Anthracene is a yellow, crystalline solid found in charcoal tar.Complete this structure for anthracene, C14H10, by including bonds andhydrogen atoms together necessary.

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Anthracene is a yellow, crystalline solid discovered in coal tar. Finish this framework for anthracene, C14H10, by adding bonds and also hydrogen atoms together necessary. If friend would like to reclaim this structure to its initial state, click on the illustration box, climate click the switch that looks favor this: What form of hybrid orbitals are utilized through carbon in anthracene? How many sigma bonds and also pi bonds space there in one anthracene molecu How countless valence electrons occupy sigma-bond orbitals and also how many

Draw the framework of anthracene and also count the number sigma bonds, variety of pi bonds, and variety of valence electrons. Also identify the hybridization the carbon atoms existing in anthracene.


Identify the hybridization of one atom by counting the number of sigma bonds.

If carbon is attached with 4 sigma bonds then the hybridization is sp3\rms\rmp^3sp3 .

If carbon is attached with 3 sigma bonds then the hybridization is sp2\rms\rmp^\rm2sp2 .

If carbon is attached v 2 sigma bonds then the hybridization is sp\rmspsp .


The molecule formula that anthracene is C14H10\rmC_\rm14\rmH_\rm10C14​H10​ .

Calculatethedoublebondequivalenceasfollows:DBE=NC+1−((NH+NCl)−NN2)ThemolecularformulaforthegivencompoundisC14H10DBE=14+1−((10+0)−02)DBE=10So,thecompoundhas10doublebonds,itindicatesthepresenceofbenzeneringintheanthracene.\beginarrayl\\\rmCalculate\,\,\rmthe\,\,\rmdouble\,\,\rmbond\,\,\rmequivalence\,\,\rmas\,\,\rmfollows:\\\\\rmDBE\,\rm = \,\rmN_\rmC\,\rm + \,\rm1\, - \left( \frac\rm(\rmN_\rmH\,\rm + \,\rmN_\rmCl\rm) - \rmN_\rmN\rm2 \right)\\\\\rmThe\,\rmmolecular\,\rmformula\,\rmfor\,\rmthe\,\rmgiven\,\rmcompound\,\rmis\,\,\rmC_14\rmH_\rm10\\\\\rmDBE\,\rm = \,14\rm + 1\, - \left( \frac\rm(10\,\rm + \,\rm0) - \rm0\rm2 \right)\\\\\rmDBE\,\rm = \,\,10\\\\\rmSo,\,\rmthe\,\rmcompound\,\rmhas\,\,10\,\rmdouble\,\rmbonds,\,\rmit\,\rmindicates\,\rmthe\,\rmpresence\,\rmof\,\rmbenzene\,\rmring\\\\\rmin\;\rmthe\;\rmanthracene.\\\endarrayCalculatethedoublebondequivalenceasfollows:DBE=NC​+1−(2(NH​+NCl​)−NN​​)ThemolecularformulaforthegivencompoundisC14​H10​DBE=14+1−(2(10+0)−0​)DBE=10So,thecompoundhas10doublebonds,itindicatesthepresenceofbenzeneringintheanthracene.​

The framework of anthracene is,

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Hybridization is sp2\rms\rmp^\rm2sp2

Number the sigma binding is 26.

Number of pi bond is 7.

Number the valence electron in sigma bond is 52 electrons.

Number the valence electrons in pi bonds is 14 electrons.

Ans:

sp2\rms\rmp^\rm2sp2 hybrid orbitals are utilized through carbon in anthracene.

26 sigma bonds and also 7 pi binding are existing in anthracene.

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The variety of valence electron in sigma bond room 52 and variety of pi electrons space 14.