Block D and F Elements

Ce, Pr, Nd, Tb and Dy are the only lanthanoids which shows + 4 O.S, so can form MO₂, but Yb does not shows +4 O.S so can't form MO₂.

Cu is the only element of 3d – series whose M²⁺ / M value is positive because of fact that low hydration enthalpy and high sublimation & ionization enthalpies.

${\left[FeC{l}_4\right]}^{2-}\Rightarrow$

$\begin{array}{l}n=4\\ \mu =\sqrt[]{4\left(4+2\right)}\\ =4.90BM\end{array}$                                                                          

[Co (C₂O₄)₃]^3- Þ Co³⁺

Co => [Ar]4s²3d⁷

Co⁺³ = [Ar]4s⁰3d⁶

  $\mu =\sqrt[]{n\left(n+2\right)}=\sqrt[]{1\left(3\right)}=1.73BM$         

VOSO₄ -> V⁴⁺ Reducing agent

Cr₂O₃ -> Amphoteric oxide

Red colour of ruby due to Cr³⁺ present in Al₂O₃

Ru⁸⁺ -> oxidizing agent

Density = $\frac{}{}$ $\frac{mass}{volume}$

Cu > Co > Fe > Cr > Zn

(1) $C{r}^{+1}=4s^{0}3d^{5}M{n}^{+2}=4s^{0}3d^5$  

(2)    $N{i}^{+2}=4s^{0}3d^{8}C{u}^{+}=4s^{0}3d^{10}$

(3) $F{e}^{+2}=4s^{0}3d^{6}C{o}^{+}=4s^{1}3d^7$  

(4)  $V^{+2}=4s^{0}3d^{3}C{r}^{+}=4s^{o}3d^5$

Cu⁺ = 3d⁹ 4s⁰

n = 1

$\mu =\sqrt[]{1*\left(1+2\right)}=\sqrt[]{3}=1.73BM$

the nearest integer is 2.

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.