Question

a) The dipole moment of HBr is 2.60  1030 C m and interatomic distance 141 pm. What is
the percentage ionic character of HBr?
b) Draw all of the stereoisomers of 2,3-butandiol, label meso compounds and pairs of
enantiomer.

Answer 1

Answer on Question #76271, Chemistry / Inorganic Chemistry

a) The dipole moment of HBr is $2.60 \cdot 10^{\wedge} - 30\mathrm{C}\cdot \mathrm{m}$ and interatomic distance $141~\mathrm{pm}$ . What is the percentage ionic character of HBr?

b) Draw all of the stereoisomers of 2,3-butandiol, label meso compounds and pairs of enantiomer.

Solution

Question a

a) To determine the percentage ionic character of HBr we should find the ratio of $\mu_{\mathrm{exp}}$ experimentally obtained to the $\mu_{\mathrm{theor}}$ mathematically calculated. When calculating $\mu_{\mathrm{theor}}$ we make an assumption that one electron from hydrogen is completely transferred to bromine.

The percentage ionic character $= \left( {\mu }_{\text{exp }}/\mu_{\text{theor }}\right) \cdot {100}\%$

Find $\mu_{\mathrm{theor}}$

$\mu_{\mathrm{theor}} = q \cdot r,$

Where $q$ is charge of an electron, $q = 1.602 \cdot 10^{-19} C$ ,

r- interatomic distance, $r = 141\mathrm{pm} = 1.41 \cdot 10^{-10}\mathrm{m}$

$\mu_{\text{theor}} = q = 1.602 \cdot 10^{-19} \, \text{C} \cdot 1.41 \cdot 10^{-10} \, \text{m} = 2.259 \cdot 10^{-29} \, \text{C} \cdot \text{m}$

The percentage ionic character $= (2.60\cdot 10^{-30}\mathrm{C}\cdot \mathrm{m} / 2.259\cdot 10^{-29}\mathrm{C}\cdot \mathrm{m})\cdot 100\% = 12\%$

Answer: $12\%$

Question b

b) Enantiomer (optical isomer) is one of two stereoisomers that are mirror images of each other that are non-superposable (not identical).

Meso isomer is a stereoisomer with an identical or superimposable mirror image i.e., a non-optically active member of a set of stereoisomers, at least two of which are optically active.

2,3-butandiol has three stereoisomers (there are four forms are shown, where C=D, D and C are the same forms):

A

B

C

D

A and B are enantiomers, C and D are the same form, C is meso form.

$A = (2R,3R) - 2,3$ -butandiol (optically active)

B= (2R,3R)-2,3-butandiol (optically active)

C=D = meso-2,3-butandiol (non-optically active, has plane of symmetry).