The correct order of decreasing first ionization energy is :

S i > A l > M g > N a

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S i > M g > A l > N a

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A l > S i > M g > N a

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M g > L i > A l > S i

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Solution

The correct option is D $S i > M g > A l > N a$
The decreasing order of first ionization energy is $S i > M g > A l > N a$ .
Across a period, ionization energy increases.
The ionization energy of Mg is higher than that of Al due to stable outer electronic configuration of $3 s^{2}$ for Mg.
The removal of electron from completely filled 3s electron requires high energy.

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Similar questions

The first ionistation enthalpies of Na, Mg, Al, and Si are in the order
(a) Na< Mg > Al < Si (b) Na> Mg > Al > Si
(c) Na< Mg< Al< Si (d) Na> Mg> Al < Si

Question 17
Arrange the following elements in the order of their decreasing metallic character
Na, Si, Cl, Mg, Al

(a) Cl > Si >Al > Mg >Na
(b) Na >Mg >Al >Si > Cl
(c) Na > Al > Mg > Cl > Si
(d) Al > Na> Si > Ca> Mg

$N a^{+} . M g^{2 +}, A l^{3 +}$ and $S i^{4 +}$ are isoelectronic. The order of their ionic size is

N a^{+}

M g^{2 +}

A l^{3 +}

and

S i^{4 +}

are isoelectronic, their ionic size will follow the order:

Q. The first ionisation energies of six

3^{r d}

period elements are 496, 578, 738, 789, 1000 and 1012

{kJ mol}^{- 1}

. The elements to which these energies correspond respectively are:

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