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Hint: Lithium has little size and enormous warmth of hydration. The hydration vitality more than remunerates the ionization vitality. This results in high oxidation potential for \[Li\].
Complete step-by-step answer:
This expands the straightforwardness with which the accompanying change happens.
\[Li(s)\to Li+{{e}^{-}}\]
Because of this, lithium is the most grounded diminishing operator in watery solutions. Undoubtedly, antacid metals are acceptable decreasing specialists attributable to inexactly held single "\[n{{s}^{1}}\]" valence electrons.
In fact, "extreme" honourable gas centre gives powerful protection and consequently it has low ionization vitality.
Further, diminishing character or the reactivity of antacid metals increments going down the gathering in vaporous state.
It is supposed to be for the most part because of the expansion in the quantity of shells among different variables.
Based on this conversation, \[Li\] is the most fragile lessening specialist and \[Cs\] is the most grounded diminishing operator (among salt metals) in vaporous basic state.
Nonetheless, for the most part we manage conduct and responses in water (fluid arrangement). Diminishing or oxidizing conduct is subsequently referenced to be needy upon or identified with the "terminal possibilities". Further, as should be obvious from action arrangement, anode potential is identified with the development/decrease of watery particles from/into metals in their stable physical state.
Thus, as a matter of course, diminishing character is estimated by oxidation potential or the inclination of metals to lose electrons to shape (solvated) particles in arrangement.
The correct answer is C.
Note: Lithium being little in size has high ionization enthalpy. Then again in light of little size it is broadly hydrated and has exceptionally high hydration enthalpy. This high hydration enthalpy repays the high vitality expected to evacuate electrons. In this manner \[Li\] tends to lose electrons in arrangement than other antacid metals. In this manner, \[Li\] is the most grounded decreasing operator.
Complete step-by-step answer:
This expands the straightforwardness with which the accompanying change happens.
\[Li(s)\to Li+{{e}^{-}}\]
Because of this, lithium is the most grounded diminishing operator in watery solutions. Undoubtedly, antacid metals are acceptable decreasing specialists attributable to inexactly held single "\[n{{s}^{1}}\]" valence electrons.
In fact, "extreme" honourable gas centre gives powerful protection and consequently it has low ionization vitality.
Further, diminishing character or the reactivity of antacid metals increments going down the gathering in vaporous state.
It is supposed to be for the most part because of the expansion in the quantity of shells among different variables.
Based on this conversation, \[Li\] is the most fragile lessening specialist and \[Cs\] is the most grounded diminishing operator (among salt metals) in vaporous basic state.
Nonetheless, for the most part we manage conduct and responses in water (fluid arrangement). Diminishing or oxidizing conduct is subsequently referenced to be needy upon or identified with the "terminal possibilities". Further, as should be obvious from action arrangement, anode potential is identified with the development/decrease of watery particles from/into metals in their stable physical state.
Thus, as a matter of course, diminishing character is estimated by oxidation potential or the inclination of metals to lose electrons to shape (solvated) particles in arrangement.
The correct answer is C.
Note: Lithium being little in size has high ionization enthalpy. Then again in light of little size it is broadly hydrated and has exceptionally high hydration enthalpy. This high hydration enthalpy repays the high vitality expected to evacuate electrons. In this manner \[Li\] tends to lose electrons in arrangement than other antacid metals. In this manner, \[Li\] is the most grounded decreasing operator.
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