Question

for ______, the 52nd electron must pair with one of the electrons in the 5p orbital.
a antimony
b molybdenum
c samarium
d tellurium

Explanation:

• Antimony (\(Sb\)) has an atomic number of \(51\). Its electron configuration is \([Kr]4d^{10}5s^{2}5p^{3}\).
• Molybdenum (\(Mo\)) has an atomic number of \(42\). Its electron configuration is \([Kr]4d^{5}5s^{1}\).
• Samarium (\(Sm\)) has an atomic number of \(62\). Its electron configuration is \([Xe]4f^{6}6s^{2}\).
• Tellurium (\(Te\)) has an atomic number of \(52\). Its electron configuration is \([Kr]4d^{10}5s^{2}5p^{4}\). In the \(5p\) orbital, when adding the \(52^{nd}\) electron, it must pair with one of the electrons already in the \(5p\) orbital (since \(5p\) can hold a maximum of \(6\) electrons and before adding the \(52^{nd}\) electron, there are \(3\) electrons in \(5p\) for \(Sb\) (\(Z = 51\)) and for \(Te\) (\(Z=52\)), the \(52^{nd}\) electron goes into the \(5p\) orbital where it pairs with an existing electron in \(5p\)).

Answer:

d. Tellurium