Question

give the noble gas shortened electron configuration for ga
1 42 33 44
give the noble gas shortened electron configuration for the only ga ion, ga+3
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: Determine atomic number of Ga

Gallium (Ga) has an atomic number of 31. The nearest noble - gas with a lower atomic number is Ar (atomic number 18).

Step2: Write electron configuration of Ga

After [Ar], we need to account for 31 - 18=13 more electrons. The electron configuration is [Ar] 4s²3d¹⁰4p¹. So for the first part: 1. C. Ar, 2. J. s², 3. CC. d¹⁰, 4. M. p¹.

Step3: Determine electron configuration of Ga³⁺

Ga³⁺ is formed by losing 3 electrons. Ga has the electron configuration [Ar] 4s²3d¹⁰4p¹. Losing 3 electrons (2 from 4s and 1 from 4p), the electron configuration of Ga³⁺ is [Ar] 3d¹⁰. So for the second part: 5. C. Ar, 6. H. s⁰, 7. CC. d¹⁰, 8. L. p⁰.

Answer:

1. C. Ar
2. J. s²
3. CC. d¹⁰
4. M. p¹
5. C. Ar
6. H. s⁰
7. CC. d¹⁰
8. L. p⁰