In a compound, nitrogen atoms (N) make cubic close packed lattic and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N ?

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Solution

Let the number of N atoms in ccp is x

$∴$ Number of tetrahedral volds = 2x

$∴$ Number of N atoms = $\frac{1}{3} \times 2 x$

$\frac{N u m b e r o f N a t o m s}{N u m b e r o f M a t o m s} = \frac{3 x}{2 x} = \frac{3}{2}$

So, the formula of the compound is $M_{2} N_{3}$

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In a compound, nitrogen atoms (N) make cubic close packed lattic and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N ?

Let the number of N atoms in ccp is x ∴ Number of tetrahedral volds = 2x ∴ Number of N atoms = 13×2x Number of N atomsNumber of M atoms=3x2x=32 So, the formula of the compound is M2N3

Let the number of N atoms in ccp is x

$∴$ Number of tetrahedral volds = 2x

$∴$ Number of N atoms = $\frac{1}{3} \times 2 x$

$\frac{N u m b e r o f N a t o m s}{N u m b e r o f M a t o m s} = \frac{3 x}{2 x} = \frac{3}{2}$

So, the formula of the compound is $M_{2} N_{3}$