaaleaacaqG4aaabaGaaeymaiaabIdaaaGccaqGybGaaeiiaiaabMgacaqGZbGaaeiiaiaab2dacaqGGaGaaeymaiaabcdacaqGWaGaaeiiaiaabobicaqGGaGaaeyEamaaBeaaleaacaqGXaaabeaakiaabccaaeaacaqGbbGaaeODaiaabwgacaqGYbGaaeyyaiaabEgacaqGLbGaaeiiaiaabggacaqG0bGaae4Baiaab2gacaqGPbGaae4yaiaabccacaqGTbGaaeyyaiaabohacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabIfacaqGGaGaaeypaiaabccadaWcaaqaaiaabIcacaqGqbGaaeyzaiaabkhacaqGJbGaaeyzaiaab6gacaqG0bGaaeyyaiaabEgacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiamaaDeaaleaacaqG4aaabaGaaeymaiaabAdaaaGccaqGybGaaeiiaiaabIhacaqGGaGaaeymaiaabAdacaqG1bGaaeykaiaabccacaqGRaGaaeiiaiaabIcacaqGXaGaaeimaiaabcdacaqGGaGaae4eGiaabccacaqGLaGaaeiiaiaab+gacaqGMbGaaeiiamaaDeaaleaacaqG4aaabaGaaeymaiaabAdaaaGccaqGybGaaeykaiaabccacaqG4bGaaeiiaiaabgdacaqG4aGaaeyDaaqaaiaabgdacaqGWaGaaeimaaaaaeaacaqGVbGaaeOCaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGXaGaaeOnaiaab6cacaqGYaGaaeiiaiaabwhacaqGGaGaaeiiaiaabccacaqG9aGaaeiiamaalaaabaGaaeikaiaabMhadaWgaaWcbaGaaeymaaqabaGccaqGGaGaaeiEaiaabccacaqGXaGaaeOnaiaabccacaqG1bGaaeykaiaabccacaqGRaGaaeiiaiaabIcacaqGXaGaaeimaiaabcdacaqGGaGaae4eGiaabccacaqG5bWaaSbaaSqaaiaabgdaaeqaaOGaaeykaiaabccacaqG4bGaaeiiaiaabgdacaqG4aGaaeiiaiaabwhaaeaacaqGXaGaaeimaiaabcdaaaaabaGaae4BaiaabkhacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeymaiaabAdacaqGYaGaaeimaiaabccacaqG1bGaaeiiaiaabccacaqG9aGaaeiiaiaabgdacaqG2aGaaeiiaiaabMhadaWgaaWcbaGaaeymaaqabaGccaqG1bGaaeiiaiaabUcacaqGGaGaaeymaiaabIdacaqGWaGaaeimaiaabccacaqG1bGaaeiiaiaabobicaqGXaGaaeioaiaabccacaqG5bWaaSbaaSqaaiaabgdaaeqaaOGaaeyDaaqaaiaab+gacaqGYbGaaeiiaiaabccacaqGXaGaaeioaiaabccacaqG5bWaaSbaaSqaaiaabgdaaeqaaOGaaeiiaiaabwhacaqGGaGaae4eGiaabccacaqGXaGaaeOnaiaabccacaqG5bWaaSbaaSqaaiaabgdaaeqaaOGaaeiiaiaabwhacaqGGaGaaeypaiaabccacaqGXaGaaeioaiaabcdacaqGWaGaaeyDaiaabccacaqGtaIaaeiiaiaabgdacaqG2aGaaeOmaiaabcdacaqGGaGaaeyDaaqaaiaab+gacaqGYbGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGYaGaaeyEamaaBaaaleaacaqGXaaabeaakiaabccacaqG1bGaaeiiaiaab2dacaqGGaGaaeymaiaabIdacaqGWaGaaeiiaiaabwhaaeaacaqGVbGaaeOCaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabMhadaWgaaWcbaGaaeymaaqabaGccaqGGaGaaeypaiaabccadaWcaaqaaiaabgdacaqG4aGaaeimaaqaaiaabkdaaaGaaeiiaiaab2dacaqGGaGaaeyoaiaabcdaaeaacaqGobGaae4BaiaabEhacaqGSaGaaeiiaiaabccacaqGqbGaaeyzaiaabkhacaqGJbGaaeyzaiaab6gacaqG0bGaaeyyaiaabEgacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiamaaDeaaleaacaqG4aaabaGaaeymaiaabAdaaaGccaqGybGaaeiiaiaab2dacaqGGaGaaeyoaiaabcdacaqGGaGaaeyjaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabcfacaqGLbGaaeOCaiaabogacaqGLbGaaeOBaiaabshacaqGHbGaae4zaiaabwgacaqGGaGaae4BaiaabAgacaqGGaWaa0raaSqaaiaabIdaaeaacaqGXaGaaeioaaaakiaabIfacaqGGaGaaeypaiaabccacaqGXaGaaeimaiaabcdacaqGGaGaae4eGiaabccacaqG5aGaaeimaiaabccacaqG9aGaaeiiaiaabgdacaqGWaGaaeiiaiaabwcacaqGGaaaaaa@AC98@

