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NCERT Exemplar Class 10 Science Solutions Chapter 5 Periodic Classification Of Elements

NCERT Exemplar Class 10 Science Solutions Chapter 5 Periodic Classification Of Elements

Edited By Sumit Saini | Updated on Sep 09, 2022 02:02 PM IST | #CBSE Class 10th
Upcoming Event
CBSE Class 10th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

Periodic Classification of Elements NCERT exemplar Class 10 Science solutions chapter 5 discusses the periodic table and its historical development. This chapter describes how the periodic table is classified. Our experienced subject matter experts have carefully prepared the NCERT exemplar Class 10 Science chapter 5 solutions to present exact and relevant responses to the students while practicing NCERT Class 10 Science questions. These NCERT exemplar Class 10 Science chapter 5 solutions develop a better understanding of the periodic classification of elements they are comprehensive in nature. The NCERT exemplar Class 10 Science solutions chapter 5 covers all the aspects within the CBSE syllabus for Class 10 chapter 5.

This Story also Contains
  1. NCERT Exemplar Class 10 Science Solutions Chapter 5-MCQ
  2. NCERT Exemplar Class 10 Science Solutions Chapter 5-Short Answer
  3. NCERT Exemplar Class 10 Science Solutions Chapter 5-Long Answer
  4. NCERT Exemplar Solutions Class 10 Science Chapter 5 Important Topics:
  5. Features of NCERT Exemplar Class 10 Science Solutions Chapter 5:

Also read - NCERT Solutions for Class 10

NCERT Exemplar Class 10 Science Solutions Chapter 5-MCQ

Question:1

Upto which element, the Law of Octaves was found to be applicable
(a) Oxygen
(b) Calcium
(c) Cobalt
(d) Potassium
Ans. (b)
Newlands octave rule was applicable only to lighter elements up to 40u therefore Newland Octave’s rules are applicable only up to calcium. After calcium, each eighth element doesn't have properties like that of the first.

Question:2

According to Mendeleev's Periodic Law, the elements were arranged in the periodic table in the order of:
(a) increasing atomic number
(b) decreasing atomic number
(c) increasing atomic masses
(d) decreasing atomic masses
Ans.(c)
Mendeleev realized that the chemical and physical properties of elements were related to their atomic mass in a 'periodic'
way, and arranged them so that groups of elements with similar properties fell into vertical columns in his table.
So, according to Mendeleev’s periodic table elements are arranged according to atomic mass.

Question:3

In Mendeleev ’s Periodic Table, gaps were left for the elements to be discovered later. Which of the following elements found a place in the periodic table later
(a) Germanium
(b) Chlorine
(c) Oxygen
(d) Silicon
Ans (a)
Chlorine, oxygen and silicon were already included in Mendeleev's periodic table.
Eka-Silicon -- Germanium
Eka-aluminum -- Gallium
were the elements that are left by Mendeleev in his periodic table.

Question:4

Which of the following statement (s) about the Modern Periodic Table are incorrect
(i) The elements in the Modern Periodic Table are arranged on the basis of their decreasing atomic number
(ii) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic masses
(iii) Isotopes are placed in adjoining group (s) in the Periodic Table
(iv) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic number
(a) (i) only
(b) (i), (ii) and (iii)
(c) (i), (ii) and (iv)
(d) (iv) only

Ans.(b)
(i) The elements in the Modern Periodic Table are arranged on the basis of their decreasing atomic number – Incorrect as the elements in periodic table are arranged according to increasing atomic number.
(ii) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic masses – Incorrect as atomic mass was the criteria for Mendeleev’s periodic table.
(iii) Isotopes are placed in adjoining group (s) in the Periodic Table – Incorrect as isotopes have same electronic configuration and thus they are placed in the same group.
(iv) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic number – Correct.
Hence option (b) is correct.

Question:5

Which of the following statements about the Modern Periodic Table is correct:
(a) It has 18 horizontal rows known as Periods
(b) It has 7 vertical columns known as Periods
(c) It has 18 vertical columns known as Groups
(d) It has 7 horizontal rows known as Groups
Ans (c)
According to Modern Periodic Table:
18 – Vertical Columns – known as Groups
7 - Horizontal Rows – known as Periods
Hence option (c) is correct.

Question:6

Which of the given elements A, B, C, D and E with atomic number 2, 3, 7, 10 and 30 respectively belong to the same period?
(a) A, B, C
(b) B, C, D
(c) A, D, E
(d) B, D, E
Ans(b)
First period – 2 elements – Atomic number 1-2
Second period – 8 elements – Atomic number 3-10
Third period – 8 elements – Atomic number 11-18
Fourth period – 16 elements – Atomic number 19-36
So, 2nd Period has elements with atomic number 3 to 10
Hence option (b) is correct.

Question:7

The elements A, B, C, D and E have atomic number 9, 11, 17, 12 and 13 respectively. Which pair of elements belong to the same group?
(a) A and B
(b) B and D
(c) A and C
(d) D and E
Ans (c)
Electronic configuration of A (atomic number 9): 2,7
Electronic configuration of B (atomic number 11): 2,8,1
Electronic configuration of C (atomic number 17): 2,8,7
Electronic configuration of D (atomic number 12): 2,8,2
Electronic configuration of E (atomic number 13): 2,8,3
Elements having same number of electrons in the outer most shell belong to same group therefore A and C belong to same group
Hence option (c) is correct.

Question:8

Where would you locate the element with electronic configuration 2,8 in the Modern Periodic Table?
(a) Group 8
(b) Group 2
(c) Group 18
(d) Group 10
Ans (c)
Electronic configuration 2, 8 means a complete octet. So the element is a noble gas.
Element which has 8 electrons in the outer most shell belong to 18th group.
Hence option (c) is correct.

Question:9

An element which is an essential constituent of all organic compounds belongs to
(a) group 1
(b) group 14
(c) group 15
(d) group 16
Ans (b)
Carbon is the essential constituent of all organic compounds.
Carbon belongs to the 14th group.
Hence option (b) is correct.

Question:10

Which of the following is the outermost shell for elements of period 2?
(a) K shell
(b) L shell
(c) M shell
(d) N shell
Ans (b)
Second period means elements having two valence shells
So, elements of period 2 has L as the outer most shell.
Hence option (b) is correct.

Question:11

Which one of the following elements exhibit maximum number of valence electrons?
(a) Na
(b) Al
(c) Si
(d) P
Ans (d)
Na – Group 1 – 1 valence electron
Al – Group 13 – 3 valence electrons
Si – Group 14 – 4 valence electrons
P – Group 15 – 5 valence electrons
Hence option (d) is correct.

Question:12

Which of the following gives the correct increasing order of the atomic radii of O, F and N ?
(a) O, F, N
(b) N, F, O
(c) O, N, F
(d) F, O, N
Ans (d)
Along a period from left to right - Number of protons increases and valence shell remains the same.
Hence radius of the atom decreases.
Correct increasing order of the atomic radii of O, F and N:
F , O , N
Hence option (d) is correct.

Question:13

Which among the following elements has the largest atomic radii?
(a) Na
(b) Mg
(c) K
(d) Ca
Ans (c)
Atomic radius increases when we move from top to bottom along a group and it decreases when it moves from left to right along a period.
So Potassium has the highest radius as it is left on the periodic table and also at the bottom of a group compared to all the elements in the same group.
Simultaneously it belongs to alkali metals which are highly electropositive and gain inert gas configuration by losing an electron.
Hence option (c) is correct

Question:14

Which of the following elements would lose an electron easily?
(a) Mg
(b) Na
(c) K
(d) Ca
Ans (c)
Atomic radius increases when we move from top to bottom along a group and it decreases when it moves from left to right along a period.
So Potassium has the highest radius as it is left on the periodic table and also at the bottom of a group compared to all the elements in the same group.
Simultaneously it belongs to alkali metals which are highly electropositive and gain inert gas configuration by losing an electron.
Hence option (c) is correct.

Question:15

Which of the following elements does not lose an electron easily?
(a) Na
(b) F
(c) Mg
(d) Al
Ans (b)
Fluorine (F) has atomic number 9
Electronic configuration: 2, 7.
Fluorine has a tendency to gain one electron to form a stable configuration with 8 electrons in the outermost shell.
Fluorine is the most electronegative element in the whole periodic table. So, it does not lose the electron that easily.
Hence option (b) is correct.

