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The electronic affinity values (in KJ molt) of 3 halogens X,Y,Z are respectively $-349,-333,-325$. Then X and Y are Z respectively

$ (a)\;F_2,Cl_2\;and\;Br_2 \\(b)\;Cl_2,F_2,Br_2 \\ (c)\;Cl_2,Br_2,F_2 \\ (d)\;Br_2,Cl_2,F_2 $

1 Answer

Solution :
The electronic affinity values (in KJ molt) of 3 halogens X,Y,Z are respectively −349,−333,−325−349,−333,−325.
Then X and Y are Z respectively
$Cl_2,F_2,Br_2 $ is correct answer.
In IB group all elements are metals.
Hence b is correct answer.
answered Mar 6, 2014 by meena.p
edited Oct 23 by meena.p
 

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