Q 1. Give formula for the following: (i) Bromide of element X of second group. (ii) Oxide of element Y of third group. (iii) Chloride of element Z of fourth group.
Solution
(i) Bromide of element X of second group = XBr2 (ii) Oxide of element Y of third group = Y2O3 (iii) Chloride of element Z of fourth group = ZCl4
Q 2. (a) Did Dobereiner’s triads also exist in the columns of Newlands law of octaves? Explain your answer. (b) What were the limitations of Dobereiner’s classification of elements?
Solution
(a) Yes, Dobereiners triads also exist in the columns of Newlands Octaves. Consider the elements lithium (Li), sodium (Na) and potassium (K) which are present in the second column of Newlands classification of elements. Now, if we start with lithium as the 1st element, then the 8th element from it is sodium and according to Newlands law of octaves, the properties of the 8th element, sodium should be similar to that of the 1st element, lithium. Again, if we take sodium as the 1st element, then the 8th element from it is potassium and according to Newlands law of octaves, the properties of the 8th element, potassium should be similar to that of the 1st element, sodium. This means that according to Newlands law of octaves, the elements lithium, sodium and potassium should have similar chemical properties. We also know that lithium, sodium and potassium form a Dobereiner’s triad having similar chemical properties. From this, we conclude that Dobereiners triads also exist in the columns of Newlands Octaves. (b) The main limitation of Dobereiner’s classification of elements was that it failed to arrange all the then known elements in the form of triads of elements having similar chemical properties. Dobereiner could identify only three triads from the elements known at that time. So, his classification of elements was not much successful. Another limitation was that Dobereiner failed to explain the relation between atomic masses of elements and their chemical properties.
Q 3. Hydrogen can be placed in group 1 and group 7 of periodic table. Why?
Solution
Hydrogen forms both positive ions like alkali metals (group 1) and negative ions like halogens (group 7). Thus, it can be placed in both group 1 and group 7. In Mendeleev”s Periodic table, the position of hydrogen was not clear.
Q 4. Give reasons for the need of classification of elements.
Solution
The continuous discovery of new elements and their compounds led to confusions. It became difficult to study, remember and recall the properties of all the elements. Thus, the need to classify and place them in certain groups was felt.
Q 5. Given below are some elements of the modern periodic table : 4Be, 9Fe, 14Si, 19K, 20Ca (i) Select the element that has one electron in the outermost shell and write its electronic configuration. (ii) Select two elements that belong to the same group. Give reasons for your answer. (iii) Select two elements that belong to the same period. Which one of the two has bigger atomic size?
Solution
(i) 19K has one electron in the outermost shell and its electronic configuration is 2, 8, 8, 1. (ii) 4Be and 20Ca belongs to same group i.e. Group 2. Electronic configuration: 4Be – 2, 2 20Ca – 2, 8, 8, 2 4Be and 20Ca have same number of valence electrons in outermost shell i.e. 2 so they belong to same group. (iii) 9F and 4Be belongs to the same period i.e. period 2. Electronic configuration: 9F – 2, 7 4Be – 2, 2 4Be has bigger atomic size than 9F because the atomic radius decreases as we move from left to right due to an increase in nuclear charge which tends to pull the electrons closer to the nucleus and hence size of atom reduces.
Q 6. How does atomic size vary in a group from top to bottom and in a period from left to right? Explain.
Solution
(i) Atomic size increases as we move from top to bottom because each time one new shell is added and electrons in the outermost shell move away from the nucleus. (ii) Atomic size decreases across the period, as electrons are added in the same shell, effective nuclear charge increases and thus, atomic size decreases.
Q 7. An element X has mass number 35 and number of neutrons 18 (a) Write the atomic number of X. (b) Give electronic configuration of X. (c) To which group and period does it belong?
Solution
(a) Atomic number = 35 – 18 = 17 (b) 2, 8, 7 (c) 17thgroup and 3rdperiod
Q 8. Name two elements whose properties were correctly predicted by Mendeleev. Mention their present day name.
