If an atom has sp3d2 hybridization in a molecule:
2 answers:
Answer : The maximum number of σ bonds that the atom can form is 6.
The maximum number of p-p bonds that the atom can form is 0.
Explanation :
As we are given that the hybridization is .
Bond pair = 6
Lone pair = 0
The number of electron density around the central atom is 6.
That means these molecule have central atoms with 0 lone pair and 6 atoms bond pair. Thus, the electronic geometry or molecular geometry of the molecule will be octahedral. For example :
The maximum number of σ bonds that the atom can form is 6.
The maximum number of p-p bonds that the atom can form is 0.
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.
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The reaction is given as:
Here, two moles of copper nitrate reacts with four moles of potassium iodide to give two moles of copper iodide, one mole of iodine and four moles of potassium nitrate.
First, calculate the number of moles of copper nitrate.
Number of moles is equal to the product of molarity and volume of solution in litre.
Number of moles =
(1 L =1000 mL)
=
Copper nitrate requires = mole of potassium iodide
= of potassium iodide
Volume of solution in litre =
Thus, volume of potassium iodide is =
=
1 L =1000 mL
Volume of potassium iodide in mL =
Hence, 0.2089 M potassium iodide consist of sufficient potassium iodide to react with copper nitrate in 3.88 mL of a 0.3842 M solution of copper nitrate .