Answer
Please see the work below.
Work Step by Step
A positive ion or cation is formed by the removal of electrons from the valence shell of the parent atom.
The charge on the ions is equal to the number of electrons removed.
a) Scandium, Sc – atomic number – 21.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{1}4s^{2}$ or $[Ar]3d^{1}4s^{2}$.
By the removal of 3 electrons from Sc, $Sc^{+3}$ is formed.
Therefore, the ground state electronic configuration of $Sc^{+3}$ is $[Ar]$.
b) Titanium, Ti – atomic number – 22.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{2}4s^{2}$ or $[Ar]3d^{2}4s^{2}$.
By the removal of 4 electrons from Ti, $Ti^{+4}$ is formed.
Therefore, the ground state electronic configuration of $Ti^{+4}$ is $[Ar]$.
c) Vanadium, V – atomic number – 23.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{3}4s^{2}$ or $[Ar]3d^{3}4s^{2}$.
By the removal of 5 electrons from V, $V^{+5}$ is formed.
Therefore, the ground state electronic configuration of $V^{+5}$ is $[Ar]$.
d) Chromium, Cr – atomic number – 24.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{5}4s^{1}$ or $[Ar]3d^{5}4s^{1}$.
By the removal of 3 electrons from Cr, $Cr^{+3}$ is formed.
Therefore, the ground state electronic configuration of $Cr^{+3}$ is $[Ar] 3d^{3}$.
e) Manganese, Mn – atomic number – 25.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{5}4s^{2}$ or $[Ar]3d^{5}4s^{2}$.
By the removal of 2 electrons from Mn, $Mn^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Mn^{+2}$ is $[Ar] 3d^{5}$.
f) Iron, Fe – atomic number – 26.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{6}4s^{2}$ or $[Ar]3d^{6}4s^{2}$.
By the removal of 2 electrons from Fe, $Fe^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Fe^{+3}$ is $[Ar] 3d^{6}$.
g) By the removal of 3 electrons from Fe, $Fe^{+3}$ is formed.
Therefore, the ground state electronic configuration of $Fe^{+3}$ is $[Ar] 3d^{5}$.
h) Cobalt, Co – atomic number – 27.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{7}4s^{2}$ or $[Ar]3d^{7}4s^{2}$.
By the removal of 2 electrons from Co, $Co^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Co^{+2}$ is $[Ar] 3d^{7}$.
i) Nickel, Ni – atomic number – 28.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{8}4s^{2}$ or $[Ar]3d^{8}4s^{2}$.
By the removal of 2 electrons from Ni, $Ni^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Ni^{+2}$ is $[Ar] 3d^{8}$.
j) Copper, Cu – atomic number – 29.
Electronic configuration - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}3d^{10}4s^{1}$ or $[Ar]3d^{10}4s^{1}$.
By the removal of 1 electron from Cu, $Cu^{+}$ is formed.
Therefore, the ground state electronic configuration of $Cu^{+}$ is $[Ar] 3d^{10}$.
k) By the removal of 2 electrons from Cu, $Cu^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Cu^{+2}$ is $[Ar] 3d^{9}$.
l)Silver, Ag – atomic number – 47.
Electronic configuration - $[Kr]4d^{10}5s^{1}$.
By the removal of 1 electron from Ag, $Ag^{+}$ is formed.
Therefore, the ground state electronic configuration of $Ag^{+}$ is $[Kr] 4d^{10}$.
m) Gold, Au – atomic number – 79.
Electronic configuration - $[Xe]4f^{14}5d^{10}6s^{1}$.
By the removal of 1 electron from Au, $Au^{+}$ is formed.
Therefore, the ground state electronic configuration of $Au^{+}$ is $[Xe]4f^{14}5d^{10}$
n) By the removal of 3 electrons from Au, $Au^{+3}$ is formed.
Therefore, the ground state electronic configuration of $Au^{+3}$ is $[Xe]4f^{14}5d^{8}$
o) Platinum, Pt – atomic number – 78.
Electronic configuration - $[Xe]4f^{14}5d^{9}6s^{1}$.
By the removal of 2 electrons from Pt, $Pt^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Pt^{+2}$ is $[Xe]4f^{14}5d^{8}$
