Past Year - 3 Mark Questions

Class 10
Chapter 5 Class 10 - Periodic Classification of Elements

## Study the data of the following three categories A. B and C.

 Category Name of the element and Atomic Mass Checking triad A Li (Atomic mass = 7) Na (Atomic mass = 23) K (Atomic mass = 39) Mean of first and last = (7 + 39)/2 = 23 Atomic mass of middle = 23 Since they are the same , it is a dobereiner’s triad. B N (Atomic mass = 14) P (Atomic mass = 31) As (Atomic mass = 74) Mean of first and last = (14 + 74)/2 = 44 Atomic mass of middle = 31 Since they are not the same , it is not a dobereiner’s triad. C B (Atomic mass = 10.8) Al (Atomic mass = 27) Ga (Atomic mass = 69.7) Mean of first and last = (10.8 + 69.7)/2 = 40.25 Atomic mass of middle = 27 Since they are not the same , it is not a dobereiner’s triad.

Therefore, category 3 is a dobereiner’s triad.

### Why did Mendeleev place elements of category  A, B and C in three different groups?

• Mendeleev arranged the elements according to
1. increasing atomic masses
2. The elements having similar chemical properties were placed together
• The elements of category A, B and C are similar among themselves ( have similar chemical properties ) but each category is different from each other.

Therefore, category A,B and C were placed in different groups.

### Is Newland law of octaves applicable to all the three categories ? Give reason to justify your answer.

• In newlands law of octave, the first and eighth elements show similar properties
• This rule is not valid for elements after Ca in the periodic table. Therefore, the newland’s law of octave is not applicable for all the three categories.