Question

1. which of the following would have the same electron configuration as neon? (0.5 points) *

se²-

o-

s²-

mg²+

ca²+

k+

Explanation:

Step1: Determine the electron number of neon

Neon (\(Ne\)) has an atomic number of \(10\), so it has \(10\) electrons.

Step2: Analyze each ion

• For \(Se^{2 - }\): Selenium (\(Se\)) has an atomic number of \(34\). \(Se^{2 - }\) has \(34 + 2=36\) electrons.
• For \(O^{-}\): Oxygen (\(O\)) has an atomic number of \(8\). \(O^{-}\) has \(8+ 1 = 9\) electrons.
• For \(S^{2 - }\): Sulfur (\(S\)) has an atomic number of \(16\). \(S^{2 - }\) has \(16+2 = 18\) electrons.
• For \(Mg^{2+}\): Magnesium (\(Mg\)) has an atomic number of \(12\). \(Mg^{2+}\) has \(12 - 2=10\) electrons.
• For \(Ca^{2+}\): Calcium (\(Ca\)) has an atomic number of \(20\). \(Ca^{2+}\) has \(20 - 2 = 18\) electrons.
• For \(K^{+}\): Potassium (\(K\)) has an atomic number of \(19\). \(K^{+}\) has \(19-1=18\) electrons.

Answer:

\(Mg^{2+}\)