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Please answer the Question with explanation, to let me study it for EXAM.

Ionic Bonding Activity In your group you will determine and demonstrate the formation of ionic compounds. Complete the table

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Answer #1

ion

pairs

valance

electron

electron

transfer

charge of ion

number of

ion

charge

balance

formula

potassium 1 lose 1 +1 3 +3 K3P
Phosphorus 3 gain 3 -3 1 -3
magnesium 2 lose 2 +2 1 +2 MgO
oxygen 6 gain 2 -2 1 -2
sodium 1 lose 1 +1 2 +2 Na2O
oxygen 6 gain 2 -2 1 -2
aluminium 3 lose 3 +3 1 +3 AlP
phosphorus 5 gain 3 -3 1 -3
Barium 2 lose 2 +2 1 +2 BaF2
fluorine 7 gain 1 -1 2 -2
Aluminium 3 lose 3 +3 2 +6 Al2O3
oxygen 6 gain 2 -2 3 -6
calcium 2 lose 2 +2 3 +6 Ca3N2
nitrogen 5 gain 3 -3 2 -6

the table is filled
let me explain 1 or 2 examples

take K3P

we need to write valance electron for each element,with the help of periodic table

for K , V.electrons=1 and for P, valance electron = 5
they need to complete their octet either by loosing or gaining of electrons
K is metal has strong tendency to loose electron.Once that one valance electron gone ,K+
will complete octet.
For P, if it gain 3 electron its octet will be completed
But one K can give only 1 electron. So P will take up 3 electrons from three K atoms
Thus for 1 P3- ion ,there must be three K+ ion .Also note that in this way charge
of each ions get balanced
three K+ ions has charge = 3[+1} = +3
one P3- ion has charge = 1{-3} =-3
Charge must be balanced when a neutral molecule formed
The from number of ions present we can write chemical formula K3P
**************************
In case of Al2O3

Al has 3 and O has 6 valance electron.
so each Al will give up 3 electrons for octet
and each O will take up 2 electrons
In order to make number of electron lost= number of electron gained [charge balance]
there must be two Al atom ; so that a total of 6 e- {2*(+3)}
and three O atom ;so that a total of 6e- again [3*(-2)}
*************
hope it is helpful


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