Chapter 7, Problem 58QP

Interpretation Introduction

Interpretation:

The ground-state electronic configuration of the given ions is to be written.

Concept introduction:

Elements that lose electrons are called cations. They have a lesser number of electrons than that of their neutral states.

Elements that gain electrons are called anions. They have a greater number of electrons than that of their neutral states.

Answer to Problem 58QP

Solution:

a) $1s^2$

b) $1s^2$

c) $1s^{2}2s^{2}2p^6$

d) $1s^{2}2s^{2}2p^6$

e) $\text{[Ne]}3s^{2}3p^6$

f) $\text{[Ne]}$

g) $\text{[Ar]}3d^{10}4s^{2}4p^6$

h) $\text{[Ar]}3d^{10}4s^{2}4p^6$

i) $\text{[Kr]}$

j) $\text{[Kr]}$

k) $\text{[Kr]}4d^{10}5s^2$

l) $\text{[Kr]}4d^{10}5s^{2}5p^6$

m) $\text{[Xe]}$

n) $\text{[Xe]4}{f}^{14}5d^{10}6s^2$

o) $\text{[Kr]5}{d}^{10}$

p) $\text{[Xe]4}{f}^{14}5d^{10}6s^2$

q) $\text{[Xe]4}{f}^{14}5d^{10}$

Explanation of Solution

a)Ion ${\text{Li}}^{\text{+}}$

The neutral state of lithium is ${\text{Li}}_3$ withelectronic configuration $1s^{2}2s^1$. ${\text{Li}}^{\text{+}}$ has oneelectronless

than its neutral state, and hence, electronic configuration of ${\text{Li}}^{\text{+}}$ ion is $1s^{2}2s^0$.

So, its ground-state electronic configuration is $1s^{2}2s^0$.

b)Ion ${\text{H}}^{-}$

Hydrogen in its neutral state has only one electron (${\text{H}}_1$) in the $s-\text{orbital}$. ${\text{H}}^{-}$ has two electrons (${\text{H}}_1$), having theelectronic configuration as $1s^2$. Hydrogen has a general tendency of losing its electron forming a cation as; ${\text{H}}^{+}$. In ${\text{H}}^{-}$ form, it forms hydrides with electropositive elements.

Its ground-state electronic configuration is $1s^2$.

c)Ion ${\text{N}}^{3-}$

The neutral state electron-arrangement of nitrogen is ${\text{N}}_7$ with the electronic configuration $1s^{2}2s^{2}2p^3$. ${\text{N}}^{3-}$ has ten electrons as threemore are added to its neutral state. The electronic configuration of ${\text{N}}^{3-}$ is $1s^{2}2s^{2}2p^6$.

So, its ground-state electronic configuration is $1s^{2}2s^{2}2p^6$.

d)Ion ${\text{F}}^{-}$

The neutral state electronic configuration of fluorine (${\text{F}}_9$) is $1s^{2}2s^{2}2p^5$. ${\text{F}}^{-}$ has ten electrons as onemore is added to its neutral state. It is isoelectronic with ${\text{N}}^{3-}$ ion.

So, its ground-state electronic configuration is $1s^{2}2s^{2}2p^6$.

e)Ion ${\text{S}}^{2-}$

The neutral state electron-arrangement of sulphur is ${\text{S}}_{16}$ with the electronic configuration $\text{[Ne]}3s^{2}3p^4$. ${\text{S}}^{2-}$ has eighteen electrons astwomore are added to its neutral state. Electronic configuration of ${\text{S}}^{2-}$ is $\text{[Ne]}3s^{2}3p^6$.

So, its ground-state electronic configuration is $\text{[Ne]}3s^{2}3p^6$.

f)Ion ${\text{Al}}^{3+}$

The neutral state of aluminium is ${\text{Al}}_{13}$ withelectronic configuration $\text{[Ne]}3s^{2}3p^1$.

${\text{Al}}^{3+}$ has ten electrons. The ground state electronic configuration of ${\text{Al}}^{3+}$ is $\text{[Ne]}3s^{0}3p^0$. It is isoelectronic with ${\text{N}}^{3-}$ and ${\text{F}}^{-}$ ions.

So, its ground-state electronic configuration is $\text{[Ne]}3s^{0}3p^0$.

g)Ion ${\text{Se}}^{2-}$

The ground state electron-arrangement of selenium is ${\text{Se}}^{2-}$ with the electronic configuration $\text{[Ar]}3d^{10}4s^{2}4p^4$. ${\text{Se}}^{2-}$ has thirty-six electrons astwo more are added to its neutral state. Electronic configuration of ${\text{S}}^{2-}$ is $\text{[Ar]}3d^{10}4s^{2}4p^6$. It is isoelectronic with ${\text{Kr}}_{\text{36}}$.

