The lanthanide ion having four unpaired elect | vedclass

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(D) To find the number of unpaired electrons in lanthanide ions, we first write their electronic configurations:$1$. $Ce (Z=58): [Xe] 4f^1 5d^1 6s^2$.

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$Ho^{3+}$

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(D) To find the number of unpaired electrons in lanthanide ions, we first write their electronic configurations:$1$. $Ce (Z=58): [Xe] 4f^1 5d^1 6s^2$. $Ce^{3+} = [Xe] 4f^1$. Unpaired electrons = $1$.$2$. $Nd (Z=60): [Xe] 4f^4 6s^2$. $Nd^{3+} = [Xe] 4f^3$. Unpaired electrons = $3$.$3$. $Tb (Z=65): [Xe] 4f^9 6s^2$. $Tb^{3+} = [Xe] 4f^8$. In $4f^8$, the electrons are filled as: $\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow, \uparrow, \uparrow$. Unpaired electrons = $6$.$4$. $Ho (Z=67): [Xe] 4f^{11} 6s^2$. $Ho^{3+} = [Xe] 4f^{10}$. In $4f^{10}$, the electrons are filled as: $\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow$. Unpaired electrons = $4$.Thus, $Ho^{3+}$ has four unpaired electrons.

The lanthanide ion having four unpaired electrons is
(Given : Atomic number of $Ce = 58, Nd = 60, Tb = 65$ and $Ho = 67$)

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The lanthanide ion having four unpaired electrons is(Given : Atomic number of $Ce = 58, Nd = 60, Tb = 65$ and $Ho = 67$)