Question

What kind of bond is present in NaBr?

Answer 1

Answer:

D Ionic bond

Explanation:

Atomic number = number of electrons = number of protons.

Atomic number of sodium is 11

So the atom contains 11 protons and 11 electrons

To find the number of neutrons we make use of the formula  

Mass number - atomic number = number of neutrons

From the periodic table we know mass number of sodium is 23

So number of neutron = 23 - 11 = 12.

When a sodium atom loses an electron it will have 11 positive protons and 10 negative electrons. Since 1 positive charge is more Na becomes $Na^+$.

Positively charged ion is called as cation

Bromine atomic number is 35 so it has 35 protons and 35 electrons.

When it gains an electrons it will have 35 positive protons and 36 negative electrons since 1 negative charge is more Br becomes $Br^-$.

Negatively charged ion is called as anion.

The electrostatic attraction between $Na^+$ and $Br^-$ due to the opposite charges gets attracted and forms a bond called ionic bond in NaBr

Look at the diagram of NaCl for example  

Answer 2

Answer:

D. Ionic bond

Explanation:

Sodium loses an electron to bromine