Question:16

Which of the following are the characteristics of isotopes of an element?
(i) Isotopes of an element have same atomic masses
(ii) Isotopes of an element have same atomic number
(iii) Isotopes of an element show same physical properties
(iv) Isotopes of an element show same chemical properties
(a) (i), (iii) and (iv) (b) (ii), (iii) and (iv)
(c) (ii) and (iii) (d) (ii) and (iv)
Ans (d)
Isotopes of an element are atoms with same atomic number and different atomic mass. Hence they have similar chemical properties as they have same electronic configuration and they have different physical properties due to different mass.
Hence option (d) is correct.

Question:17

Arrange the following elements in the order of their decreasing metallic character Na, Si, Cl, Mg, Al
(a) Cl > Si >Al > Mg >Na
(b) Na >Mg >Al >Si > Cl
(c) Na > Al > Mg > Cl > Si
(d) Al > Na> Si > Ca> Mg
Ans (b)
Metallic character decreases when we move from left to right in a period
Na > Mg > Al > Si > Cl
Hence option (b) is correct.

Question:18

Arrange the following elements in the order of their increasing non-metallic character Li, O, C, Be, F
(a) F < O < C < Be < Li
(b) Li < Be < C < O< F
(c) F < O < C < Be < Li
(d) F < O < Be < C < Li

Ans (b)
Metallic character decreases when we move from left to right in a period
Li < Be < C < O < F
Hence option (b) is correct.

Question:19

What type of oxide would Eka– aluminium form?
(a) EO3
(b) E3 O2
(c) E2 O3
(d) EO
Ans (c)
Eka-aluminum is the element Gallium
Gallium has a valency of 3. Hence it forms an oxide having molecular formula E2O3.
Hence option (c) is correct.

Question:20

Three elements B, Si, and Ge are
(a) metals
(b) non-metals
(c) metalloids
(d) metal, non-metal, and metalloid respectively
Ans (c)
Boron is a chemical element with the symbol B and atomic number 5. It is a metalloid.
Silicon is a chemical element with the symbol Si and atomic number 14. It is a metalloid.
Germanium is a chemical element with the symbol Ge and atomic number 32. It is a metalloid.
Hence, Boron, Silicon, and Germanium are metalloids which are placed diagonally in periodic table.
Hence option (c) is correct.

Question:21

Which of the following elements will form an acidic oxide?
(a) An element with atomic number 7
(b) An element with atomic number 3
(c) An element with atomic number 12
(d) An element with atomic number 19
Ans (a)
Non Metals form an acidic oxide and Metal forms basic oxides
Element with atomic number 7: Nitrogen - Non-metal
Element with atomic number 3: Lithium - Metal
Element with atomic number 12: Magnesium - Metal
Element with atomic number 19: Potassium - Metal
So element with atomic number 7 forms acidic oxide.
Hence option (a) is correct.

Question:22

The element with atomic number 14 is hard and forms acidic oxide and a covalent halide. To which of the following categories does the element belong?
(a) Metal
(b) Metalloid
(c) Non-metal
(d) Left-hand side element
Ans (b)
Element with atomic number 14 – Silicon (Si)
It is hard and brittle.
Silicon forms acidic oxides such as silicon dioxide which is weakly acidic.
Non-metals need to gain electrons to become stable. When non-metals react with halogens, they can share electrons and form covalent bonds.
Si has a tendency to react with halogens and form covalent bonds.
Hence, it is a metalloid (shows characteristics of both metals and non metals)
Hence option (b) is correct.

Question:23

Which one of the following depict the correct representation of atomic radius(r) of an atom?

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Ans (b)
Atomic radius is the distance between the centre of the nucleus to the outer most shell which contains electrons.
We can see that in figure (ii) and (iii), correct representation is shown.
Hence option (b) is correct.

Question:24

Which one of the following does not increase while moving down the group of the periodic table?
(a) Atomic radius
(b) Metallic character
(c) Valence
(d) Number of shells in an element
Ans (c)
Atomic radius increases down a group
Metallic character increases down a group
The number of electrons in the outer most shell are same in a group hence valence electrons and valency of all the elements in the group remains the same when we move from top to bottom.
Number of shells in an element increases down a group
Hence option (c) is correct.

