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Periodic classification class 10 notes,test paper, Board Questions for SA – II
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Periodic classification class 10 Questions from: Summative Assessment – II
In 1789, Lavoisier first attempted to classify the elements into two divisions namely Metals and Non-metals.
Gist of Lesson for Quick Revision (By JSUNIL)
Question: If an element A is a member of group 14, write the formulae of its chloride and oxide. Predict the nature of bonding in the compound formed. Give reason for your answers.
Ans: A member of group 14 have valecy = 4 So, there is Covalent bonding
Chloride - Acl4 Belong to Gr.14 valency is 4
Oxide – AO2
Question: Among O2- ion and Mg+2 ion, which one have larger size and why?
Answer: Mg+2 has more number of proton and experience higher nuclear forces than that of O2- . Therefore the size of O2- is larger than that of Mg+2
Question: Explain how the tendency to gain electron changes from left to right across the period.
Ans: From left to right across the period atomic size reducess . As a result, effective nuclear charge increases and hence the tendency to gain electrons(electronegativity) increases
Question: Explain how the tendency to gain electrons change on moving down a group.
Ans: On moving down the group, atomic size increases . As a result, effective nuclear charge decreases and hence the tendency to gain electrons(electronegativity) decreases
Q. What happens to the metallic character of the elements as we move in a period from left to right in the periodic table? Give reason.
Ans: Metallic character of the element decreases along a period . In period atomic size decrease,As a result, effective nuclear charge increases and hence the tendency to loose electrons(electroposivity) decreases .
Question: Cl (17) is surrounded by F(9), S(16), Ar(18) and Br(35) in the Modern Periodic Table.
(a) Which amongst them would have same number of
(i) Shells as Chlorine
(ii) Valence electrons as chlorine
(b) Which amongst them would have chemical properties similar to chlorine ?
Ans: (a) (i) S and Ar (ii) F and Br (b) F and Br
Question: (a) Discuss any two achievements of Modern Periodic Table.
(b) Discuss any two limitations of Mendeleev’s classification.
(c) How were these removed in the Modern Periodic Table (any one)
Ans:
(a) (i) Prediction of properties of elements could be made with more precision.
(ii) Elements were sequentially arranged in increasing order of atomic number. (no reversals were required)
(iii) Number of elements between two elements was countable.
(b) (i) Position of Hydrogen (ii) Positioning of Isotopes (iii) Position exchange of Ni and Co. (iv) Variable increase in mass from one element to other.
(c) When the elements are arranged according to their atomic numbers on the basis of modern periodic law, all the anomalies (defects) of Mendeleev’s classification disappear. For example, Position of isotopes: All the isotopes of an element have the same number of protons, so their atomic number is also the same. Since all the isotopes of an element have the same atomic number, they can be put at one place in the same group of the periodic table
Q. Give one example of each:
(a) Metal having valency 2.
(b) Non metal having valency 2.
(c) Element with completely filled outermost shell.
(d) Element with three shells, having 4 electrons in the outermost shell.
Ans: (a) Mg (b) Oxygen (c) he (iv) silicon
Q.From the following elements: Be(4),F(19),K(19),Ca(20) [2015]
(i) Select the element having one element in outermost shell
(ii)two elements of the same group
(iii)write the formula of and mention the nature of the compound formed by the union of K and X(2,8,7)
Ans: (i) K / Potassium. (ii) Be and Ca. (iii) the formula of the compound KX or KCl , Ionic / Electrovalent.
Q. From the following elements : Na,Mg and Al , which one (i) large atomic size (ii)is least reactive. [2015]
Ans: (i) i) Na / Sodium. Reason – The atomic size decreases from left to right due to the increase in the nuclear charge
(ii) Al / Aluminium. Reason – The tendency to lose electrons decreases from left to right.
Q.Write the main aim of classifying elements. Name the basic property of elements used in the development of Modern Periodic Table. State the Modern Periodic Law. On which side (part) of the Modern Periodic Table do you find metals, metalloids and non-metals ?[2015]
Ans: The main aim of classifying elements :For systematic and simplified study of elements and their compounds. ½
Basic property of elements used in the development of Modern Periodic Table : Atomic Number. ½
Modern periodic Law: The properties of elements are a periodic function of their atomic number. ½
Metals are found on the left side and centre of the Modern Periodic Table. ½
Metalloids are found in a zig-zag manner between the metals and the nonmetals. ½
Non-metals are found on the right side of the Modern Periodic Table. ½
Q.The atomic number of an element ‘X’ is 20.
(i) Determine the position of the element ‘X’ in the periodic table.
(ii) Write the formula of the compound formed when ‘X’ reacts/combines with another element ‘Y’ (atomic number 8).
