The melting points of four solids A, B, C and D are $78^{o} C, 213^{o} C, 100^{o} C$ and $154^{o} C .$ Arrange them in increasing order of their interparticle forces of attraction.

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Solution

Here $A = 78^{o} C, B = 213^{o} C, C = 100^{o} C, D =^{o} C, 154^{o} C$ have melting points.
As stronger the faces of attraction between the particles more will be melting point and vice-versa.
Hence,
Order of their inter-particle forces of attraction is $B > D > C > A$ maximum melting point, so maximum inter particles forces of attraction.

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The melting points of four solids A, B, C and D are 78oC,213oC,100oC and 154oC. Arrange them in increasing order of their interparticle forces of attraction.

Here A=78oC,B=213oC,C=100oC,D=oC,154oC have melting points.As stronger the faces of attraction between the particles more will be melting point and vice-versa.Hence,Order of their inter-particle forces of attraction is B>D>C>A maximum melting point, so maximum inter particles forces of attraction.

The melting points of four solids A, B, C and D are $78^{o} C, 213^{o} C, 100^{o} C$ and $154^{o} C .$ Arrange them in increasing order of their interparticle forces of attraction.