Strontium, Sr, will lose 2 valence electrons when forming an ionic bond.
The correct answer and explanation is:
The statement is correct. Strontium (Sr) will lose 2 valence electrons when forming an ionic bond.
Strontium is an alkaline earth metal found in Group 2 of the periodic table. Elements in this group have two electrons in their outermost shell, called valence electrons. These valence electrons determine how an atom interacts chemically with other atoms.
When strontium forms an ionic bond, it tends to lose its two valence electrons to achieve a stable electron configuration. This stable configuration resembles that of the nearest noble gas, which has a full outer shell. By losing two electrons, strontium forms a positively charged ion called a cation with a charge of +2, denoted as Sr²⁺.
This loss of electrons happens because metals like strontium have relatively low ionization energy, meaning it requires less energy to remove their valence electrons. Once strontium loses its two outer electrons, its electron configuration matches that of krypton, the noble gas preceding strontium, resulting in increased stability.
In ionic bonding, this Sr²⁺ ion then attracts negatively charged ions (anions) such as chloride (Cl⁻), which gain electrons to complete their own outer shells. The electrostatic attraction between the positively charged Sr²⁺ and negatively charged anions holds the compound together.
This process is typical for alkaline earth metals, which commonly form ions with a +2 charge. For example, calcium (Ca) and magnesium (Mg) also lose two electrons to form Ca²⁺ and Mg²⁺ ions, respectively.
In summary, strontium loses two valence electrons to form a stable Sr²⁺ ion during ionic bonding, allowing it to achieve a noble gas electron configuration and form ionic compounds through electrostatic attraction with anions.