If covalent radius of A is 0.99 A0 and bond length is 1.46 A0 Calculate the covalent radius of B. XA and XB are 3.5 and 2.5 respectively.

 (A) 0.35 A⁰              (B) 0.56 A⁰              (C) 1.98A⁰               (D) 10.48A⁰

The covalent radius of B is calculated by the relation given by Stevenson & Schoemaker, given as:

For a diatomic Hetero molecule:

Bond Length (l_A-B) = r_A + r_B- 0.09(X_A-X_B)

Where X_A= Electronegativity of more electronegative atom

Where X_B= Electronegativity of less electronegative atom

Bond Length=1.46 A⁰ = 0.99 A⁰ + r_B - 0.09(3.5-2.5)

 Hence r_B = 0.56 A⁰ 

Solved Questions:

(1) The electronegativity of F and H are 4.0 and 2.1 respectively. The percentage ionic character in H and F bond is.

(2) A given compound AB whose electronegative difference is 1.9 . Atomic radius of A and B are 4 and 2 Angstroms the distance between A and B means dA-B is ?

(3) The C–C single bond length is 1.54 Angstroms and that of Cl–Cl is 1.98 Angstroms. If the electronegativity of Cl and C are 3.0 and 2.5 respectively, the C–Cl bond-length will be equal to ?

(4) Stevenson & Schromaker Equation: Determination covalent radius of Heterogeneous Molecules.

(5) How to calculate percentage (%) ionic character in covalent compounds?