Assume that in CHX 3- only the CX bond is polar and it has 20 - ionic character- Also- CHX 3 is a perfect tetrahedral i-e- - HCX - XCX - 109- 28- If the C-H bond length is 150 pm- then find dipole moment in Debye of CHX 3 - Given-cos 180-109- 28-1-3 -

Percent ionic character = observed dipole moment (in D)/4.8 ×bond length(in ∘A)× 100 Dipole moment due to bond CX =4.8 × 1.5 × 0.20=1.44D cos α=cos(180 1 ...

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Standard XII

Chemistry

Mathematics of Dipole Moment

Assume that i...

Question

Assume that in $C H X_{3}$ , only the CX bond is polar and it has $20 %$ ionic character. Also, $C H X_{3}$ is a perfect tetrahedral i.e. $∠ H C X = ∠ X C X = 109^{\circ} 28^{^{'}}$ . If the C-H bond length is 150 pm, then find dipole moment (in Debye) of $C H X_{3}$ .(Given: $c o s (180 - 109^{\circ} 28^{^{'}}) = \frac{1}{3}$ ) .

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Solution

$Percent ionic character = \frac{o b s e r v e d d i p o l e m o m e n t (i n D)}{4.8 \times bond length (i n \circ A)} \times 100$

Dipole moment due to bond CX $= 4.8 \times 1.5 \times 0.20 = 1.44 D$

$c o s α = c o s (180 - 109^{\circ} 28^{^{'}}) = \frac{1}{3}$ , So dipole moment of the molecule
$μ = (3 \times c o s α) \times 1.44 = 1.44 D$

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Assume that in CHX3, only the CX bond is polar and it has 20% ionic character. Also, CHX3 is a perfect tetrahedral i.e. ∠HCX=∠XCX=109∘28′. If the C-H bond length is 150 pm, then find dipole moment (in Debye) of CHX3.(Given:cos(180−109∘28′)=13) .

Percent ionic character=observed dipole moment(in D)4.8×bond length(in∘A)×100 Dipole moment due to bond CX =4.8×1.5×0.20=1.44D cos α=cos(180−109∘28′)=13, So dipole moment of the molecule μ=(3×cos α)×1.44=1.44D