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Question

Find out the number of electrons, protons, and neutrons present in sodium and chloride ions.


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Solution

Given:

Sodium ion (Na+)

Chloride ion (Cl-)

Formula Used:

  1. Atomic number of element=Number of electrons=Number of protons
  2. Mass number of element=Number of protons+Number of neutrons

For Sodium (Na):

Atomic number=11

Mass number=23

Sodium ion (Na+) is formed when Sodium atom loses one electron.

Thus, there is no change in the number of protons and neutrons but the number of electrons decreases by one.

Calculate the number of electrons in Sodium ion (Na+):

Number of electrons in Sodium ion=11-1

Number of electrons in Sodium ion=10

Calculate the number of protons in Sodium ion (Na+):

Number of protons=Atomic number of element

Number of protons in Sodium ion=11

Calculate the number of neutrons in Sodium ion (Na+):

Mass number of element=Number of protons+Number of neutrons

Thus,

Number of neutrons=Mass number of element-Number of protons

Number of neutrons in Sodium ion=23-11

Number of neutrons in Sodium ion=12

Thus, the number of electrons, protons, and neutrons present in Sodium ion are 10,11 and 12 respectively.

For Chlorine (Cl):

Atomic number=17

Mass number=35

Chloride ion (Cl-) is formed when Chlorine atom gains one electron.

Thus, there is no change in the number of protons and neutrons but the number of electrons increases by one.

Calculate the number of electrons in Chloride ion (Cl-):

Number of electrons in Chloride ion=17+1

Number of electrons in Chloride ion=18

Calculate the number of protons in Chloride ion (Cl-):

Number of protons=Atomic number of element

Number of protons in Chloride ion=17

Calculate the number of neutrons in Chloride ion (Cl-):

Mass number of element=Number of protons+Number of neutrons

Thus,

Number of neutrons=Mass number of element-Number of protons

Number of neutrons in Chloride ion=35-17

Number of neutrons in Chloride ion=18

Thus, the number of electrons, protons, and neutrons present in Chloride ion are 18,17 and 18 respectively.


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