Q.12 If Z = 3, what would be the valency of the element? Also, name the element.

Answer

Z represents the atomic number of an element. Atomic number 3 is of the element Lithium.

Electronic configuration of Lithium is 2,1. In order to complete its duplet it will lose its valence electron therefore, its valency is 1.

Composition of the nuclei of two atomic species X and Y are given as under

X Y
Protons 6 6
Neutrons 6 8

 

 

 

 

 

Q.13 Give the mass numbers of X and Y. What is the relation between the two species?

Answer

We know that

Mass number = Number of protons + Number of neutrons

Mass number of X = 6 + 6 = 12

Mass number of Y = 6 + 8 = 14

X and Y have the same atomic number but different mass numbers therefore; X and Y are the Isotopes.

Q.14 For the following statements, write T for True and F for False.

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.

(c) The mass of an electron is about 1/2000 times that of proton.

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Answer

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons. (F)

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral. (F)

(c) The mass of an electron is about 1/2000 times that of proton. (T)

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. (T)

Q.15 Which one of the following is a correct electronic configuration of sodium?

(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.

Answer

2,8,1

Q.16 Complete the following table.

Atomic number Mass number Number of neutrons Number of protons Number of electrons Name of the atomic species
9 – 10 – – –
16 32 – – – Sulphur
– 24 – 12 – –
– 2 – 1 – –
– 1 0 1 0 –

Answer

Atomic number Mass number Number of neutrons Number of protons Number of electrons Name of the atomic species
9 19 10 9 9 Fluorine
16 32 16 16 16 Sulphur
12 24 12 12 12 Magnesium
1 2 1 1 1 Deuterium
1 1 0 1 0 Protium

Q.17 Rutherford’s alpha-particle scattering experiment was responsible for the discovery of

(a) Atomic Nucleus (b) Electron

(c) Proton (d) Neutron

Answer

Atomic nucleus

Q.18 Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.

Answer

Isotopes of an element have different number of neutrons.

Q.19 Number of valence electrons in Cl– ion are:
(a) 16
(b) 8
(c) 17
(d) 18

Answer

8

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FAQs (Frequently Asked Questions)

Students can find the NCERT Solutions for Class 9 Science Chapter 4 on the Extramarks website. The solutions explain this chapter in simple words for better understanding. Students can register on the website and get access to the study materials.

 

A variety of questions are covered in the NCERT textbook and the NCERT Exemplar books. These include; multiple-choice questions, match the following, true or false and descriptive types of questions. Students should practise these various formats of problems in Class 9 Science, which will help them increase their knowledge of critical topics and finally leverage higher marks in their examination.