Question

give the noble gas shortened electron configuration for se 1 42 33 44 give the noble gas shortened electron configuration for the only se ion, se - 2 remember that ion formation usually results in completely full or completely empty subshells. 5 46 37 48 a. he b. ne c. ar d. kr e. xe f. rn g. uuo h. s^0 i. s^1 j. s^2 k. s^3 l. p^0 m. p^1 n. p^2 o. p^3 p. p^4 q. p^5 r. p^6 s. d^0 t. d^1 u. d^2 v. d^3 w. d^4 x. d^5 y. d^6 z. d^7 aa. d^8 bb. d^9 cc. d^10 dd. f^0 ee. f^1 ff. f^2 gg. f^3 hh. f^4 ii. f^5 jj. f^6 kk. f^7 ll. f^8 mm. f^9 nn. f^10 oo. f^11 pp. f^12 qq. f^13 rr. f^14

Explanation:

Step1: Identify noble - gas core for Se

Selenium (Se) has an atomic number of 34. The noble - gas that comes before Se is Ar (atomic number 18). So the noble - gas core is [Ar].

Step2: Determine remaining electron configuration for Se

After using the Ar core, we have 34 - 18=16 electrons left. The electron configuration after Ar is $4s^{2}3d^{10}4p^{4}$.

Step3: Identify noble - gas core for $Se^{2 -}$

When Se gains 2 electrons to form $Se^{2 -}$, it has a total of 36 electrons. The noble - gas with 36 electrons is Kr. So the noble - gas core is [Kr].

Step4: Determine electron configuration for $Se^{2 -}$

Since $Se^{2 -}$ has a full outer shell, its electron configuration after Kr is $4s^{2}3d^{10}4p^{6}$.

Answer:

1. C. Ar, J. s2, CC. d10, P. p4
2. D. Kr, J. s2, CC. d10, R. p6