What is the formula of a magnetic oxide of cobalt used in recording tapes that crystallises with cobalt atoms occupying one-eighth of the tetrahedral holes and onehalf of the octahedral holes in a closest packed array of oxide ions?

Question

What is the formula of a magnetic oxide of cobalt used in recording tapes that crystallises with cobalt atoms occupying one-eighth of the tetrahedral holes and onehalf of the octahedral holes in a closest packed array of oxide ions? (A) Co2O3 (B) CoO2 (C) Co3O4 (D) CoO

Tardigrade Admin · Accepted Answer

Tetrahedral holes = 8, Octahedral holes = 4 Cobalt atoms = (8/8) = 1 Oxide ions =(4/2) = 2 Thus, formula is CoO2.

Q. What is the formula of a magnetic oxide of cobalt used in recording tapes that crystallises with cobalt atoms occupying one-eighth of the tetrahedral holes and onehalf of the octahedral holes in a closest packed array of oxide ions?

$C o_{2} O_{3}$

$C o O_{2}$

$C o_{3} O_{4}$

$C o O$

Tetrahedral holes $= 8$ , Octahedral holes $= 4$
Cobalt atoms $= \frac{8}{8} = 1$
Oxide ions $= \frac{4}{2} = 2$
Thus, formula is $C o O_{2}$ .

Q. What is the formula of a magnetic oxide of cobalt used in recording tapes that crystallises with cobalt atoms occupying one-eighth of the tetrahedral holes and onehalf of the octahedral holes in a closest packed array of oxide ions?

Co2​O3​

CoO2​

Co3​O4​

CoO

Tetrahedral holes =8, Octahedral holes =4 Cobalt atoms =88​=1 Oxide ions =24​=2 Thus, formula is CoO2​.

$C o_{2} O_{3}$

$C o O_{2}$

$C o_{3} O_{4}$

$C o O$

Tetrahedral holes $= 8$ , Octahedral holes $= 4$
Cobalt atoms $= \frac{8}{8} = 1$
Oxide ions $= \frac{4}{2} = 2$
Thus, formula is $C o O_{2}$ .