What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigo - nal planar arrangement of electron pairs? A linear ar - rangement of electron pairs? How many unhybridized p atomic orbitals are present when a central atom exhibits tetrahedral geometry? Trigonal planar geometry? Linear geometry? What are the unhybridized p atomic orbitals used for?

Question

Jimena

Chemistry

17 March, 22:07

What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigo - nal planar arrangement of electron pairs? A linear ar - rangement of electron pairs? How many unhybridized p atomic orbitals are present when a central atom exhibits tetrahedral geometry? Trigonal planar geometry? Linear geometry? What are the unhybridized p atomic orbitals used for?

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Trace · Answer 1

sp³;

sp²;

sp;

None;

One;

Two;

They're used to pi bonds.

Explanation:

The central atom in a molecule is generally the one that can make a greater number of bonds. The covalent bonds are made by the sharing of electrons, and, for that, the electron must be alone in the orbital.

To explain this, the hybridization theory was created, which states that, the orbitals are joined to form hybrids ones, and so, by the Hund's law, the electrons are alone in them.

The sigma bonds are done in the hybrids orbitals, and at the pure orbitals, the pi bonds are done. The lone pair of electrons are at a pure orbital. So, to know the hybridization of the central atom, we must know how many sigma bonds it does, and it will be the number of hybrids orbitals (each orbital may have two electrons, thus each bond are done in one orbital).

Double bonds and triple bonds have always only one sigma bond, so the number of sigma bonds is equal to the number of bonds, it's not necessary to know if they are simple, double or triple.

When the arrangement is tetrahedral, the central atom does 4 bonds, so it has 4 sigma bonds, and 4 hybrids orbitals (one of s and three for p), does its hybridization is sp³. Because exists only 3 p orbitals, there are no unhybridized p orbitals in this case.

When the arrangement is trigonal, the central atom does 3 bonds, so it has 3 hybrids orbitals (one of s and two of p), thus the hybridization is sp². So there are one unhybridized p orbitals.

When the arrangement is linear, the central atom does 2 bonds, so it has 2 hybrids orbitals (one of s and one of p), thus the hybridization is sp. So, there are two unhybridized p atoms.

As stated before, the unhybridized p orbitals are used to pi bonds.