Explanation:
It is known that atomic number of carbon is 6 and its electronic configuration is . This means that in its neutral state it contains 2 electrons in its s-orbital and 2 electrons in its p-orbital.
After excitation there will be one electron present in its s-orbital and three electrons present in p-orbital.
Therefore, after the hybridization there will be in total 2 sp hybrid orbitals, 2 p-orbitals and zero s-orbital.
Answer : The correct option is, (a) 0.44
Explanation :
First we have to calculate the concentration of .
Now we have to calculate the dissociated concentration of .
The balanced equilibrium reaction is,
Initial conc. 1.0 M 0
At eqm. conc. (1.0-x) M (2x) M
As we are given,
The percent of dissociation of =
= 28.0 %
So, the dissociate concentration of =
The value of x = = 0.28 M
Now we have to calculate the concentration of at equilibrium.
Concentration of = 1.0 - x = 1.0 - 0.28 = 0.72 M
Concentration of = 2x = 2 × 0.28 = 0.56 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the equilibrium constant for the reaction is, 0.44