Gadolinium (atomic number $=64$ ) is a member of $4 f$ series. It's electronic configuration in +3 oxidation…
Gadolinium (atomic number $=64$ ) is a member of $4 f$ series. It's electronic configuration in +3 oxidation state is [Xe] $4 f^7$. What is the ground state electronic configuration of gadolinium?
$[\mathrm{Xe}] 4 f^{10}$
$[\mathrm{Xe}] 4 f^8 6 s^2$
$[\mathrm{Xe}] 4 f^7 5 d^3$
$[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2$
Solution
The atomic number of gadolinium is 64 . Electronic configuration is [Xe] $4 f^7 5 d^1 6 s^2$ $\mathrm{Gd}^{3+}=[\mathrm{Xe}] 4 f^7 5 d^0 6 s^0$
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