Question:25

On moving from left to right in a period in the periodic table, the size of the atom.
(a) increases
(b) decreases
(c) does not change appreciably
(d) first decreases and then increases
Ans (b)
When move from left to right in a period the atomic radius decreases as the number of electrons are increasing in the outer most shell
which increases the nuclear forces among them.
Hence option (b) is correct.

Question:26

Which of the following set of elements is written in order of their increasing metallic character?
(a) Be Mg Ca
(b) Na Li K
(c) Mg Al Si
(d) C O N
Ans (a)
When we move from top to bottom in group the metallic nature increases as the atomic size increases due to increase in the number of shells.
Hence metallic character:
Ca > Mg > Be
K >Na > Li
Metallic character goes on decreasing while going from left to right in a period.
Hence metallic character:
Mg > Al > Si
C> N >O
Hence option (a) is correct.

NCERT Exemplar Class 10 Science Solutions Chapter 5-Short Answer

Question:27

The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as? Give one example of such a set of elements.
Answer:

Such an arrangement of elements is called triad, as in Dobereiner’s Triads.
Doberenier’s Triads: It is the arrangement of elements as a set containing 3 elements where mass of
middle element is equal or close to the average of other two elements.
Y= \frac{X+Z}{2}
For example,
Lithium (Atomic mass 7), Sodium (Atomic mass 23) and Potassium (Atomic mass 39)
Average atomic mass of Li and K = \frac{39+7}{2}= 23

Question:28

Elements have been arranged in the following sequence on the basis of their increasing atomic masses.
F, Na, Mg, Al, Si, P, S, Cl, Ar, K
(a) Pick two sets of elements which have similar properties.
(b) The given sequence represents which law of classification of elements
Answer:

(a) SET 1 – F, Cl
SET 2 – Na, K
(b)The law of octaves states that every eighth element has similar properties when the elements are arranged in the increasing order of their atomic masses.
Based on Newlands octaves rule the above elements can be arranged as Follows:
F Na Mg Al Si P S Cl K Ar

Question:29

Can the following groups of elements be classified as Dobereiner's triad ?
(a) Na, Si, Cl
(b) Be, Mg, Ca
Atomic mass of Be 9; Na 23; Mg 24; Si 28; Cl 35; Ca 40
Explain by giving reason.
Answer:

Doberenier’s Triads: It is the arrangement of elements as a set containing 3 elements where mass of middle element is equal or close to the average of other two elements.
(a)(23) ; Si(28) ; Cl(35)
Atomic Mass of Si= \frac{23+35}{2}= \frac{58}{2}
= 29
Silicon mass is close to the average of Sodium and chlorine but the above three elements are not dobereiner’s traits as the properties of the elements don’t match.
(b)Be(9) ; Mg(24) ; Ca(40)
Atomic mass of Mg= \frac{9+40}{2}= 24\cdot 5
Magnesium mass is close to the average of Beryllium and calcium and all the three
elements have common properties therefore the above three elements are a dobereiner’s triads.

Question:30

In Mendeleev ’s Periodic Table the elements were arranged in the increasing order of their atomic masses. However, cobalt with atomic mass of 58.93 amu was placed before nickel having an atomic mass of 58.71 amu. Give reason for the same.
Answer:

In Mendeleev ’s Periodic Table the elements were arranged in the increasing order of their atomic masses. But cobalt with atomic mass of 58.93 amu was placed before nickel having an atomic mass of 58.71 amu because
a)The properties of cobalt are similar to rhodium and iridium
b)The properties of nickel are similar to palladium and platinum
And in order to have the elements with similar properties under one group cobalt is placed before nickel even if its mass is greater than that of nickel.

Question:31

“Hydrogen occupies a unique position in Modern Periodic Table”. Justify the statement.
Answer:

Hydrogen is placed separately in periodic table because:
a) Considering the properties, it is similar to alkali metals
b) Based on electronic configuration it is one electron short for inert gas configuration which is similar to halogens
c) Some of the properties do not resemble with alkali metals like the nature of oxides. Hydrogen oxide (H2O) is neutral but alkali
metal oxides are basic and halogen oxides are acidic.

