Question

Which of the following have same electronic configuration ?

• A. ${\mathrm{Na}}^{+},{\mathrm{Mg}}^{++}$
• B. ${\mathrm{Mg}}^{2+},{\mathrm{Al}}^{3+},{\mathrm{Cl}}^{-}$
• C. ${\mathrm{Cl}}^{-},{\mathrm{Na}}^{+}$
• D. ${\mathrm{Na}}^{+},{\mathrm{Ca}}^{++},{\mathrm{Cl}}^{-}$

Answer 1

The correct option is A

${\mathrm{Na}}^{+},{\mathrm{Mg}}^{++}$

Explanation for the correct option

Option A

• ${\mathrm{Na}}^{+},{\mathrm{Mg}}^{++}$have the same electronic configuration.
• If an atom will lose one electron, it will gain one positive charge.
• If an atom will gain one electron, it will gain one negative charge.
• Name of chemical symbol $\mathrm{Na}$ = $\mathrm{Sodium}$
• Atomic number of Sodium = $11$
• Electronic configuration of Sodium = $2,8,1$
• Here ion ${\mathrm{Na}}^{+}$ is formed by the donation of one electron.
• Electronic configuration of ${\mathrm{Na}}^{+}$ = $2,8$
• Name of chemical symbol $\mathrm{Mg}$ = $\mathrm{Magnesium}$
• Atomic number of Magnesium = $12$
• Electronic configuration of Magnesium = $2,8,2$
• ${\mathrm{Mg}}^{++}$ ion is formed by the donation of 2 electrons.
• Electronic configuration of ${\mathrm{Mg}}^{++}$= $2,8$
• Hence ${\mathrm{Na}}^{+},{\mathrm{Mg}}^{++}$ have the same electronic configuration.

Explanation for the incorrect option

Option B

• Name of chemical symbol $\mathrm{Mg}$ = $\mathrm{Magnesium}$
• Atomic number of Magnesium = $12$
• Electronic configuration of Magnesium = $2,8,2$
• ${\mathrm{Mg}}^{++}$ ion is formed by the donation of 2 electrons.
• Electronic configuration of ${\mathrm{Mg}}^{++}$= $2,8$
• Name of chemical symbol $\mathrm{Al}$ = $\mathrm{Aluminium}$
• Atomic number of Aluminium = $13$
• Electronic configuration of Aluminium = $2,8,3$
• ${\mathrm{Al}}^{3+}$ ion is formed by the donation of 3 electrons.
• Electronic configuration of ${\mathrm{Al}}^{3+}$=$2,8$.
• Name of chemical symbol $\mathrm{Cl}$ = $\mathrm{Chlorine}$
• Atomic number of Chlorine = $17$
• Electronic configuration of Chlorine = $2,8,7$
• ${\mathrm{Cl}}^{-}$ ion is formed by gaining one electron.
• Electronic configuration of ${\mathrm{Cl}}^{-}$= $2,8,8$
• Hence${\mathrm{Mg}}^{++}$, ${\mathrm{Al}}^{3+},{\mathrm{Cl}}^{-}$does not have the same electronic configuration.

Option C

• Name of chemical symbol $\mathrm{Cl}$ = $\mathrm{Chlorine}$
• Atomic number of Chlorine = $17$
• Electronic configuration of Chlorine = $2,8,7$
• ${\mathrm{Cl}}^{-}$ ion is formed by gaining one electron
• Electronic configuration of ${\mathrm{Cl}}^{-}$= $2,8,8$
• Name of chemical symbol $\mathrm{Na}$ = $\mathrm{Sodium}$
• Atomic number of Sodium = $11$
• Electronic configuration of Sodium = $2,8,1$
• Here ion ${\mathrm{Na}}^{+}$ is formed by the donation of one electron.
• Electronic configuration of ${\mathrm{Na}}^{+}$ = $2,8$
• Hence ${\mathrm{Cl}}^{-},{\mathrm{Na}}^{+}$does not have the same electronic configuration.

Option D

• Name of chemical symbol $\mathrm{Na}$ = $\mathrm{Sodium}$
• Atomic number of Sodium = $11$
• Electronic configuration of Sodium = $2,8,1$
• Here ion ${\mathrm{Na}}^{+}$ is formed by the donation of one electron.
• Electronic configuration of ${\mathrm{Na}}^{+}$ = $2,8$
• Name of chemical symbol $\mathrm{Ca}$ = $\mathrm{Calcium}$
• Atomic number of Calcium = $20$
• Electronic configuration of Calcium= $2,8,8,2$
• Here ion ${\mathrm{Ca}}^{++}$ is formed by the donation of two-electron.
• Electronic configuration of ${\mathrm{Ca}}^{++}$ = $2,8,8$
• Name of chemical symbol $\mathrm{Cl}$ = $\mathrm{Chlorine}$
• Atomic number of Chlorine = $17$
• Electronic configuration of Chlorine = $2,8,7$
• ${\mathrm{Cl}}^{-}$ ion is formed by gaining one electron.
• Electronic configuration of ${\mathrm{Cl}}^{-}$= $2,8,8$
• Hence ${\mathrm{Na}}^{+},{\mathrm{Cl}}^{-},{\mathrm{Ca}}^{++}$does not have the same electronic configuration.

Conclusion: Hence option A is correct.