Home | | Chemistry 11th std | Atomic radius

Chapter: 11th Chemistry : UNIT 3 : Periodic Classification of Elements

Atomic radius

Atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

Atomic radius

 

Atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

 

It is not possible to measure the radius of an isolated atom directly. Except for noble gases, usually atomic radius is referred to as covalent radius or metallic radius depending upon the nature of bonding between the concerned atoms.


 

Covalent radius

It is one-half of the internuclear distance between two identical atoms linked together by a single covalent bond. Inter nuclear distance can be determined using x-ray diffraction studies.


Example:

The experimental internuclear distance in Cl2 molecule is 1.98 Ã…. The covalent radius of chlorine is calculated as below.


The formation of covalent bond involves the overlapping of atomic orbitals and it reduces the expected internuclear distance. Therefore covalent radius is always shorter than the actual atomic radius.

The covalent radius of individual atom can also be calculated using the internuclear distance (dA-B) between two different atoms A and B. The simplest method proposed by Schomaker and Stevenson is as follows.

dA-B = rA + rB - 0.09 (χA-χB)

where Ï‡A and Ï‡B are the electronegativities of A and B respectively  in Pauling units
 Here Ï‡A Ï‡and radius is in Ã….

Let us calculate the covalent radius of hydrogen using the experimental d value is 1.28 Ã… and the covalent radius of chlorine is 0.99 Ã…. In pauling scale the electronegativity of chlorine and hydrogen are 3 and 2.1 respectively.

dH-Cl = rH + rCl - 0.09 ( χCl - χH)

1.28 = rH + 0.09 - 0.09 (3 - 2.1)

1.28 = rH + 0.09 - 0.09 (0.9)

1.28 = rH + 0.09 - 0.081

1.28 = rH + 0.909

∴ rH = 1.28 - 0.909 = 0.317 Ã…


Metallic radius

It is defined as one-half of the distance between two adjacent metal atoms in the closely packed metallic crystal lattice.

For example, the distance between the adjacent copper atoms in solid copper is 2.56 Ã… and therefore the metallic radius of copper is


The metallic radius can be calculated using the unit cell length of the metallic crystal. You will study the detailed calculation procedure in XII standard solid state unit.

 

Periodic Trends in Atomic Radius

 

Variation in Periods

Atomic radius tends to decrease in a period. As we move from left to right along a period, the valence electrons are added to the same shell. The simultaneous addition of protons to the nucleus, increases the nuclear charge, as well as the electrostatic attractive force between the valence electrons and the nucleus. Therefore atomic radius decreases along a period.

 

Effective nuclear charge

In addition to the electrostatic forces of attraction between the nucleus and the electrons, there exists repulsive forces among the electrons. The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. It is approximated by the below mentioned equation.

Zeff = Z – S

Where Z is the atomic number and 'S' is the screening constant which can be calculated usi ng Slater's rules as described below.

 

Step 1 :

 

Write the electronic configuration of the atom and rearrange it by grouping ns and np orbitals together and others separately in the following form.

 

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p)...

 

Step 2 :

 

Identify the group in which the electron of interest is present. The electron present right to this group does not contribute to the shielding effect.

 

Each of the electrons within the identified group (denoted by 'n') shields to an extent of 0.35 unit of nuclear charge. However, it is 0.30 unit for 1s electron.

 

Step 3 :

 

Shielding of inner shell electrons.

If the electron of interest belongs to either s or p orbital,

i) each electron within the (n-1) group shields to an extent of 0.85 unit of nuclear charge, and

ii) each electron within the (n-2) group (or) even lesser group (n-3, (n-4) etc...completely shields i.e. to an extent of 1.00 unit of nuclear charge.

 

If the electron of interest belongs to d or f orbital, then each of electron left of the group of electron of interest shields to an extent of 1.00 unit of nuclear charge.

 

Step 4 :

 

Summation of the shielding effect of all the electrons gives the shielding constant 'S'

Example: Let us explain the calculation of effective nuclear charge on 4s electron and 3d electron in scandium. The electronic configuration of scandium is 1s2, 2s2, 2p6, 3s 2, 3p6, 4s2, 3d1. we can rearrange as below.


Zeff = Z – S i.e.= 21- 18  âˆ´Zeff = 3

Calculation of effective nuclear charge on 3d electron


∴ Zeff = Z – S i.e. =21 – 18 âˆ´ Zeff = 3


Variation in Group

In the periodic table, the atomic radius of elements increases down the group. As we move down a group, new shells are opened to accommodate the newly added valence electrons. As a result, the distance between the centre of the nucleus and the outermost shell containing the valence electron increases. Hence, the atomic radius increases. The trend in the variation of the atomic radius of the alkali metals down the group os shown below.



Activity 3.1

Covalent radii (in Ã…) for some elements of different groups and periods are listed below. Plot these values against atomic number. From the plot, explain the variation along a period and a group.

2nd group elements : Be (0.89), Mg (1.36), Ca (1.74), Sr (1.91) Ba(1.98)

17th group elements : F (0.72), Cl (0.99), Br (1.14), I (1.33)

3rd Period elements : Na(1.57), Mg(1.36), Al (1.25), Si(1.17), P(1.10), S(1.04), Cl(0.99)

4th period elements : K(2.03), Ca(1.74), Sc(1.44), Ti(1.32), V(1.22), Cr(1.17), Mn(1.17), Fe(1.17), Co(1.16), Ni(1.15), Cu(1.17), Zn(1.25), Ga(1.25), Ge(1.22), As(1.21), Se(1.14), Br(1.14)


Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Chemistry : UNIT 3 : Periodic Classification of Elements : Atomic radius |


Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.