Question

give the noble gas shortened electron configuration for hf
1 62 43 54
give the noble gas shortened electron configuration for the only hf ion, hf+4
remember that ion formation usually results in completely full or completely empty subshells.
5 66 47 58
a. he b. ne c. ar d. kr e. xe f. rn g. uuo h. s0
i. s1 j. s2 k. s3 l. p0 m. p1 n. p2 o. p3 p. p4
q. p5 r. p6 s. d0 t. d1 u. d2 v. d3 w. d4 x. d5
y. d6 z. d7 aa. d8 bb. d9 cc. d10 dd. f0 ee. f1 ff. f2
gg. f3 hh. f4 ii. f5 jj. f6 kk. f7 ll. f8 mm. f9 nn. f10
oo. f11 pp. f12 qq. f13 rr. f14

Explanation:

Step1: Determine noble - gas core for Hf

Hafnium (Hf) has an atomic number of 72. The noble - gas that comes before Hf is Xenon (Xe) with an atomic number of 54. So the noble - gas core is [Xe].

Step2: Determine remaining electron configuration for Hf

After using the Xe core, we have 72 - 54=18 electrons left. The electron configuration of the remaining electrons is $6s^{2}4f^{14}5d^{2}$.

Step3: Determine noble - gas core for $Hf^{4 +}$

When Hf loses 4 electrons to form $Hf^{4+}$, it loses the 2 electrons from the 6s orbital and 2 electrons from the 5d orbital. The noble - gas core for $Hf^{4+}$ is also [Xe].

Step4: Determine remaining electron configuration for $Hf^{4 +}$

After losing 4 electrons, all the outer - shell electrons are removed. The electron configuration of $Hf^{4+}$ is $[Xe]4f^{14}5d^{0}6s^{0}$.

Answer:

1. E. Xe, J. s2, EE. f14, U. d2
2. E. Xe, H. s0, EE. f14, S. d0