Question:32

Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.
Answer:

Eka-Silicon means Germanium (Ge) which belongs to group 4 (carbon group) whose valency is 4
So chloride of germanium is : GeCl4
Eka-aluminum means Gallium (Ga) which belongs to group 3 (Boron group) whose valency is 3
So chloride of Gallium is : GaCl3

Question:33

Three elements A, B and C have 3, 4 and 2 electrons respectively in their outermost shell. Give the group number to which they belong in the Modern Periodic Table. Also, give their valencies.
Answer:

i) Element A has 3 electrons in the outer most shell therefore its valency is 3. So element A can be one of the following: B, Al, Ga, In , Tl
ii) Element B has 4 electrons in the outer most shell therefore its valency is 4. So element B can be one of the following: C, Si, Ge, Sn, Pb
iii) Element C has 2 electrons in the outer most shell therefore its valency is 2. So element C can be one of the following: Be, Mg, Ca, Sr, Ba, Ra

Question:34

If an element X is placed in group 14, what will be the formula and the nature of bonding of its chloride?
Answer:

Elements belonging to group 14 will have an electronic configuration: 2 8 4
So, it has 4 electrons in the outer most shell therefore its valency is 4
Thus, the chloride of element X will be XCl4

Question:35

Compare the radii of two species X and Y. Give reasons for your answer.
(a) X has 12 protons and 12 electrons
(b) Y has 12 protons and 10 electrons
Answer:

Species X:
K L M
2 8 2
Which has three shells.
Species Y:
K L
2 8
Which has two shells.
Since element X has 3 shells and Y has only 2 shells, we can make a conclusion the element X has larger radius compared to Y.

Question:36

Arrange the following elements in increasing order of their atomic radii.
(a) Li, Be, F, N
(b) Cl, At, Br, I
Answer:

(a) Li, Be, F, N
All the elements above belong to same period (2nd Period) and when we move from left to right in a period, the atomic radius decreases.
F<N<Be<Li
(b) Cl, At, Br, IAll the elements above belong to same group (17th Group) and
when we move from top to bottom in a group the atomic radius increases.
Cl <Br < I< At

Question:37

Identify and name the metals out of the following elements whose electronic configurations are given below.
(a) 2, 8, 2
(b) 2, 8, 1
(c) 2, 8, 7
(d) 2, 1
Answer:

Electronic Configuration

Valence electrons

Metal / Non-Metal

Atomic Number

Name of the elements

2,8,2

2

Metal

2+8+2=12

Magnesium

2,8,1

1

Metal

2+8+1=11

Sodium

2,8,7

7

Non-Metal

2+8+7=17

Chlorine

2,1

1

Metal

2+1=3

Lithium

Question:38

Write the formula of the product formed when the element A (atomic number 19) combines with the element B (atomic number 17). Draw its electronic dot structure. What is the nature of the bond formed?
Answer:

Electronic configuration of A(21) : 2 8 8 1
Electronic Configuration of B(17): 2 8 7
A is an electropositive element so it tends to lose an electron to attain noble gas configuration on the other hand B is an electronegative element
which takes an electron and gains noble gas configuration.
Based on atomic number A is potassium and B is Chlorine.
Electron dot structure:

Ionic bond will be formed.

Question:39

Arrange the following elements in the increasing order of their metallic character: Mg, Ca, K, Ge, Ga
Answer:

Group No : 1 2 13 14
3rd Period : Mg
4th Period : K Ca Ga Ge
When we move from left to right in a period the metallic nature decreases and all the nonmetals are present in p-block.
On the other hand, when me move from top to bottom in a group the metallic character increases. So,
taking all these trends into consideration:
Ge<Ga<Mg<Ca<K

Question:40

Identify the elements with the following property and arrange them in increasing order of their reactivity
(a) An element which is a soft and reactive metal
(b) The metal which is an important constituent of limestone
(c) The metal which exists in liquid state at room temperature
Answer:

a) Sodium (Na) and Potassium(K) belong to alkali metals which are highly reactive due to their electro positive nature and these elements are so reactive that these are kept in kerosene so that they won’t react with oxygen or moisture in air.
b) Limestone is CaCO3: Calcium is the only metal in its molecule. Therefore, calcium is the most important metal in limestone.
c) Mercury is the only metal which exists as liquid in room temperature.
Order of reactivity Mercury (Hg) < Calcium (ca) <sodium (Na)

