What are the atoms involved in Sodium bromide?
The correct answer and explanation is :
Sodium bromide (NaBr) is a chemical compound composed of two atoms: sodium (Na) and bromine (Br). These atoms combine to form an ionic bond, where sodium donates an electron, and bromine accepts it, resulting in the formation of ions.
Atoms Involved in Sodium Bromide:
- Sodium (Na): Sodium is a metal from the alkali metals group on the periodic table. It is highly reactive and has a single electron in its outermost electron shell. To achieve a stable electron configuration (similar to the noble gases), sodium tends to lose this electron, becoming a positively charged ion (Na⁺).
- Bromine (Br): Bromine is a halogen, located in Group 17 of the periodic table. It has seven electrons in its outermost shell and needs one more electron to achieve a full outer shell, like the nearest noble gas, argon. Bromine accepts an electron from sodium, forming a negatively charged ion (Br⁻).
Explanation of the Bonding in Sodium Bromide:
Sodium bromide forms through an ionic bond. When sodium reacts with bromine, sodium atoms lose their single valence electron to become Na⁺ ions, and bromine atoms gain that electron to become Br⁻ ions. The resulting oppositely charged ions (Na⁺ and Br⁻) attract each other due to electrostatic forces, forming a stable ionic compound, NaBr.
This ionic bond gives sodium bromide its characteristic properties. The compound is typically found as a solid crystal at room temperature and dissolves readily in water. When dissolved, it dissociates into Na⁺ and Br⁻ ions, making it a good conductor of electricity in solution. Sodium bromide is used in various applications, such as in photography, medicine (as a sedative), and as a disinfectant in swimming pools.
In summary, sodium bromide consists of sodium (Na) and bromine (Br) atoms, with sodium forming a positive ion (Na⁺) and bromine forming a negative ion (Br⁻), held together by an ionic bond.