Explanation
Zn, Cd, Hg have (n – 1) d10 ns2 as general electronic config.
AgCl does not react with NaNO3.
Na2CdCl4 contain Cd2+ which has 4d10 configuration and is thus colourless.
CuF2 contains Cu2+ (3d9) ion which is coloured.
Elect. Config. of Gd = [Xe] 4f1 5d1 6s2
The 3d – series strats from Z = 21 – 30.
(4) Hg2Cl2
Any FeO is oxidised to Fe2O3 So that it may not be lost as slag. Then Fe2O3 is reduced by CO to Fe metal.
AgF is soluble in water and the solubility decreases as the size of halogen increases down the group.
Fe3+ + 3SCN– → Fe(SCN)3 (blood red colour)
In this reaction Ag+ in Ag2O is reduced to Ag metal.
Cu (Z =29)=3d10 4s1 outer electronic configuration.
In pyrolusite (MnO2), the oxidation state of Mn is +4 ; in K2MnO4, the oxidation state of Mn is + 6 .
In the formation of complexes, the transition metal atom will act as a Lewis acid (electron acceptor) and the ligand will act as a Lewis base (electron donor).
Gd = –4f75d16s2; Gd3+ = 4f7
2AgNO3 + Na2S2O3 → Ag2S2O3 + 2NaNO3
Ag2S2O3 + H2O → H2SO4 + Ag2S(Black)
4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H2O
Zn + 2NaOH → Na2ZnO2 + H2.
As Cu is more electropositive than Ag, Cu metal will dissolve in AgNO3 solution giving a blue solution.
(n – 2)f1 – 14 (n – 1)s2 p6 d0–1 ns2 is the general electronic configuration of lanthanides.
As there are 5 ( 3 in d and 2 in s orbitals) valence electrons the atom belongs to group 5 th of the periodic table.
10 ml of 1 M KMnO4 oxidises 10 ml of 5 M FeSO4 in acidic medium.
Ionic size decreases from La3+ to Lu3+ due to lanthanide contraction
Due to lanthanoid contraction, the ionic radii of Lu3+ decrease from La3+ to Lu3+. Thus, the lowest value (0.85 Å) is the ionic radius of Lu3+.
Please disable the adBlock and continue. Thank you.