The order of decrease in atomic radii for $B e, N a$ & $M g$ is :

N a > M g > B e

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M g > N a > B e

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B e > N a > M g

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B e > M g > N a

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Solution

The correct option is A $N a > M g > B e$
While going down the group size increases due to the addition of a shell. Therefore, the size of Mg is greater than that of Be.

On moving from left to right in a period size decreases due to increase in nuclear charge. Therefore, the size of Na is greater than that of Mg.

Option A is correct.

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Similar questions

Q. The correct order of atomic radii for the following isoelectronic species

F^{-}, O^{2 -}, N a^{+}, M g^{2 +}

is:

Q. Arrange the following elements in the increasing order of their metallic character.

S i, B e, M g, N a, P

The correct arrangement of increasing order of atomic radius among Na, K, Mg, Rb is:

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

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The order of decrease in atomic radii for Be,Na & Mg is :

The order of decrease in atomic radii for $B e, N a$ & $M g$ is :