In this chapter, students will come to know about early attempts at the classification of elements, Newlands Law Of Octaves, Mendeleev’s Periodic Table, the modern periodic table, metallic and non-metallic properties. At the end students are supposed to attempt question and answer exercise, multiple choice questions and group activity.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 5 – Periodic Classification of Elements solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 5 – Periodic Classification of Elements Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

Periodic Classification of Elements Class 10 Science In Text Book Questions

Question 1.
Did Doberiener’s triads also exist in the columns of Newlands’ octaves ? Compare and find out.
Answer:
Yes, some of the Doberiener’s triads did exist in the columns of Newlands’ octaves. For example,
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Question 2.
What were the limitations of Doberiener’s triads ?
Answer:
Doberiener was in a position to identify three triads. It could not apply to all the elements known at that time. Therefore, the classification was not so useful.

Question 3.
What were the limitations of Nawlands’ Law of Octaves ?
Answer:

  1. Actually this classification was successful only upto the element calcium. After that, every eighth element did not possess the same properties as by the element lying above it in the same group. For example, the elements cobalt (Co) and nickel (Ni) placed below chlorine had different properties. Same was the case with copper (Cu) placed after potassium in the same group.
  2. Newland committed another mistake. He placed two elements in the same slot in a particular group. For example, Co and Ni in the first group after chlorine. Similarly, elements cerium (Ce) and lanthanum (La) were placed after yitterium (Y) in the same group. Newland could not offer any explanation for such an arrangement.
  3. Newland somehow thought that only 56 elements existed in nature and no more elements were likely to be discovered. But this belief ultimately proved to be wrong.
  4. When noble gas elements were discovered at a later stage, their inclusion in the table disturbed the entire arrangement.

Question 4.
Use Mendeleev’s periodic table to predict the formulae for the oxides of the elements : K, C, Al, Si, Ba.
Answer:
Oxygen is a member of group VTA in Mendeleev’s periodic table. Its valency is 2. Similarly, the valencies of all the elements listed can be predicted from their respective groups. This can help in writing the formulae of their oxides.

  1. Potassium (K) is a member of group IA. Its valency is 1. Therefore, the formula of its oxide is K2O.
  2. Carbon (C) is a member of group IVA. Its valency is 4. Therefore, the formula of its oxide is C2O4 or CO2.
  3. Aluminium (Al) belongs to groups IIIA and its valency is 3. The formula of the oxide of the element is Al2O3.
  4. Silicon (Si) is present in group IVA after carbon. Its valency is also 4. The formula of its oxide is Si2O4 or SiO2
  5. Barium (Ba) belongs to group IIA and the valency of the element is 2. The formula of the oxide of the element is Ba2O2 or BaO.

Question 5.
Besides gallium which two other elements have since been discovered that fill the gaps left by Mendeleev in creating his periodic table ?
Answer:
Two other elements are scandium (Sc) and germanium (Ge). In their gaps, the elements with names Eka-boron and Eka-silicon were placed.

Question 6.
What was the criteria used by Mendeleev in creating his periodic table ?
Answer:
Mendeleev used atomic masses of the elements as the criteria for creating his periodic table. In this table, the elements were arranged in order of increasing atomic masses.

Question 7.
Why do you think that the noble gases should be placed in a separate group ?
Answer:
In the Mendeleev’s periodic table, the elements have been arranged in the different groups on the basis of valency. For example, the elements placed in group I (IA and IB) have Valency equal to one. Same is the
case with the elements placed in other groups. Since the noble gas elements He, Ne, Ar, Kr, Xe, and Rn have zero valency, they could not find a place in Mendeleev’s periodic table. These have been placed in a separate group called zero group in the periodic table. Please note that the noble gas elements were not a part of the Mendeleev’s periodic table. They were added later on.

Question 8.
How could the Modern Periodic table remove various anomalies of Mendeleev’s periodic table ?
Answer:
We have already studied that the Mendeleev’s periodic table is based on the atomic masses of the elements whereas Modern Periodic Table takes into account their atomic numbers. Since the properties of the elements are linked with the electronic configuration of their atoms (i.e., atomic number), this means that Modern Periodic Table is better than the Mendeleev’s Periodic Table. The important advantages are listed.

  1. The position of the elements in the periodic table are linked with their electronic configuration.
  2. Each group is an independent group and the idea of sub-groups has been discarded.
  3. One position for all the isotopes of an element is justified since the isotopes have the same atomic number. For example, the three isotopes of the element hydrogen e., protium, deuterium and tritium have atomic number one (Z = 1). Similarly, two isotopes of chlorine i.e. Cl-35 and Cl-37 are placed in the same slot since they have same atomic number (Z = 17).
  4. The positions of certain elements which were earlier misfits in the Mendeleev’s periodic table  are now justified because it is based on   atomic numbers of the elements.  For example, Ar precedes K because its atomic number (18) is less than that of K (19).
  5. It is quite easy to remember and reproduce.

Question 9.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice ?
Answer:
Magnesium (Mg) belongs to group 2 known as Alkaline Earth Family. The two other elements belonging to the same group are calcium (Ca) and strontium (Sr). The basis of choice is the electronic distribution in the valence shell of these elements. All of them have two electrons each.

