Question

part c
ti
express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. for example, he2s^22p^2 should be entered as he2s^22p^2.
part d
te
express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. for example, he2s^22p^2 should be entered as he2s^22p^2.

Explanation:

Step1: Find Xe electron configuration

Xenon (Xe, atomic number 54) is the noble gas preceding Tellurium (Te, atomic number 52). The condensed form uses Xe as the core.

Step2: Add valence electrons for Te

Te has 52 electrons; Xe has 54, so Te has 2 fewer electrons than Xe. The valence orbitals fill as $5s^24d^{10}5p^4$.

Step3: Combine core and valence

Condensed form: $[Xe]5s^24d^{10}5p^4$
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Step1: Find Kr electron configuration

Krypton (Kr, atomic number 36) is the noble gas preceding Titanium (Ti, atomic number 22). The condensed form uses Kr as the core.

Step2: Add valence electrons for Ti

Ti has 22 electrons; Kr has 36, so Ti has 14 fewer electrons than Kr. The valence orbitals fill as $4s^23d^2$.

Step3: Combine core and valence

Condensed form: $[Kr]4s^23d^2$

Answer:

Part C: $[Kr]4s^23d^2$
Part D: $[Xe]5s^24d^{10}5p^4$