The correct order of atomic size of $N a, B e$ and $M g$ is:

B e > M g > N a

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N a > M g > B e

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M g > N a > B e

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N a = M g > B e

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Solution

The correct option is C $N a > M g > B e$
As we move down the group, the atomic size increases.
$∴ M g > B e$ . And along the period, the atomic size decreases towards the right, hence $N a > M g$ .

Thus final order is $N a > M g > B e$ .

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Similar questions

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

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

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

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

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

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

Q. The order of decrease in atomic radii for

B e, N a

M g

is :

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The correct order of atomic size of Na,Be and Mg is:

The correct option is C Na>Mg>BeAs we move down the group, the atomic size increases.∴Mg>Be. And along the period, the atomic size decreases towards the right, hence Na>Mg. Thus final order is Na>Mg>Be.

The correct order of atomic size of $N a, B e$ and $M g$ is:

The correct option is C $N a > M g > B e$
As we move down the group, the atomic size increases.
$∴ M g > B e$ . And along the period, the atomic size decreases towards the right, hence $N a > M g$ .

Thus final order is $N a > M g > B e$ .