(iii) What would be the nature (acidic or basic) of the compound formed ? Justify your answer.[2015]
Ans: Electronic configuration : 2 ,8 . 8 , 2 ½
i) ‘X’ is present in the 2nd group and 4th period of the periodic table. ½ , ½ ii) XY ½
iii) Basic because X is a metal and the oxides of metals are basic in nature. (Y , Atomic number= 8 , oxygen ) ½ , ½
Q.Four elements P, Q, R and S belong to the third period of the Modern Periodic Table and have respectively 1, 3, 5 and 7 electrons in their outermost shells.
Write the electronic configurations of Q and R and determine their valences.
Write the molecular formula of the compound formed when P and S combine.
Ans: Electronic configuration of Q : 2 , 8 , 3 Valency of Q : 3 Electronic configuration of R : 2 , 8 . 5
Valency of R : 8 – 5 = 3 Electronic config. of P : 2 , 8 , 1 Electronic config of S : 2 , 8 , 7
Q.Two elements ‘P’ and ‘Q’ belong to the same period of the modern periodic table and are in Group-1 and Group-2 respectively. Compare their following characteristics in tabular form :
(a) The number of electrons in their atoms (b) The sizes of their atoms
(c) Their metallic characters (d) Their tendencies to lose electrons
(e) The formula of their oxides (f) The formula of their chlorides [2015]
Ans: Property P Q
(a) No. of electrons in the atom 3 4
11 12
19 20
(b) Size of the atom Bigger Smaller
(c) Metallic character More metallic Less metallic
(d) Tendency to lose electrons More Less
(e) Formula of oxides P2O QO
(f) Formula of chlorides PCl QCl2
Q.Taking the example of an element of atomic number 16, explain how the electronic configuration of the atom of an element relates to its position in the modern periodic table and how valency of an element is calculated on the basis of its atomic number.[2015]
Ans:Electronic configuration of element with atomic no. 16 is 2,8,6.
Since the no. of valence electrons is 6, the group no. will be 10 + 6 = 16.
Valency of the element will be 8- valence electrons ie 8 – 6 = 2.
Q. Why was it necessary to change the basis of classification from atomic mass to atomic number ?
Ans: This is because atomic no. of every element is unique but atomic mass may be same like sotopes.
Q. A quiz contest was being held in the school for chemistry students. The quiz-master said :
An element has the electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) Which of the elements N, F, P and Ar shows similarity with this element ?
(c) We daily use a compound of this element in our food. What is that ?
(d) A compound of this element causes hardness of water. What is that ?
Ans: (a) Atomic number=17 (b) F - as it belongs to the same group as the element(Cl)
(c)NaCl-Common Salt (d)Ca(HCO3)2 causes temporary hardness of water.
In 1789, Lavoisier first attempted to classify the elements into two divisions namely Metals and Non-metals.
Gist of Lesson for Quick Revision (By JSUNIL)
Question: If an element A is a member of group 14, write the formulae of its chloride and oxide. Predict the nature of bonding in the compound formed. Give reason for your answers.
Ans: A member of group 14 have valecy = 4 So, there is Covalent bonding
Chloride - Acl4 Belong to Gr.14 valency is 4
Oxide – AO2
Question: Among O2- ion and Mg+2 ion, which one have larger size and why?
Answer: Mg+2 has more number of proton and experience higher nuclear forces than that of O2- . Therefore the size of O2- is larger than that of Mg+2
Question: Explain how the tendency to gain electron changes from left to right across the period.
Ans: From left to right across the period atomic size reducess . As a result, effective nuclear charge increases and hence the tendency to gain electrons(electronegativity) increases
Question: Explain how the tendency to gain electrons change on moving down a group.
Ans: On moving down the group, atomic size increases . As a result, effective nuclear charge decreases and hence the tendency to gain electrons(electronegativity) decreases
Q. What happens to the metallic character of the elements as we move in a period from left to right in the periodic table? Give reason.
Ans: Metallic character of the element decreases along a period . In period atomic size decrease,As a result, effective nuclear charge increases and hence the tendency to loose electrons(electroposivity) decreases .
Question: Cl (17) is surrounded by F(9), S(16), Ar(18) and Br(35) in the Modern Periodic Table.
(a) Which amongst them would have same number of
(i) Shells as Chlorine
(ii) Valence electrons as chlorine
(b) Which amongst them would have chemical properties similar to chlorine ?
Ans: (a) (i) S and Ar (ii) F and Br (b) F and Br
Question: (a) Discuss any two achievements of Modern Periodic Table.
(b) Discuss any two limitations of Mendeleev’s classification.
(c) How were these removed in the Modern Periodic Table (any one)
Ans:
(a) (i) Prediction of properties of elements could be made with more precision.
(ii) Elements were sequentially arranged in increasing order of atomic number. (no reversals were required)
(iii) Number of elements between two elements was countable.
(b) (i) Position of Hydrogen (ii) Positioning of Isotopes (iii) Position exchange of Ni and Co. (iv) Variable increase in mass from one element to other.
