Answer
Please see the work below.
Work Step by Step
When a positive ion or cation is formed, electrons are removed from the valence shell of the parent atom.
When a negative ion or anion is formed, electrons are added from the valence shell of the parent atom.
The charge on the ions is equal to the number of electrons removed or added.
a. Li - $1s^{2} 2s^{1}$.
By the removal of 1 electron from Li, $Li^{+}$ is formed.
Therefore, the ground state electronic configuration of $Li^{+}$ is $1s^{2}$.
b. H - $1s^{1}$.
By the addition of 1 electron to H, $H^{-}$ is formed.
Therefore, the ground state electronic configuration of $H^{-}$ is $1s^{2}$.
c. N - $1s^{2} 2s^{2}2p^{3}$.
By the addition of 3 electrons to N, $N^{-3}$ is formed.
Therefore, the ground state electronic configuration of $N^{-3}$ is $1s^{2} 2s^{2}2p^{6}$.
d. F - $1s^{2} 2s^{2}2p^{5}$.
By the addition of 1 electron to F, $F^{-}$ is formed.
Therefore, the ground state electronic configuration of $F^{-}$ is $1s^{2} 2s^{2}2p^{6}$.
e. S - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{4}$.
By the addition of 2 electrons to S, $S^{-2}$ is formed.
Therefore, the ground state electronic configuration of $S^{-2}$ is $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{6}$.
f. Al - $1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{1}$.
By the removal of 3 electrons from Al, $Al^{+3}$ is formed.
Therefore, the ground state electronic configuration of $Al^{+3}$ is $1s^{2} 2s^{2}2p^{6}$.
g. Se - $[Ar] 3d^{10} 4s^{2}4p^{4}$.
By the addition of 2 electrons to Se, $Se^{-2}$ is formed.
Therefore, the ground state electronic configuration of $Se^{-2}$ is $[Ar] 3d^{10}4s^{2} 4p^{6}$.
h. Br - $[Ar] 4s^{2}3d^{10} 4p^{5}$.
By the addition of 1 electron to Br, $Br^{-}$ is formed.
Therefore, the ground state electronic configuration of $Br^{-}$ is $[Ar] 4s^{2}3d^{10} 4p^{6}$.
i. Rb - $[Kr] 5s^{1}$.
By the removal of 1 electron from Rb, $Rb^{+}$ is formed.
Therefore, the ground state electronic configuration of $Rb^{+}$ is $[Kr]$.
j. Sr - $[Kr] 5s^{2}$.
By the removal of 2 electrons from Sr, $Sr^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Sr^{+2}$ is $[Kr]$.
k. Sn - $[Kr] 4d^{10} 5s^{2}5p^{2}$.
By the removal of 1 electron from Sn, $Sn^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Sn^{+2}$ is $[Kr] 4d^{10} 5s^{2}$.
l. Te - $[Kr] 4d^{10}5s^{2} 5p^{4}$.
By the addition of 2 electrons to Se, $Te^{-2}$ is formed.
Therefore, the ground state electronic configuration of $Te^{-2}$ $[Kr] 4d^{10} 5s^{2}5p^{6}$.
m. Ba - $[Xe] 6s^{2}$.
By the removal of 2 electrons from Ba, $Ba^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Ba^{+2}$ is $[Xe]$.
n. Pb - $[Xe]4f^{14}5d^{10}6s^{2}6p^{2}$.
By the removal of 2 electrons from Pb, $Pb^{+2}$ is formed.
Therefore, the ground state electronic configuration of $Pb^{+2}$ is $[Xe]4f^{14}5d^{10}6s^{2}$.
o. In - $[Kr] 4d^{10}5s^{2} 5p^{1}$.
By the removal of 3 electrons from In, $In^{+3}$ is formed.
Therefore, the ground state electronic configuration of $In^{+3}$ is $[Kr] 4d^{10}$.
p. Tl - $[Xe] 4f^{14}3d^{10}6s^{2}6p^{1}$.
By the removal of 1 electron Tl, $Tl^{+}$ is formed.
Therefore, the ground state electronic configuration of $Tl^{+}$ is $[Xe]4f^{14}5d^{10}6s^{2}$.
q. Tl - $[Xe] 4f^{14}3d^{10}6s^{2}6p^{1}$.
By the removal of 3 electrons from Tl, $Tl^{+3}$ is formed.
Therefore, the ground state electronic configuration of $Tl^{+3}$ is $[Xe]4f^{14}5d^{10}$.
