Match List-I with List-II
|
List-I (Electronic configuration) |
List-II (Ion) |
|
(A) $[Ar] 3d^{10}$ |
(I) $Ce^{4+}$ |
|
(B) $[Xe]$ |
(II) $Cu^+$ |
|
(C) $[Rn]$ |
(III) $Th^{4+}$ |
|
(D) $[Xe] 4f^{14} 5d^1$ |
(IV) $Lu^{2+}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
|
List-I (Electronic configuration) |
List-II (Ion) |
|
(A) $[Ar] 3d^{10}$ |
(II) $Cu^+$ |
|
(B) $[Xe]$ |
(I) $Ce^{4+}$ |
|
(C) $[Rn]$ |
(III) $Th^{4+}$ |
|
(D) $[Xe] 4f^{14} 5d^1$ |
(IV) $Lu^{2+}$ |
(A) $[Ar]3d^{10}$ $\rightarrow$ (II) $Cu^+$: Copper ($Z=29$) has a neutral configuration of $[Ar]3d^{10}4s^1$. Removing one electron from the $4s$ orbital to form the $Cu^+$ ion leaves the filled $3d$ subshell.
(B) $[Xe]$ $\rightarrow$ (I) $Ce^{4+}$: Cerium ($Z=58$) has a neutral configuration of $[Xe]4f^1 5d^1 6s^2$. To form the $Ce^{4+}$ ion, all four valence electrons are removed, leaving the stable noble gas configuration of Xenon.
(C) $[Rn]$ $\rightarrow$ (III) $Th^{4+}$: Thorium ($Z=90$) is an actinide with a neutral configuration of $[Rn]6d^2 7s^2$. Losing four electrons results in the noble gas configuration of Radon.
(D) $[Xe]4f^{14}5d^1$ $\rightarrow$ (IV) $Lu^{2+}$: Lutetium ($Z=71$) has a neutral configuration of $[Xe]4f^{14}5d^1 6s^2$. Removing the two $6s$ electrons to form $Lu^{2+}$ leaves the $4f^{14}5d^1$ configuration.