Calculate the percentage ionic character of Si-H bond in `SiH_4` assuming Pauling electronegativity values of Si and H to be 1.8 and 2.1 respectively.

Explanation

Percentage ionic character, `P=18(X_B–X_A)^(1.4)`.
Substituting electronegativity values of Si and H
Percentage ionic character of Si-H bond, `P=18 (2.1-1.8)^(1.4) = 18 (0.3)^(1.4) = 3.336`
Taking logarithms: `log 18 + log (0.3)^(1.4) = log 18+ 1.4 log 0.3`
1.2552 + 1.4(-0.5228) = 1.2552 - 0.7319 = 0.5233
Taking antilogarithms: Antilog 0.5233 = 3.34
Percentage ionic character of Si-H bond, P= 3.34