Students should pay attention to the following important questions topics for class 10 science chapter 5 – Periodic Classification of Elements:
- Triads of Döbereiner
- The Law of Octaves of Newlands
- Ordering Chaos – Mendeléev’s Periodic Table
- Modern Periodic Table Positions of Elements
- Modern Periodic Table Trends
These are some of the important questions topic for class 10 science chapter 5 – Periodic Classification of Elements from an examination point of view.
Class 10 Science Important Questions Chapter 5 – Periodic Classification of Elements
Q1. Why did Mendeleev leave some gaps in the Periodic table?
Mendeleev left some gaps in the periodic table for yet to be discovered elements. Mendeleev predicted the properties of these elements on the basic of their positions. For example, he predicted the properties of gallium (eka-aluminium) and germanium (eka-silicon) which were unknown at that time.
Q2. An element ‘X’ is forming an acidic oxide. Its position in modern periodic table will be
(a) group 1 and period 3
(b) group 2 and period 3
(c) group 13 and period 3
(d) group 16 and period 3.
(d) : As the element X forms an acidic oxide, hence ‘X is a non-metal. Hence, X is sulphur.
Q3. Write any one difference in the electronic configurations of group 1 and group 2 elements.
Group 1 elements have one electron in their outermost shell while group 2 elements have two electrons in their outermost shell.
Q4. Write the atomic numbers of two elements ‘X’ and ‘Y’ having electronic configurations 2, 8, 2 and 2, 8, 6 respectively.
Electronic configuration of X = 2, 8, 2
∴ Atomic number = 2+ 8 + 2 = 12 Similarly,
Electronic configuration of Y = 2, 8, 6
∴ Atomic number = 2 + 8 + 6 = 16
Q5. Out of the three elements P, Q and R having atomic numbers 11, 17 and 19 respectively, which two elements will show similar properties and why?
Atomic number of P = 11 Electronic configuration of P = 2, 8,1 Electronic configuration of Q(17) = 2, 8, 7 and for R(19) = 2, 8, 8, 1
Thus, from electronic configurations of P and R, it is observed that they belong to group 1 as both have one valence electron and have valency equal to 1. Thus, P and R will have similar properties.
Q6. How can it be proved that the basic structure of the Modern Periodic Table is based on the electronic configuration of atoms of different elements?
Electronic configuration of an element decides its position in the Modern periodic table.
Let’s take an example of sodium (Na).
Atomic number of sodium = 11
Thus, electronic configuration of Na = 2, 8, 1 As Na contains 1 electron in its outermost shell, it belongs to group 1. Sodium contains 3 shells so, it belongs to period number 3.
Thus, we can conclude that
Group number = Number of valence electrons
(When valence electrons are 1 and 2) and group number = 10 + valence electrons
(When valence electrons are 3 and above) Period number = Number of shells in which electrons are filled.
Q7. 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(nine)?
Valency of an element is determined by the number of electrons present in its outermost shell. For elements having outermost electrons 1 to 4, valencies are equivalent to their respective valence electrons.
For elements having outermost electrons 5 to 8, valency is calculated as;
Valency = 8 – (Number of valence electrons)
For element having atomic number = 9
Electronic configuration = 2, 7
Valency = 8 – 7 = 1
Q8. Can the following groups of elements be classified as Dobereiner’s triad?
(a) Na, Si, CI
(b) Be, Mg, Ca
Atomic mass of Be 9; Na 23; Mg 24; Si 28; C| 35; Ca 40
Explain by giving a suitable reason.
(a) Na, Si, and Cl are not a Dobereiner’s triad. Although the atomic mass of silicon (Si) is the average of atomic masses of sodium (Na) and chlorine (Cl), but these elements do not possess similar properties. Hence, it can’t be classified as a Dobereiner’s triad.
23 (Na) + 35 (Cl) / 2 = 29 (Si)
(b) Be, Mg, and Ca is a Dobereiner’s triad. They have similar properties, and the atomic masses of magnesium (Mg) is approximately the average of the atomic mass of Be and Ca.
9 (Be) + 40 (Ca) / 2 = 24.5 (Si)
Q9. Identify and name the metals from the following elements whose electronic configurations are given below.
(a) 2, 8, 2
(b) 2, 8, 1
(c) 2, 8,7
(d) 2, 1
(a) Magnesium; It is a metal.
(b) Sodium; It is a metal.
(c) Chlorine; It is a nonmetal.
(d) Lithium; It is a metal.
Q10. 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.
Based on 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 among them.
(c) Name the element with the smallest atomic size
(d) Name the element which is a metalloid
(e) Name the element that shows maximum valency.
(a) The most electropositive element among them is Lithium.
(b) The most electronegative element among them is Fluorine.
(c) The element with the smallest atomic size is Fluorine.
(d) Boron (5) is a metalloid.
(e) Carbon shows maximum valency.