CBSE Class 10

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

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Multiple Choice Questions

Question 1.
Upto which element, the Law of Octaves was found to be applicable ?
(a) Oxygen
(b) Calcium
(c) Cobalt
(d) Potassium.
Answer:
(b). Law was found to be applicable upto the element calcium.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

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.
Answer:
(c).

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.
Answer:
(a). The element was called Eka-silicon.

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 numbers
(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 numbers
(a) (i) only
(b) (i), (ii) and (iii)
(c) (i), (ii) and (iv)
(d) (iv) only.
Answer:
(b). Statements (i), (ii) and (iii) are all in 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.
Answer:
(c).

Question 6.
Which of the given elements A, B, C, D and E with atomic numbers 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.
Answer:
(b). The elements B (Z = 3), C (Z = 7) and D(Z = 10) belong to the same period. It is second period.

Question 7.
The elements A, B, C, D and E have atomic numbers 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.
Answer:
(c). Elements A (Z = 9) and C (Z = 17) belong to the same group. It is a halogen family (group 17).

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.
Answer:
(c). It is a noble gas element Neon (Ne) present in group 18.

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.
Answer:
(b). The element is carbon. It belongs to group 14.

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
Answer:
(b). In second (2) period, the electrons are filled in second shell also known as L-shell.

Question 11.
Which one of the following elements exhibits maximum number of valence electrons ?
(a) Na
(b) Al
(c) Si
(d) P
Answer:
(d). The element phosphorus (P) has five electrons (2, 5) in the valence shell.

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.
Answer:
(d). It is the correct order. These elements are present in second period.

Question 13.
Which among the following elements has the largest atomic radius ?
(a) Na
(b) Mg
(c) K
(d) Ca.
Answer:
(c). The element potassium (K) present in group 1 has the largest atomic radius.

Question 14.
Which of the following elements would lose an electron easily ?
(a) Mg
(b) Na
(c) K
(d) Ca.
Answer:
(c). The element potassium (K) with maximum size would lose electron easily.

Question 15.
Which of the following elements does not lose an electron easily ?
(a) Na
(b) F
(c) Mg
(d) Al.
Answer:
(b). The element fluorine (F) with smallest size does not lose electron easily.

Question 16.
Which of the following are the characteristics of isotopes of an element ?
(i) Isotopes of an element have same atomic mass.
(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).
Answer:
(d).

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.
Answer:
(b). The metallic character of the elements decreases along a period. These elements are present in third period.

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.
Answer:
(b). The non-metallic character of the elements increases along a period. These elements are present in second period in the order Li, Be, C, O, F.

Question 19.
What type of oxide would Eka-aluminium form ?
(a) EO₃
(b) E₃O₂
(c) E₂O₃
(d) EO.
Answer:
(c). Is the correct answer.

Question 20.
Three elements B, Si and Ge are
(a) metals
(b) non-metals
(c) metalloids
(d) metal, non-metal and metalloid respectively.
Answer:
(c). is the correct answer. These are also called semi-metals and possess the characteristics of both metals and non-metals.

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.
Answer:
(a). The element with atomic number (Z) = 7 is nitrogen. It forms acidic oxides such as N₂O₃, N₂O₅ etc.
All other elements are metals and they form basic oxides.

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.
Answer:
(b). The element with atomic number (Z) = 14 is silicon (Si). It is a metalloid.

Question 23.
Which one of the following depicts 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).
Answer:
(b). The atom (ii) has only one shell (K-shell). No electrons are present in the other shells. Therefore, the arrow represents correct atomic radius.
The arrow in atom (iii) also represents the correct atomic radius of the element.

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.
Answer:
(c). In a group, the valence or valency does not change since all the elements present have same valence shell configuration.

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
Answer:
(b). Atomic size decreases along a period. However, the noble gas atom is an exception. It has very large size.

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
Answer:
(a). These elements belong to group 2. These are written in increasing order of their size. Since the metallic character increases down the group, the order is the correct.

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Short Answer Questions

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:
The arrangement is known as Dobereiner’s triad. For example, Calcium (Ca), Strontium (Sr) and Barium (Ba).

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) The elements that have similar properties belong to the same group. From the list of elements available, elements which belong to same group are : .
Na, K (Alkali metals) ; F, Cl (Halogens)
(b) The sequence is according to Newland’s law of octaves.

Question 29.
Can the following groups of elements be classified as Dobereiner’s triad ?
(a) Na, Si, Cl
(b) Be, Mg, Ca
Answer:
Atomic mass of Be = 9; Na = 23; Mg = 24; Si = 28; Cl = 35; Ca = 40 Explain by giving reason.
(a) No, these elements cannot be classified as triads because these do not have same properties. However, the atomic mass of Si (28) is almost the mean of the atomic masses of elements Na (23) and Cl (35).
Note : The Dobereiner’s triad is meaningful only if the elements present in the triad have almost identical properties.
(b) Yes, these elements can be classified as triads. These belong to the same group (2) and have almost identical properties. The atomic mass of Mg (24) is almost the mean of the atomic masses of the elements Be (9) and Ca(40) i.e., 9 + 40 = 49/2 = 24.5.

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:
This is regarded as a defect in the Mendeleev’s periodic table. However, the main reason for making the arrangement was that the elements with similar characteristics must be grouped together. In this case,
• Cobalt (Co) should be in the company of the elements Rhodium (Rh) and Iridium (Ir).
• Nickel (Ni) should be in the company of the elements Palladium (Pd) and Platinum (Pt).

