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The first , second, third and fourth ionization energies of a given element are $80,2.43,3.66$ and $25.03 \;MJmol$ respectively

$ (a)\; \text {Boron} \\(b)\;\text {Carbon}\\ (c)\;\text {Aluminium}\\ (d)\;\text {Ditrogen} $
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1 Answer

Boron is correct.
Hence a is the correct answer.
answered Mar 6, 2014 by meena.p
 

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