The lowest value of $(I E)_{1}$ among the following:
$B, S c, A l, B r, M g, P b$

A l

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S c

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B r

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P b

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Solution

The correct option is A $A l$
$M g$ and $A l$ belongs to III period and are most metallic in nature among all but $A l$ is still more metallic than $M g$ even after lying down than $M g$ in the period. The reason is the electronic configuration. $M g$ has completely filled outermost orbital ( $3 s^{2}$ ) whereas $A l$ has one electron in 3p orbital so it will be easier for $A l$ to lose an electron to acquire stable configuration.

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Z n

P b

A l

E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

Z n

P b

A l

E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

Z n

P b

A l

E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

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