Question

The dipole measured for HI is 0.380 D. The bond distance is 161 pm. What is the percent ionic character of the HI bond?

Answer 1

Answer:

Percent ionic character of HI bond is 4.91%.

Explanation:

Given Data:

Measured Dipole = 0.380D

bond distance = d = 161pm = 1.61*10^-8 cm

Calculation:

% ionic character is determined by following equation:

% ionic= (dipole measured/dipole calculated)*100

Now,

$Dipole(calc)=qd$

$Dipole(calc)= (1.6*10^{-19}*3*10^{9})esu *1.61*10^{-8}cm$

(In above step 3*10^8 is multiplied to convert coulomb into esu)

$Dipole(calc)=7.728*10^{-18} esu*cm$

As,

$10^{-18}esu*cm= 1D$

So,

$Dipole(calc)=7.728D$

Now we can % ionic character using above equation:

%ionic=(0.380D/7.728D)*100

% ionic character=4.91%

Answer 2

Using the ratio formula which relates the measured and calculated dipole to the percentage ionic character, the percentage ionic character of the HI bond is 4.92%

Given the Parameters :

• Measured dipole = 0.380D
• Bond distance = 161 pm = 1.61 × 10^-8 m
• $1 \times 10^{-18} esu cm = 1D$

The percentage ionic character is defined as :

• (Measured dipole / calculated dipole) × 100%

Calculated dipole :

• qd
• Where q = magnitude of charges ; d = distance apart

Calculated dipole = $(1.6 \times 10^{-19})(1.61 \times 10^{-8}) = 2.576 \times 10^{-27} C\: cm$

To convert to esu cm:

$(2.576 \times 10^{-27})(3 \times 10^{9} = 7.728 \times 10^{-18} = 7.728D$

Hence,

% ionic character = $\frac{0.380D}{7.728D} \times 100$% = 0.0492%

Therefore, the percentage ionic character is 4.92%

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