From elementary molecular orbital theory we can deduce the electronic configuration of the singly positive nitrogen molecular ion as :
1s2 , 1s2 , 2s2 , 2s2 , 2p4 , 2p1
1s2 , 1s2 , 2s2 , 2s2 , 2p2 , 2p3
1s2 , 1s2 , 2s2 , 2s2 , 2p3 , 2p2
1s2 , 1s2 , 2s2 , 2s2 , 2p2 , 2p4
A solution contains 25% H2O , 25% C2H5OH and 50% CH3COOH by mass .The mole fraction of H2O would be :
0.25
2.5
0.502
5.03
If the solubility of PbCl2 at 25°C is 6.3 x 10-2 mol/L , its solubility product is :
1 x 10-6
1 x 10-3
1.1 x 10-6
1.1 x 10-5
B.
1 x 10-3
PbCl2 completely ionised in the solution as
PbCl2 Pb2+ + 2Cl-
i.e , 1 mole of PbCl2 in the solution gives 1 mole of Pb2+ ion and 2 moles of Cl- ions .
Now , as the solubility of PbCl2
= 6.3 x 10-2 mol/L
[Pb2+] = 6.3 x 10-2 mol/L
and [Cl-] = 2 X 6.3 X 10-2 mol/L
= 12.6 x 10-2 mol/L
Ksp for PbCl2 = [Pb2+][Cl-]2
= (6.3 x 10-2) x (12.6 x 10-2)2
= 1 x 10-3
A double bond connecting two atoms .There is a sharing of :
2 electrons
1 electron
4 electrons
All electrons