The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is nearly

Question

The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is nearly (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

% ionic character of a bond is determined with the help of Hanny-Smith's equation. % ionic characters =16(Δ EN )+3.5(Δ EN )2 Where, Δ EN = difference of electronegativity =4.0-1.2=2.80 So, % ionic character =1.6(2.80)+3.5(2.80)2=72.24 %

Q. The values of electronegativity of atom $A$ and $B$ are $1.20$ and $4.0$ respectively. The percentage of ionic character of $A - B$ bond is nearly

$%$ ionic character of a bond is determined with the help of Hanny-Smith's equation.
$%$ ionic characters $= 16 (Δ EN) + 3.5 (Δ EN)^{2}$
Where, $Δ EN =$ difference of electronegativity
$= 4.0 - 1.2 = 2.80$
So, $%$ ionic character
$= 1.6 (2.80) + 3.5 (2.80)^{2} = 72.24%$

Q. The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A−B bond is nearly

% ionic character of a bond is determined with the help of Hanny-Smith's equation. % ionic characters =16(ΔEN)+3.5(ΔEN)2 Where, ΔEN= difference of electronegativity =4.0−1.2=2.80 So, % ionic character =1.6(2.80)+3.5(2.80)2=72.24%

Q. The values of electronegativity of atom $A$ and $B$ are $1.20$ and $4.0$ respectively. The percentage of ionic character of $A - B$ bond is nearly

$%$ ionic character of a bond is determined with the help of Hanny-Smith's equation.
$%$ ionic characters $= 16 (Δ EN) + 3.5 (Δ EN)^{2}$
Where, $Δ EN =$ difference of electronegativity
$= 4.0 - 1.2 = 2.80$
So, $%$ ionic character
$= 1.6 (2.80) + 3.5 (2.80)^{2} = 72.24%$