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Alecsey [184]
3 years ago
6

If an atom has sp3d2 hybridization in a molecule:

Chemistry
2 answers:
mojhsa [17]3 years ago
5 0

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 sp^3d^2.

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 : SF_6

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.

never [62]3 years ago
4 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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Answer:

1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm

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An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

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The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
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