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a
AlF3>MgF2
b
Li3N>Li2O
c
NaCl > LiF
d
TiC > ScN
answer is C.
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Detailed Solution
L.E. ∝q+⋅q−r++r−(1) AlF3>MgF2 [Charge on cation] (2) Li3N>Li2O [Charge on anion] (3) NaCl>LiF [Size of cation and anion] (4) TiC > ScN [Charge on cation and anion]