Question 31.
“Hydrogen occupies a unique position in Modern Periodic Table”. Justify the statement.
Answer:
The position of the element hydrogen is still not clear even in the Modern Periodic Table.
• In electronic configuration, it resembles alkali metals of group 1. All of them have only one electron in the valence shell. Actually, hydrogen has only one shell (K-shell) which has one electron. However, it is not a metal whereas alkali-metals are metallic in nature.
• In characteristics, it resembles halogens of group 17. For example, like halogens hydrogen is a non-metal and diatomic as well.
It has been therefore, decided to assign hydrogen a unique position in the Modern or Long Form. Periodic Table. It is placed at the top in group 1 of alkali metals. However, it is not a member of that group.*

Question 32.
Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the ^elements predicted by Mendeleev.
Answer:
Eka-aluminium represents gallium (Ga) with valency three and Eka-silicon is for germanium (Ge) with valency four. The formulae of their respective chlorides are GaCl₃ and GeCl₄.

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:
The electrons present in the outermost shell are also known as valence electrons. The desired information is given :

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:
The element X present in group 14 has four valence electrons in its atom. It can complete its octet by sharing four valence electrons with the electrons of other atoms. Therefore, it will form covalent bonds with the four atoms of chlorine. The formula of the chloride of the element X is.

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:
The available information makes it clear that :
(a) Species X with equal number of protons (12) and electrons (12) is an atom.
(b) Species Y with two electrons less (10) than the number of protons (12), is a divalent cation y²⁺ of species X.
Therefore, the radius of species X is more as compared to that of y²⁺.
Note : The radius of cation is always less than that of atom while that of anion is more.

Question 36.
Arrange the following elements in increasing order of their atomic radii.
Answer:
(a) All the elements belong to the same period (second). The atomic size or radius decreases along a period. Therefore, increasing order of atomic radii is : F < N < Be < Li
(b) The listed elements are present in group 17 (halogen family). The atomic size or radius increases down
a group. Therefore, the correct increasing order of atomic radii is : 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:
(a) Electronic configuration : 2, 8, 2 (Mg is a metal)
(b) Electronic configuration : 2, 8, 1 (Na is a metal)
(c) Electronic configuration : 2, 8, 7 (Cl is a non-metal)
(d) Electronic configuration : 2, 1 (Li is a metal)

Question 38.
Write the formula of the compound 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 ? (CBSE 2013)
Answer:
Electronic configuration of element A (Z = 19) is 2, 8, 8, 1
Electronic configuration of element B(Z = 17) is 2, 8, 7.
The element A has one valence electron while the element B has seven electrons in its valence shell. One electron gets transferred from the atom of element A to the atom of element B. As a result of electron transfer, an ionic bond is formed.

Question 39.
Arrange the following elements in the increasing order of their metallic character. ,
Mg, Ca, K, Ge, Ga
Answer:
In general, the metallic character increases down the group and decreases along a period. From the relative positions of the elements in the periodic table, the increasing order of metallic character is :
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 metal which is soft and reactive
(b) The metal which is an important constituent of limestone
(c) The metal which exists in liquid state at room temperature
Answer:
(a) The metal may be either Na or K.
(b) The metal is Ca.
(c) The metal is Hg.

Question 41.
The increasing order of reactivity of the metals is :
Hg < Ca < Na < K.
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 (Group 1 and Period 3) or Potassium (Group 1 and Period 4)
(b) Phosphorus (Group 15 and Period 3). It shows variable valencies 3 and 5 and is stored under water.
(c) Carbon (Group 14 and Period 2)
(d) Helium (Group 18 and Period 1)
(e) Aluminium (Group 13 and Period 3)

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Long Answer Questions

Question 42.
An element placed in 2nd group and 3rd Period of the Periodic Table, burns in the 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) The element is magnesium (Mg).
(b) The electronic configuration is 2, 8, 2.
(c) Magnesium burns in oxygen (air) to form magnesium oxide which is of basic nature.
2Mg(s) + O₂(g) ————-> 2MgO(s)
(d) Magnesium hydroxide is formed.
MgO(s) + H₂O(aq) ————> Mg(OH)₂(s)

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:
Element X (Z = 17) is chlorine while element Y (Z = 20) is calcium.
(a) Chlorine (Cl) is a member of group 17 and period 3. Calcium (Ca) is present in group 2 and period 4.
(b) Chlorine is a typical non-metal while calcium is a metal.
(c) The oxide of calcium is calcium oxide (CaO). It is a basic oxide.

Question 44.
Atomic number of a few elements are : 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:
(a) The elements are : Neon (Z = 10), Calcium (Z = 20), Nitrogen (Z = 7) and Silicon (Z = 14)
(b) Group numbers : Neon (18), Calcium (2), Nitrogen (15), Silicon (14).
(c) Periods : Neon (2), Calcium (4), Fluorine (2), Silicon (3).
(d) Electronic Configuration : Neon (2, 8) ; Calcium (2, 8, 8, 2) Nitrogen (2, 5) ; Silicon (2, 8, 4)
(e) Valency : Neon (zero) ; Calcium (2) ; Nitrogen (3) ; Silicon (4)

Question 45.
(a) In this ladder 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, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Br.
(b) Group 1 — H, Li, Na, K
Group 2 — Be, Mg, Ca
Group 13 — B, A1
Group 14 — C, Si
Group 15 — N, P
Group 16 — O, S
Group 17 — Cl, Br
Group 18 — He, Ne, Ar

Question 46.
Complete the following cross word puzzle
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 the 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.

                             (Cross-Puzzle)
Answer:

                               (Puzzle Solved)

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) Germanium (Ge) for Eka-silicon and gallium (Ga) for Eka-aluminium
(b) Germanium (Group 14 and Period 4)
Gallium (Group 13 and Period 4)
(c) Germanium (Metalloid) ; Gallium (Metal)
(d) Germanium (Z = 32), Four valence electrons (2, 8, 18, 4)
Gallium (Z = 31) ; Three valence electrons (2, 8, 18, 3).

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:
(a) Lithium (Z = 3) is the most electropositive element,
(b) Fluorine (Z = 9) is the most electronegative element.
(c) Fluorine (Z = 9) has the smallest atomic size.
(d) Boron (Z = 5) is a metalloid.
(e) Carbon (Z = 6) shows the maximum valency (4). However, the element nitrogen (Z = 7) can show valency 5 in some compounds (e.g., N₂O₅).

