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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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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

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The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

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K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

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The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

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