Question

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

Answer 1

Ionic Radius vs. Atomic Radius

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

Answer 2

Because effective nuclear charge decreases.

Explanation:

Atomic radii  is the distance from the center of the nucleus to the valence shell.

Florine (F) has atomic number of 9 and thus contain 9 protons and 9 electrons.  

$F:9:1s^22s^22p^5$

$F^-$ has 10 electrons and 9 protons. Now 9 protons present in the nucleus can not influence 10 electrons towards itself, the effective nuclear charge decreases, the valence electrons are less tightly held by the nucleus and thus the size increases.

$F^-:10:1s^22s^22p^6$

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