Block D and F Elements

$M{n}^{+2}\to groupIV\left(M{n}^{+2},C{o}^{+2},Z{n}^{+2},N{i}^{2+}\right)$

$A{s}^{+3}\to groupIIB\left(A{s}^{+3},A{s}^{+5},S{b}^{+3},S{b}^{+5},S{n}^{+2},S{n}^{+4}\right)$

$\begin{array}{l}C{u}^{+2}\to groupIIA\left(C{u}^{+2},P{b}^{+2},H{g}^{+2},C{d}^{+2},B{i}^{+3}\right)\\ A{\mathcal{l}}^{+3}\to groupIII\left(F{e}^{+3},A{\mathcal{l}}^{+3},C{r}^{+3}\right)\end{array}$

Here A is CH₃ - CH₂ – CH₂ – COO – CH₂ – CH₃

The given sequence of reaction is:

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-OH\underset{}{\overset{Oxidation}{\to }}C{H}_3-C{H}_2-C{H}_2-COOH$

Nucleophilic addition of sodium hydrogen sulphite to aldehyde or ketone is a;

$NaHS{O}_3?N{a}^{+}+HS{O}_3^{?}$                

So; nucleopilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.

Addition of hydrogen cyanide;

Final product is cyanohydrin.

La⁺² (z = 57) = 5d¹

Ce⁺² (z = 58) = 4f²

Nd⁺² (z = 60) = 4f⁴

Yb⁺² (z = 70) = 4f¹⁴

Yb⁺² has no unpaired electrons thus diamagnetic in nature.

Non-linear and symmetrical Cr-O-Cr bond due to resonance.

One water molecule is associated with hydrogen bond.

Ans. = 1

$\underset{\left(z=93\right)}{Np}\Rightarrow 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$

$4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{4}6d^1$

Total number of f-electrons = 14 + 4 = 18 electrons

Ans. = 18

$2KMn{O}_4\stackrel{\Delta }{\to }\underset{\left(green\right)}{\overset{+6}{K_{2}Mn{O}_4}}+\underset{\left(black\right)}{Mn{O}_2}+O_2$

Mn⁶⁺ has one unpaired electron so paramagnetic and has green colour.

MnO -> Antiferromagnetism

O₂ -> Paramagnetism

NaCl -> Diamagnetism

Fe₃O₄ -> Ferrimagnetism

V₂O₃ – Basic

CrO – Basic

Metal oxides are generally basic in nature but if more electronegative element are connected through central atom then deficiency of electron occurs on central atom and it behaves as acidic in nature.