Among the following, electropositivity is very high for:
(A) Al
(B) Ge
(C) Li
(D) Ba
Answer
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Hint: Electropositivity is defined as the tendency of an atom to give or donate electrons from the positively charged cation. This property is mainly exhibited by the metallic elements, especially alkaline earth metals.
Complete answer:
Let us understand what electropositivity is. Electropositivity is defined as the tendency of an atom to give or donate electrons from the positively charged cation. It depends on ionization enthalpy, that is if the electropositivity is high the lesser will be the ionizing enthalpy. When this enthalpy is less the electron is removed easily and hence the electropositivity is increased. Hence we have to find out which of the following elements have low ionisation enthalpy.
Let us now see which of the elements are placed and find out the correct option;
Li is placed in the $ 2nd $ period and $ 1st $ group, Al belongs to $ 3rd $ period and $ 13th $ group, Ge belongs to $ 4th $ period and $ 14th $ group, and Ba belongs to $ 6th $ period and $ 2nd $ group. As we know along the period the ionization energy increases and it decreases down the group. So according to the positing and property of the elements we can see that Ba is having the least ionising enthalpy.
Hence the correct option is D .i.e. BARIUM is the most electropositive element out of all.
Note:
Ge or germanium is a metalloid, Al or aluminium is a metal with very low electropositive group $ 2 $ . Electropositivity increases down the group, Li or lithium lies on the top of group $ 1 $ . Electropositivity is the opposite of electronegativity.
Complete answer:
Let us understand what electropositivity is. Electropositivity is defined as the tendency of an atom to give or donate electrons from the positively charged cation. It depends on ionization enthalpy, that is if the electropositivity is high the lesser will be the ionizing enthalpy. When this enthalpy is less the electron is removed easily and hence the electropositivity is increased. Hence we have to find out which of the following elements have low ionisation enthalpy.
Let us now see which of the elements are placed and find out the correct option;
Li is placed in the $ 2nd $ period and $ 1st $ group, Al belongs to $ 3rd $ period and $ 13th $ group, Ge belongs to $ 4th $ period and $ 14th $ group, and Ba belongs to $ 6th $ period and $ 2nd $ group. As we know along the period the ionization energy increases and it decreases down the group. So according to the positing and property of the elements we can see that Ba is having the least ionising enthalpy.
Hence the correct option is D .i.e. BARIUM is the most electropositive element out of all.
Note:
Ge or germanium is a metalloid, Al or aluminium is a metal with very low electropositive group $ 2 $ . Electropositivity increases down the group, Li or lithium lies on the top of group $ 1 $ . Electropositivity is the opposite of electronegativity.
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