Question:41

Properties of the elements are given below. Where would you locate the following elements in the periodic table?
(a) A soft metal stored under kerosene
(b) An element with variable (more than one) valency stored under water.
(c) An element which is tetravalent and forms the basis of organic chemistry
(d) An element which is an inert gas with atomic number 2
(e) An element whose thin oxide layer is used to make other elements
corrosion resistant by the process of “anodising”
Answer:

a) Sodium (Na) and Potassium (K) belong to alkali metals which are highly reactive due to their electropositive nature and these elements are so reactive that these are kept in kerosene so that they won’t react with oxygen or moisture in air. Sodium is found in group 1 period 3 and potassium is found in group 1 period 4.
b) Phosphorous shows variable valency of 3 and 5. It is stored under water as it is highly
reactive when it comes in contact with air. Phosphorous is placed in group 15 period 3.
c) Carbon is tetravalent and it is the basis for all the organic compounds in nature. Carbon is placed in group 14 and period 2.
d) Helium is the inert gas with atomic number 2 and it is the lightest inert gas. Helium is placed in group 18 and period 1.
e) Anodizing is the process of making a thin layer of Al2­O3 on other elements to protect from corrosion. Aluminum is the main metal in that thin layer, it is placed in group 13 and period 3.

NCERT Exemplar Class 10 Science Solutions Chapter 5-Long Answer

Question:42

An element is placed in 2nd Group and 3rd Period of the Periodic Table, burns in presence of oxygen to form a basic oxide.
(a) Identify the element
(b) Write the electronic configuration
(c) Write the balanced equation when it burns in the presence of air
(d) Write a balanced equation when this oxide is dissolved in water
(e) Draw the electron dot structure for the formation of this oxide
Answer:

a) Element is in 2nd Group and 3rd Period that means it belongs to Alkali Earth Metals and it is Magnesium (Mg).
b) Atomic Number of Magnesium is 12 whose electronic configuration is:
K L M
2 8 2
c) When Magnesium is burnt in the presence in air it will form Magnesium Oxide.
2Mg\left ( s \right )+O_{2}\left ( g \right )\overset{Heat}{\rightarrow}2MgO\left ( s \right )
d) When Magnesium Oxide is dissolved in water it will form Magnesium Hydroxide.
2MgO\left ( s \right )+2H_{2}O\left ( l \right )\rightarrow 2Mg\left ( OH \right )_{2}
e) Electronic Configuration of Magnesium: 2 8 2
Electronic Configuration of Oxygen: 2 6
Electron Dot Structure:

Question:43

An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide.
(a) Where in the periodic table are elements X and Y placed?
(b) Classify X and Y as metal (s), non-metal (s) or metalloid (s)
(c) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed
(d) Draw the electron dot structure of the divalent halide
Answer:

a) Element X has atomic number 17 (Chlorine) and its electronic configuration is 2 8 7
It has 7 valence electrons, so it belongs to group 17 and it has 3 shells which means it belongs to 3rd Period.
Element Y has atomic number 20 (Calcium) and its electronic configuration is 2 8 8 2
It has 2 valence electrons and it belongs to group 2 and it has 4 shells which means it belongs to 4th Period.
b) As element X has 7 electrons in the outer most shell it is short of one
electron for inert gas configuration. So, X will prefer accepting an electron hence it is a Non-Metal. Whereas element Y has 2 electrons in the outermost shell which means this element prefers to lose electrons and reach inert gas configuration. So, element Y is going to be a metal.
c) Since Y is a metal, and all the metallic oxides are basic in nature. Oxygen being a non-metal, the bonding between them is going to be Ionic in nature (Electropositive (Metals) and electronegative (Non-metals) elements will form Ionic Bond).
d) Electronic Configuration of Calcium: 2 8 8 2
Electronic Configuration of Chlorine: 2 8 7
Electron Dot Structure of Calcium Chloride CaCl2

Question:44

Atomic number of a few elements are given below 10, 20, 7, 14
(a) Identify the elements
(b) Identify the Group number of these elements in the Periodic Table
(c) Identify the Periods of these elements in the Periodic Table
(d) What would be the electronic configuration for each of these elements?
(e) Determine the valency of these elements
Answer:

Group Number is Electrons in the outer most shell plus 10 as long as the electrons in the outermost shell is greater than 2.
Period Number is the number of shells that are filled by electrons.
Valency is the least number of electrons an element loses or gains to have an inert gas configuration

