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Question

When the following five anions are arranged in order of decreasing ionic radius, the correct sequence is?


A

Se2,I,Br,O2,F

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B

I,Se2,O2,Br,F

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C

Se2,I,Br,F,O2

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D

I,Se2,Br,O2,F

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Solution

The correct option is D

I,Se2,Br,O2,F


Lets calculate number of electrons for each one of them.

Se2 36 electrons (Valence n = 4)

I 54 electrons (Valence n = 5)

Br 36 electrons (Valence n = 4)

O2 10 electrons (Valence n = 2)

F 10 electrons (Valence n = 2)

I is biggest in size

Down the group from Top to Bottom, Atomic Radius increases because the principal quantum number (n) increases and the valance electrons arefarther from the nucleus.

Se2 and Br are isoelectronic, but n = 4 which is greater than n = 2, so they both will have greater radii than O2 and F

From these two

Se2 > Br

Now, try to remember from the videos, for isoelectronic species, the more cationic charge on the atom, lesser will be then ionic radius.

More the ionic charge on the atom, more will be the ionic radius.

Now, In option (a), the correct order will be Mg > Li > Be

Because Mg has principal quantum number (n) = 3 and Be gas n = 2

Among Li and Be, Li > Be because along a period from Left to Right, atomic radius increased

Now, In option (b)

H+ < Li+ < H

For H+, there are no electrons, so no atomic radius. In case of Li+ and H, Li+ < H,

Here, Na+ has more charge than others, therefore, Na+ is the smallest in size.Now, for O2 and F

O2 > F

Now, try to remember from the videos, for isoelectronic species, the more cationic charge on the atom, lesser will be then ionic radius.

More the ionic charge on the atom, more will be the ionic radius.

Now, In option (a), the correct order will be Mg > Li > Be

Because Mg has principal quantum number (n) = 3 and Be gas n = 2

Among Li and Be, Li > Be because along a period from Left to Right, atomic radius increased

Now, In option (b)

H+ < Li+ < H

For H+, there are no electrons, so no atomic radius. In case of Li+ and H,

Li+ < H, Here, Na+ has more charge than others, therefore, Na+ is the smallest in size.

Therefore,

I > Se2 > Br > O2 > F

Chemistry is so simple !!!


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