Solution
Two elements whose properties were correctly predicted by Mendeleev are: Eka-aluminium – Gallium Eka-silicon – Germanium
Q 9. Lithium, sodium and potassium are all elements that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements? Explain.
Solution
Yes, there is a similarity in the atoms of lithium, sodium and potassium elements. All these elements have a similar electronic configuration having one electron each in their valence shells. The electronic configurations of lithium, sodium and potassium are given below: Lithium: 2, 1 Sodium: 2, 8, 1 Potassium: 2, 8, 8, 1
Q 10. Group → Period ↓ 1 2 13 14 15 16 17 18 I a j II b e g h k III c f i l IV d Consider the table given above and answer the following questions: (i) Name the most reactive metal. (ii) How many shells does ‘d’ have? (iii) Name the element (s) having valency 2. (iv) How many valence electrons does ‘j’ have? (v) Which is more non – metallic, ‘h’ or ‘i’? (vi) The atom of which element is bigger in size, ‘e’ or ‘h’?
Solution
(i) Element ‘b’ is the most reactive. (ii) ‘d’ has 4 shells. (iii) Elements ‘e’ and ‘g’ have valency 2. (iv) ‘j’ has zero valence electrons. (v) Element ‘h’ is more non-metallic. (vi) Element ‘e’ is bigger in size.
Q 11. Account for the following: (a) How is metallic character of an element expressed? (b) Noble gases are placed in a separate group.
Solution
(a) The metallic character of an element is expressed in terms of electron releasing tendency of its atom. As a result, a positive ion is formed. Metals are hence electropositive in nature. (b) Noble gases like helium, neon and argon were discovered very late because they are very inert and present in extremely low concentrations in our atmosphere. Hence, they were placed in a new separate group without disturbing the existing order.
Q 12. Explain why the number of elements in the third period is 8.
Solution
According to the 2n2 rule, the maximum number of electrons in the third period = 2 x (3)2 = 18. But, the last shell cannot accommodate more than 8 electrons so, the number of electrons in third period is 8. Hence, the number of elements is also 8.
Q 13. An element ‘M’ has atomic number 12. (a) Write its electronic configuration. (b) State the group to which ‘M’ belongs. (c) Is ‘M’ a metal or a non-metal? (d) Write the formula of its oxide.
Solution
(a) The electronic configuration of M is 2, 8, 2. (b) M belongs to 2nd group. (c) M is a metal. (d) MO
Q 14. (a) How does the tendency of an element to lose electrons change in a group? (b) How does this tendency change in a period?
Solution
(a) On moving down a group, the number of shells increases at each succeeding element. As a result, the valence electrons lies farther away from the nucleus. Therefore, the force of attraction between nucleus and the electrons decreases and tendency to lose electrons increases down the group. (b) As we move left to right in a period, number of shells remain same but number of valence electrons increases. As a result, effective nuclear charge increases and the tendency to lose electrons decreases.
Q 15. Account for the following: (a) Elements of group 18 are called zero valent. (b) Elements in a group of periodic table have similar chemical properties.
Solution
(a) Elements of group 18 have their outermost shell completely filled. So, they neither gain electron nor lose any electron and hence have zero valency. Thus, they are called zero valent. (b) Chemical properties of elements are determined by the number of valence electrons. Since in a group, all the elements have same number of valence electrons, thus they exhibit similar chemical properties.
Q 16. a. How does valency vary in a group on going from top to bottom? b. How does the number of valence electrons vary in a period on going from left to right and from top to bottom in a group?
Solution
(a) Valency remains same on moving from top to bottom in a particular group. This is because the outermost electronic configuration of all the elements in a group remains same. (b) On moving from left to right in a period, the number of valence electrons in elements increases from 1 to 8. The elements in a period have consecutive atomic numbers. On moving down a group in the periodic table, the number of valence electrons in the elements remains the same.
