Pbi2
What is the compensation of PBN on P2?
2+
PB has 2+ so you can draw two of the two with 1.
As I said, K has a +1 charge. And in Formula A you see only one which I have added to K. This means I also have to be 1 to end the 11K charge.
Pbi2
Pbi2
Usually, you usually know some kind of payload to make such laws.
Pb is the load of x.
Since P2 is a connection, there is no charge. The total charge is 0.. In addition, the charge of the iodide ion is 1.
x + 2 (1) = 0
x = +2
Use the same mtd to charge the iodide ion.
If the charge of the iodide ion is y,
y + 1 = 0
j = 1.
As you do more, you can see the charge immediately. All my wishes! :)
First, see that the charge is nothing more than an oxidation state.
Do not count by OT coordination number.
There are two types of synchronization, first and second, which are only necessary for complex balances.
Non-oxidation, however, is the amount or charge that occurs to an element in the base case, for example, lead only.
The charge of PB on P2 is +2. Since the charge of single halide ions cancels out the total charge of two iodine ions in P2, PB is charged with +2.
For similar I, if the charge on K is +1, then the charge on me will be 1.
Try to cancel or balance the charge, the amount of both charges may be e.
The charge of pb is x.
Since P2 is a connection, there is no charge. The total charge is 0.. In addition, the charge of the iodide ion is 1.
x + 2 (1) = 0
x = +2
Use the same mtd to load the iodide.
If the charge of the iodide ion is y,
y + 1 = 0
j = 1.
Pbi2
Pbi2
What is the charge of Pb ion on P2? 3
Can you tell me where you got it from?
And if the charge of ion K is 1+, then what is the charge of iodide ion in KI?
please explain
Thank you very much!!!!
2+
Pb has 2+ so you can drag two I's each with 1.
As I said, K has a +1 charge. And in the AI formula you see only one which I add to K. This means that I also have to be 1 to eliminate the +1K charge.
Pbi2
Pbi2
To create this type of QNS, you usually know the payload of a species.
PBX is a burden.
As P2 is a connection, there is no charge. The total charge is 0. Also, the charge of iodide ion is 1.
x + 2 (1) = 0
x = +2
Use the same MTD to charge the iodide ion.
If the charge of the iodide ion is y,
y + 1 = 0
j = 1.
As you do more, you can see the charge immediately. All my wishes! :)
First, notice that the charge is nothing more than an oxidation state.
Do not count by OT coordination number.
There are two types of coordination numbers, pri and sec, which are only needed to balance the complex.
However, non-oxidation is the amount or charge that occurs in the primary case of an element, for example, only lead;
The charge of Pb on P2 is +2. Since the charge of a single halide anion cancels out the total charge of two iodine ions in P2, Pb charges +2.
Similarly for KI, if the charge on K is +1, then the charge on I will be 1.
Try to cancel or balance the charge, the combination of the two charges may be e.
The charge of pb is x.
As P2 is a connection, there is no charge. The total charge is 0. Also, the charge of iodide ion is 1.
x + 2 (1) = 0
x = +2
Use the same MTD to load the iodide.
If the charge of the iodide ion is y,
y + 1 = 0
j = 1.
