May/June 2007
Paper 04
Q-2 c)
Q-3 b)
Q-4
more doubts coming from me soon
Ans 2.c- part i:
Gas A:
16 = O
17 = OH
18 = H2O
Therefore, it'll be H2O
Gas B:
14 = N
16 = O
28 = N2 (14+14)
30 = NO (14+16)
44 = N2O (14+14+16)
Therefore, it'll be N2O
part ii: NH4NO3 ? N2O + 2H2
Ans3.b- Not so sure.
(i) Pb is in two oxidation states: +2 and +4
The ratio will be 1:2
(ii) Pb3O4 ? 3PbO + ½O2
(iii) Pb3O4 + 4HNO3 ? 2Pb(NO3)2 + PbO2 + 2H2O
(iv) Pb2+ more basic than Pb4+
Ans4- The d-orbital will be split in two ways: along the axis and in between the axes.
(ii) Ligands have lone pair of electrons which is how they form a covalent bond with the transition metal ion. The electrons in the orbitals pointing towards the ligands have higher energy.
c(i)- Accoring to graph and data given:
C= red and D=blue
c(ii)- More energy less wavelength. Therefore, at C.
Hope that's cleared.