Pauling developed an equation relating bond and the electronegativity difference. |xA-xB| of the bonded atoms A and B viz: percentage ionic character =16(D x)-3.5(D x)2. Calculate (D x)=|xA-xB| for 15 % ionic character.

Question

Pauling developed an equation relating bond and the electronegativity difference. |xA-xB| of the bonded atoms A and B viz: percentage ionic character =16(D x)-3.5(D x)2. Calculate (D x)=|xA-xB| for 15 % ionic character. (A) 0.6 (B) 0.5 (C) 0.8 (D) 1.2

Tardigrade Admin · Accepted Answer

35( Dx )2-160 Dx +150=0 so Dx =(160 ± √25600-21000/70) =(160 ± √4600/70)=1.2

Q. Pauling developed an equation relating bond and the electronegativity difference. $∣ x_{A} - x_{B} ∣$ of the bonded atoms $A$ and $B$ viz : percentage ionic character $= 16 (D x) - 3.5 (D x)^{2}$ . Calculate $(D x) = ∣ x_{A} - x_{B} ∣$ for $15%$ ionic character.

0.6

0.5

0.8

1.2

$35 (D x)^{2} - 160 D x + 150 = 0$
so $D x = \frac{160 \pm 25600 - 21000}{70}$
$= \frac{160 \pm 4600}{70} = 1.2$

Q. Pauling developed an equation relating bond and the electronegativity difference. ∣xA​−xB​∣ of the bonded atoms A and B viz : percentage ionic character =16(Dx)−3.5(Dx)2. Calculate (Dx)=∣xA​−xB​∣ for 15% ionic character.

0.6

0.5

0.8

1.2

35(Dx)2−160Dx+150=0 so Dx=70160±25600−21000​​ =70160±4600​​=1.2

Q. Pauling developed an equation relating bond and the electronegativity difference. $∣ x_{A} - x_{B} ∣$ of the bonded atoms $A$ and $B$ viz : percentage ionic character $= 16 (D x) - 3.5 (D x)^{2}$ . Calculate $(D x) = ∣ x_{A} - x_{B} ∣$ for $15%$ ionic character.

$35 (D x)^{2} - 160 D x + 150 = 0$
so $D x = \frac{160 \pm 25600 - 21000}{70}$
$= \frac{160 \pm 4600}{70} = 1.2$