K

 L M N

O

Mg (Z =12)

 2 8 2 — —

Ca (Z = 20)

 2 8 8 2

—
Sr (Z = 38) 2 8 18 8

2

Question 10.
Name :
(a) three elements that have a single electron in their outermost shells.
(b) three elements that have two electrons in their outermost shells.
(c) three elements with filled outermost shells.
Answer:
(a) Lithium, sodium, potassium (Alkali metals present in group 1)
(b) Beryllium, magnesium, calcium (Alkaline earth metals present in group 2)
(c) Helium, neon, argon (Noble gases present in group 18).

Question 11.
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements ?
(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common ?
Answer:
(a) The atoms of all these elements have one electron each in their valence shell. That is why, these elements are placed in the group 1 known as alkali metal group. The electronic configurations of these elements are given :

K

 L M

N

Li (Z = 3)

 2 1 — —

Na (Z = 11)

 2 8 1

—
K (Z = 19) 2 8 8

1

All the three elements evolve hydrogen gas on reacting with water
2 Li + 2 H2O ———-> 2 LiOH + H2
2 Na + 2 H2O ————> 2 NaOH+ H2
2 K + 2 H2O ————> 2 KOH + H2
Apart from this, all the elements happen to be the first elements of their respective periods. For example,

  • Second period starts from lithium (Li)
  • Third period starts from sodium (Na)
  • Fourth period starts from potassium (K).

(b) Both these elements have completely filled shells. Helium (Z = 2) has two electrons in the only shell (K shell). The other element neon (Z = 10) has both K and L shells fully filled (2, 8). Because of the filled shells, the atoms of these elements do not have any desire to take part in chemical combination and they have been placed together in the same group known as group 18 or zero group.

Question 12.
In the modern periodic table, which are the metals among the first ten elements ?
Answer:
Metals among the first ten elements are lithium (Li) and beryllium (Be) . These are placed towards the left of the table.

Period

 Group 1 Group 2 Group 13 Group 14 Group 15

Group 16

1

2

 —

—

 —

Be

 —

—

 —

—

 —

—

 —

—
3

4

 —

—

 —

—

 —

Ga

 —

Ge

 —

As

—

Se

Question 13.
By considering their position in the periodic table, which one of the following elements would you expect to have maximum metallic characteristics ?
Answer:
Before identifying the metallic character from the list of the elements, we must remember two points :

  • The metallic character of an element is related to the electron releasing tendency of its atoms. Greater the tendency, more will be the metallic character.
  • In general, metallic character of the elements increases down the group and decreases along a period.
    With the help of the Modern Periodic Table, let us identify the group and period to which these elements

Since the metallic character increases down a group and decreases along a period, the obvious choice is between two elements. These are Be (beryllium) present in group 2 and Ga (gallium) present in group 13. Now, the size of Ga is very big as compared to that of Be. Actually, the element Ga has three shells since it belongs to period 4 while the element Be has only two shells as it is a member of period 2. This means that the element Ge has a greater electron releasing tendency of its atom as compared to the element Be. Therefore, Ge has more metallic character, rather maximum metallic character among the elements that are listed.

Periodic Classification of Elements Class 10 Science NCERT End Exercise

Question 1.
Which of the following statements is not correct about the trends while going from left to right across the periodic table ?
(a) The elements become less metallic in nature
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c). The atoms lose their valence electrons with difficulty and not easily. This is on account of the reason that

  • nuclear charge increases from left to the right since the atomic number of the elements gradually increases.
  • with the increase in nuclear charge, the force binding the electrons with the nucleus increases. Therefore, the atoms lose their valence electrons with difficulty.

Question 2.
Element X forms a chloride with the formula XCI2 which is a solid with high melting point. X would most likely to be in the same group of the periodic table as :
(a) Na
(b) Mg
(c) Al
(d) Si.
Answer:
(b). The formula of the chloride of the element is XCl2. This means that the valency of the element X is 2 since chlorine is monovalent. The element with valency 2 is expected to be present in group 2 to which magnesium (Mg) belongs.

Question 3.
Which element has :
(a) two shells, both of which are completely filled with electrons ?
(b) the electronic configuration 2, 8, 2 ?
(c) a total of three shells with four electrons in the valence shell ?
(d) a total of two shells with three electrons in the valence shell ?
(e) twice as many electrons in the second shell as in the first shell ?
Answer:
(a) The elements with completely filled shells are noble gas elements and they belong to group 18. Since the element has two shells ; it must be present in second period and is neon (Ne) with electronic configuration 2, 8.
(b) The electronic configuration suggests that the element belongs to third period and second group. It is therefore, magnesium (Mg).
(c) The element with three shells is present in third- period. Since it* has four electrons in the valence
shell, it must belong to group 14 and is silicon (Si) with electronic configuration 2, 8, 4.
(d) The element with two shells is expected to be present in the second period. With three electrons in
the valence shell, it must belong to group 13 and is boron (B) with electronic configuration 2, 3.
(e) The element has only two shells. The first shell can have a maximum of two electrons. The second shell has four electrons which is twice the number of electrons present in the first shell. Therefore, the electronic configuration of element is 2, 4. It is carbon with atomic number (Z) equal to 6.

