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The dipole moment (mu) of HBr (a polar covalent molecule) is 0.851 D (debye), and its percent ionic character is 12.6 %. Estima...Asked by Summer
The dipole moment (μ) of HBr (a polar covalent molecule) is 0.851D (debye), and its percent ionic character is 12.6% . Estimate the bond length of the H−Br bond in picometers.
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Notes for conversion:
1D= 3.34*10^-30 C*m & Q= 1.6*10^-19 C
The equation we will be using is μ=Q*r
-we need to rearrange this equation to solve for the bond length (r) making the equation now: r=μ/Q
The 0.851D needs to be converted to C*m. So we make a ratio now and multiply 0.851D=3.34*10^-30 C*m/ 1 D = 2.84*10^-30 C*m
(multiply 0.851 times 3.34*10^-30)
We need to also change out the Q (since it is in percent form to the actual amount) 100% Q= 1.6*10^-19 C
.126(which was 12.6%) times 1.6*10^-19 = 2.02*10^-20 C
now we can solve with our μ (2.84*10^-30) and Q (2.02*10^-20) using our equation r=μ/Q which will equal 1.41*10^-10 m
we need to make this into picometers now. 1m= 10^12pm
= 1.41*10^-10m * 10^12pm/1m = 141 pm
(if you need to answer this in 3 sig figs it is 14*10^1 pm)
1D= 3.34*10^-30 C*m & Q= 1.6*10^-19 C
The equation we will be using is μ=Q*r
-we need to rearrange this equation to solve for the bond length (r) making the equation now: r=μ/Q
The 0.851D needs to be converted to C*m. So we make a ratio now and multiply 0.851D=3.34*10^-30 C*m/ 1 D = 2.84*10^-30 C*m
(multiply 0.851 times 3.34*10^-30)
We need to also change out the Q (since it is in percent form to the actual amount) 100% Q= 1.6*10^-19 C
.126(which was 12.6%) times 1.6*10^-19 = 2.02*10^-20 C
now we can solve with our μ (2.84*10^-30) and Q (2.02*10^-20) using our equation r=μ/Q which will equal 1.41*10^-10 m
we need to make this into picometers now. 1m= 10^12pm
= 1.41*10^-10m * 10^12pm/1m = 141 pm
(if you need to answer this in 3 sig figs it is 14*10^1 pm)
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