Question

What is the electronic configuration of $Na$ and $Cl$ before the formation of $NaCl$ ?

• A. $Na$: $\left[Ne\right]3s^1$; $Cl$: $\left[Ne\right]$ $3s$${}^{2}3p^5$
• B. $Na$: $\left[Ne\right]2s^2$; $Cl$: $\left[Ne\right]3s^{2}3p^6$
• C. $Na$: $\left[Ne\right]2s^2$; $Cl$: $\left[Ne\right]3s^{2}3p^4$
• D. $Na$: $\left[Ne\right]2s^1$; $Cl$: $\left[Ne\right]3s^{2}3p^6$

Answer 1

The correct option is A $Na$: $\left[Ne\right]3s^1$; $Cl$: $\left[Ne\right]$ $3s$${}^{2}3p^5$

In writing the electron configuration for sodium the first two electrons will go in the 1s orbital. Since $1s$ can only hold two electrons the next $2$ electrons for sodium go in the $2s$ orbital. The next six electrons will go in the $2p$ orbital. The $p$ orbital can hold up to six electrons. We'll put six in the $2p$ orbital and then put the remaining electron in the $3s$. Therefore the sodium electron configuration will be $1s^{2}2s^2{2p}^{6}3s^1=\left[Ne\right]3s^1$

In writing the electron configuration for Chlorine the first two electrons will go in the 1s orbital. Since $1s$ can only hold two electrons the next $2$ electrons for Chlorine go in the $2s$ orbital. The next six electrons will go in the $2p$ orbital. The $p$ orbital can hold up to six electrons. We'll put six in the $2p$ orbital and then put the next two electrons in the $3s$. After filling $3s$ we'll move to the $3p$ where we'll place the remaining five electrons. Therefore the Chlorine electron configuration will be $1s^{2}2s^2{2p}^{6}3s^{2}3p^5=\left[Ne\right]3s^{2}3p^5$

Note: [$1s^{2}2s^2{2p}^6=\left[Ne\right]$] is the electronic configuration of noble gas Ne.