Atomic Number

(d)

Electronic Configuration

(b)

Group Number

(c)

Period Number

(e) Valency

(a) Element

10

2,8

(8+10=18)

2

0

Neon

20

2,8,8,2

2

4

2

Calcium

7

2,5

15

2

3

Nitrogen

14

2,8,4

14

3

4

Silicon

Question:45

Complete the following cross word puzzle (Figure 5.1)

Across:
(1) An element with atomic number 12.
(3) Metal used in making cans and member of Group 14.
(4) A lustrous non-metal which has 7 electrons in its outermost shell.
Down:
(2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene.
(5) The first element of second Period
(6) An element which is used in making fluorescent bulbs and is second member of Group 18 in the Modern Periodic Table
(7) A radioactive element which is the last member of halogen family.
(8) Metal which is an important constituent of steel and forms rust when exposed to moist air.
(9) The first metalloid in Modern Periodic Table whose fibres are used in making bullet-proof vests
Answer:

Across
1) Element with atomic number 12: Magnesium
3) Metal used in making cans and member of Group 14: Tin
4) A lustrous non-metal which has 7 electrons in its outermost shell: Iodine
Down
2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene: Sodium
5) The first element of second Period: Lithium
6) An element which is used in making fluorescent bulbs and is second member of Group 18 in the Modern Periodic Table: Neon
7) A radioactive element which is the last member of halogen family: Astatine
8) Metal which is an important constituent of steel and forms rust when exposed to moist air: Iron
9) The first metalloid in Modern Periodic Table whose fibres are used in making bullet-proof vests: Boron

Question:46

(a) In this ladder (Figure 5.2) symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic number in the Periodic Table.
(b) Arrange them in the order of their group also.

Answer:

a)H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca
b)

1

2

13

14

15

16

17

18

H







He

Li

Be

B

C

N

O

F

Ne

Na

Mg

Al

Si

P

s

Cl

Ar

K

Ca








Question:47

Mendeleev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
(a) Name the elements which have taken the place of these elements
(b) Mention the group and the period of these elements in the Modern Periodic Table.
(c) Classify these elements as metals, non-metals or metalloids
(d) How many valence electrons are present in each one of them?
Answer:

a)
Eka-Silicon---------- Germanium
Eka-aluminum---------- Gallium
b)
Germanium is in group 14, Period 4
Gallium is in group 13, Period 4
c)
Gallium is a Metal
Germanium is Metalloid
d)
Ga lies in group 13 which means it has 3 valence electrons.
Ge lies in group 14 which means it has 4 valence electrons.

Question:48

(a) Electropositive nature of the element(s) increases down the group and decreases across the period
(b) Electronegativity of the element decreases down the group and increases across the period
(c) Atomic size increases down the group and decreases across a period (left to right)
(d) Metallic character increases down the group and decreases across a period.
On the basis of the above trends of the Periodic Table, answer the following about the elements with atomic numbers 3 to 9.
(a) Name the most electropositive element among them
(b) Name the most electronegative element
(c) Name the element with smallest atomic size
(d) Name the element which is a metalloid
(e) Name the element which shows maximum valency.
Answer:

Atomic Number

3

4

5

6

7

8

9

Name

Lithium

Beryllium

Boron

Carbon

Nitrogen

Oxygen

Fluorine

Symbol

Li

Be

B

C

N

O

F

a) Lithium (atomic number 3) is the most electropositive element in 2nd Period as it is the left most element in the period
d) Fluorine (atomic number 9) is the most electronegative element in 2nd Period as it is the right most element in the period
c) Element with smaller atomic size is fluorine (atomic number 9) as the atomic size decreases as we move from left to right in a period
d) Boron (atomic number 5) is the metalloid in 2nd Period.
e) Carbon (atomic number 6) has the maximum valency of 4.

Question:49

An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
(a) Identify the element X
(b) Write the electronic configuration of X
(c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals?
(d) What would be the nature (acidic/ basic) of oxides formed?
(e) Locate the position of the element in the Modern Periodic Table.
Answer:

(a) X is Sulphur
Atomic number of Sulphur = 16
(b) Electronic configuration of Sulphur = 2, 8, 6
(c)
2FeSO_{4}\left ( s \right )+Heat\rightarrow Fe_{2}O_{3}\left ( s \right )+SO_{2}\left ( g \right )+SO_{3}\left ( g \right )
(d) Oxides of Sulphur (SO2, SO3) are acidic in nature.
(e) Sulphur belongs to Group 16, Period 3 in the Modern Periodic Table.

