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

The correct option is;

4 percent ionic, 96 percent covalent, 222 pm

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;

Where:

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

Therefore;

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.

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