What is the hybridization about the nitrogen center in "NCl"_2^(+)NCl+2?

1 Answer
Truong-Son N.
Apr 26, 2017

"N"N, from the periodic table, has 55 valence electrons, since its electron configuration ends in 2s^2 2p^32s22p3. Each "Cl"Cl has 77 valence electrons. The charge of +1+1 then means that there is one less electron than in "NCl"_2NCl2. That accounts for a total of 5 + 2 xx 7 - 1 = 185+2×7−1=18 valence electrons.

  • Distributing 33 lone pairs on each chlorine atom, we use 1212 of these electrons.
  • Making two bonds off of nitrogen uses 44 more.
  • The remaining 22 become a lone pair.

But then nitrogen only has 6 valence electrons. So, instinctively, the next best thing is to make a second "N"-"Cl"N−Cl bond...

Since "Cl"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:

  • "7 valence" - "6.5 owned" = +0.57 valence−6.5 owned=+0.5 on each chlorine atom
  • "5 valence" - "5 owned" = 05 valence−5 owned=0 on the nitrogen atom

which adds up to a total charge of +1+1 as expected. As a check, each atom has 88 valence electrons:

  • 2+2+1+2+1/2(2) = 82+2+1+2+12(2)=8 total around each chlorine atom
  • 2+2+2+1/2(2)+1/2(2) = 82+2+2+12(2)+12(2)=8 total around the nitrogen atom

Assuming this is the correct structure, this indicates a hybridization of color(blue)(sp^2)sp2 since it has three electron groups.

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