The atomic orbitals of the central atom mix to form hybrid orbitals are one s and three p.
<h3>Which atomic orbitals are used to form hybrid orbitals?</h3>
Pauling supposed that in the presence of four hydrogen atoms, the s and p orbitals form four equivalent combinations which he called hybrid orbitals.
<h3>How many bonds does PF5?</h3>
Phosphorus pentafluoride has 5 regions of electron density around the central phosphorus atom (5 bonds, no lone pairs).
The resulting shape is a trigonal bipyramidal in which three fluorine atoms occupy equatorial and two occupy axial positions.
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The delocalized cloud of π electrons in benzene is formed by the overlap of 6<u> p-orbitals</u>.
So, option B is correct one.
In the case of hydrocarbons, delocalisation occurs in benzene rings , where a hexagon of six carbon atoms has decalized electrons spread over the whole ring.
All of the carbon atoms in the benzene rings are 
orbitals around the ring produces a framework of six sigma bonds, while the unhybridized p-orbitals which are perpendicular to this plane over in side-to-side fashion to form three pi-bonds.
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D or c I’m not sure maybe c sorry if it’s weonggg
Answer:
A 10% (v/v) methanol solution can be made by adding 10 mL of methanol to 100 mL of water, while a 10% (m/m) methanol solution can be made by adding 10 g of methanol to 90 g of water
Explanation:
This percents are a sort of concentrations:
v/v means volume of solute in 100 mL of solution
m/m means mass of solute in 100 g of solution
Certainly in solutions that can be made by two liquids, we don't consider additive, the volumes.
In m/m, as the percent are the grams of solute in 100 g of solution, we consider the mass of water (generally the solvent) as 100 g - x % of solute
Mass of solution = Mass of solvent + Mass of solute
