In alkaline earth metal lose 2 electron because present in 2nd group in period and acquire noble gas configuration.

Alkaline earth metals are the elements in Group 2 of the periodic table, which includes elements such as beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). These elements have two electrons in their outermost energy level or valence shell.

To acquire noble gas configuration, alkaline earth metals need to lose these two electrons from their valence shell. Noble gases, such as helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn), have completely filled valence shells, which makes them highly stable and unreactive.

To attain a noble gas configuration, alkaline earth metals aim to have the same electron configuration as the nearest noble gas in the periodic table, which is the previous noble gas element in the same period. For example, beryllium (Be) aims to have the same electron configuration as helium (He), magnesium (Mg) aims to have the same electron configuration as neon (Ne), and so on.

The noble gas configuration for helium (He) is 1s², and for neon (Ne) it is 2s²2p⁶. By losing their two valence electrons, alkaline earth metals can achieve the same electron configuration as the nearest noble gas. This corresponds to forming a +2 cation, as the loss of two negatively charged electrons results in a positive charge.

Therefore, the correct answer is option 'B', which states that alkaline earth metals lose 2 electrons to acquire noble gas configuration.