# What is the hybridization about the nitrogen center in "NCl"_2^(+)?

Apr 26, 2017

$\text{N}$, from the periodic table, has $5$ valence electrons, since its electron configuration ends in $2 {s}^{2} 2 {p}^{3}$. Each $\text{Cl}$ has $7$ valence electrons. The charge of $+ 1$ then means that there is one less electron than in ${\text{NCl}}_{2}$. That accounts for a total of $5 + 2 \times 7 - 1 = 18$ valence electrons.

• Distributing $3$ lone pairs on each chlorine atom, we use $12$ of these electrons.
• Making two bonds off of nitrogen uses $4$ more.
• The remaining $2$ become a lone pair.

But then nitrogen only has 6 valence electrons. So, instinctively, the next best thing is to make a second $\text{N"-"Cl}$ bond... Since $\text{Cl}$ is slightly more electronegative, this is not extremely favorable. So, it is a bit better to delocalize like this, using one valence electron from each chlorine instead of two from one: This gives for formal charge:

• $\text{7 valence" - "6.5 owned} = + 0.5$ on each chlorine atom
• $\text{5 valence" - "5 owned} = 0$ on the nitrogen atom

which adds up to a total charge of $+ 1$ as expected. As a check, each atom has $8$ valence electrons:

• $2 + 2 + 1 + 2 + \frac{1}{2} \left(2\right) = 8$ total around each chlorine atom
• $2 + 2 + 2 + \frac{1}{2} \left(2\right) + \frac{1}{2} \left(2\right) = 8$ total around the nitrogen atom

Assuming this is the correct structure, this indicates a hybridization of $\textcolor{b l u e}{s {p}^{2}}$ since it has three electron groups.