Ing D.docx
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1. Demuestre que el cociente mínimo de catión a anión para un número de coordinación de 4 es 0,225.
AB = √a2 + a2
AB = 2ra AB = 2√a
a√2 = 2ra
a=
2ra √2
“a” Queda en función de los lados del cubo con la diagonal que relaciona el anión y el catión. 2
(AEF)2 = (AB) + (FB)2
AEF = 2(ra + rc )
AB = 2ra FB = a =
2 ra √2
Reemplazando 𝐀𝐄𝐅 = 𝟐(𝐫𝐚 + 𝐫𝐜 ) 2ra 2 ) √2
[ 2 (ra + rc )]2 = (2ra )2 + ( 2 ra + 2rc = √6ra2 rc ra
=
√6−2 2
rc ra
2 (ra+ rc ) = √(4ra )2 +
2 rc = ra √6 − 2ra
= 0,225
(4ra )2 2
2rc = ra (√6 − 2)
AB = √a2 + a2
AB = 2ra AB = 2√a
a√2 = 2ra
a=
2ra √2
“a” Queda en función de los lados del cubo con la diagonal que relaciona el anión y el catión. 2
(AEF)2 = (AB) + (FB)2
AEF = 2(ra + rc )
AB = 2ra FB = a =
2 ra √2
Reemplazando 𝐀𝐄𝐅 = 𝟐(𝐫𝐚 + 𝐫𝐜 ) 2ra 2 ) √2
[ 2 (ra + rc )]2 = (2ra )2 + ( 2 ra + 2rc = √6ra2 rc ra
=
√6−2 2
rc ra
2 (ra+ rc ) = √(4ra )2 +
2 rc = ra √6 − 2ra
= 0,225
(4ra )2 2
2rc = ra (√6 − 2)
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