
Which of the following species is neither hypervalent nor hypovalent?
A.${\text{ClO}}_4^ - $
B.${\text{B}}{{\text{F}}_3}$
C.${\text{SO}}_4^{2 - }$
D.${\text{CO}}_3^{2 - }$
Answer
563.4k+ views
Hint: Hypervalent compound means one which contains more than eight electrons in the valence shell. Hypovalent compound means one which possesses less electrons than an octet in its valence shell. This abnormal behaviour can be seen only in covalent bonding, that is, by the sharing of electrons.
Complete step by step solution:
In option A, the given compound is ${\text{ClO}}_4^ - $.
In chlorine, the third shell is the valence shell. Thus chlorine possesses vacant d orbital. It extends its covalency to seven using the vacant d orbitals. It forms double bonds with three oxygen atoms and a single bond with the fourth. As a result, it has 14 valence electrons and is hypervalent.
In option B, the given compound is ${\text{B}}{{\text{F}}_3}$
Boron has three valence electrons and cannot increase its covalency. After forming bonds with the chlorine atoms, it still has only six valence electrons and is hypovalent.
In option C, the given compound is ${\text{SO}}_4^{2 - }$
In sulphur, the third shell is the valence shell. Thus it possesses a vacant d orbital. It extends its covalency to six using the vacant d orbitals. It forms double bonds with two oxygen atoms and a single bond each with the other two. As a result, it has 12 valence electrons and is hypervalent.
In option D, the given compound is ${\text{CO}}_3^{2 - }$.
Carbon in the given ion has eight valence electrons after bonding with the oxygen atoms and is neither hypervalent nor hypovalent.
Hence, the correct option is option D.
Note:
We know that, in hypervalent compounds, the central atom exceeds the number of electrons more than an octet. It is possible in only those cases, if the central atom has vacant d orbitals present in its valence shell. Thus, we can say that only elements present in the third period or above can be hypervalent. These elements increase their covalency by increasing the number of unpaired electrons by exciting electrons to the empty d orbitals.
Complete step by step solution:
In option A, the given compound is ${\text{ClO}}_4^ - $.
In chlorine, the third shell is the valence shell. Thus chlorine possesses vacant d orbital. It extends its covalency to seven using the vacant d orbitals. It forms double bonds with three oxygen atoms and a single bond with the fourth. As a result, it has 14 valence electrons and is hypervalent.
In option B, the given compound is ${\text{B}}{{\text{F}}_3}$
Boron has three valence electrons and cannot increase its covalency. After forming bonds with the chlorine atoms, it still has only six valence electrons and is hypovalent.
In option C, the given compound is ${\text{SO}}_4^{2 - }$
In sulphur, the third shell is the valence shell. Thus it possesses a vacant d orbital. It extends its covalency to six using the vacant d orbitals. It forms double bonds with two oxygen atoms and a single bond each with the other two. As a result, it has 12 valence electrons and is hypervalent.
In option D, the given compound is ${\text{CO}}_3^{2 - }$.
Carbon in the given ion has eight valence electrons after bonding with the oxygen atoms and is neither hypervalent nor hypovalent.
Hence, the correct option is option D.
Note:
We know that, in hypervalent compounds, the central atom exceeds the number of electrons more than an octet. It is possible in only those cases, if the central atom has vacant d orbitals present in its valence shell. Thus, we can say that only elements present in the third period or above can be hypervalent. These elements increase their covalency by increasing the number of unpaired electrons by exciting electrons to the empty d orbitals.
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