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

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(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 $27$).

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

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