First slide

Trends in Atomic Properties & Physical Properties

Question

A, B, C, D , E and F are the elements belonging to Second group of Modern periodic table. The correct trend of reducing power for these elements in aqueous solution is A = F > C > D > E>B. The element ‘A’ in the sequence is

Moderate

A

Ba
B

Ra
C

Be
D

Either 1 or 2

Solution

A) Atomic radius…Be<Mg<Ca<Sr<Ba<Ra;
B) IP₁…..Be>Mg>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) $E_{M^{2 +} / M}^{°}$ values are negative.
Most negative for Ra=Ba and least negative for Be.
J) $E_{{Ra}^{2 +} / Ra}^{°} (most negative) = E_{{Ba}^{2 +} / Ba}^{°} < E_{{Sr}^{2 +} / Sr}^{°} < E_{{Ca}^{2 +} / Ca}^{°} < E_{{Mg}^{2 +} / Mg}^{°} < E_{{Be}^{2 +} / Be}^{°} (least negative)$
Best reducing agent is Ra=Ba, best oxidant is Be²⁺;
K) Reducing power…Be<Mg<Ca<Sr<Ba=Ra ;
L) Oxidizing power… Be²⁺> Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺ = Ra²⁺ ;

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