Chemistry - GENERAL CHEMISTRY Question with Solution | TestHub

In Isoelectronic species, number of electrons should be same
Number of electrons present in${\text{Li}}^{\text{+}}\text{= 3}-\text{1 = 2}$
Number of electrons present in${\text{Na}}^{\text{+}}\text{= 11}-\text{1 = }\text{10}$
(A)${\mathrm{F}}^{-}=9+1=10{\mathrm{e}}^{-}$
$\mathrm{L}{\mathrm{i}}^{+}=3-1=2{\mathrm{e}}^{-}$
$\mathrm{N}{\mathrm{a}}^{+}=11-1=10{\mathrm{e}}^{-}$
$\mathrm{M}{\mathrm{g}}^{+2}=12-2=10{\mathrm{e}}^{-}$
(B)${\mathrm{N}}^{-3}=7+3=10{\mathrm{e}}^{-}$
$\mathrm{L}{\mathrm{i}}^{+}=3-1=2{\mathrm{e}}^{-}$
$\mathrm{M}{\mathrm{g}}^{+2}=12-2=10{\mathrm{e}}^{-}$
$\mathrm{N}{\mathrm{a}}^{+}=11-1=10{\mathrm{e}}^{-}$
(C)${\text{F}}^{-}\text{= 9+1 = 10}$
${\text{O}}^{\text{2}-}\text{= 8+2 = 10}$
${\text{N}}^{\text{3}-}\text{= 7+3 = 10}$
$\mathrm{M}{\mathrm{g}}^{2+}=12-2=10$
(d)$\mathrm{L}{\mathrm{i}}^{+}=3-1=2{\mathrm{e}}^{-}$
$\mathrm{N}{\mathrm{a}}^{+}=11-1=10{\mathrm{e}}^{-}$
${\mathrm{O}}^{-2}=8+2=10{\mathrm{e}}^{-}$
${\mathrm{F}}^{-}=9+1=10{\mathrm{e}}^{-}$
Hence,${\mathrm{N}}^{3-},{\mathrm{O}}^{2-},{\mathrm{F}}^{-},\mathrm{N}{\mathrm{a}}^{+}$are isoelectronic.