Structure of the atom Worksheet-5

Structure of the atom Worksheet-5

 

  1. There are 15 protons and 16 neutrons in the nucleus of an element. Calculate its atomic number and mass number. How will you represent the element?

 

  1. The symbol of oxygen atom is 8O16

(i) What is the name given to the number of protons in the nucleus of the atom?

(ii) What is the name given to the number of protons plus number of neutrons in the nucleus of the atom?

 

  1. The atom of an element has 9 protons, 9 electrons and 10 neutrons.

(i) What is the atomic number of the element?

(ii) What is the mass number of the element?

(iii) Name the element and give its electronic configuration.

(iv) Predict the valeney of the element.

 

  1. The element sulphur has atomic number 16, and mass number 32. State the number of protons, electrons and neutrons in it. Give the arrangement of the electrons in different energy shells. What is the valency of the element?

 

  1. An ion M2+ contains 10 electrons and 12 neutrons. What is the atomic number and mass number of the element M? Name the element.

 

  1. The element boron occurs in nature as two isotopes having atomic masses 10 u and 11 u. What are the percentage abundances of these isotopes in a sample of boron having average atomic mass of 10.8 u?

 

  1. The relative atomic mass of copper is 63.5 u. It exists as two isotopes which are 29Cu63 and 29Cu65.

Calculate the percentage of each present in it.

 

Answer:

  1. Atomic number of the element = No. of protons = 15

Mass number of the element = No. of protons (15) + No. of neutrons (16)

          = 15 + 16 = 31

Representation of the element = 15X31

The element with atomic number (Z = 15) is phosphorus (P). It is represented as 15P31

 

  1. (i) No. of protons in the nucleus of the atom is known as atomic number.

 ∴ Atomic no. of oxygen (Z) = 8

(ii) No. of protons in the nucleus plus no. of neutrons in the nucleus of an atom is known as mass number.

 ∴ Mass no. of oxygen (A) = 8 + 8 = 16

 

  1. (i) The atomic no. of element = No. of protons = 9

(ii) The mass no. of element = No. of protons + No. of neutrons

          = 9 + 10

          = 1

(iii) The element with Z = 9 is fluorine (F).

 Its electronic configuration : 2, 7.

(iv) The valiancy of fluorine is 1 and is calculated as 8 – 7 = 1.

 

  1. No. of protons = Atomic number =16

No. of electrons = No. of protons =16

No. of neutrons = Mass no. — no. of protons

          = 32 – 16 = 16

Electronic arrangement = K(2), L(8), M(6)

Valency of the element = 8 – 6 = 2

 

  1. No. of electrons in M2+ ion = 10

Atomic number of atom M = 10 + 2 = 12

No. of protons in atom M = 12

Mass number of atom M = No. of protons + No. of neutrons =12 +12 = 24

The element M with atomic number 12 is magnesium (Mg).

 

  1. Let the percentage of B-10 isotope = x

∴ the percentage of B – 11 isotope = 100 – x

The average atomic mass of boron

         

But the given average atomic mass of boron = 10.8 u

         

10x + 1100 – 11x = 10.8 × 100 = 1080

X = 1100 – 1080 = 20

∴ Percentage Abundance of B-10 isotope = 20%

Abundance of B – 11 isotope = 80%

 

  1. Let the percentage of 29Cu63 isotope = x

∴ The percentage of 29Cu65 isotope = 100 – x

From the above data, the relative atomic mass of Cu

         

But the given relative atomic mass of Cu = 63.5 u

         

63x + 6500 – 65x = 6350

–2x = 6350 – 6500 = –150

          2x = 150

or      x = 75u

Percentage of 29Cu63 isotope = 75%

Percentage of 29Cu65 isotope = 25%.