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8.1     Write down the electronic configuration of:

               (vi)\; Lu^{2+}

 

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The atomic number of lutetium is 71 and the previous noble element is Xe (xenon)
the electronic configuration of Lu^{2+}=[Xe]^{54}4f^{14}5d^{1}6s^{0}

Posted by

manish

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[Xe]4f14 

Posted by

Diptangshu Nath

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