Question

which is the noble - gas notation for lead (pb)?
○ rn6s²4f¹⁴5d¹⁰6p²
○ rn6s²5d¹⁰6p²
○ xe6s²4f¹⁴5d¹⁰6p²
○ xe6s²5d¹⁰6p²

Explanation:

1. First, determine the atomic number of lead (Pb), which is 82.
2. The noble gas before lead in the periodic table is xenon (Xe), with an atomic number of 54. So the noble - gas core should be [Xe].
3. Now, we need to fill the electrons after the Xe core. The electron configuration of lead is built by filling the orbitals in the order of energy levels. The electron configuration of lead is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}4f^{14}5s^{2}5p^{6}5d^{10}6s^{2}6p^{2}\). After the Xe core (\([Xe]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}\)), we have the remaining electrons: \(4f^{14}5d^{10}6s^{2}6p^{2}\)? Wait, no. Wait, the order of filling is \(6s\), then \(4f\), then \(5d\), then \(6p\). Wait, the correct electron configuration after Xe (\(Z = 54\)) for Pb (\(Z=82\)): the number of electrons after Xe is \(82 - 54=28\). Let's calculate the electrons in each sub - shell: \(6s^{2}\) (2 electrons), \(4f^{14}\) (14 electrons), \(5d^{10}\) (10 electrons), \(6p^{2}\) (2 electrons). \(2 + 14+10 + 2=28\). So the noble - gas notation is \([Xe]6s^{2}4f^{14}5d^{10}6p^{2}\).

Answer:

[Xe]6s²4f¹⁴5d¹⁰6p² (the third option)