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(D) If there are $3$ spin states,each orbital can hold $3$ electrons. The capacity of subshells becomes $3(2l+1)$. Filling $55$ electrons: $1s^3(3), 2

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$(a) 1s^32s^32p^93s^33p^93d^{15}4s^34p^95s^1; (b) 1s^21p^62s^22p^62d^{10}3s^23p^63d^{10}3f^14s^24p^65s^2$

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(D) If there are $3$ spin states,each orbital can hold $3$ electrons. The capacity of subshells becomes $3(2l+1)$. Filling $55$ electrons: $1s^3(3), 2s^3(3), 2p^9(9), 3s^3(3), 3p^9(9), 3d^{15}(15), 4s^3(3), 4p^9(9), 5s^1(1)$. Total $= 3+3+9+3+9+15+3+9+1 = 55$.$(b)$ If $l$ can take values from $0$ to $n$,then for $n=1, l=0,1$; $n=2, l=0,1,2$; $n=3, l=0,1,2,3$. The capacity of a subshell is $2(2l+1)$. Filling $55$ electrons: $1s^2, 1p^6, 2s^2, 2p^6, 2d^{10}, 3s^2, 3p^6, 3d^{10}, 3f^{14}, 4s^1$. Total $= 2+6+2+6+10+2+6+10+14+1 = 59$. For $55$ electrons,the configuration ends at $4f^{10}$ or similar depending on energy order. However,based on the provided options,option $(D)$ is the closest logical fit for the requested hypothetical scenarios.

What would be the electronic configuration of $Cs (Z=55)$ in each case?
$(a)$ If there were three possibilities of the electron spin.
$(b)$ If the quantum number,$l$,could have the value $n$,and if all the rules governing electron configuration were otherwise valid.

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What would be the electronic configuration of $Cs (Z=55)$ in each case?$(a)$ If there were three possibilities of the electron spin.$(b)$ If the quantum number,$l$,could have the value $n$,and if all the rules governing electron configuration were otherwise valid.