
For each of the following pairs, indicate which of the two species is of large size
(a)$F{e^{2 + }}\;or\;F{e^{3 + }}$
(b)$M{g^{2 + }}\;or\;C{a^{2 + }}$
Answer
573.3k+ views
Hint: The size of a species depends upon its outer shell, or the valence electron. In a periodic table when the number of shells increases the size of the species also increases.
Complete step by step solution:
(a)$F{e^{2 + }}\;or\;F{e^{3 + }}$: The two ionic radii of $Fe$, $F{e^{2 + }}$and $F{e^{3 + }}$ we can compare their size by taking the valence shells.
We know that $Fe$ has 26 electrons in their electron configuration. And $F{e^{2 + }}$ and $F{e^{3 + }}$ has 24 and 23 electrons in their electron configuration.
Now let's look into the electron configuration of $Fe$
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}$
From this we are removing two electron to get, $F{e^{2 + }}$and their electron configuration becomes
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}$
The two electrons from the 4s shell are removed. And for $F{e^{3 + }}$ we are removing one more electron from the 3d shell, hence the electron configuration becomes
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}$
Now, coming to their size. Sizes are determined by the number of electrons and protons present in an atom. When the number of protons increases they exert more force on the electron and pull them to the centre, making them small in radius. $Fe$ has 26 protons and electrons each. In $F{e^{2 + }}$, the number of electrons are decreased by 2 and In $F{e^{3 + }}$ number of electrons are decreased by 3. Hence the increase in protons is larger in $F{e^{3 + }}$therefore, they are in smaller size. That means $F{e^{2 + }}$ is larger among both.
(b)$M{g^{2 + }}\;or\;C{a^{2 + }}$: Here also we can see the electron configuration of both.
$Mg$ has 12 electrons in its electron configuration. Hence it has 10 electron in $M{g^{2 + }}$
Therefore, the electron configuration is
$1{s^2}2{s^2}2{p^6}$
$C{\text{a}}$ has 20 electrons, hence $C{a^{2 + }}$ has 18 electrons
Therefore, the electron configuration is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
We can clearly see that they have an extra valence shell in them. Therefore clearly $C{a^{2 + }}$ has larger radius, hence the larger size.
Note: In a neutral atom, we can say that its ionic radii and atomic radii are the same but when an atom loses electrons to form positively charged cation its ionic radius is smaller than atomic radius. Also when an atom gain an electron to have negative charge anion its ionic radius is greater than it atomic radius
Complete step by step solution:
(a)$F{e^{2 + }}\;or\;F{e^{3 + }}$: The two ionic radii of $Fe$, $F{e^{2 + }}$and $F{e^{3 + }}$ we can compare their size by taking the valence shells.
We know that $Fe$ has 26 electrons in their electron configuration. And $F{e^{2 + }}$ and $F{e^{3 + }}$ has 24 and 23 electrons in their electron configuration.
Now let's look into the electron configuration of $Fe$
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}$
From this we are removing two electron to get, $F{e^{2 + }}$and their electron configuration becomes
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}$
The two electrons from the 4s shell are removed. And for $F{e^{3 + }}$ we are removing one more electron from the 3d shell, hence the electron configuration becomes
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}$
Now, coming to their size. Sizes are determined by the number of electrons and protons present in an atom. When the number of protons increases they exert more force on the electron and pull them to the centre, making them small in radius. $Fe$ has 26 protons and electrons each. In $F{e^{2 + }}$, the number of electrons are decreased by 2 and In $F{e^{3 + }}$ number of electrons are decreased by 3. Hence the increase in protons is larger in $F{e^{3 + }}$therefore, they are in smaller size. That means $F{e^{2 + }}$ is larger among both.
(b)$M{g^{2 + }}\;or\;C{a^{2 + }}$: Here also we can see the electron configuration of both.
$Mg$ has 12 electrons in its electron configuration. Hence it has 10 electron in $M{g^{2 + }}$
Therefore, the electron configuration is
$1{s^2}2{s^2}2{p^6}$
$C{\text{a}}$ has 20 electrons, hence $C{a^{2 + }}$ has 18 electrons
Therefore, the electron configuration is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
We can clearly see that they have an extra valence shell in them. Therefore clearly $C{a^{2 + }}$ has larger radius, hence the larger size.
Note: In a neutral atom, we can say that its ionic radii and atomic radii are the same but when an atom loses electrons to form positively charged cation its ionic radius is smaller than atomic radius. Also when an atom gain an electron to have negative charge anion its ionic radius is greater than it atomic radius
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