So, its ground-state electronic configuration is $\text{[Ar]}3d^{10}4s^{2}4p^6$.

h) Ion ${\text{Br}}^{-}$

The ground state electron-arrangement of bromine is ${\text{Br}}_{35}$ with the electronic configuration of $\text{[Ar]}3d^{10}4s^{2}4p^5$. ${\text{Br}}^{-}$ has thirty-six electrons asone more added to its neutral state. Electronic configuration of ${\text{Br}}^{-}$ is $\text{[Ar]}3d^{10}4s^{2}4p^6$. It is isoelectronic with ${\text{Kr}}_{\text{36}}$.

So, its ground-state electronic configuration is $\text{[Ar]}3d^{10}4s^{2}4p^6$.

i) Ion ${\text{Rb}}^{+}$

The neutral state of rubidium is ${\text{Rb}}_{37}$ withelectronic configuration $\text{[Kr]}5s^1$. ${\text{Rb}}^{+}$ has one electron less than its neutral state, and hence, its electronic configurationis $\text{[Kr]}5s^0$. It is isoelectronic with ${\text{Kr}}_{\text{36}}$.

So, its ground-state electronic configuration is $\text{[Kr]}$.

j)Ion ${\text{Sr}}^{2+}$

The neutral state of strontium is ${\text{Sr}}_{38}$ with the electronic configuration $\text{[Kr]}5s^2$. ${\text{Sr}}^{2+}$ ion has two electrons less than its neutral state with an electronic configuration $\text{[Kr]}$. It is isoelectronic with ${\text{Kr}}_{\text{36}}$ and ${\text{Rb}}^{+}$.

So, its ground-state electronic configuration is $\text{[Kr]}$.

k) Ion ${\text{Sn}}^{\text{2+}}$

The neutral state of stannous is ${\text{Sn}}_{50}$ with the electronic configuration $\text{[Kr]}4d^{10}5s^{2}5p^2$. ${\text{Sn}}^{\text{2+}}$ ion hastwo electrons less than its neutral state with an electronic configuration $\text{[Kr]}4d^{10}5s^{2}5p^0$.

So, its ground-state electronic configuration is $\text{[Kr]}4d^{10}5s^2$.

l)Ion ${\text{Te}}^{2-}$

${\text{Te}}^{2-}$ has two more electrons than its neutral state of ${\text{Te}}_{52}$ with its ground state electronic configuration as $\text{[Kr]}4d^{10}5s^{2}5p^4$. The electronic configuration of ${\text{Te}}^{2-}$ is $\text{[Kr]}4d^{10}5s^{2}5p^6$. It is isoelectronic with the noble gas ${\text{Xe}}_{54}$.

So, its ground-state electronic configuration is $\text{[Kr]}4d^{10}5s^{2}5p^6$.

m)Ion ${\text{Ba}}^{2+}$

The neutral state of barium is ${\text{Ba}}_{56}$ with the electronic configuration $\text{[Xe]}6s^2$. ${\text{Ba}}^{2+}$ ion hastwo electrons less than its neutral state with theelectronic configuration $\text{[Xe]}6s^0$. It is isoelectronic with ${\text{Xe}}_{54}$ and ${\text{Te}}^{2-}$.

So, its ground-state electronic configuration is $\text{[Xe]}$.

n)Ion ${\text{Pb}}^{2+}$

The neutral state of lead is ${\text{Pb}}_{82}$ with the electronic configuration $\text{[Xe]4}{f}^{14}5d^{10}6s^{2}6p^2$. ${\text{Pb}}^{2+}$ ion has two electrons less than its neutral state. Its electronic configurationis $\text{[Xe]4}{f}^{14}5d^{10}6s^{2}6p^0$.

So, its ground-state electronic configuration is $\text{[Xe]4}{f}^{14}5d^{10}6s^2$.

o)Ion ${\text{In}}^{3+}$

The neutral state of indium is ${\text{In}}_{49}$ with the electronic configuration $\text{[Kr]}4d^{10}5s^{2}5p^1$. ${\text{In}}^{3+}$ ion has three electrons less than its neutral state with an electronic configuration $\text{[Kr]5}{d}^{10}5s^{0}5p^0$.

So, its ground-state electronic configuration is $\text{[Kr]5}{d}^{10}$.

p)Ion ${\text{Tl}}^{+}$

The neutral state of thallium is ${\text{Tl}}_{81}$ having theelectronic configuration $\text{[Xe]4}{f}^{14}5d^{10}6s^{2}6p^1$. ${\text{Tl}}^{+}$ ion has one electron less than its neutral state with an electronic configuration $\text{[Xe]4}{f}^{14}5d^{10}6s^2$ with a total of eighty electrons.

So, its ground-state electronic configuration is $\text{[Xe]4}{f}^{14}5d^{10}6s^2$.

q)Ion ${\text{Tl}}^{3+}$

The neutral state of thallium is ${\text{Tl}}_{81}$ having the electronic configuration $\text{[Xe]4}{f}^{14}5d^{10}6s^{2}6p^1$. ${\text{Tl}}^{3+}$ ion has three electrons less than its neutral state with an electronic configuration $\text{[Xe]4}{f}^{14}5d^{10}$, having a total of seventy-eight electrons.

So, its ground-state electronic configuration is $\text{[Xe]4}{f}^{14}5d^{10}$.