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) The available information suggests that the element X is sulphur.
(b) Electronic configuration of S(Z = 16) 2, 8, 6

(d) Fe₂O₃ (basic oxide), SO₂(acidic oxide), SO₃(acidic oxide)
(e) Sulphur is a member of group 16 and period 3.

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. What type of bonds is formed in it ?
Answer:
(a) The available information suggests that the element X is nitrogen (N) and exists in diatomic form as N,
Electronic configuration of N(Z = 7) ; 2, 5. It has five valence electrons.

Question 51.
Which group of elements could be placed in Mendeleev’s Periodic Table without disturbing the original order ? Give reason.
Answer:
Group of noble gases (called zero group) could be placed in the Mendeleev’s Periodic Table with out disturbing the original order.
Reasons : Elements present in a group have same valency in the Mendeleev’s Periodic Table. Based on their electronic configuration, the members of noble gas family have zero valency. That is why they are called inert gases. They could be easily placed in the Mendeleev’s Periodic Table as a separate group without disturbing the arrangement of other elements.

Question 52.
Give an account of the process adopted by Mendeleev for the classification of elements. How did he arrive at “Periodic Law” ?
Answer:

1. The basis of classification of elements adopted was atomic masses of the elements. The elements were arranged in order of increasing atomic masses.
2. Elements with similar properties were kept in a particular group in order of increasing atomic masses.
3. Elements placed in a particular group (or sub group) were having same valency.

Note : Please note that when Mendeleev arranged the elements in the Periodic Table only 63 elements were known. Even noble gas or inert gas elements were adjusted at a later stage. Moreover, many groups were left in the table.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements helpful to you.

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HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements

Question 1.
Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound. This compound is added in a small amount to almost all vegetable dishes during cooking Oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neutral. Based on the above information answer the following questions :

1. To which group or period of the periodic table do the listed elements belong ?
2. What would be the nature of compound formed by a combination of elements B and F ?
3. Which two elements could definitely be metals ?
4. Which one of the eight elements is most likely to be found in gaseous state at room temperature ?
5. If the number of electrons in the outermost shell of elements C and G be 3 and 7 respectively, write the formula of the compound formed by the combination of C and G.

Answer:
The clue for the answer is given by the compound which is added to almost all vegetables during cooking. It is sodium chloride (NaCl). Both these elements belong to third period :

1. All the elements belong to third period. They have three shells (K, L, M) and the number of electrons vary from 1 to 8. These belong to different groups which are listed.
   
2. The element ‘B’ is Ca and ‘F’ is S. They combine to form CaS. It is an ionic compound.
3. The elements A, B are definite metals because the elements present in group 1 and 2 are all metals.
4. The element H is a noble gas element. It is most likely to be found in the gaseous state at room temperature.
5. The electronic configuration of the element C(Z = 13) is 2, 8, 3 while that of G(Z = 17) is 2, 8, 7. The formula of the compound formed by their combination is CG₃. It is actually AlCl₃ and is formed as a result of electron sharing.
   

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
The following table shows the position of six elements A, B, C, D, E and F in the period table.

Using the above table answer the following questions :
(a) Which element will form only covalent compounds ?
(b) Which element is a metal with valency 2 ?
(c) Which element is a non-metal with valency of 3 ?
(d) Out of D and E, which one has more atomic radius and why ?
(e) Write a common name for the family of elements C and F.
Answer:
(a) The element ‘E’ present in group 14 is a non-metal. Its name is silicon (Si) and the compounds of the element are only covalent.
(b) The element ‘D’ present in group 2 is a metal known as magnesium (Mg). It exhibits valency 2 in its compounds.
(c) The elements ‘B’ present in group 15 is a non-metal. It is nitrogen (N) and exhibits valency 3 in its compounds.
(d) The element ‘D’ has more atomic radius than the element ‘E’ as the atomic size decreases along a period.
(e) The elements ‘C’ and ‘F’ present in group 18 belong to a family known as noble gases.

Question 3.
Two elements X and Y belong to group 1 and 2 respectively in the same period. Compare them with respect to :
(a) the number of valence electrons
(b) valency
(c) metallic character
(d) size of the atoms
(e) formulae of their oxides and chlorides.
Answer:
(a) The valence electrons present in element X (group 1) and element Y (group 2) are 1 and 2 respectively.
(b) The valency of the element X is one while that of the element Y is two.
(c) Metallic character decreases along a period. This means that the element X is more metallic as compared to element Y.
(d) Atomic size decreases along a period. As a result, the element Y has a smaller size than the element X.
(e) For element X : oxide (X₂O), chloride (XCl).
For element Y : oxide (YO) and chloride (YCl₂).

Question 4.
Atoms of seven elements A, B, C, D, E, F and G have a different number of electronic shells but have the same number of electrons in their outermost shells. The elements A and C combine with chlorine to form an acid and common salt respectively. The oxide of element A is liquid at room temperature and is a neutral substance while the oxides of the remaining six elements are basic in nature. Based on the above information, answer the following questions given ahead :

1. What could the element A be ?
2. Will elements A to G belong to the same period or same group of the periodic table ?
3. Write the formula of the compound formed by the reaction of the element A with oxygen.
4. Show the formation of the compound by a combination of element C with chlorine with the help of electronic structure.
5. What would be the ratio of number of combining atoms in a compound formed by the combination element A with carbon ?
6. Which one of the given elements is likely to have the smallest atomic radius ?

Answer:
Since all the seven elements have same number of electrons in their outer most shells, this means that they belong to the same group. From the available information, it becomes clear that the group is of alkali metals (group 1). Hydrogen (A) is also included. The elements present are;

Let us answer the questions asked

1. The element A’ is hydrogen (H)
2. The elements A’ to ‘G’ belong to the same group and not same period
3. The formula of compound is H₂O
4. Compound between ‘C’ and chlorine (Cl) is sodium chloride (NaCl) formed by the transference of one electron.
   
