Write down the electronic configuration of:

(i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+


(ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+


(i) The atomic number of Cr is 24 and the electronic configuration is [Ar] 3d54s1


When 3 electrons are removed, it becomes Cr3+. The electronic configuration of Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or [Ar]3d3


(ii) The atomic number of Pm is 61 and the electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f5 5s2 5p6 6s2 or [Xe] 4f56s2. When 3 electrons are removed, it becomes Pm+3, having the electronic configuration,


Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4


Or, [Xe]544f4


(iii) The atomic number of Cu is 29 and has the electronic configuration of [Ar] 3d104s1. On removing one electron, the Cu+is obtained with the electronic configuration, 1s2 2s2 2p6 3s2 3p6 3d10 Or [Ar] 3d10


(iv)The atomic number of Ce is 58 and has the electronic configuration of [Xe] 4f15d16s2. When the valence electrons are removed, the Ce4+ ion with configuration, 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Or [Xe]is obtained.


(v) The atomic number of Co is 27 and has the electronic configuration of [Ar] 3d74s2. When 2 electrons are removed from s-orbital, it becomes Co2+with electronic configuration, 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar] 3d7.


(vi)The atomic number of Lu is 71 and has the electronic configuration of [Xe] 4f145d16s2. When 2 electrons are removed, Lu2+with the configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f 14 5d 1 Or, [Xe] 4f145d1


(vii) The atomic number of Mn is 25 and has the electronic configuration of [Ar] 3d54s2 . When 2 electrons are removed, Mn2+ ion is obtained and has an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10 .


(viii) The atomic number of Th is 90 and has the electronic configuration of [Rn] 6d2 7s2 . When 4 electrons are removed, the electronic configuration becomesTh4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f 14 5d 10 6s 2 6p 6 Or, [Rn]


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