Q 17. (a) The atomic number has been chosen as the basis for classifying elements. Why? (b) By considering their positions in the Periodic Table, which one of the following elements would you expect to have maximum metallic character? Na, Mg, Al.
Solution
(a) Atomic number decides the electronic configuration of the elements which is responsible for the chemical properties of the elements. (b) Na will have maximum metallic character since metallic character decreases on going from left to right in a period.
Q 18. Elements A, B, C, D, E have following electronic configurations: A: 2, 3 B: 2, 8, 3 C: 2, 8, 5 D: 2, 8, 7 E: 2, 8, 8, 2 (i) Which of these belong to same group? (ii) Which of these belong to same period?
Solution
(i) Element A and B belong to same group since they have the same number of valence electrons. (ii) Element B, C and D belong to same period since they have same number of shells in their atoms.
Q 19. Name the element which has twice as many electrons in its second shell as in its first shell. Write its electronic configuration also.
Solution
Carbon has twice as many electrons in its second shell as in its first shell. K-shell- 2 electrons L-shell- 4 electrons
Q 20. Consider the following elements: 20Ca, 8O, 18Ar, 16S, 4Be, 2He Which of the above elements would you expect to be: (i) Very stable (ii) In group 2 of the periodic table (iii) In group 16 of the periodic table. Give reasons.
Solution
(i) 18Ar and 2He (Noble gases) are very stable because of completely filled shells. (ii) 20Ca and 4Be belong to group 2 since they have two shells only. (iii) 8O and 16S belong to group 16 since they both have 6 valence electrons in their outermost shell.
Q 21. (a) Why physical properties and chemical properties of an element are periodic function of their atomic numbers? (b) How does Modern Periodic Law justifies one position for isotopes?
Solution
(a) Physical and chemical properties of elements are determined by their atomic numbers which is equal to the number of electrons. By using the number of electrons, we can find out the number of valence electrons and hence the physical and chemical properties. (b) According to Modern Periodic Law, elements are arranged in the Modern Periodic Table in the increasing order of their atomic numbers. Isotopes have the same atomic number and different atomic mass. So, though they have different atomic masses still they are given the same position in the Modern Periodic Table.
Q 22. Three elements A, B and C have atomic numbers 7, 8 and 9 respectively. (a) What would be their positions in the modern periodic table (Mention group and period both)? (b) Arrange A, B and C in the decreasing order of their size. (c) Which one of the three elements is most reactive and why?
Solution
(a) Position of A = 15th group, 2nd period Position of B = 16th group, 2nd period Position of C = 17th group, 2nd period (b) A > B > C since atomic size decreases as we move from left to right along a period. (c) C is most reactive as it has smallest size, so it can easily gain electrons.
Q 23. Arrange the following elements in the decreasing order of metallic character: (i) Si, Mg, Na, P (ii) B, Al, Mg, K
Solution
On going down a group, metallic character of elements increases. On moving from left to right in a period, metallic character of elements decreases. (i) Na > Mg > Si > P (ii) K > Mg > Al > B
Q 24. A part of modern periodic table is given below. Answer the following questions based on this table. H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar (a) Why do H, Li and Na show similar properties? (b) Atomic size of Mg is bigger than Be. Why? (c) Why are He, Ne and Ar called noble gases? (d) Write a common name of the family to which F and Cl belong. (e) Write the trend of non – metallic character in the horizontal row from Na to Cl. (f) How does the atomic size vary as we move from Li to F in the second period of the periodic table?
Solution
(a) H, Li and Na show similar properties because they have one electron in their valence shell and belong to same group. (b) Atomic size of Mg is bigger than Be because Mg consists of three shells whereas Be consist of 2 shells. This increases the distance between outermost electrons and the nucleus. (c) He, Ne and Ar are called noble gases because their outermost shell is complete and their combining capacity is zero i.e. they are least or less reactive. (d) Halogen family (e) Non-metallic character increases from Na to Cl. (f) Atomic size decreases as we move from Li to F in the second period of the periodic table.
