Question

give the noble gas shortened electron configuration for zr
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give the noble gas shortened electron configuration for the only zr ion, zr+4
remember that ion formation usually results in completely full or completely empty subshells.
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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 Zr's atomic number

Zirconium (Zr) has an atomic number of 40.

Step2: Determine the noble - gas core

The noble gas before Zr with the closest and lower atomic number is Kr (atomic number 36). So the noble - gas core is [Kr].

Step3: Fill the remaining electrons

After using Kr as the core, we have 40 - 36=4 electrons left. These electrons fill the 5s and 4d orbitals. The electron configuration is [Kr]5s$^2$4d$^2$.

Step4: Consider the Zr$^{4 + }$ ion

Zr$^{4 + }$ has lost 4 electrons. Starting from [Kr]5s$^2$4d$^2$, after losing 4 electrons, the electron configuration becomes [Kr]5s$^0$4d$^0$.

Answer:

1. D. Kr, J. s$^2$, U. d$^2$
2. D. Kr, H. s$^0$, S. d$^0$