$C_{60}$ (bucky ball) is cubic closest packed (face Centred cubic) in its crystalline form. If you insert potassium atoms into all the tetrahedral and octahedral holes of the $C_{60}$ structure, the formula would become $K_{x} C_{60}$ . What is the value of $x$ ?

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Solution

The correct option is C 3
$C_{60}$ forms fcc unit cell. If N represents the number of $C_{60}$ present in one unit cell, then the number of octahedral voids in one unit cell will be N and the number of tetrahedral voids in one unit cell will be 2N. Total number of voids in one unit cell will be $N + 2 N = 3 N$ . This is equal to the number of K atoms.
Thus, for N $C_{60}$ units in one unit cell , there will be 3N K atoms.
For 1 $C_{60}$ unit, there will be 3 K atoms.
The formula will be $K_{3} C_{60} = K_{x} C_{60}$ or $x = 3$ .

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C60 (bucky ball) is cubic closest packed (face Centred cubic) in its crystalline form. If you insert potassium atoms into all the tetrahedral and octahedral holes of the C60 structure, the formula would become KxC60. What is the value of x?

$C_{60}$ (bucky ball) is cubic closest packed (face Centred cubic) in its crystalline form. If you insert potassium atoms into all the tetrahedral and octahedral holes of the $C_{60}$ structure, the formula would become $K_{x} C_{60}$ . What is the value of $x$ ?