(c) When the elements are arranged according to their atomic numbers on the basis of modern periodic law, all the anomalies (defects) of Mendeleev’s classification disappear. For example, Position of isotopes: All the isotopes of an element have the same number of protons, so their atomic number is also the same. Since all the isotopes of an element have the same atomic number, they can be put at one place in the same group of the periodic table
Q. Give one example of each:
(a) Metal having valency 2.
(b) Non metal having valency 2.
(c) Element with completely filled outermost shell.
(d) Element with three shells, having 4 electrons in the outermost shell.
Ans: (a) Mg (b) Oxygen (c) he (iv) silicon
Q.From the following elements: Be(4),F(19),K(19),Ca(20) [2015]
(i) Select the element having one element in outermost shell
(ii)two elements of the same group
(iii)write the formula of and mention the nature of the compound formed by the union of K and X(2,8,7)
Ans: (i) K / Potassium. (ii) Be and Ca. (iii) the formula of the compound KX or KCl , Ionic / Electrovalent.
Q. From the following elements : Na,Mg and Al , which one (i) large atomic size (ii)is least reactive. [2015]
Ans: (i) i) Na / Sodium. Reason – The atomic size decreases from left to right due to the increase in the nuclear charge
(ii) Al / Aluminium. Reason – The tendency to lose electrons decreases from left to right.
Q.Write the main aim of classifying elements. Name the basic property of elements used in the development of Modern Periodic Table. State the Modern Periodic Law. On which side (part) of the Modern Periodic Table do you find metals, metalloids and non-metals ?[2015]
Ans: The main aim of classifying elements :For systematic and simplified study of elements and their compounds. ½
Basic property of elements used in the development of Modern Periodic Table : Atomic Number. ½
Modern periodic Law: The properties of elements are a periodic function of their atomic number. ½
Metals are found on the left side and centre of the Modern Periodic Table. ½
Metalloids are found in a zig-zag manner between the metals and the nonmetals. ½
Non-metals are found on the right side of the Modern Periodic Table. ½
Q.The atomic number of an element ‘X’ is 20.
(i) Determine the position of the element ‘X’ in the periodic table.
(ii) Write the formula of the compound formed when ‘X’ reacts/combines with another element ‘Y’ (atomic number 8).
(iii) What would be the nature (acidic or basic) of the compound formed ? Justify your answer.[2015]
Ans: Electronic configuration : 2 ,8 . 8 , 2 ½
i) ‘X’ is present in the 2nd group and 4th period of the periodic table. ½ , ½ ii) XY ½
iii) Basic because X is a metal and the oxides of metals are basic in nature. (Y , Atomic number= 8 , oxygen ) ½ , ½
Q.Four elements P, Q, R and S belong to the third period of the Modern Periodic Table and have respectively 1, 3, 5 and 7 electrons in their outermost shells.
Write the electronic configurations of Q and R and determine their valences.
Write the molecular formula of the compound formed when P and S combine.
Ans: Electronic configuration of Q : 2 , 8 , 3 Valency of Q : 3 Electronic configuration of R : 2 , 8 . 5
Valency of R : 8 – 5 = 3 Electronic config. of P : 2 , 8 , 1 Electronic config of S : 2 , 8 , 7
Q.Two elements ‘P’ and ‘Q’ belong to the same period of the modern periodic table and are in Group-1 and Group-2 respectively. Compare their following characteristics in tabular form :
(a) The number of electrons in their atoms (b) The sizes of their atoms
(c) Their metallic characters (d) Their tendencies to lose electrons
(e) The formula of their oxides (f) The formula of their chlorides [2015]
Ans: Property P Q
(a) No. of electrons in the atom 3 4
11 12
19 20
(b) Size of the atom Bigger Smaller
(c) Metallic character More metallic Less metallic
(d) Tendency to lose electrons More Less
(e) Formula of oxides P2O QO
(f) Formula of chlorides PCl QCl2
Q.Taking the example of an element of atomic number 16, explain how the electronic configuration of the atom of an element relates to its position in the modern periodic table and how valency of an element is calculated on the basis of its atomic number.[2015]
Ans:Electronic configuration of element with atomic no. 16 is 2,8,6.
Since the no. of valence electrons is 6, the group no. will be 10 + 6 = 16.
Valency of the element will be 8- valence electrons ie 8 – 6 = 2.
Q. Why was it necessary to change the basis of classification from atomic mass to atomic number ?
Ans: This is because atomic no. of every element is unique but atomic mass may be same like sotopes.
Q. A quiz contest was being held in the school for chemistry students. The quiz-master said :
An element has the electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) Which of the elements N, F, P and Ar shows similarity with this element ?
(c) We daily use a compound of this element in our food. What is that ?
(d) A compound of this element causes hardness of water. What is that ?
Ans: (a) Atomic number=17 (b) F - as it belongs to the same group as the element(Cl)
(c)NaCl-Common Salt (d)Ca(HCO3)2 causes temporary hardness of water.
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