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Electronic configuration of Deuterium is

$(a)\;1s^2\qquad(b)\;1s^1\qquad(c)\;1s^22s^2\qquad(d)\;1s^22s^4$

1 Answer

Electronic configuration of Deuterium is $1s^1$
Hence (b) is the correct answer.
answered Jan 22, 2014 by sreemathi.v
 
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