If the dipole moment of AB molecule is given by 1.2 D and A-B the bond length is 1 Å then % ionic character of the bond is [Given: 1 debye =10-18 esu. cm ]

Question

If the dipole moment of AB molecule is given by 1.2 D and A-B the bond length is 1 Å then % ionic character of the bond is [Given: 1 debye =10-18 esu. cm ] (A) 75 (B) 50 (C) 60 (D) 25

Tardigrade Admin · Accepted Answer

μ text Cal =4.8 × 10-10 e.s. u × 10-8 cm =4.8 × 10-18 esu cm =4.8 D Hence % ionic character =(μ text experimental /μ text caculated ) × 100 =25 %

Q. If the dipole moment of $A B$ molecule is given by $1.2 D$ and $A - B$ the bond length is $1 Å$ then $%$ ionic character of the bond is
[Given: 1 debye $= 1 0^{- 18}$ esu. $c m$ ]

75

50

60

25

$μ_{Cal} = 4.8 \times 1 0^{- 10}$ e.s. $u \times 1 0^{- 8} c m$
$= 4.8 \times 1 0^{- 18}$ esu cm
$= 4.8 D$
Hence $%$ ionic character $= \frac{μ _{experimental}}{μ _{caculated}} \times 100 = 25%$

Q. If the dipole moment of AB molecule is given by 1.2D and A−B the bond length is 1Å then % ionic character of the bond is [Given: 1 debye =10−18 esu. cm ]

75

50

60

25

μCal ​=4.8×10−10 e.s. u×10−8cm =4.8×10−18 esu cm =4.8D Hence % ionic character =μcaculated ​μexperimental ​​×100=25%

Q. If the dipole moment of $A B$ molecule is given by $1.2 D$ and $A - B$ the bond length is $1 Å$ then $%$ ionic character of the bond is
[Given: 1 debye $= 1 0^{- 18}$ esu. $c m$ ]

$μ_{Cal} = 4.8 \times 1 0^{- 10}$ e.s. $u \times 1 0^{- 8} c m$
$= 4.8 \times 1 0^{- 18}$ esu cm
$= 4.8 D$
Hence $%$ ionic character $= \frac{μ _{experimental}}{μ _{caculated}} \times 100 = 25%$