Question 4.
(a) Which property do all elements in the same column of the periodic table as boron have in common ?
(b) Which property do all elements in the same column of the periodic table as fluorine have in common ?
Answer:
(a) The element boron (B) is the first member of group (also called column) 13. It has three electrons in
the valence shell (2, 3). The other elements included in the same column are aluminium (Al),
gallium (Ga), indium (In) and thalium (Tl). They too have three electrons in the valence shell of their atoms. Just like boron, these elements also show a valency of 3 in their compounds.
(b) The element fluorine (F) is the first member of group (also called column) 17. It has seven electrons in the valence shell (2, 7). The other members present in the same group known as halogen family are chlorine (Cl), bromine (Br), iodine (I) and astatine (At). They have also seven electrons in the valence shell of their atoms. Like fluorine, they all show a valency of 1 in their compounds.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element ?
(b) To which of the following elements would it be chemically similar ? (Atomic numbers are given in parentheses).
N(7), F(9), P(15), Ar(18).
Answer:
(a) The atomic number of the element is 17 (2 + 8 + 7 = 17).
(b) It would be chemically similar with fluorine (F) which has also 7 electrons in valence shell (2, 7)

Question 6.
The position of three elements A, B and C in the periodic table are shown below :

Group 16

Group 17

–

 A
B

C

(a) State whether A is metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B ?
(d) Which type of ion, cation or anion will be formed by the element A ?
Answer:
(a) Group 17 represents halogen family. All the elements included in the family are Therefore, element A is a non-metal.
(b) Reactivity of non-metals is generally due to the electron accepting tendency of their atoms. Down the group, the atomic size increases. Therefore, the attraction of the nucleus for the outside electrons decreases. This means that down the group of non-metals, reactivity decreases. Thus, the element C is less reactive than the element A.
(c) Atomic size of the elements decreases along a period. The elements B and C are present in the same period. Since C is placed after B, the size of the element C is less than that of B.
(d) The elements A, as pointed out earlier is a non-metal which belongs to group 17. It has seven valence electrons (2, 8, 7). In order to have the configuration of the nearest noble gas element, it will take up one electron and change to anion i.e., A– ion.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write their electronic configuration. Which of these will be more electronegative and why ? (CBSE 2012)
Answer:
The electronic configurations of the two elements are :
Nitrogen (Z = 7) 2, 5 ;
Phosphorus (Z = 15) 2, 8, 5
Since the size of nitrogen is small as compared to phosphorus, it has a greater tendency to take up electrons. It is therefore, more electronegative than phosphorus.

Question 8.
How does the electronic configuration of an atom relate to its position in the modern periodic table ?
(CBSE Delhi 2011)
Answer:
The modern periodic table is based upon atomic numbers of the elements. Since electronic configurations of the elements depend upon their atomic numbers, this means that the periodic table is based on the electronic configurations of the elements. For example, all the alkali metals have one electron each in their valence shell. These are placed in group 1. Similarly, the alkaline earth metals with two electrons in their valence shell are placed in group 2 and so on.

Question 9.
In the modern periodic table, calcium (Z = 20) is surrounded by the elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium ?
(CBSE All India 2011)
Answer:
Only those elements are placed in the same group in which the gaps of atomic numbers are : 8, 8, 18, 18, 32. If we look at the atomic numbers of the elements that are listed, it becomes clear that the elements with atomic numbers 12, 20 (Ca), 38 fit into this pattern. They are placed in the same group and have also similar physical and chemical properties.

Question 10.
Compare and contrast the arrangement of elements in Mendeleevs periodic table and the modern periodic table.
Answer:
The main points of distinction between Mendeleevs periodic table and Modern periodic table are as follows :

Mendeleev’s Periodic Table

Modern Periodic Table
1. It regards atomic masses of the elements as the basis of classification. It regards atomic number of the elements as the basis of classification.
2. No separate positions or slots have been allotted to the isotopes of an element since they have different atomic masses. There is no need for the separate slots for the isotopes since they have the same atomic numbers.
3. No justification is made for placing hydrogen in group IA along with alkali metals. Justification has been made for placing hydrogen along with alkali metals in group 1 since both hydrogen and alkali metals have one valence electron.
4. Except for the elements in group VIII, the remaining groups have been divided with sub-groups A and B. There are no sub-groups and all groups one independent in nature.
5. Position of certain elements based on their atomic masses are misfits. For example, the element cobalt (atomic mass = 58-9) has been its placed ahead of nickel (atomic mass = 58-7) Modern periodic table is free from such anomalies. The element cobalt is placed before nickel since its atomic number (27) is less than that of nickel (28).
6. Electronic configurations and properties of the elements can not be predicted from their positions to the table. Both electronic configuration and certain properties of the elements can be predicted from their positions in the periodic table.
7. It is not very systematic and is difficult to remember. It is quite systematic and is easy to remember.

NCERT Solutions for Class 10 Science Chapter 5 – Periodic Classification of Elements

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