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Solid sodium chloride is mixed with solid potassium dichromate and concentrated sulfuric acid is added. A reddish-brown gaseous product X is formed. How many d-orbitals participate in the hybridization of X?

1. 1
2. 2
3. 3
4. 4

Correct answer: 1

Solution

The reaction is: K2Cr2O7 + 4NaCl + 6H2SO4 → 2CrO2Cl2 (chromyl chloride) +... Chromyl chloride (CrO2Cl2) is the reddish-brown fuming gas. Chromium in CrO2Cl2 is in the +6 oxidation state. The structure has Cr bonded to 2 oxygen atoms (double bonds) and 2 chlorine atoms, with no lone pairs on Cr, giving 4 electron domains. The hybridization is sp3 (some sources describe it as involving one d orbital giving sp2d or dsp2, but the tetrahedral geometry of Cr with 4 sigma bonds and pi bonds via d-orbitals means one d orbital is used in hybridization). Cr: [Ar] 3d5 4s1 in ground state; in CrO2Cl2 the hybridization involves sp3 but Cr uses one 3d orbital for the pi bonding with O. The hybridization of Cr is sp3 with one d-orbital participating (d2sp3 would be wrong here; it is sp3 with 1 d orbital contribution). The answer is 1 d-orbital involved in hybridization.

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