Block D and F Elements

In Ce & Eu (+3) oxidation state is more stable

$?C{e}^{+4}+e^{?}?C{e}^{+3}\left(Reduction\right)$     so CeO₂ is oxidising agent

$E{u}^{+2}?E{u}^{+3}+e?\left(oxidation\right)$ so EuSO₄ is reducing agent

Cu, Zn, Ni

Composition of german silver [Cu-50%, Zn-20%, Zi-30%]

Given are the oxide of alkali and alkaline earth metals which are ionic in nature.

Simple oxide are Li₂O, CaO, MgO and K₂O.

Peroxide is Na₂O₂ and superoxide is KO₂.

All simple oxides are diamagnetic as it has no unpaired electron.

$2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$

IUPAC nomenclature of element with atomic no. 103 is uniltrium.

Uses of catalyst is very specific for a particular reaction.

$C{e}^{3+}\to \left[Xe\right]4f^{1}5d^0$

$C{e}^{4+}\to \left[Xe\right]4f^{0}5d^0$  (Noble gas configuration)

$Mn{O}_4^{-2}\to A+B$  

Oxidation state of Mn in B < A 

$Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

B is MnO₂

$\therefore$  Oxidation state of Mn = +4

${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

  $\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$  

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

$=3.87\approx 4$  

$X\underset{H_{2}S{O}_4}{\overset{Conc}{\to }}$ Brown fumes + Brown ring test with FeSO₄/H₂SO₄

$x\underset{HCl}{\overset{H_{2}S}{\to }}Y\left(ppt\right)\underset{HN{O}_3}{\overset{Conc}{\to }}dissolved\underset{N{H}_{4}OH}{\overset{}{\to }}$ Deep blue colour solution

In cation analysis of Cu⁺ ions, precipitate formed is CuS on treating with H₂S and HCl which dissolved in HNO₃ and produced blue colour complex solution ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{++}$ in NH₄OH. Brown fumes and brown ring performance given by nitrate sample. Hence salt is Cu (NO₃)₂.

$NiC{l}_2+NaCN\underset{O.A.}{\overset{strong}{\to }}{\left[Ni{\left(CN\right)}_6\right]}^{2-}$

Complex has Ni⁴⁺ and strong ligand, hence following are the metal ion electronic configuration

Change of unpaired electron = 2