logo

Tuesday, June 1, 2010

Periodic Properties of the Elements (ATOMIC AND IONIC RADIUS)

Measures of atomic radius: Unlike a ball, an atom doesn't have a fixed radius.
Why? Because it is impossible to determine exactly the location of an electron around the nucleus, so atomic radius can't be measured

Again: Why it is impossible to determine exactly the location of an electron around the nucleus
To know click HERE

The left hand diagram shows bonded atoms. The atoms are pulled closely together and so the measured radius is less than if they are just touching. This is what you would get if you had metal atoms in a metallic structure, or atoms covalently bonded to each other. The type of atomic radius being measured here is called the metallic radius or the covalent radius depending on the bonding.
AND
The right hand diagram shows what happens if the atoms are just touching. The attractive forces are much less, and the atoms are essentially "unsquashed".
The distance between the nuclei of two touching atoms (bonded atoms) known as: Bond length.
So:
Bond length can be defined as:
(it is the distance between the nuclei of two bonded atoms)
As you can see from the diagrams, the same atom could be found to have a different radius depending on what was around it.
Atomic radius can be defined as:
{it is half the distance between the centers of two similar atoms in diatomic molecule}
Examples (1): if bond length in H2 is 0.6 Angstrom, and bond length of HCl is 1.29 Angstrom. Find covalent radius of Cl2 atom
Bond length in HCl = radius of H + radius of Cl2
Bond length in HCl = 0.5 bond length in H2 + radius of Cl2
1.29 = 0.5 × 0.6 + radius of Cl2
1.29 = 0.3 + radius of Cl2
Radius of Cl2 = 1.29 – 0.3 = 0.99 Angstrom
Examples (2): in water molecule bond length is 0.96 Angstrom, H2 is 0.6 Angstrom. Find covalent radius for O2 atom
Bond length in H2O = radius of H2 + radius of O2
Bond length in H2O = 0.5 bond length in H + radius of O2
0.96 = 0.5 × 0.6 + radius in O2
0.96 = 0.3 + radius in O2
Radius in O2 = 0.96 – 0.3 = 0.66 Angstrom
Examples (3): Find bond length in NH3, given that bond length in H2 is 0.6 Angstrom, N2 is 1.4 A0
Bond length in NH3 = radius in N2 + radius in H2
Bond length in NH3 = 0.5 bond length in N2 + radius in H2
Bond length in NH3 = 0.5 × 1.4 + 0.5 × 0.6
Bond length in NH3 = 0.7 + 0.3 = 1 Angstrom
Examples (4): If sum of bond length in CH4 is 4.28 Angstrom, bond length in H2 is 0.6 Angstrom. Find atomic radius in C atom.
Bond length in CH4 = radius in C + radius in H2
Bond length in CH4 = radius in C + 0.5 × bond length in H2
4.28 ÷ 4 = radius in C + 0.5 × 0.6
1.07 = radius in C + 0.3
Radius in C = 1.07 – 0.3 = 0.77 Angstrom
Problems
1.Calculate the bond length in fluorine molecule provided that the bond length of hydrogen fluoride molecule is 0.94 Angstrom and the bond length in hydrogen molecule is 0.6 Angstrom
2.Calculate the atomic radius of carbon atom providing the bond length between the chlorine molecules (Cl – Cl) is 1.98 Angstrom and the bond length between carbon and chlorine atoms (C – Cl) is 1.76 Angstrom

1 comments:

JeffreyShockley said...

The latest news from the Texas Education Agency is available through news releases, online correspondence, mailing lists, and other posted information. solution manual test bank

Post a Comment