Question

1. choose the correct electron configuration for the element. refer to the periodic table.

he
2s²
ne 1s²
xe 2s²
1s²

1. choose the correct electron configuration for the element. refer to the periodic table.

po
xe 6s²
xe 6s²4f¹⁴5d¹⁰6p²
xe 6s²4f¹⁴5d¹⁰6p⁴
xe 6s²4f¹⁴5d⁷

Explanation:

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• Helium (He) has an atomic number of 2.
• The electron configuration starts from the lowest - energy orbital \(1s\).
• For \(n = 1\), the \(s\) - orbital (\(l=0\)) can hold a maximum of \(2(2l + 1)=2\) electrons.
• So, the electron configuration of He is \(1s^{2}\).
• Option \(2s^{2}\) is incorrect because the \(1s\) orbital fills before the \(2s\) orbital.
• Options \([Ne]1s^{2}\) and \([Xe]2s^{2}\) are incorrect because Ne (\(Z = 10\)) and Xe (\(Z=54\)) are much larger atoms than He, and using their noble - gas cores for He is wrong.

• Polonium (Po) has an atomic number \(Z = 84\).
• The noble - gas core for Po is \([Xe]\) (\(Z = 54\)).
• After the \(Xe\) core (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}\)), we start filling the \(6s\) orbital (\(2\) electrons: \(6s^{2}\)), then the \(4f\) orbitals (\(14\) electrons: \(4f^{14}\)), then the \(5d\) orbitals (\(10\) electrons: \(5d^{10}\)), and finally the \(6p\) orbitals.
• For Po, the number of electrons after the \(Xe\) core is \(84 - 54=30\). After filling \(6s^{2}4f^{14}5d^{10}\) (which accounts for \(2 + 14+10 = 26\) electrons), we need to fill \(6p\) orbitals. Since \(p\) orbitals can hold a maximum of \(6\) electrons and we have \(30-(2 + 14+10)=4\) electrons left for the \(6p\) orbitals.
• Option \([Xe]6s^{2}\) is incorrect as it only accounts for \(2\) electrons after the \(Xe\) core.
• Option \([Xe]6s^{2}4f^{14}5d^{10}6p^{2}\) is incorrect because it only has \(2\) electrons in the \(6p\) orbital.
• Option \([Xe]6s^{2}4f^{14}5d^{7}\) is incorrect as it does not follow the correct filling order and does not account for the correct number of electrons.

Answer:

\(1s^{2}\)

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