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$Be^{2+}$ is isoelectronic with :

$(a)\;Mg^{2+}\qquad(b)\;Na^+\qquad(c)\;Li^+\qquad(d)\;H^+$

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Number of electrons $Be^{2+}=2$
Number of electrons $Li^+=2$
So these are isoelectronic.
Hence (c) is the correct option.
answered Nov 27, 2013 by sreemathi.v
 
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