Question

an atom of element x has one more shell of electrons than an atom of beryllium, but it has one less valence electron than beryllium. which element is element x?
○ sodium (na)
○ boron (b)
○ magnesium (mg)
○ lithium (li)

Explanation:

Step1: Analyze Beryllium's Electron Configuration

Beryllium (Be) has an atomic number of 4. Its electron configuration is $1s^2 2s^2$. So, it has 2 electron shells and 2 valence electrons (electrons in the outermost shell, which is the 2nd shell here, with 2 electrons).

Step2: Determine Requirements for Element X

• Shells: Element X has one more shell than Be. Be has 2 shells, so X should have $2 + 1 = 3$ electron shells.
• Valence Electrons: Element X has one less valence electron than Be. Be has 2 valence electrons, so X should have $2 - 1 = 1$ valence electron.

Step3: Analyze Each Option's Electron Configuration

• Sodium (Na): Atomic number 11. Electron configuration: $1s^2 2s^2 2p^6 3s^1$. Number of shells: 3 (matches the shell requirement). Valence electrons: 1 (matches the valence electron requirement).
• Boron (B): Atomic number 5. Electron configuration: $1s^2 2s^2 2p^1$. Shells: 2 (does not match, needs 3 shells). Valence electrons: 3 (does not match, needs 1).
• Magnesium (Mg): Atomic number 12. Electron configuration: $1s^2 2s^2 2p^6 3s^2$. Shells: 3 (good). Valence electrons: 2 (needs 1, so no).
• Lithium (Li): Atomic number 3. Electron configuration: $1s^2 2s^1$. Shells: 2 (needs 3, so no). Valence electrons: 1 (but shells don't match).

Answer:

A. sodium (Na)