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(B) To determine the block of an element,we look at the subshell in which the last electron enters:$1$. For $Z = 37$ (Rubidium,$Rb$): The electronic c

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$A-S, B-R, C-P, D-Q$

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(B) To determine the block of an element,we look at the subshell in which the last electron enters:$1$. For $Z = 37$ (Rubidium,$Rb$): The electronic configuration is $[Kr] 5s^1$. Since the last electron enters the $s-$orbital,it belongs to the $s-$block $(S)$.$2$. For $Z = 42$ (Molybdenum,$Mo$): The electronic configuration is $[Kr] 4d^5 5s^1$. Since the last electron enters the $d-$orbital,it belongs to the $d-$block $(R)$.$3$. For $Z = 34$ (Selenium,$Se$): The electronic configuration is $[Ar] 3d^{10} 4s^2 4p^4$. Since the last electron enters the $p-$orbital,it belongs to the $p-$block $(P)$.$4$. For $Z = 92$ (Uranium,$U$): The electronic configuration is $[Rn] 5f^3 6d^1 7s^2$. Since the last electron enters the $f-$orbital,it belongs to the $f-$block $(Q)$.Therefore,the correct matching is $A-S, B-R, C-P, D-Q$.

Column-$I$	Column-$II$
$A. Z = 37$	$P. p-$ block
$B. Z = 42$	$Q. f-$ block
$C. Z = 34$	$R. d-$ block
$D. Z = 92$	$S. s-$ block

List-$I$	List-$II$
$(1)$ Active non-metal with high electron gain enthalpy.	$(A)$ $Na$
$(2)$ Soft metal with low ionization enthalpy.	$(B)$ $Sb$
$(3)$ Metalloid that forms ${M_2}{O_3}$ type oxide.	$(C)$ $He$
$(4)$ Chemically inert gas.	$(D)$ $Cl$

Match the following elements with their respective blocks in the periodic table based on their atomic number $(Z)$:
Column-$I$ Column-$II$
$A. Z = 37$ $P. p-$ block
$B. Z = 42$ $Q. f-$ block
$C. Z = 34$ $R. d-$ block
$D. Z = 92$ $S. s-$ block

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Match the following elements with their respective blocks in the periodic table based on their atomic number $(Z)$:Column-$I$Column-$II$$A. Z = 37$$P. p-$ block$B. Z = 42$$Q. f-$ block$C. Z = 34$$R. d-$ block$D. Z = 92$$S. s-$ block