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Select the atoms that are likely to lose electrons to form cations:

A. Fluorine (F)
B. Magnesium (Mg)
C. Sodium (Na)
D. Sulfur (S)

User Gulbrandr
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1 Answer

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Final answer:

Magnesium (Mg) and Sodium (Na) are likely to lose electrons to form cations, while Fluorine (F) and Sulfur (S) are likely to gain electrons to form anions. The correct options are B & C.

Step-by-step explanation:

The atoms that are likely to lose electrons to form cations are Magnesium (Mg) and Sodium (Na). This is because both magnesium and sodium are metals and elements that typically lose electrons to form positively charged ions, known as cations.

Magnesium, being in group 2 of the periodic table, will lose two electrons to have the same number of electrons as the previous noble gas, neon, forming a Mg2+ ion.

Similarly, sodium, an alkali metal in group 1, will lose one electron to form a Na+ ion with the electron configuration of the previous noble gas, neon.

In contrast, Fluorine (F) and Sulfur (S) are more likely to gain electrons to form anions. Fluorine, with seven electrons in its valence shell, gains one electron to become fluoride (F−), while sulfur would gain two electrons to form a sulfide anion (S2−) as it is in group 16 and needs two electrons to complete its valence shell. The correct options are B & C.

User Rousseauo
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