Question

Calculate the percent ionic, the percent covalent, and the bond length (in picometers) of a chemical bond between phosphorus and selenium.

Phosphorus—atomic radius: 109 pm; covalent radius: 106 pm; ionic radius: 212 pm.
Selenium—atomic radius: 122 pm; covalent radius: 116 pm; ionic radius: 198 pm.

98 percent ionic, 2 percent covalent, 410 pm
4 percent ionic, 96 percent covalent, 222 pm
2 percent ionic, 98 percent covalent, 222 pm
96 percent ionic, 4 percent covalent, 410 pm

Answer 1

The correct option is;

4 percent ionic, 96 percent covalent, 222 pm

Step-by-step explanation:

The parameters given are;

Phosphorus:

Atomic radius = 109 pm

Covalent radius = 106 pm

Ionic radius = 212 pm

Electronegativity of phosphorus = 2.19

Selenium:

Atomic radius = 122 pm

Covalent radius = 116 pm

Ionic radius = 198 pm

Electronegativity of selenium= 2.55

The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;

Using Pauling's alternative electronegativity difference method, we have;

`\% \, Ionic \ Character = \left [18* (\bigtriangleup E.N.)^(1.4) \right ] \%`

Where:

Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36

Therefore;

`\% \, Ionic \ Character = \left [18* (0.36)^(1.4) \right ] \% = 4.3 \%`

Hence the percentage ionic character = 4.3% ≈ 4%

the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%

The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;

The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.

The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.