NCERT Exemplar ProblemsPERIODIC CLASSIFICATION OF ELEMENTS
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Multiple Choice Questions:
Q-1: Up to which element, the law of octaves was found to be applicable?
(a) Oxygen (b) Calcium (c) Cobalt (d) Potassium
Solution: (b)
Newlands’s law of octaves was applicable only to lighter elements having atomic masses upto 40 u i.e., upto calcium. After calcium, every eighth element did not possess properties similar to that of the first element.
Q-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
Mendeleev’s periodic law states that, the physical and chemical properties of elements are a periodic function of their atomic masses.
Solution: (c)
According to Mendeleev’s periodic law, the elements were arranged in the periodic table in the order of increasing atomic masses.
Q-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
Solution: (a)
Mendeleev’s left some gaps in the periodic table for those elements which were not known at that time.
Germanium element found a place in the periodic table later and Mendeleev’s predictions were found to be remarkably correct.
Q-4: Which of the following statement(s) about the modern periodic table are incorrect?
(i) Their 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) Their elements in the modern periodic table are arranged on the basis of their increasing atomic number.
(a) Only (i) (b) (i), (ii) and (iii) (c) (i), (ii) and (iv) (d) Only (iv)
In the modern periodic table or in the long form of the periodic table, the elements are arranged in increasing order of their atomic numbers. This table was prepared by Bohr and is based upon the electronic configuration of elements.
Solution: (b)
Only statement 4 is correct. All the elements in the modern periodic table are arranged on the basis of their increasing atomic number. All the isotopes can be placed at one place in the same group of the periodic table.
Q-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 rows known as periods
(c) It has 18 vertical columns known as groups
(d) It has 7 horizontal rows known as groups
Solution: (c)
Modern periodic table has 18 vertical columns, known as groups and 7 horizontal rows, known as periods.
Q-6:Which of the given elements A, B, C, D and E with atomic number 2, 3, 7, 10 and 30 respectively belong to the same period?
(a) A, B, C (b) B, C, D
(c) A, D, E (d) B, D, E
The number of elements in any period is fixed by the maximum number of electrons that can be accommodated in that particular shell. The maximum number of electrons that can be accommodated in any shell is given by the formula where, n is the number of the given shell starting form the nucleus.
Solution: (b)
2nd period contains elements with atomic number 3(Li), 7(N), 10(Ne).
Since, 2nd period has elements having atomic number 3 to 10.
Q-7:The elements A, B, C D and E have atomic number 9, 11, 17, 12 and 13 respectively. Which pair of elements belong to the same group?
(a) A and B (b) B and D
(c) A and C (d) D and E
Solution: (c)
Electronic configuration of A (atomic number 9) is 2, 7.
Electronic configuration of B (atomic number 11) is 2, 8, 1
Electronic configuration of C (atomic number 17) is 2, 8, 7
Electronic configuration of D (atomic number 12) is 2, 8, 2
Electronic configuration of E (atomic number 13) is 2, 8, 3
Elements which differ in atomic number by 8 i.e., 9 (A fluorine) and 17
(Cl chlorine) lie in the same group i.e., group 17 (halogens).
Q-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
:
Group 18 elements are known as noble gases or inert gases. They are helium, neon argon, krypton etc. All the noble gases have 8 electrons in their outermost shell (except helium which has only 2 electrons) in the K valence shell. Thus, the valence shell of all the noble gases are completely filled with electrons.
Solution: (c)
Element with electronic configuration 2, 8 is an inert gas i.e., neon and hence, belongs to group 18,
Q-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
Solution: (b)
Constituent of all organic compounds is carbon. It belongs to group 14.
Q-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
Solution: (b)
The elements of 2nd period involve the filling of 2nd shell i.e., L shell. Because in period 2, there are two shell, K and L.
Q-11: Which one of the following elements exhibit maximum number of valence electrons?
(a) Na (b) Al (c) Si (d) PT
Amongst representative elements, the number of valence electrons increases as the group number increases from left to right.
Solution: (d)
Thus, Na (group 1) has one, Al (group 13) has three (13-10), SI (group 14) has four (14-10) and P (group 15) has five (15-10) valence electrons. Therefore, P has maximum number of valence electrons i.e., 5 (maximum among the given).
Q-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, NTAtomic radii is the distance between the centre of the nucleus and the outermost shell which contains electrons in an isolated system
Solution: (d)
As we move from left to right in a period, the atomic number of each succeeding element increases by 1. So, the electrons are attracted close to the nucleus and hence, the atomic size decreases, along period from left to right.
Therefore, the increasing order of radii is F, O, N as atomic number of F, O and N are 9.8 and 7 respectively. Due to which F belongs to 17th group. O belongs to 16th group and N belongs to 15th group.
Q-13:Which among the following elements has the largest atomic radii?
(a) Na (b) Mg (c) K (d) Ca
Thinking Process:
Atomic radii increases on moving down the groups as new shell of electrons is added at each succedding element.
Solution: (c)
Na and K are the same group and K is below than Na, so K will have higher atomic radius i.e., K > Na. In a period on moving from left to right, atomic radius decreases.
Since, K and Ca are in the same period and K is in 1 group and Ca is in 2 group, so atomic radius of K will be more than Ca, i.e., K > Ca.
Also Na and Mg are in the same period, but Na belongs to 1 group and Mg belong to 2 group, so atomic radius of Na is more than Mg, i.e., Na > Mg. Thus, if we take all these together we get K > Na > Mg and K > Ca > Mg. Hence we can say the atomic radius of K is largest.
Q-14:Which of the following elements would lose an electron easily?
(a) Mg (b) Na (c) K (d) Ca
Solution: (c)
Electronic configuration of Mg (atomic number 12)2, 8, 2.
EC of Na (atomic number 11) 2, 8, 1
E. configuration of K (atomic number 19) 2, 8, 8, 1
Electronic config of Ca (atomic number 20) 2, 8, 8, 2
From the above electronic configuration it is clear that K and Na will lose electron easily to achieve stable configuration. But out of K and Na, K will lose electron more easily because the force of attraction on valence electron of K is least among the given.
In other words, larger the atomic radius of an element more easily it can lose its valence electron. Among all these, K has the largest atomic radius, therefore, it can lose its valance electron more easily.
For atomic size refer to Q.13.
Q-15: Which of the following elements does not lose an electron easily?
(a) Na (b) F (c) Mg (d) Al
:
Elements of 2nd period have smaller size than those of the corresponding elements of the 34d period. Further in a period, halogens have the smallest size. F belongs to 2nd period and rest three to 3rd period.
Solution: (b)
Smaller the atomic radius of an element, more difficult is lose electron. Among Na, F, Mg and F has the smallest size. Also it contains 7 electrons in its valence shell and it is most electronegative element. Hence, it does not lose an electron easily rather it gains an electron easily.