Question

Consider the following elements: Po, B, C, B for all of the following a) Give the shorthand electron configuration for each Po [Xe] Gs?, 4th Seppo 6py Xe] 65° 464 Salo 683 cs De] 65' Ba [xe se b) Which of the above elements is the largest Explain: c) Which of the above elements is the most paramagnetic Explain: d) Which of the above elements has the highest ionization energy? Explain:

Answer 1

vhus | PO = Electronic Configuration [Xe] 65², 4914 5d 6p4 604 [Irlalo 2 unpaired e- Bi= [XeJ 662,48", 5dº6p3 Da s 3 unfaired & umferinde cs = [Xe] 65 as I inpound e- Ba = [x] 4s | Ô unfores en 6p3 2 b) Laugest dementa es Atomic sadius decreases from left to right within a period. This caused by the increase in the number of protons and electron avroy a heriod. One proton has a greater effect than one election, this elections are hülled touarede nucleus veduting In Smaller radies.

C) mest hacamagnetic element = Bi because Bi has 3 unpaired election. It is highest unpaired e compare to them, 80. nee knalle Meldon fee haramagnetic behevioue de Unpairede, more unpaired é make hana - magnetic beheeran d) highest Ponization Energy = Bi. because Bi haws half filled oubitat, which is relatively most utable configuration among them. Thank yeau, May it help you.