
Which end of the bonds $ Si-Cl $ , $ At-Br $ , $ Hg-P $ are positively charged?
A. $ Si,\ At,\ P $
B. $ Cl,\ Br,\,P $
C. $ Cl,\ At,\ Hg $
D. $ Si,\ At,\ Hg $
Answer
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Hint: Generally, elements in the modern periodic table are divided into two categories i.e., electropositive elements and electronegative elements. The elements which have a tendency to donate electrons and form positive charged ions are termed as electropositive, whereas the elements which have a tendency to accept electrons and form anions are known as electronegative elements.
Complete answer:
We know that on moving from left to right in a period, the atomic size of elements decreases, so the valence shell comes closer to the nucleus due to which the force of attraction between nucleus and electrons increases. Thus, the tendency to accept the electrons also increases. So, electronegativity increases on moving from left to right in a period.
On moving top to bottom in a group, the atomic size of elements increases, due to which there is a decrease in effective nuclear charge i.eComparatively less force of attraction exists between nucleus and electrons of valence shell electrons. Thus, the tendency to accept electrons decreases and hence, electronegativity of elements decreases down the group.
Hence, we can conclude that an element in a compound which is comparatively less electronegative will show electropositive character and forms will be considered as the positive end of the compound.
Therefore, for the given compounds:
$ Si-Cl\Rightarrow $ Silicon is less electronegative than chlorine and hence, it will be the positive end of the bond.
$ At-Br\Rightarrow $ Astatine is less electronegative than chlorine and hence, it will be the positive end of the bond.
$ Hg-P\Rightarrow $ Mercury is less electronegative than chlorine and hence, it will be the positive end of the bond.
Thus, option (D) is the correct answer.
Note:
It is important to note that both electropositive and electronegative are comparing terms i.e., we cannot always consider an element to show a specific behaviour. It depends on the type of element it is bonded with. For example, a hydrogen atom shows electropositive nature and donates an electron in $ {{H}_{2}}O $ while in $ NaH $ it behaves like an electronegative element.
Complete answer:
We know that on moving from left to right in a period, the atomic size of elements decreases, so the valence shell comes closer to the nucleus due to which the force of attraction between nucleus and electrons increases. Thus, the tendency to accept the electrons also increases. So, electronegativity increases on moving from left to right in a period.
On moving top to bottom in a group, the atomic size of elements increases, due to which there is a decrease in effective nuclear charge i.eComparatively less force of attraction exists between nucleus and electrons of valence shell electrons. Thus, the tendency to accept electrons decreases and hence, electronegativity of elements decreases down the group.
Hence, we can conclude that an element in a compound which is comparatively less electronegative will show electropositive character and forms will be considered as the positive end of the compound.
Therefore, for the given compounds:
$ Si-Cl\Rightarrow $ Silicon is less electronegative than chlorine and hence, it will be the positive end of the bond.
$ At-Br\Rightarrow $ Astatine is less electronegative than chlorine and hence, it will be the positive end of the bond.
$ Hg-P\Rightarrow $ Mercury is less electronegative than chlorine and hence, it will be the positive end of the bond.
Thus, option (D) is the correct answer.
Note:
It is important to note that both electropositive and electronegative are comparing terms i.e., we cannot always consider an element to show a specific behaviour. It depends on the type of element it is bonded with. For example, a hydrogen atom shows electropositive nature and donates an electron in $ {{H}_{2}}O $ while in $ NaH $ it behaves like an electronegative element.
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