5. The compound formed between A (H) and carbon is methane (CH₄). In this the two elements are present in the ratio of 1 : 4
6. The element A (hydrogen) has the smallest atomic radius since it is the first element of the group.

Hope given HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements helpful to you.

If you have any doubts, please comment below. We try to provide online math tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Multiple Choice Questions

Question 1.
Carbon exists in the atmosphere in the form of
(a) carbon monoxide only
(b) carbon monoxide and carbon dioxide in traces
(c) carbon dioxide only
(d) coal gas.
Answer:
(b). Both CO and CO₂ gases are present in traces in the atmosphere.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Which of the following statements are usually correct for carbon compounds ? These
(i) are good conductors of heat and electricity
(ii) are poor conductors of electricity
(iii) have strong forces of attraction between their molecules
(iv) do not have strong forces of attraction between their molecules
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(d). Carbon compounds are of covalent nature. That is why these are poor conductors of electricity and do not have strong forces of attraction.

Question 3.
A molecule of ammonia (NH₃) has
(a) only single bonds
(b) only double bonds
(c) only triple bonds
(d) two double bonds and two single bonds.
Answer:
(a). All bonds are single covalent in nature.

Question 4.
Buckminster fullerene is an allotropie form of
(a) nitrogen
(b) sulphur
(c) carbon
(d) tin
Answer:
(c).

Question 5.
Which of the following are correct chain isomers of butane ?

(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(c). Isomer (i) has a straight chain while (ii) is a branched chain isomer.

Question 6.

In the above reaction, alkaline KMnO₄ acts as
(a) reducing agent
(b) oxidising agent
(c) catalyst
(d) dehydrating agent.
Answer:
(b). Alkaline KMnO₄ is also known as Baeyer’s reagent. It acts as an oxidising agent.

Question 7.
Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats. This is an example of
(a) Addition reaction
(b) Substitution reaction
(c) Displacement reaction
(d) Oxidation reaction
Answer:
(a). Oils are unsaturated in nature. By the addition of hydrogen, they become saturated.

Question 8.
In which of the following compounds, —OH is the functional group ?
(a) Butanone
(b) Butanol
(c) Butanoic acid
(d) Butanal
Answer:
(b). —OH is the functional group in butanol which is an alcohol.

Question 9.
The soap molecule has a
(a) hydrophilic head and a hydrophobic tail
(b) hydrophobic head and a hydrophilic tail
(c) hydrophobic head and a hydrophobic tail
(d) hydrophilic head and a hydrophilic tail
Answer:
(a). ‘Head’ is a polar group and is attacked towards H₂O molecules of water. ‘Tail’ is a long chain of hydrocarbons and is water repellent.

Question 10.
Which of the following is the correct representation of electron dot structure of nitrogen ?

Answer:
(d). Both the N atoms have a complete octet after electron sharing.

Question 11.
Structural formula of ethyne is

Answer:
(a).

Question 12.
Identify the unsaturated compounds from the following
(i) Propane
(ii) Propene
(iii) Propyne
(iv) Chloropropane
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (ii) and (iii)
Answer:
(d). Both propene and propyne are unsaturated hydrocarbons.

Question 13.
Chlorine reacts with saturated hydrocarbons at room temperature in the
(a) absence of sunlight
(b) presence of sunlight
(c) presence of water
(d) presence of hydrochloric acid
Answer:
(b). These reactions are called photochemical reactions.

Question 14.
In the soap micelles
(a) the ionic end of soap is on the surface of the cluster while the carbon chain is in the interior of the cluster.
(b) ionic end of soap is in the interior of the cluster and the carbon chain is out of the cluster,
(c) both ionic end and carbon chain are in the interior of the cluster
(d) both ionic end and carbon chain are on the exterior of the cluster
Answer:
(a).

Question 15.
Pentane has the molecular formula C₅H₁₂. h has
(a) 8 covalent bonds
(b) 10 covalent bonds
(c) 16 covalent bonds
(d) 14 covalent bonds
Answer:
(c).

Question 16.
Structural formula of benzene is

Answer:
(c). The carbon atom skeleton in benzene has alternate single and double bonds

Question 17.
Ethanol reacts with sodium and forms two products. These are :
(a) sodium ethanoate and hydrogen
(b) sodium ethanoate and oxygen
(c) sodium ethoxide and hydrogen
(d) sodium ethoxide and oxygen
Answer:
(c).

Question 18.
The correct structural formula of butanoic acid is

Answer:
(d).

Question 19.
Vinegar is a solution of
(a) 30% — 40% acetic acid in alcohol
(b) 5% — 8% acetic acid in alcohol
(c) 5% – 8% acetic acid in water
(d) 15% – 20% acetic acid in water
Answer:
(c).

Question 20.
Mineral acids are stronger acids than carboxylic acids because
(i) mineral acids are completely ionised
(ii) carboxylic acids are completely ionised
(iii) mineral acids are partially ionised
(iv) carboxylic acids are partially ionised
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(a). Mineral acids like HCl are completely ionised in solvent like water whereas carboxylic acids such as CH₃COOH are only partially ionised.

Question 21.
Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms e.g. hydrogen. After the formation of four bonds, carbon attains the electronic configuration of
(a) helium
(b) neon
(c) argon
(d) krypton
Answer:
(b). The compound formed is methane (CH₄). In this, carbon atom has a complete octet and configuration of neon which is a noble gas element.

Question 22.
The correct electron dot-structure of a water molecule is

Answer:
(c).

Question 23.
Which of the following is not a straight chain hydrocarbon ?

Answer:
(d). Please note that the continuous chains of carbon atoms whether straight or bent are not branched in nature.

