Chemistry, 13.11.2019 07:31 anna8352

In carbon tetra-chloride, ccl4, what is true about the charge distribution? select one:
a. a. the carbon will be more negative than the chlorine.
b. b. the chlorine will be more negative than the carbon.
c. c. the chlorine and carbon would be equally negative.
d. d. the atoms will become positive and negative ions.
e. e. it is impossible to predict how the electrons will distribute themselves in this type of bond.

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Answer from: shellxavier1

Option B, The chlorine will be more negative than the carbon

There is a theorized number assigned to every atom known as electro-negativity.

Basically, it just means how strong an atom can pull on electrons. If an atom has a larger electronegativity number than the other, that means it pulls on electrons stronger.

(Just a side note, the general trend for electronegativities is that it increases from left to right in a row and decreases down a period)

Also remember that when you gain an electron, you become more negative. When you lose an electron, you become more positive.

We can apply these two facts to this question.

Even if you didn't know that chlorine is more to the right of the periodic table than carbon (Remember my side note about the trend!), you can also infer that chlorine is stronger than Carbon.

Chlorine is a halogen, making it one of the most reactive elements in the periodic table. It has 7 valence electrons, and only needs 1 more, making it extremely reactive because it's craving that one last electron.

Carbon? Not so much. It needs to either gain or lose 4 electrons, making it less reactive than Chlorine.

Knowing this, when Carbon and Chlorine form a covalent bond, Chlorine will likely pull on the electrons in the bond more, thus making it more negative (since it has more of the electrons!)

Carbon on the other hand, becomes positive, since it loses more of the electrons.

That's all there is to it! If you have any questions, don't be afraid to ask.

Good luck!

Answer from: Quest

u i forgot the answer?

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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