Question

Bond length of A - A bond is 124 pm and bond length of B-B bond is 174 pm. The bond lengths
(in pm) of A-B bond in AB molecule is equal 70 X of percentage of ionic charactereft-B
bond is 19.5% then X value is​

Answer 1

Given that,

Bond length of A-A=124 pm

Bond length of B-B=174 pm

Ionic character = 19.5%

Bond length of A-B = 70X

We need to calculate the value of $r_{A}$

Using  formula for $r_{A}$

$r_{A}=\dfrac{bond\ length\ of\ A-A}{2}$

Put the value into the formula

$r_{A}=\dfrac{124}{2}$

$r_{A}=62\ pm$

We need to calculate the value of $r_{B}$

Using  formula for $r_{B}$

$r_{B}=\dfrac{bond\ length\ of\ B-B}{2}$

Put the value into the formula

$r_{B}=\dfrac{174}{2}$

$r_{B}=87\ pm$

We need to calculate the electronegativity

Using formula of ionic character

$ionic\ character=16\times(\Delta E.N)+3.5(\Delta E.N)^2$

Put the value into the formula

$19.5=16\times(\Delta E.N)+3.5(\Delta E.N)^2$

Suppose, $E.N=x$

Then, $19.5=16x+3.5x^2$

$3.5x^2+16x-19.5=0$

$x= 1, -5.57$

We take positive value of x.

So, The electronegativity is 1.

We need to calculate the value of X

Using formula of bond length

$bond\ length\ of\ A-B=r_{A}+r_{B}-9(\Delta E.N)$

Put the value into the formula

$70X=62+87-9\times1$

$X=\dfrac{62+87-9}{70}$

$X=2$

Hence, The value of X is 2.