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A, B, C, D, E and F are the elements belonging to Second group of Modern periodic table. The correct trend of Ionization potential for these elements is  A > B > C > D > E > F. The element ‘A' in the sequence is

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a
Be
b
Mg
c
Ca
d
Sr

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detailed solution

Correct option is A

A) Atomic radius…BeMg>Ca>Sr>Ra>Ba;C) EN…..Be>Mg>Ca=Sr>Ba>Ra;D) MP…Be>Ca>Sr>Ba>Ra>Mg ;E) BP…Be>Ba>Ra>Ca>Sr>Mg ;F) Density…..Ra>Ba>Sr>Be>Mg>Ca ;G) Abundance in Earth's crust….Ca>Mg>Ba>Sr>Be>Ra ;H) Hydration Enthalpy….Be2+>Mg2+>Ca2+>Sr2+>Ba2+I) EM2+/M° values are negative.Most negative for Ra=Ba and least negative for Be.J) ERa2+/Ra°(most negative)=EBa2+/Ba° Mg2+ > Ca2+ > Sr2+ > Ba2+ = Ra2+ ;

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