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प्रश्न
KF crystallizes in fcc structure like sodium chloride. calculate the distance between K+ and F− in KF.
(Given: density of KF is 2.48 g cm−3)
उत्तर
Density of KF 2.48 g cm−3
ρ = `"nM"/("a"^3"N"_"A")`
n = 4 (For fcc)
ρ = `(4 xx 58)/("a"^3 xx 6.023 xx 10^23)`
2.48 = `(4 xx 58)/("a"^3 xx 6.023 xx 10^23)`
a3 = `(4 xx 58)/(2.48 xx 6.023 xx 10^23)`
a3 = `232/(14.93 xx 10^23)`
a3 = 15.54 × 10−23
a = 537.5 pm
d = `"a"/sqrt2` (For fcc)
d = `537.5/1.414`
d = 380.12 pm
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