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]