The atomic number of the element having the m | vedclass

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Question

(A) To find the element with the maximum number of unpaired electrons,we write the electronic configuration for each option in the ground state:$A) \

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) To find the element with the maximum number of unpaired electrons,we write the electronic configuration for each option in the ground state:$A) \ Z = 15 \ (P): 1s^2 2s^2 2p^6 3s^2 3p^3$. The $3p$ subshell has $3$ unpaired electrons.$B) \ Z = 10 \ (Ne): 1s^2 2s^2 2p^6$. All electrons are paired ($0$ unpaired electrons).$C) \ Z = 12 \ (Mg): 1s^2 2s^2 2p^6 3s^2$. All electrons are paired ($0$ unpaired electrons).$D) \ Z = 8 \ (O): 1s^2 2s^2 2p^4$. The $2p$ subshell has $2$ unpaired electrons.Comparing the number of unpaired electrons,$Z = 15$ has the maximum number ($3$ unpaired electrons). Therefore,the correct option is $A$.

$Set$	$(n)-(l)-(m)-(s)$
$(i)$	$(3)-(2)-(1)-(+1/2)$
$(ii)$	$(2)-(2)-(1)-(+1/2)$
$(iii)$	$(4)-(3)-(-2)-(-1/2)$
$(iv)$	$(1)-(0)-(-1)-(-1/2)$
$(v)$	$(3)-(2)-(3)-(+1/2)$

The atomic number of the element having the maximum number of unpaired electrons (in the ground state) is ........

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Consider the following sets of quantum numbers:
$Set$ $(n)-(l)-(m)-(s)$
$(i)$ $(3)-(2)-(1)-(+1/2)$
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