Match the following:List-IList-IIA. (Atomic n | vedclass

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(A) The electronic configurations of the given elements are:$A$. $Z = 112$ (Copernicium): $[Rn] 5f^{14} 6d^{10} 7s^2$. The last electron enters the $d

Vedclass · Accepted Answer

$A-III, B-II, C-I, D-IV$

Vedclass · Answer

(A) The electronic configurations of the given elements are:$A$. $Z = 112$ (Copernicium): $[Rn] 5f^{14} 6d^{10} 7s^2$. The last electron enters the $d$-orbital $(III)$.$B$. $Z = 116$ (Livermorium): $[Rn] 5f^{14} 6d^{10} 7s^2 7p^4$. The last electron enters the $p$-orbital $(II)$.$C$. $Z = 88$ (Radium): $[Rn] 7s^2$. The last electron enters the $s$-orbital $(I)$.$D$. $Z = 100$ (Fermium): $[Rn] 5f^{12} 7s^2$. The last electron enters the $f$-orbital $(IV)$.Thus,the correct match is $A-III, B-II, C-I, D-IV$.

List-$I$	List-$II$
$A$. (Atomic number,$Z = 112$)	$I$. $s$
$B$. (Atomic number,$Z = 116$)	$II$. $p$
$C$. (Atomic number,$Z = 88$)	$III$. $d$
$D$. (Atomic number,$Z = 100$)	$IV$. $f$

Match the following:
List-$I$ List-$II$
$A$. (Atomic number,$Z = 112$) $I$. $s$
$B$. (Atomic number,$Z = 116$) $II$. $p$
$C$. (Atomic number,$Z = 88$) $III$. $d$
$D$. (Atomic number,$Z = 100$) $IV$. $f$

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Match the following:List-$I$List-$II$$A$. (Atomic number,$Z = 112$)$I$. $s$$B$. (Atomic number,$Z = 116$)$II$. $p$$C$. (Atomic number,$Z = 88$)$III$. $d$$D$. (Atomic number,$Z = 100$)$IV$. $f$