Question 24.
Which among the following are unsaturated hydrocarbons ?

(a) (i) and (iii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv).
Answer:
(c). Both (ii) and (iv) are unsaturated in nature.

Question 25.
The IUPAC name of the compound
CH₃—CH₂—CHO is
(a) Propanal
(b) Propanone
(c) Ethanol
(d) Ethanal
Answer:
(a).

Question 26.
The heteroatoms present in CH₃—O—CH₂—CH₂ (Br) are
(i) oxygen
(ii) carbon
(iii) hydrogen
(iv) bromine
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(d). Both oxygen (O) and bromine (Br) are heteroatoms. Please remember that apart from C and H atoms, all other atoms present in an organic compound are hetero atoms.

Question 27.
Which of the following does not belong to the same homologous series ?
(a) CH₄
(b) C₂H₆
(c) C₃H₈
(d) C₃H₆.
Answer:
(d). It is an alkene while the rest are all alkane molecules in nature.

Question 28.
The first member of alkene family is
(a) ethyne
(b) ethene
(c) propyne
(d) ethane
Answer:
(b).

Question 29.
Which of the following represents saponification

Answer:
(d). This is an example of saponification reaction. All other options are not correct.

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Short Answer Questions

Question 30.
Draw the electron dot structure of ethyne and also draw its structural formula
Answer:

Question 31.
Write the names of the following compounds :

Answer:
(a) Pentanoic acid
(b) But-l-yne
(c) Heptanal
(d) Pentan-l-ol

Question 32.
Identify and name the functional groups present in the following compounds.


Answer:

Question 33.
A compound X is formed by the reaction of a carboxylic acid C₂H₄O₂ and an alcohol in presence of a few drops of H₂SO₄. The alcohol on oxidation with alkaline KMnO₄ followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of (a) carboxylic acid,
(b) alcohol and
(c) the compound X. Also write the reaction. (CBSE Sample Paper 2017)
Answer:
The available information suggests that the alcohol which gives the same carboxylic acid upon oxidation has two carbon atoms. It is, therefore ethanol (C₂H₅OH). The structures of the different compounds are :

Question 34.
Why are detergents better cleansing agents than soaps ? Explain.
Answer:
A substance capable of removing grease and dirt from any fabric or body is called detergent. The detergents are of two types Le. soapy and non-soapy detergents. The soapy detergents are soaps whereas the non-soapy detergents are synthetic detergents or simply detergents. Although both are cleansing agents, they differ in chemical composition. In the present chapter, we shall briefly discuss the composition and cleansing action of soaps and synthetic detergents.
Soaps are the sodium and potassium salts of long chain fatty acids with general formula RCOONa or RCOOK. The acids present have the formula RCOOH where R may have following values.

These fatty acids exist as triesters of glycerol which is a trihydric alcohol. The triesters are also called triglycerides or simply glycerides and are the constituents of edible oils and fats. These are of animal and vegetable origin e.g. castor oil, linseed oil or soyabean oil. Chemically the triglycerides are formed as a result of esterification reaction.

Question 35.
Identify the functional groups present in the following compounds

Answer:
(a) >C = O
(b) —COOH
(c) —CHO
(d) —OH.

Question 36.
How is ethene prepared from ethanol ? Give the reaction involved in it.
Answer:
From ethanol: This method involves the slow oxidation of a dilute solution of ethanol (10-15 per cent) by oxygen present in air in the presence of an enzyme acetobactor.

The acid obtained is in the form of dilute solution called vinegar.
We have also studied under ethanol that it gets oxidised to ethanoic acid in the presence of dilute solution of alkaline KMnO₄ or acidified K₂Cr₂O₇.

Question 37.
Intake of small quantity of methanol can be lethal. Comment.
Answer:
From methanol: These days ethanoic acid is manufactured by the reaction between methanol and carbon monoxide in the presence of iodine-rhodium (I₂ — Rh) catalyst mixture.

Question 38.
A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation for the reaction involved.
Answer:
The gas evolved is hydrogen. The balanced chemical equation for the reactibn is :

Question 39.
Ethene is formed when ethanol at 443 K is heated with excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction ? Write the balanced chemical equation of this reaction.
Answer:
Concentrated sulphuric acid acts as a dehydrating agent in the reaction. Ethene is formed as the

Question 40.
Carbon, Group (14) element in the Periodic Table, is known to form compounds with many elements. Write an example of a compound formed with
(a) chlorine (Group 17 of Periodic Table)
(b) oxygen (Group 16 of Periodic Table)
Answer:
(a) The compound is carbon tetrachloride (CCl₄)
(b) The compound is carbon dioxide (CO₂)

Question 41.
In electron dot structure, the valence shell electrons are represented by crosses or dots.
(a) The atomic number of chlorine is 17. Write its electronic configuration
(b) Draw the electron dot structure of chlorine molecule.
Answer:
(a) 2, 8, 7

Question 42.
Catenation is the ability of an atom to form bonds with other atoms of the same element. It is exhibited by both carbon and silicon. Compare the ability of catenation of the two elements. Give reasons.
Answer:
The tendency to show catenation is very small in case of Si as compared to C although both the elements belong to the same group (14). This is because of greater atomic size of Si atom (118 pm) than that of carbon (77 pm). As a result, the strength of Si—Si bond is less as compared to that of C—C bond. This means that lesser number of Si atoms can be linked to each other by covalent bonds as compared to carbon atoms.

Question 43.
Unsaturated hydrocarbons contain multiple bonds between the two C-atoms and show addition reactions. Give the test to distinguish ethane from ethene.
Answer:
Distinction between ethane and ethene can be done with the help of bromine water test. Whereas ethene dicolourises the yellow colour of bromine water, ethane does not.

Question 44.
Match the reactions given in Column (A) with the names given in column (B).

