Question

If The dipole moment is given by charge x bond length (qxd) then, the DM of HF should have been less than the DM of HI (due to smaller bond length). But DM of HF=1.9D & DM of HI=0.4D..Why??

Answer 1

greater differences in electronegativity between H and F than in H and I

Answer 2

But the formula does not include the difference in electronegativities, and the charges for both F and I are same

Answer 3

Periodic table trends:
Electronegativity increases down the row and up a column.
You are aware that, both, I and F, are on the 7th column (7 valence e), F is higher on the table than I and is more electronegative (most electronegative element actually) and it pulls electrons closer to it than Iodine would and thus the difference between H and F is greater.

Answer 4

yeah but since F tends to pull e- closer , the bond length decreases..and since DM is charge x bond length..won't the DM be low for HF (as bond length is reduced due to pulling of e-)

Answer 5

H-F .. 3.98-2.18 = 1.8
H-I.. 2.66-2.18 = 0.48

say bond lengths were idk 100 pm and 300pm respectively (i made these up)

(1.8)100 = 180
0.48(300)=144

Greater differences in electronegativity

Answer 6

so when you take the charge..don't you have to take it as the charge on cation and anion??? why do you take charge as difference in electronegativities...??

Answer 7

you mean H has +1 and F has -1? those are oxidation states

those do not define dipole moments

Answer 8

No i mean.. isn't the charge on both F and I -1 in HF and HI??? and if that is the case than the only factor deciding DM will be bond length...and the bond length is smaller in case of HF (due to higher electronegativity) so shouldn't the DM of HF be lesser than that of HI??

Answer 9

thats not the charge one speaks of in terms of dipole moments

Answer 10

the charge is the difference of electronegativity between the two elements

Answer 11

ohk Thank u

Answer 12

no prob!