Hey there!
No of hybrid orbitals , H = ( V +S - C + A ) / 2
Where H = no . of hybrid orbitals
V = Valence of the central atom = 5
S = No . of single valency atoms = 4
C = No . of cations = 1
A = No . of anions = 0
For PCl4 +
Plug the values we get H = ( 5+4-1+0) / 2
H = 4 ---> sp3 hybridization
sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations
Answer C
Hope that helps!
C7H14 + 10.5 O2 -> 7 CO2 + 7 H2O
Or, if whole numbers must be used:
2 C7H14 + 21 O2 -> 14 CO2 + 7 H2O