Answer:
(a)—(iv) ;
(b)—(i) ;
(c)—(ii) ;
(d)—(iii)

Question 45.
Write the structural formulae of all the isomers of hexane.
Answer:

Question 46.
What is the role of metal or reagents written on arrows in the given chemical reactions?

Answer:
(a) Nickel (Ni) acts as a hydrogenation catalyst for the reaction
(b) Cone. H₂SO₄ removes a molecule of H₂O from the reaction mixture and acts as a dehydrating agent.
(c) Alkaline KMnO₄ acts as an oxidising agent and oxidises ethanol to ethanoic acid.

NCERT Exemplar Solutions for Class 10 Science Chapter 4Long Answer Questions

Question 47.
A salt X is formed and a gas is evolved when ethanoic acid reacts with sodium hydrogen carbonate. Name the salt X and the gas evolved. Describe an activity and draw the diagram of the apparatus to prove that the evolved gas is the one which you have named. Also, write chemical equation of the reaction involved.
Answer:
The salt X formed in the reaction is sodium ethanoate. The gas evolved is carbon dioxide gas. For the activity and chemical reaction,

Question 48.
(a) What are hydrocarbons ? Give examples.
(b) Give the structural differences between saturated and unsaturated hydrocarbons with two examples of each.
(c) What is a functional group ? Give examples of four different functional groups.
Answer:
(a) Hydrocarbons are the organic compounds containing only carbon and hydrogen atoms as their constituents. These may be alkanes, alkenes and alkynes.

(b) Saturated hydrocarbons or alkanes contain either C—C or C—H bonds in their molecules. These are represented by the general formula C_nH_2n+2 For example,

Unsaturated hydrocarbons contain either atleast one >C = C< bond or triple —C ≡ C— bond in their molecules. These may be either alkenes or alkynes in nature. The general formula of alkenes is C_nH_2n while that of alkynes is C_nH_2n-2 for example,

(c) For the definition of functional group and example,

Question 49.
Name the reaction which is commonly used in the conversion of vegetable oils to fats. Explain the reaction in detail.
Answer:
The reaction is known as catalytic hydrogenation. For details, constult section 4.15.

Question 50.
(a) Write the formula and draw electron dot structure of carbon tetrachloride.
(b) What is saponification ? Write the reaction involved in this process.
Answer:

(b) For saponification reaction,

The reaction is known as saponification reaction because it is the basis for the formation of soap.

Question 51.
Esters are sweet-smelling substances and are used in making perfumes. Suggest some activity and the reaction involved for the preparation of an ester with well labelled diagram.
Answer:
Esters as pointed, are pleasant smelling compounds. These are therefore, commonly used as flavouring agents and also in perfumes. When an ester is reacted with water in the presence of a dilute acid like dilute HCl, acid and alcohol are formed as the product. The reaction is called ester hydrolysis.

Ester hydrolysis is the reverse of esterification reaction.
When an ester is reacted with an aqueous solution of base like NaOH or KOH, the product is an alcohol and salt of the acid. For example,

The reaction is known as saponification reaction because it is the basis for the formation of soap.

Question 52.
A compound C (molecular formula, C₂H₄O₂) reacts with Na-metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, C₃H₆O₂). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A.
Identify C, R, A, S and write down the reactions involved.
Answer:
From the available information, it is evident that

1. Compound C with molecular formula C₂H₄O₂ is ethanoic acid (CH₃COOH)
2. The compound R is sodium ethanoate and has formula CH₃COONa
3. Since the compound S has only three carbon atoms (C₃H₆O₂) and has been formed by the action of an alcohol on compound C (C₂H₄O₂), this means that the alcohol A has only one carbon atom. It is methanol (CH₃OH).
4. The compound S with a sweet smell is methyl ethanoate with formula CH₃COOCH₃.

The chemical reactions involved are as follows :

Question 53.
Look at given figure and answer the following questions :

(a) What change would you observe in the calcium hydroxide solution taken in tube B ?
(b) Write the reaction involved in test tubes A and B respectively.
(c) If ethanol is used instead of ethanoic acid, would you expect the same change ?
(d) How can a solution of lime water be prepared in the laboratory ?
Answer:
(a) It would become milky.

In test tube B, calcium hydroxide reacts with carbon dioxide to form calcium carbonate which is milky in colour.

(c) No, it would be different. No chemical reaction is possible between ethanol and sodium carbonate.
(d) Lime w’ater is prepared by keeping suspension of calcium hydroxide overnight in a beaker. The solution is decanted and is transferred to another beaker. It contains traces of calcium hydroxide and is called lime water.

Question 54.
How would you bring about the following conversions ? Name the process and write the reaction involved.
(a) ethanol to ethene.
(b) Methanol to Ethanoic acid. Write the reactions.
Answer:
(a) From ethanol: This method involves the slow oxidation of a dilute solution of ethanol (10-15 per cent) by oxygen present in air in the presence of an enzyme acetobactor.

The acid obtained is in the form of dilute solution called vinegar.
We have also studied under ethanol that it gets oxidised to ethanoic acid in the presence of dilute solution of alkaline KMnO₄ or acidified K₂Cr₂O₇.
(b) From methanol: These days ethanoic acid is manufactured by the reaction between methanol and carbon monoxide in the presence of iodine-rhodium (I₂ — Rh) catalyst mixture.

Question 55.
Draw the possible isomers of the compound with molecular formula C₃H₆O and also give their electron dot structures.
Answer:
Two isomers are possible for the molecular formula C₃H₆O. These are known as functional isomers.

Question 56.
Explain the given reactions with the examples :
(a) Hydrogenation reaction
(b) Oxidation reaction
(c) Substitution reaction
(d) Saponification reaction
(e) Combustion reaction
Answer:
Hydrogenation of oils : The reaction is extremely useful in the hydrogenation of vegetable oils also called edible oils e.g. ground nut oil, cotton seed oil etc. These are also called cooking oils and are unsaturated in the sense that their molecules contain atleast one C=C bond in their structures. Upon passing hydrogen gas through oil in the presence of nickel catalyst, the double bond changes to single bond. As a result, the unsaturated oil changes to solid fat which is of saturated nature. Vegetable ghees such as Dalda, are of saturated nature and are formed by catalytic hydrogenation reaction.

