Question

choose the electron configuration for sodium (na) in both longhand notation and noble - gas notation. a link to the periodic table is provided for you. longhand notation: 1s²2s²2p⁶3s²3p⁴ 1s²2s²2p⁶3s¹ 1s²2s²2p⁶3s²3p⁶4s²3d³ 1s²2s²2p³ noble - gas notation: ar3s¹ he2s²2p⁶3s¹ ne3s²3p⁴ ne3s¹

Explanation:

Sodium (Na) has an atomic number of 11. In longhand notation, electrons are filled in orbitals following the Aufbau principle. The first two electrons go in the 1s orbital ($1s^{2}$), the next two in 2s ($2s^{2}$), six in 2p ($2p^{6}$), and the last one in 3s ($3s^{1}$), giving $1s^{2}2s^{2}2p^{6}3s^{1}$. For noble - gas notation, we use the noble gas core that precedes sodium in the periodic table. Neon (Ne) has an atomic number of 10, so we write [Ne] to represent its electron configuration, and then add the remaining electron in the 3s orbital for sodium, resulting in [Ne]$3s^{1}$.

Answer:

Longhand notation: $1s^{2}2s^{2}2p^{6}3s^{1}$
Noble - gas notation: [Ne]$3s^{1}$