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The bond length in an HI molecule is 1.61 Å and the measured dipole moment is 0.44 D. What is the magnitude (in units of e) of the negative charge on I in HI?

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Okay, let's think through this step-by-step:

* The HI bond length is given as 1.61 Å

* The measured dipole moment is 0.44 D

* We need to find the magnitude of the negative charge on I in units of e

To find the charge, we can use the formula:

μ = q*d

Where:

μ is the dipole moment (0.44 D)

q is the magnitude of charge

d is the HI bond length (1.61 Å = 1.61 x 10-10 m)

Plugging in the values:

0.44 = q * 1.61 x 10-10

q = 0.44 / (1.61 x 10-10)

q = 0.273 x 1010 C

Charges are generally expressed in units of elementary charge (e):

1 e = 1.602 x 10-19 C

Therefore, converting 0.273 x 1010 C to units of e:

q = (0.273 x 1010 C) / (1.602 x 10-19 C/e)

q = 0.170 e

Therefore, the magnitude of the negative charge on I in HI is 0.170 e.

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