Oxidation reaction :

• Loss of hydrogen is known as oxidation.
• Gain of oxygen is known as oxidation.

Therefore, it is an oxidation reaction.
Substitution Reactions : Substitution reactions are also called replacement reactions and you are quite familiar with these. In organic compounds, particularly the saturated hydrocarbons (or alkanes), these reactions are very common. One or more hydrogen atoms in the molecule of alkane such methane get substituted by chlorine atoms when the reaction is carried with chlorine in the presence of ultraviolet sun light.

Combustion reaction :
We have seen that all combustion reactions are basically oxidation reactions carried in the presence of air or oxygen. It is not necessary that the reactants may burn during combustion.

• Blue flame signifies complete combustion of the fuel.
• Yellow sooty flame signifies incomplete combustion of the fuel.

Question 57.
An organic compound A on heating with concentrated H₂SO₄ forms a compound B which on addition of one mole of hydrogen in presence of Ni forms a compound C. One mole of compound C on combustion forms two moles of CO₂ and three moles of H₂O. Identify the compounds A, B and C and write the chemical equations for the reactions involved.
Answer:
Since one mole of compound C on combustion forms two moles of CO₂ and three moles of H₂O the compound C is a hydrocarbon with formula C₂H₆. It is ethane. The compound B which forms C₂H₆ upon addition of hydrogen is ethene (C₂H₄). The organic compound A which forms ethene upon acidic dehydration is ethanol. The chemical equations for the reactions involved are :

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds helpful to you. If you have any doubts, please comment below. We try to provide online math tutoring for you.

Value Based Questions in Science for Class 10 Chapter 4 Carbon and Its Compounds

Question 1.
Kamla and Reema are best friends. On one evening Kamla went to the house of Reema and found her working in the kitchen. The gas burner was emitting yellow flame instead of blue flame. Kamla immediately asked Reema to put off the gas. She helped her in cleaning the fine holes of the gas burner with a needle. The entire operation took about fifteen minutes. The gas was now ignited and there was a blue flame. Please read the above narration and answer the following questions :

1. Why was burner emitting yellow flame ?
2. What was the purpose of cleaning the holes ?
3. In what way Kamla helped Reema ?
4. What lesson can we learn from this ?

Answer:

1. The holes of the burner were blocked due to soot or some oily material released during cooking. The combustion was incomplete and the gas burnt with yellow flame.
2. By cleaning the holes, the deposits were removed and the combustion was now complete.
3. By doing this operation, there was a saving of fuel. This also helped in checking pollution.
4. We should always keep gas burners and engines of scooters or cars clean so that the combustion is complete and there is no wastage of any fuel. This also checks pollution problem.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Mohan and Sohan were working in a factory and were good friends. On one evening Sohan received a phone . call from Mohan that his eye sight had become very dim all of a sudden. Sohan immediately rushed to his place and came to know that this had occurred after consuming some alcohol. Sohan took him to an eye specialist. He checked his vision properly and gave some antibiotic drops. The specialist kept him under observation and by next morning, he was fully cured.

1. Why did eye sight of Mohan become dim ?
2. How was he cured ?
3. What is the value associated with episode ?

Answer:

1. Mohan had consumed adultrated liquor poisoned either with methyl alcohol or copper sulphate.
2. Antibiotic given by the specialist neutralised the effect of poisoning. As a result, Mohan regained his sight.
3. Timely help by Sohan saved Mohan from getting completely blind. This was need a service rendered by one friend to the other.

Question 3.
Teacher asked Hema to perform test for unsaturation in the laboratory for ethylene gas. She took some chlorine water in a tube and passed the vapours of the gas. Nothing happened Teacher asked her to pass the vapours of the gas into bromine water. The yellow colour of the gas immediately discharged.

1. What was the mistake committed by Hema ?
2. How did teacher help her ?
3. Write chemical equation for the reaction.

Answer:

1. Chlorine water has no colour. Therefore, on passing ethylene gas the colour of chlorine water did not discharge.
2. Bromine water is yellow in colour. When ethylene gas was passed through bromine water, its colour got discharged. This is the test for unsaturation.
3. The chemical equation for the reaction is :
   

Question 4.
A patient was suffering from high blood pressure and high cholesterol.He went to the specialist. He enquired about his eating habits. The patient told the doctor that he consumes both dalda ghee and desi ghee and also drinks milk with full fat. The doctor asked the patient to immediately stop these and instead use vegetable oil and also drink fat free milk.

1. What was wrong with the eating habits of the patient ?
2. How did doctor help him ?

Answer:

1. Both dalda/desi ghee and milk full of fats are very harmful to our body. The fats get deposited and lead to high chrolestrol level and high blood pressure. This may ultimately result in either paralysis or death.
2. Doctor gave the patient correct advice. The diet should be free from fats as much as possible particularly in the present set up when there is lack of exercise.

Question 5.
Ethanol, commonly called alcohol is an excellent solvent, is used in medicines and is an important chemical compound involved in synthesis of many chemical compounds. However in spite of its benefits to man, its impact on social behaviour has always been questioned. Media has often shown abnormal behaviour of people while drunk. It is considered as a curse in the lives of those who are addicted to alcohol – Alcoholic’ people are not only lowering their metabolism and affecting Central Nervous System, they are also a threat to the lives of others. Anger and rude behaviour are some of its ill effects.

1. Comment on the statement – ‘Should production of alcohol be banned’ give three valid reasons to justify.
2. As a student what initiative would you take in the common concern of‘Save Life, Do not Drink’. Give two suggestions.

