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2 years ago

Can some atoms exceed the limits of the octet rule in bonding? If so, give an example.

Chemistry

1 answer:

harkovskaia [24]2 years ago

6 0

Answer:

Yes. Example: Sulfur hexafluoride (SF₆) molecule

Explanation:

According to the octet rule, elements tend to form chemical bonds in order to have 8 electrons in their valence shell and gain the stable s²p⁶ electronic configuration.

However, this rule is generally followed by main group elements only.

Exception: SF₆ molecule

In this molecule, six fluorine atoms are attached to the central sulfur atom by single covalent bonds.

Each fluorine atom has 8 electrons in their valence shells. Thus, it follows the octet rule.

Whereas, there are 12 electrons around the central sulfur atom in the SF₆ molecule. Therefore, sulfur does not follow the octet rule.

Therefore, the SF₆ molecule is known as a hypervalent molecule or expanded-valence molecule.

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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

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Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

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Answer:

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Explanation:

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Answer:

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Explanation:

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Answer:

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