The gaseous metallic chloride (MCl type) molecule has a measured dipole moment of 9.0 D, which indicates that it is a very polar molecule. The separation between the nuclei in this molecule is 2.25 × 10-8 cm. Calculate the percentage ionic character in MCl molecule. [1 eV =1.6 × 10-19 J , C =4.8 × 10-10 esu ]

Question

The gaseous metallic chloride (MCl type) molecule has a measured dipole moment of 9.0 D, which indicates that it is a very polar molecule. The separation between the nuclei in this molecule is 2.25 × 10-8 cm. Calculate the percentage ionic character in MCl molecule. [1 eV =1.6 × 10-19 J , C =4.8 × 10-10 esu ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Dipole moment of compound would have been completely ionic =(4.8 × 10-10 esu )(2.25 × 10-8 cm ) =4.8 × 2.25 D So % ionic character =(9/4.8 × 2.25) × 100 % =83.33 %

Dipole moment of compound would have been completely ionic =(4.8×10−10esu)(2.25×10−8cm) =4.8×2.25D So % ionic character =4.8×2.259​×100% =83.33%

Dipole moment of compound would have been completely ionic
$= (4.8 \times 1 0^{- 10} es u) (2.25 \times 1 0^{- 8} c m)$
$= 4.8 \times 2.25 D$
So $%$ ionic character $= \frac{9}{4.8 \times 2.25} \times 100%$
$= 83.33%$