The first ionisation energy of $N a, M g, A l$ and $S i$ are in the order of:

N a < M g > A l < S i

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N a > M g > A l > S i

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N a < M g < A l > S i

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N a > M g > A l < S i

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Solution

The correct option is A $N a < M g > A l < S i$
$M g$ requires greater ionization energy than $A l$ due to its configuration. It is more difficult to remove an electron from $3 s$ orbital than from $3 p$ orbital since s-orbitals have greater penetration power. Moreover, $M g$ has a stable electronic configuration with full-filled $3 s$ orbital.
Among $N a$ and $M g$ , $M g$ has completely filles $s -$ orbital, hence, ionization energy of $M g$ is greater than $N a$ .

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The first ionistation enthalpies of Na, Mg, Al, and Si are in the order
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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

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:

N a^{+}

M g^{2 +}

A l^{3 +}

and

S i^{4 +}

are isoelectronic, their ionic size will follow the order:

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The first ionisation energy of Na,Mg,Al and Si are in the order of:

The first ionisation energy of $N a, M g, A l$ and $S i$ are in the order of: