Block D and F Elements

Using, µ = √n (n+2) B.M, n = number of unpaired electrons.
? Ti? ³ = 4s? 3d¹, unpaired electron = 1
µ = 1.73 B.M
? V? ² = 4s? 3d³, unpaired electron = 3
µ = 3.87 B.M
? Sc? ³ = 4s? 3d? , unpaired electron = 0
µ = 0 B.M

$C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$

Dichromate ion converted to chromate ion in basic medium and oxidation number of Cr in $Cr{O}_4^{2-}$ is +6.

[Mn₂ (CO)₁₀]

Bridging ligand (CO) is (0)

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