The density of sodium chloride at `25^(@)C "is" 2.163 xx 10 ^(3) " kg m" ^(-3)` When X -rays rom a palladium target having waveleth of 58.1 pm are used, the (200) reflection of sodium chloride occurs at an angle of ` 5.90^(@)` . How many `Na^(+) and Cl^(-)` ions are present in the unit cell ? ( Molar mass of NaCl = 58.5 `" mol" ^(-1) sin 5.9 ^(@) = 0.1028)`

Options

A. `2Na^(+) , 4Cl^(-)`
B. `3Na^(+) , 1Cl^-`
C. `4Na^+, 4Cl^-`
D. `4Na^+, 2Cl^-`

Answer

C

Explanation

We have `n lamda = 2d sin theta `
for n ` = 1 , d_200 = (lamda)/(2 sin theta) = (58.1"pm")/(2 sin 5.9^@) = 282` pm
Now ,` d_200 = (a)/(sqrt(h^2 + k^2 + l^2) )`
282(pm) ` = (a)/(sqrt(2^2 + 0^2 + 0^2) ) ` , a = 564pm
thus using the equation
`rho = (zM)/(NV) = (zM)/(Na^3)`
` therefore z = (rho Na^3)/(M) = (2.163 xx 10^3 xx (6.022 xx 10^23) xx (564 xx 10^(-12) )^3)/(58.443 xx 10^(-3)) = 3.999 = 4`
thus , a unit cell contains 4`Na^+` and `4Cl^-` ions as expected