Q 25. (a) What were the achievements of Mendeleev’s periodic table? (b) What were the basic properties of elements considered by Mendeleev for their classification in the periodic table?
Solution
(a) The achievements of Mendeleev’s periodic table were: Mendeleev kept some blank spaces in the periodic table for the elements that were yet to be discovered. Predicted element Actual element discovered later Eka-boron Scandium Eka-aluminium Gallium Eka-silicon Germanium He also predicted properties of some elements even before their discovery and were later found to be correct. Property Eka-aluminium Gallium Atomic mass 68 69.7 Formula of oxide E2O3 Ga2O3 Formula of chloride ECl3 GaCl3 Mendeleev’s periodic table could accommodate noble gases when they were discovered. (b) Mendeleev took the formulae of the oxides and hydrides formed by the elements as the basic properties of elements for their classification in the periodic table.
Q 26. Write one property of hydrogen which makes it resemble with (a) Alkali metals (b) Halogens.
Solution
Resemblance with alkali metals: Hydrogen has the same outermost electronic configuration as that of alkali metals. Resemblance with halogens: Hydrogen exists as diatomic molecule as halogens.
Q 27. Define atomic radius of an element. How does it vary along the period and group?
Solution
The distance between the centre of the nucleus and the outermost orbit of the atom is called the atomic radius of the atom of an element. Variation along the period: – The atomic radius decreases as we move from left to right along the period. This is due to an increase in nuclear charge which tends to pull the electrons closer to the nucleus and reduces the size of the atom. Variation along the group: – The atomic radius increases down the group. This is because new shells are being added as we go down the group. This increases the distance between the outermost electrons and the nucleus so the atomic size increases inspite of the increase in nuclear charge.
Q 28. How can the valency of an element be determined if its electronic configuration is known? What will be the valency of an element of atomic number 9?
Solution
An element’s valency is determined by the number of electrons in its outer shell. Hence, the number of valence electrons obtained from the electronic configuration of the element gives the valency i.e. the number of electrons lost, gained or shared by the element to attain the noble gas configuration. The valency of an element of atomic number 9 would be 1 since the number of valence electrons in its outermost shell is 7 so it needs only one electron to attain the noble gas configuration.
Q 29. (a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements? If yes, write the similarity. (b) Helium is a non-reactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Solution
(a) Yes, the atoms of all the three elements lithium, sodium and potassium have one electron each in their outermost shells. (b) Both helium (He) and neon (Ne) have filled outermost shells. Helium has a duplet in its K shell, while neon has an octet in its L shell.
Q 30. Which physical and chemical properties of the elements were used by Mendeleev in creating his periodic table? List two observations which posed a challenge to Mendeleev’s periodic table.
Solution
The creation of Mendeleev’s periodic table was based upon certain physical and chemical properties: Physical properties: The atomic masses of the elements were taken into account and the elements were arranged in order of increasing atomic masses. This influenced some of their physical properties like melting points, boiling points, density etc. Chemical properties: The distribution of the elements into different groups was linked with formation of hydrides by combining with hydrogen and formation of oxides by combining with oxygen. This is linked with the valency of the elements. The two main observations which posed challenge to Mendeleev’s periodic table are as follows: (i) Position of isotopes: Since the isotopes of an element differ in their atomic masses, they must be assigned separate positions in the periodic table. (ii) Anomalous positions of some elements: In the Mendeleev’s periodic table, certain elements with higher atomic masses precede or placed before the elements with lower atomic masses. For example: the elements Argon (At. mass = 39.9) is placed before the element Potassium (At. Mass = 39.1).
Q 31. An element X has atomic number 19. (a) Write its electronic configuration. (b) To which group of the modern periodic table does it belong? (c) State the nature of the compound formed by element X with chlorine. (d) Write the valency of element X.
Solution
(a) Electronic configuration of X = 2,8,8,1 (b) Group1 (c) ‘X’ will form ionic (electrovalent) compound with chlorine. (d) Valency of X= +1
Q 32. An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide. (a) What will be the formula of the halide formed by X and Y? (b) What is the position of elements X and Y in periodic table? (c) What will be the nature of oxide of element Y. Identify the nature of bonding in the compound formed.
Solution
X – non-metal Y – metal (a) Since it is a divalent halide, so formula of halide is YX2 (b) X: Position- Group 17, period – 3 Y: Position – Group2, period – 4 (c) Basic oxide – YO, ionic bond will be formed between Y (metal) and oxygen (non-metal).
Q 33. Properties of some elements are given below. Identify in each case the element in the periodic table to which it belongs. (a) A soft metal stored under kerosene. (b) An element with variable valency stored under water. (c) An element which is tetra valent and forms the basis of organic chemistry. (d) An element which is an inert gas with atomic number 2. (e) A metal which burns with brilliant light when ignited. (f) An element which is yellow solid at room temperature that shows catenation and allotropy.
Solution
(a) Sodium (b) Phosphorus (c) Carbon (d) Helium (e) Magnesium (f) Sulphur
Q 34. Ria and Rama are students of Class-X. Ria is very much organized and maintained. All teachers love her. She earns a great respect in the class where as Rama is un-organised and always faces a lot of problems in handling life situations. (i) In your opinion how does organization skill help in daily life? (ii) How can you relate the above fact with the chapter classification of elements. (iii) How classification of elements help us in studying them properly?
Solution
(i) Organisation makes our life simple, easy and systematic. (ii) As classification and organization help us in our daily life in the same way classification of elements has made the study of elements easier and simple. (iii) We can know about the properties of elements because of this classification. Associated Value: The learner will be motivated to practice an organized life with discipline.
Q 35. (a)State Modern periodic law. (b)How, (i) tendency to lose electron, (ii) Valency of the elements varies along the period from left to right? Give reasons to justify your answers.
Solution
(a)Modern periodic law states that “properties of elements are periodic functions of their atomic numbers”. (b) (i) Tendency to lose electrons decreases because the effective charge acting on the valence shell electron increases. (ii) Valency increase from 1 to 4 and then decrease to zero. Group No. 1, 2, 13, 14, 15, 16, 17, 18 valency 1, 2, 3, 4, 3, 2, 1, 0
Q 36. In the modern periodic table, the element Calcium (atomic number = 20) is surrounded by elements with atomic number 12, 19, 21 and 38. Which of these elements has physical and chemical properties resembling those of Calcium and why?
Solution
Ca: Electronic configuration is 2, 8, 8, 2 The physical and chemical properties of elements with atomic number 12 and 38 will resemble with calcium. This is because they all belong to the second group and all of them have two electrons in their valence shell.
Q 37. Why are isotopes of an element having different atomic masses placed at the same position in the periodic table?
Solution
All isotopes of an element have same number of protons and hence they have the same atomic number. Hence, they can be placed at the same position in the periodic table.
Q 38. Where are metals, non-metals and metalloids located in the periodic table?
Solution
(i) metals – left (ii) non -metals – right (iii) metalloids – middle
Q 39. The position of three elements A, B and C in the periodic table are shown below: Group 16 Group 17 – – – B – A – C (a) State whether A is a metal or non – metal. (b) State whether C is more reactive or less reactive than A. (c) Will C be larger or smaller in size than B? (d) Which type of ion, cation or anion, will be formed by element A?
Solution
(a) A – non-metal (b) C – less reactive than A (c) C – smaller in size than B (d) A – will form anion
Q 40. A part of the Periodic Table has been shown below. Group → 1 2 13 14 15 16 17 18 Period ↓ 1 2 A C E G 3 B D F Answer the following questions on the basis of position of elements in the above table. (i) Which element is a noble gas? Give reason. (ii) Which element is most electronegative? Give reasons. (iii) Write the electronic configuration of (i) B and (ii) E.
Solution
(i) G – It is a noble gas and has zero valency. (ii) E is the most electronegative element because of its smallest atomic size and more electron affinity. (iii) B – 2, 8, 1 E – 2, 7