Question

which of the following matches the electronic configuration of a diamagnetic element?
a. (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1})
b. (mathrm{rn} 7 s^{2} 5 f^{14} 6 d^{8})
c. (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{4})
d. (mathrm{xe} 6 s^{2} 4 f^{14} 5 d^{10} 6 p^{6})
e. (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1})

Explanation:

Diamagnetic elements have all electrons paired.

• For option A: In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\), the \(3p\) sub - shell has one unpaired electron (\(3p^{1}\)).
• For option B: In \([Rn]7s^{2}5f^{14}6d^{8}\), the \(6d\) sub - shell has \(8\) electrons. Since \(d\) sub - shell has \(5\) orbitals (\(n = 3\) to \(n- l+1\) for \(d\) orbitals, \(l = 2\)), and using Hund's rule, there are \(2\) unpaired electrons in \(6d^{8}\) (\(6d^{8}=6d^{5 + 3}\), after filling \(5\) half - filled orbitals, \(3\) more electrons pair up in \(3\) of the \(5\) orbitals, leaving \(2\) unpaired).
• For option C: In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\), the \(4p\) sub - shell has \(4\) electrons. Since \(p\) sub - shell has \(3\) orbitals, there are \(2\) unpaired electrons (\(4p^{4}=4p^{3+1}\), after filling \(3\) half - filled orbitals, \(1\) more electron pairs up in one of the \(3\) orbitals, leaving \(2\) unpaired).
• For option D: In \([Xe]6s^{2}4f^{14}5d^{10}6p^{6}\), all sub - shells (\(6s\), \(4f\), \(5d\), \(6p\)) have paired electrons. \(s\) sub - shell (\(l = 0\)) has \(1\) orbital (\(2\) electrons - paired), \(f\) sub - shell (\(l=3\)) has \(7\) orbitals (\(14\) electrons - all paired as \(14 = 2\times7\)), \(d\) sub - shell (\(l = 2\)) has \(5\) orbitals (\(10\) electrons - \(2\times5\) paired), \(p\) sub - shell (\(l = 1\)) has \(3\) orbitals (\(6\) electrons - \(2\times3\) paired).
• For option E: In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}\), the \(5s\) sub - shell has one unpaired electron.

Answer:

D. \([Xe]6s^{2}4f^{14}5d^{10}6p^{6}\)