Answer:
In favour of negative response :

1. Ethanol is used as a solvent in the laboratory.
2. Ethanol is used in the synthesis of large number of compounds.
3. Ethanol is a constituent of many drugs.

In favour of positive response :

1. Ethanol is habit forming.
2. Excessive intake of ethanol is highly injurious to the body.
3. Excessive consumption of ethanol is the major cause of crimes in our society.
4. It is a big strain on the family budget. Many families are ruined.

Initiatives :

1. Students must take a pledge not to drink alcohol.
2. Students must create awareness in the society by organising skits/seminars and also through charts.

Question 6.
School going children generally bring tiffins and they eat food during the break. One category of students marked ‘A’ carry paranthas, butter and pickels while the other category of students marked ‘B’ bring chapatties, vegetables, salad and fruits .Whereas students A’ donot like to share their food, students ‘B’ would like to share it.

1. Which acid is present in pickels ?
2. Which group of students bring healthy food and why ?
3. Which group of students bring unhealthy food and why ?
4. Which group of students have better value system and why ?

Answer:

1. Acetic acid is present in pickels
2. Students marked ‘B’ bring healthy food since it has more neutrition values and it is easy to digest.
3. Students marked A’ bring unhealthy food. Since it is quite difficulty to digest it and leads to obacity.
4. Students marked ‘B’ have better value system. Sharing of food at school level will make them better citizens at a later stage in their life.

Question 7.
Cough syrups, generally contain alcohol. Some people are habitual of drinking ‘alcohol’. Instead of drinking ‘alcohol’, they have started using cough syrups which contains alcohol and cause addiction. To solve this problem, government is thinking to ban cough syrups.

1. What is an alcohol ?
2. Should production of cough syrups be banned ?
3. As a student ‘what initiative would you take to make people aware of harmful effects of taking cough syrups unnecessarily’. Give two suggestions.

Answer:

1. Alcohol is chemically ethyl alcohol and its formula is C₂H₅OH.
2. No, production of all cough syrups cannot be banned since they given relief from cough and cold. However, they should be sold strictly if prescribed by the doctor on a proper prescription.
3. We students must acquaint people around us the harmful effects of alcohols by holding seminars and by door to door convassing.

Hope given Value Based Questions in Science for Class 10 Chapter 4 Carbon and Its Compounds helpful to you.

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HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 1.
An organic compound A of molecular formula C₂H₄ on reduction gives another compound B of molecular formula C₂H₆. B on reaction with chlorine in the presence of sunlight gives C of molecular formula C₂H₅Cl.
(a) Name the compounds A, B and C.
(b) Write chemical equation for the conversion of A to B and name the type of reaction.
Answer:
The compound A of molecular formula C₂H₄ is an alkene. Upon reduction with hydrogen, it gives ‘B’ of molecular formula C₂H₆. The compound ‘B’ upon chlorinadon gives ‘C’ of molecular formula C₂H₅Cl.

The reaction to called addition reaction.

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Name the functional group of organic compounds that can be hydrogenated. With the help of suitable example, explain the process of hydrogenation mentioning the conditions of the reaction and any one change in physical property with the formation of the product. Name any one natural source of organic compounds that are hydrogenated.
Answer:
The functional group which can be easily hydrogenated is
The family is known as alkenes. The hydrogenation reaction can be carried by heating a member of the family (e.g. ethene) with hydrogen in the presence of a catalyst like nickel (Ni) For example.

Ethene is an unsaturated hydrocarbon while ethane is of saturated nature.
Edible oils such as coconut oil, olive oil, peanut oil etc. contain atleast one
in their molecules. These are regarded as unsaturated compounds. Upon hydrogenation, these get converted into fats which no longer contain any

Question 3.
An organic compound A’ of molecular formula C₂H₆O on oxidation with dilute alkaline KMnO₄ solution gives an acid ‘B’ with the same number of carbon atoms. Compound A’ is often used for sterilization of skin by doctors. Name the compound. Write the chemical equation involved in the formation of ‘B’ from A.
Answer:
The compound ‘B’ should contain a —COOH group as it is an acid. Since it has only two carbon atoms, the other carbon atom must represent CH₃ group. Thus, compound ‘B’ is ethanoic acid (CH₃COOH). The compound A used for the sterilization of skin by doctors is ethanol C₂H₅OH (C₂H₆O). The chemical reaction involved in the oxidation by dilute alkaline KMnO₄ solution also called Baeyers reagent is :

Question 4.
A to F are the structural formulae of some organic compounds :

(i) Give the letters which represent the same family.
(ii) Give the letters which do not represent hydrocarbons.
(iii) Flow can ‘C’ be converted into A ?
Answer:
(i) Letters ‘B’ and ‘D’ represent the family of alkynes.
(ii) Letters ‘E’ and ‘F’ donot represent any hydrocarbon.
(iii) ‘C’ can be converted into ‘A’ by passing hydrogen (H₂) in the presence of Ni at 473 K.

Question 5.
(a) A test tube contains a brown liquid in it. The colour of the liquid remains the same when methane is passed through it but it disappears when ethene is passed. Suggest the name of the liquid brown in colour. Give the chemical equation involved.
(b) The formula of an ester is C₃H₇COOC₂H₅. Write the formulae of the acid and alcohol from which the ester is prepared.
Answer:
(a) The brown liquid seems to be bromine dissolved in water. Methane (CH₄) is a saturated hydrocarbon and does not react with bromine. Ethene (C₂H₄) being unsaturated in nature, decolourises bromine and its colour therefore, disappears.

(b) In an ester, the left side in the molecular formula containing C₃H₇CO is derived from the acid while the right side having OC₂H₅ is from the alcohol. This means that the acid and alcohol participating in the ester are C₃H₇COOH and C₂H₅OH respectively. The formation of ester may be shown as follows :

Hope given HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds helpful to you. If you have any doubts, please comment below. We try to provide online math tutoring for you.