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It's K2O.

K has a charge of +1
O has a charge of -2

The answer is potassium oxide right? (since one of them is positive and the other negative)

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SO THIS IS MY QUESTION:

then P Cl3

P has a charge of -3
Cl has a charge of -1

is the answer
#1 - phosphorous trichloride?
#2 - phosphorous chloride?

I got #1...

But my teacher's answer key said #2. Why?
(There were no prefixes for NEGATIVE balanced compounds. is this allowed?)

2006-09-24 16:52:45 · 4 answers · asked by gogogo 3 in Science & Mathematics Chemistry

Doesn't Cl have 2? not 3?

2006-09-24 17:01:00 · update #1

WAIIIIITTTTTTT!!!!

I think P has +5 charge... That must explain why the answer had no prefixes...

2006-09-24 17:24:20 · update #2

4 answers

The answer from rhsaunder is correct; either term may be used.

It is a bit confusing to compare PCl3 to K2O.
K2O has ionic bonds.
PCl3 has covalent bonds.

You note that P wants to be +5. Actually, it wants to be either +5 or -3, and it is easier to get to -3. That is, it is easier to add 3 electrons than to get rid of 5.

But how does P gain 3 electrons when each of the 3 Cl atoms wants to gain 1 electron?
Answer: BY SHARING (i.e., by covalent bonds).
So each Cl shares one of its electrons with P.
And P shares three of its electrons, one with each of the Cl's.
(In practice it's not that neat, but it works conceptually.)

2006-09-24 17:32:32 · answer #1 · answered by actuator 5 · 0 0

In the first part, potassium oxide. In the second, either term is acceptable. Trichoride is technically correct, but the use of chloride is acceptable because there is no ambiguity: the only chloride is the trichloride. The oxidation states are P = +3, Cl = -1.

2006-09-25 00:18:48 · answer #2 · answered by Anonymous · 0 0

Because #2 is known to be have 3 Cl. You just have to memorize the rule.

2006-09-24 23:56:27 · answer #3 · answered by Brian 3 · 0 0

Its potassium oxide.

There is a phosphorous pentachloride and a phosphorous trichloride, maybe its a mistake.

2006-09-25 00:04:10 · answer #4 · answered by Anonymous · 0 0

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