Question:50

An element X of group 15 exists as diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell.
(a) Identify the element X. How many valence electrons does it have?
(b) Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it?
(c) Draw the electron dot structure for ammonia and what type of bond is formed in it?
Answer:

(a) X is nitrogen (atomic number 7)
Electronic configuration: 2, 5
Valence electrons: 5
(b)


(c) Ammonia is NH3
N atom has the following Lewis structure:

It has three unpaired electrons, each of which can make a covalent bond by sharing electrons with an H atom.
Electron dot structure for ammonia:

It forms a covalent bond.

Question:51

Which group of elements could be placed in Mendeleev's Periodic Table without disturbing the original order? Give reason.
Answer:

The noble gases could have been placed in a separate group without disturbing the group as they are found in atmosphere with very little concentrations and have inert properties.
When Mendeleev gave periodic table, none of the noble gases were discovered. When they were later discovered, they were placed in a separate group called as Zero group.

Question:52

Give an account of the process adopted by Mendeleev for the classification of elements. How did he arrive at “Periodic Law”?
Answer:

When Mendeleev was working with the periodic table, only 63 elements were known to mankind. He focused on the masses of the elements as well as the nature of the compounds formed when these reacted with oxygen and hydrogen. He grouped the elements which had similar oxide and hydride formations under same group and arranged them according to their atomic masses.
Elements with similar properties were arranged in a group.
Mendeleev observed that elements were automatically arranged in the order of increasing atomic masses

NCERT Exemplar Solutions Class 10 Science Chapter 5 Important Topics:

The chapter on Periodic Classification of Elements in NCERT exemplar Class 10 Science solutions chapter 5 covers the below-mentioned topics:

  • Classification of the periodic table.
  • NCERT exemplar Class 10 Science solutions chapter 5 discusses octave law and its limitation in the periodic table
  • Classification to understand basic properties of similar group elements.
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NCERT Class 10 Science Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Science Solutions Chapter 5:

These Class 10 Science NCERT exemplar chapter 5 solutions provide an understanding of periodic table and how it is classified. We will also learn that Mendeleev had left some spaces for those matters which can be discovered later. These elaborate solutions can be used by the students to understand and hone the concepts of Periodic classification of elements. The Class 10 Science NCERT exemplar solutions chapter 5 Periodic classification of elements are sufficient to develop a strong base in order to solve problems given in other books such as Chemistry question bank, NCERT Class 10 Science, S. Chand by Manjit Kaur and Lakhmir Singh, et cetera.

It is highly suggested to explore the NCERT exemplar Class 10 Science solutions chapter 5 pdf download feature to keep the solutions handy when attempting the problems of NCERT exemplar Class 10 Science chapter 5 in offline mode.

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Frequently Asked Questions (FAQs)

1. Q1. How elements are arranged in the modern periodic table?

A1. In the modern periodic table, elements are arranged in increasing order of atomic number.

2. Q2. Orbits are also known as shells what are the standard nomenclatures of these shells?

A2. Shells are those orbits in which electrons revolve around the nucleus. Every orbit has limitation on number of electrons it can have. These orbits or shells are named as KLMN shells

3. Q3. What is atomic radius?

A3. The radius of the last orbit where electron can exist in any substance is called atomic radius. This radius is very large in comparison with nuclear radius.

4. Q4. Will these problems of exemplar require additional knowledge of Periodic classifications of elements?

A4. NCERT Exemplar class 10 science solutions chapter 5 equip the student with a multidimensional approach to the problems and understand the concept of Periodic classifications of elements.

5. Q5. Is the chapter Periodic classifications of elements important for Board examinations?

A5. The chapter of Periodic classifications of elements is important for Board examinations as it holds around 5-7% weightage of the whole paper.

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

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  1. Starting Point:

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    • Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
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If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer
sai raja 01 September,2024
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Read Complete Answer
SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

Read Complete Answer
Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Read Complete Answer
sadafreen356 03 July,2024
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Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

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\; K\;
Option 3)

zero\;

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K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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