Complete the hybridization and bonding scheme for xef4.
1 answer:
<span>Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons</span>
total ve = 8 + 28 = 36 ve
<span>36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)</span>
<span>28 - 4(6) = 4
(We assign the remaining electrons to F atoms)</span>
<span>4 - 2(2) = 0
(Therefore 4 electrons left => we have 2 lone pairs)</span>
The steric number = No. of σ bonds + #lone pairs
= 4 σ bonds + 2 lone pairs = 6 => d²sp³ (6 hybrid orbitals)
<span>4 bonds + 2 lone pairs => square planar</span>
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Answer:
At equilibrium, reactants predominate.
Explanation:
For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:
Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.