MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trend in Atomic Radius

A FCC crystal...

Question

A FCC crystal of $O^{2 -}$ (oxide) ion has ${M n}^{n +}$ ions in all octahedral voids. Then value of n is
(1) 4
(2) 1
(3) 2
(4) 3

Open in App

Solution

In FCC crystal, no. of oxide ions per unit cell.
In FCC, Number of close packed spheres = number of octahedral voids
$∴$ Total no. of octahedral voids = 4
Number of $M^{n +}$ particles = 4
As all O.V. are occupied by $M^{n +}$ .
$∴$ Formula of compound is $M_{4} O_{4}$
4n + (-2)4 = 0
n = $\frac{8}{4}$
n = 2
Hence, option (3) is correct.

Suggest Corrections

thumbs-up

10

Similar questions

Q. In a crystal oxide ions are arranged in fcc and

A^{2 +}

1 / 8^{t h}

of the tetrahedral voids, and ions

B^{3 +}

1 / 2

of the octahedral voids. Calculate the formula of the oxide.

Q. Oxide ion forming a fcc lattice in which

F e^{2 +}

ions are occupying

1 / 8^{t h}

of tetrahedral void and

F e^{3 +}

ions are occupying

\frac{1}{2}

of the octahedral void. What will be the formula of the compound?

Q. Aluminium oxide crystallizes with aluminium ions in 2/3 of the line octahedral and 1/3 octahedral holes remaining vacant in a closest packed array of oxide ions. What is the formula of aluminium oxide?

Q. Consider the ions

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

. Strongest oxidizing ion is ?

A^{n +}

{A O}_{3}^{-}

M n O_{4}^{-}

{M n}^{2 +}

in acidic solution. Given that

2.68 \times 10^{- 3}

A^{n +}

1.61 \times 10^{- 3}

M n O_{4}^{-}

n

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Size of an Atom or Ion

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Trend in Atomic Radius

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon