What atomic or hybrid orbitals make up the sigma bond between Cl and F in chlorine trifluoride, ClF3
1 answer:
Answer:
sp³d¹ hybridization
Explanation:
Given Cl as central element with three F substrates ...
The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.
Valence bond theory predicts the following during bonding:
Cl:[Ne]3s²3p²p²p¹3d⁰
=> [Ne]3s²p²p¹p¹d¹
=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹
giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.
Note the following images:
Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.
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Answer : The for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation :
Reduction :
Now we have to calculate the Gibbs free energy.
Formula used :
where,
= Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the for this reaction is, -88780 J/mole.