Give the number of electrons in the species $H_{2}^{+}$,$H_{2}$,and $O_{2}^{+}$.
(N/A) $H_{2}^{+}$:
Number of electrons in $H$ atom is $1$. Total electrons in $H_{2} = 1 + 1 = 2$. Since it is a cation $(H_{2}^{+})$,remove $1$ electron. Total electrons = $2 - 1 = 1$.
$H_{2}$:
Number of electrons in $H$ atom is $1$. Total electrons in $H_{2} = 1 + 1 = 2$.
$O_{2}^{+}$:
Number of electrons in $O$ atom is $8$. Total electrons in $O_{2} = 8 + 8 = 16$. Since it is a cation $(O_{2}^{+})$,remove $1$ electron. Total electrons = $16 - 1 = 15$.
Number of electrons in $H$ atom is $1$. Total electrons in $H_{2} = 1 + 1 = 2$. Since it is a cation $(H_{2}^{+})$,remove $1$ electron. Total electrons = $2 - 1 = 1$.
$H_{2}$:
Number of electrons in $H$ atom is $1$. Total electrons in $H_{2} = 1 + 1 = 2$.
$O_{2}^{+}$:
Number of electrons in $O$ atom is $8$. Total electrons in $O_{2} = 8 + 8 = 16$. Since it is a cation $(O_{2}^{+})$,remove $1$ electron. Total electrons = $16 - 1 = 15$.
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