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Question 7.3.3: Write electron configurations for the following elements, in......

Write electron configurations for the following elements, in spdf notation and orbital box notation. Identify the element as paramagnetic or diamagnetic.
a. Zn (Do not use the noble gas notation.)
b. Sm (Use the noble gas notation.)

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You are asked to write electron configurations in both spdf and orbital box notation and to identify the element as paramagnetic or diamagnetic.
You are given the identity of the element.

a. Zinc is element 30. Use the filling order to write the electron configuration, keeping in mind the maximum number of electrons that can be accommodated in a subshell.

spdf notation: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}

orbital box notation:

\underset{1s}{\fbox{↑↓}}\,\, \underset{2s}{\fbox{↑↓}}\, \, \underset{2p}{\fbox{↑↓}\fbox{↑↓}\fbox{↑↓}}\,\, \underset{3s}{\fbox{↑↓}}\,\, \underset{3p}{\fbox{↑↓}\fbox{↑↓}\fbox{↑↓}} \,\,\underset{4s}{\fbox{↑↓}}\,\, \underset{3d}{\fbox{↑↓}\fbox{↑↓}\fbox{↑↓}\fbox{↑↓}\fbox{↑↓}}

Zinc is diamagnetic because its electron configuration shows no unpaired electrons.
b. Samarium is element 62. Use the symbol [Xe] to represent the first 54 electrons in the electron configuration.

pdf notation: [Xe]6s^{2}4f^{6}

orbital box notation: [Xe]\underset{6s}{\fbox{↑↓}}\,\,\underset{4f}{\fbox{↑ }\fbox{↑ }\fbox{↑ }\fbox{↑ }\fbox{↑ }\fbox{↑ }\huge\fbox{} }

Samarium is paramagnetic because its electron configuration shows six unpaired electrons.

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