Hello people! Mdm Lao only gave me one copy of the class test answer, so I post here then you all can see.
7(c).valence shell electronic configuration of Y+: ns2 np4
.................................................................of E+"ns2np4
2 nd I.E. is to remove one of the unpaired np electrons from Y+ but one of the paired np electron from E+. Due to the intrelectronic repulsion between paired electrons in the same subshell, the 2nd IE of E is lower than that of Y.
(d). Lower. Valence shell electronic configuration of X2+: ns2np1 and .......of W2+: ns2
Because 3rd IE is to remove electron from a higher energy subshell(i.e. np) in X2+ while in W2+, electron is removed from a lower energy subshell (ns), so the 3rd IE of X is lower than that of W.
All the best for monday's lecture test!! Jiayou!
Wang Xiao =)
Subscribe to:
Post Comments (Atom)
Oh no i juz realise there is a mistake in the first line. the valence shell electronic configuration of Y+ should be ns2 np3
ReplyDelete