Answer
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Hint: To find the structure of ions we must know the electronic configuration of the atoms that ions possess. Valence electrons are used in making the structure of ions. Valence electrons are the electrons which are present in the outermost shell of an atom.
Complete step by step answer:
$ClO_3^ - $ is chlorate anion. In this compound oxidation state of chlorine is $ + 5$ (oxidation state of oxygen is $ - 2$, on whole molecule charge is $ - 1$ and on computing these values we get oxidation state of chlorine as $ + 5$). To draw the Lewis structure of an ion we should know the electronic configuration of atoms that are present in the ion. Structure is drawn with the help of valence electrons of atoms. In a given anion atoms are oxygen and chlorine. Electronic configuration of $Cl$ is $1{s^2}2{s^2}2{p^6}3{s^3}3{p^4}$ . There are $7$ electrons in the outermost shell of chlorine. This means there are $7$ valence electrons in $Cl$ atom. Electronic configuration of oxygen is $1{s^2}2{s^2}2{p^4}$. This means there are $6$ electrons in the valence shell of oxygen. Total valence electrons from three oxygen atoms and one chlorine atom are $7 + \left( {3 \times 6} \right) = 25$. Due to negative charge there will be one more valence electron. So total valence electrons will be $25 + 1 = 26$. In this anion central atom is chlorine (as there is only one atom of chlorine). Now we will draw a lone pair around the central atom.
Similarly draw lone pairs of oxygen atoms.
Number of electron pairs can be found by dividing total valence electrons with two. So total electron pairs are $\dfrac{{26}}{2} = 13$. Chlorine is a central atom so it should be bound with other atoms also which are present in ions. It is shown as
Out of total $13$ electron pairs $3$ are drawn. To draw other bond pairs draw remaining $10$ electron pairs around atoms. These are shown below:
Now add charges on each atom. there will be $ - 1$ charge on each oxygen atom (oxygen makes two bonds but there is only one bond) and $ + 2$ charge on chlorine atom (chlorine makes one bond but there are three bonds). Structure will look like:
Due to these charges ion is highly unstable. To increase the stability two oxygen atoms will make bonds with chlorine and the remaining negative charge on one oxygen atom will be in resonance with the structure. Final structure of ion will look like:
This is the final structure of $ClO_3^ - $ ion.
Note:
Resonance structures are sets of lewis structures. These structures describe the delocalization of electrons in an ion or a molecule. These structures are also called canonical structures. Delocalization of charge makes the structure more stable.
Complete step by step answer:
$ClO_3^ - $ is chlorate anion. In this compound oxidation state of chlorine is $ + 5$ (oxidation state of oxygen is $ - 2$, on whole molecule charge is $ - 1$ and on computing these values we get oxidation state of chlorine as $ + 5$). To draw the Lewis structure of an ion we should know the electronic configuration of atoms that are present in the ion. Structure is drawn with the help of valence electrons of atoms. In a given anion atoms are oxygen and chlorine. Electronic configuration of $Cl$ is $1{s^2}2{s^2}2{p^6}3{s^3}3{p^4}$ . There are $7$ electrons in the outermost shell of chlorine. This means there are $7$ valence electrons in $Cl$ atom. Electronic configuration of oxygen is $1{s^2}2{s^2}2{p^4}$. This means there are $6$ electrons in the valence shell of oxygen. Total valence electrons from three oxygen atoms and one chlorine atom are $7 + \left( {3 \times 6} \right) = 25$. Due to negative charge there will be one more valence electron. So total valence electrons will be $25 + 1 = 26$. In this anion central atom is chlorine (as there is only one atom of chlorine). Now we will draw a lone pair around the central atom.
Similarly draw lone pairs of oxygen atoms.
Number of electron pairs can be found by dividing total valence electrons with two. So total electron pairs are $\dfrac{{26}}{2} = 13$. Chlorine is a central atom so it should be bound with other atoms also which are present in ions. It is shown as
Out of total $13$ electron pairs $3$ are drawn. To draw other bond pairs draw remaining $10$ electron pairs around atoms. These are shown below:
Now add charges on each atom. there will be $ - 1$ charge on each oxygen atom (oxygen makes two bonds but there is only one bond) and $ + 2$ charge on chlorine atom (chlorine makes one bond but there are three bonds). Structure will look like:
Due to these charges ion is highly unstable. To increase the stability two oxygen atoms will make bonds with chlorine and the remaining negative charge on one oxygen atom will be in resonance with the structure. Final structure of ion will look like:
This is the final structure of $ClO_3^ - $ ion.
Note:
Resonance structures are sets of lewis structures. These structures describe the delocalization of electrons in an ion or a molecule. These structures are also called canonical structures. Delocalization of charge makes the structure more stable.
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