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CBSE Class 9 Answered

(a) How are electrons arranged around the nucleus in an atom. (b) If an atom of an element has atomic number 11 and mass number 23, find the number of protons, electrons and neutrons in its atoms.                                                                    OR (a) The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes begin mathsize 12px style straight X presubscript 8 presuperscript 16 space and space straight X presubscript 8 presuperscript 18 end style in the sample? (b) Complete the following table. Elements Atomic Number Mass Number Protons Neutrons Electrons A 11 - - 12 - B - 35 - - 17
Asked by Topperlearning User | 04 Apr, 2016, 11:01: AM
answered-by-expert Expert Answer

(a) The arrangement of electrons in different orbits around the nucleus in an atom is called the electronic configuration of the element.

     The following general rules used to write the electronic configuration of an element are:

     (i) The shells of energy levels are represented by the circles. Electrons are filled in order of increasing energy levels of shells. e.g. first K, then L, then M and so on.

     (ii) The maximum number if electrons which can be accommodated in any shell is '2n2' where 'n' is the shell number.

     (iii) The outermost shell cannot accommodate more than 8 electrons, even if it has the capacity to accommodate more electrons.

(b)Atomic number = 11

    Mass number = 23

    Hence,

    Number of protons = 11

    number of electrons = 11

    Mass number = P + N

                  23  = 11 + N

                   N   = 23 - 11 = 12

                                                                       OR

(a)Atomic mass of element X = 16.2 u

Let the % of isotope begin mathsize 12px style straight X presubscript 8 presuperscript 16 end style be Z.

Let the % of isotope begin mathsize 12px style straight X presubscript 8 presuperscript 18 end style be (100 - Z).

Hence,

begin mathsize 12px style Average space atomic space mass space of space straight X space space equals space fraction numerator percent sign space of space straight X presubscript 8 presuperscript 16 space straight x space 16 space plus space percent sign space of space straight X presubscript 8 presuperscript 18 space straight x space 18 over denominator 100 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 16.2 space equals space fraction numerator straight Z space straight x space 16 space plus space left parenthesis 100 space minus space straight Z right parenthesis space straight x space 18 over denominator 100 end fraction space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 2 straight Z space straight u space space equals space 180 space straight u
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight Z space equals space 180 over 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight Z space equals 90 percent sign end style

So, % of begin mathsize 12px style straight X presubscript 8 presuperscript 16 end style = 90%

      % of begin mathsize 12px style straight X presubscript 8 presuperscript 18 end style = 10%

Elements

Atomic Number

Mass Number

Protons

Neutrons

Electrons

A

11

23

11

12

11

B

17

35

17

18

17

 

Answered by | 04 Apr, 2016, 01:01: PM
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