Complete the following table:
`|{:("Particle","Mass No.","Atomic No.","Protons","Neitrons","Electrons"),("Nitrogen atom",-,-,-,7,7),("Calcium ion",-,20,-,20,-),("Oxygen atom",16,8,-,-,-),("Bromide ion",-,-,-,45,36):}|`

Explanation

For nitrogen atom.
No. of electron `=7` (given)
No. of neutrons `=7` (given)
`:.` No. of protons `=Z=7` ( :. Atom is electrically neutral)
Atomic number `=Z=7`
Mass No. (A)=No. of protons + No. of neutrons `=7+7=14`
For calcium ion.
No. of netrons `=20` (Given)
Atomic No. (Z)`=20` (Given)
`:.` No. of protons `=Z=20`,
But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the equation `Cararr Ca^(2+)+2e^(-)` but the composition of the nucleus remains unchanged.
`:.` No. of electrons in calcium ion `=20-2=18`
Mass number (A)= No. of protons + No. of neutrons `=20+20=40`
For oxygen atom.
Mass number (A) = No. of protons + No. of neutrons `=16` (Given)
Atomic No. `(Z)=8` (Given)
No. of protons `=Z=8`,
No. of electrons `=Z=8`
No. of neutrons `=A-Z=16-8=8`
For bromide ion.
No. of neutrons `=45` (given)
No. of electrons `=36` (given)
But in the formation of bromide ion, one electron is gained by nuclear part according to equation `Br+e^(-) rarr Br^(-),` But the composition of nucleus remains ubchanged.
`:.` No. of protons in bromide ion = No. of electrons in bromine atom `=36-1=35`
Atomic number (Z) = No. of protons `=35`
Mass number (A)= No. of neutrons + No. of protons `=45+35=80`