Question

Give reason ionic radius of f- is more than atomic radius of f class11

Answer 1

Metals - the atomic radius of a metal is generally larger than the ionic radius of the same element.

Why? Generally, metals loose electrons to achieve the octet. This creates a larger positive charge in the nucleus than the negative charge in the electron cloud, causing the electron cloud to be drawn a little closer to the nucleus as an ion.

Non-metals - the atomic radius of a non-metal is generally smaller than the ionic radius of the same element.

Why? Generally, non-metals loose electrons to achieve the octet. This creates a larger negative charge in the electron cloud than positive charge in the nucleus, causing the electron cloud to 'puff out' a little bit as an ion.

Cations (or anions) of different elements are said to be **isoelectronic** if they both have the same electronic configuration e.g. K+ and Ca 2+

For isoelectronic cations, the more positive the ionic charge, the smaller the ionic radius.

For isoelectronic anions, the more negative the ionic charge, the larger the ionic radius.


Sample problem:

List the following species in order of increasing size: Rb, Rb+, F, F-, Te.


Answer: First look in which periods the element is located. Rb and Te are in period 5, and F is in period 2.

We know that Rb+ is smaller than Rb because it loses an electron and has one less shell. Another reasoning is that since Rb+ has the same number of protons but less electrons, the electrons are held more tightly in Rb+. We know that Te is smaller that Rb because of the trend of decreasing radius to the right within a period. But since Te has one shell more than Rb+ is it bigger.

so far we have Rb+, Te, Rb

We know that F and F- are both smaller than the rest because they have less shells.

Now between F and F-: F- has one more electron, which causes the radius to be bigger.


FINAL ANSWER: F, F-, Rb+, Te, Rb

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