In a compound AB, electro negativity difference between A and B is 1.9 . Atomic radius A and B are 4 overset° A and 2 overset° A . The distance between A and B atoms means dA - B is

Question

In a compound AB, electro negativity difference between A and B is 1.9 . Atomic radius A and B are 4 overset° A and 2 overset° A . The distance between A and B atoms means dA - B is (A) 6.72 overset° A (B) 5.82 overset° A (C) 6.9 overset° A (D) 7.5 overset° A

Tardigrade Admin · Accepted Answer

dA - B = rA + rB - 0.09 (. xA - xB .) = 6 - 0.09 (. 1.9 .) ≈ 5.829

Q. In a compound AB, electro negativity difference between A and B is $1.9$ . Atomic radius A and B are $4 A^{\circ}$ and $2 A^{\circ}$ . The distance between A and B atoms means $d_{A - B}$ is

$6.72 A^{\circ}$

$5.82 A^{\circ}$

$6.9 A^{\circ}$

$7.5 A^{\circ}$

$d_{A - B} = r_{A} + r_{B} - 0.09 (x_{A} - x_{B}) = 6 - 0.09 (1.9) \approx 5.829$

Q. In a compound AB, electro negativity difference between A and B is 1.9 . Atomic radius A and B are 4A∘ and 2A∘ . The distance between A and B atoms means dA−B​ is

6.72A∘

5.82A∘

6.9A∘

7.5A∘