Answer:
a. the maximum number of σ bonds that the atom can form is 4
b. the maximum number of p-p bonds that the atom can form is 2
Explanation:
Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.
The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.
This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.
And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.
Whats your question? id love to help but I’m not sure I understand
Answer:
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Given: Kp = 
Temperature = 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
T = (25 + 273.15) K = 298.15 K
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (2)-(2+1) = -1
Thus, Kc is:
Answer:
∆H = -763kJ
Explanation:
Using Hess's law we can determine the ΔH of a reaction from the sum of similar reactions. Using the reactions:
<em>(1) </em>2PCl3(l) → 2P(s) + 3Cl2(g) ∆H = -640 kJ
<em>(2) </em>2P(s) + 5Cl2(g) → 2PCl5(s) ∆H = -886 kJ
The sum of (1)/2 + (2)/2 gives:
(1) / 2 = PCl3(l) → P(s) + 3/2Cl2(g) ∆H = -640 kJ/2 = -320kJ
(2) / 2 = P(s) + 5/2Cl2(g) → PCl5(s) ∆H = -886 kJ/2 = -443kJ
PCl3(l) + Cl2(g) → PCl5(s) ∆H = -320kJ - 443kJ =
<h3>∆H = -763kJ
</h3>
Charle's law states that, v1 / t1 = v2 / t2 .
<span>Hence,v(volume) is directly proportional to t(temperature). </span>
<span>Thus, the volume decreases.</span>
