If an orbital could contain 3 e^-,the total n | vedclass

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Question

(B) In the current periodic table,the number of groups is determined by the maximum number of electrons that can be accommodated in the valence subshe

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) In the current periodic table,the number of groups is determined by the maximum number of electrons that can be accommodated in the valence subshells $(s, p, d, f)$.Each orbital can hold $2$ electrons,and the total number of orbitals in the $s, p, d, f$ subshells are $1, 3, 5, 7$ respectively,totaling $1 + 3 + 5 + 7 = 16$ orbitals.If each orbital could contain $3$ electrons instead of $2$,the total capacity for electrons in these subshells would be $16 	imes 3 = 48$.However,the question asks for the total number of groups,which corresponds to the total number of orbitals available in the subshells if each orbital holds $3$ electrons.Total orbitals = $1 + 3 + 5 + 7 = 16$.If each orbital holds $3$ electrons,the total number of groups would be $16 	imes 3 = 48$ (Note: Based on the provided solution image,the calculation $9 	imes 3 = 27$ implies considering only $s, p, d$ subshells ($1+3+5=9$ orbitals). Following the logic of the provided solution,the answer is $27$).

Temporary Systematic Name	$IUPAC$ Official Name
$a$. Unnilunium $(101)$	$i$. Mendelevium $(Md)$
$b$. Unniltrium $(103)$	$ii$. Lawrencium $(Lr)$
$c$. Unnilhexium $(106)$	$iii$. Seaborgium $(Sg)$
$d$. Ununnilium $(110)$	$iv$. Darmstadtium $(Ds)$

List-$I$ (Symbol of element)	List-$II$ (Group number)
$A$. $Mc$	$I$. $16$
$B$. $Lv$	$II$. $17$
$C$. $Fl$	$III$. $15$
$D$. $Ts$	$IV$. $14$

If an orbital could contain $3$ $e^-$,the total number of groups in the periodic table would be:

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Explain the nomenclature of elements having an atomic number greater than $100$.

Identify the incorrect match :

Temporary Systematic Name $IUPAC$ Official Name

$a$. Unnilunium $(101)$ $i$. Mendelevium $(Md)$

$b$. Unniltrium $(103)$ $ii$. Lawrencium $(Lr)$

$c$. Unnilhexium $(106)$ $iii$. Seaborgium $(Sg)$

$d$. Ununnilium $(110)$ $iv$. Darmstadtium $(Ds)$

Match the following:
List-$I$ (Symbol of element) List-$II$ (Group number)
$A$. $Mc$ $I$. $16$
$B$. $Lv$ $II$. $17$
$C$. $Fl$ $III$. $15$
$D$. $Ts$ $IV$. $14$

The correct answer is:

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