Final answer:
Sodium's electron configuration in longhand notation is 1s²2s²2p⁶3s¹, and its noble gas configuration is [Ne] 3s¹. The [Ne] represents the 10 core electrons with the same configuration as the noble gas neon, and 3s¹ represents the valence electron. Upon ionization, sodium's configuration matches that of neon, giving it a stable noble gas electron arrangement.
Step-by-step explanation:
The electron configuration for sodium (Na), which is element number 11, is represented in longhand notation as 1s²2s²2p⁶3s¹. This notation indicates that sodium has two electrons in the 1s subshell, two in the 2s subshell, six in the 2p subshell, and one electron in the 3s subshell - its outermost or valence electron. The noble gas preceding sodium is neon (Ne), which has an electron configuration identical to the inner-shell electrons of sodium. Therefore, sodium's noble gas configuration can be written in shorthand as [Ne] 3s¹, where [Ne] represents the core electrons with the configuration of neon, and 3s¹ shows the valence electron for sodium.
It is important to note that when sodium forms a cation (Na+), it loses its single valence electron, resulting in an electron configuration identical to that of neon. Thus, the formation of a cation leaves sodium with a complete octet, reflecting the stable electron configuration of a noble gas.