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  4. 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

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

NTA AbhyasNTA Abhyas 2020Classification of Elements and Periodicity in Properties

Solution:

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