JEE Main & Advanced Chemistry Classification of Elements and Periodicity in Properties / तत्त्वों का वर्गीकरण एवं गुणों में आवर्ति Question Bank Topic Test - Periodic Table

A compound AB whose electro negativity difference is 1.9. Atomic radius of A and B are 4 and \[2\,\overset{\text{o}}{\mathop{\text{A}}}\,\]. The distance between A & B means \[{{d}_{A-B}}\] is

A) \[6.72\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

B) \[5.82\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

C) \[6.9\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

D) \[7.5\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

Correct Answer: B

Solution :

[b] Given \[{{r}_{A}}=4\overset{\text{o}}{\mathop{\text{A}}}\,,{{r}_{B}}=2\overset{\text{o}}{\mathop{\text{A}}}\,,\Delta x=1.9\]

By the formula \[{{d}_{A-B}}={{r}_{A}}+{{r}_{B}}-.09\,(\Delta x)\]

\[=4+2-.09\times 1.9=6-0.171=5.82\overset{\text{o}}{\mathop{\text{A}}}\,\]