Question

match the elements with their correct noble gas notation configurations. chlorine, radium, arsenic, flourine. ne3s²3p⁵, rn7s², ar4s²3d¹⁰4p³, he2s²2p⁵

Explanation:

Noble - gas notation uses the nearest previous noble - gas element in brackets and then writes the remaining electron configuration. Chlorine (Cl) has 17 electrons. The nearest previous noble gas is neon (Ne). After Ne (10 electrons), there are 7 more electrons, so its configuration is $[Ne]3s^{2}3p^{5}$. Radium (Ra) has 88 electrons. The nearest previous noble gas is radon (Rn). After Rn (86 electrons), there are 2 more electrons in the 7s orbital, so its configuration is $[Rn]7s^{2}$. Arsenic (As) has 33 electrons. The nearest previous noble gas is argon (Ar). After Ar (18 electrons), there are 15 more electrons, which fill the 4s, 3d, and 4p orbitals as $4s^{2}3d^{10}4p^{3}$, so its configuration is $[Ar]4s^{2}3d^{10}4p^{3}$. Fluorine (F) has 9 electrons. The nearest previous noble gas is helium (He). After He (2 electrons), there are 7 more electrons, so its configuration is $[He]2s^{2}2p^{5}$.

Answer:

Chlorine - $[Ne]3s^{2}3p^{5}$
Radium - $[Rn]7s^{2}$
Arsenic - $[Ar]4s^{2}3d^{10}4p^{3}$
Fluorine - $[He]2s^{2}2p^{5}$