Block D and F Elements

Kindly go through the solution

 $C{r}^{+3}+NaOH\left(dil\right)\to C{r}_2{O}_3.{\left(H_{2}O\right)}_n$

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

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

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.

According to the John-Tollen distortion, there is a unsymmetrical filling of eq set of Cu⁺= ions but there is symmetrical filling of t₂g set of ions.

Following are the $\Delta H°$  hydration of given ions

Cr²⁺ - 1926 KJ/mole

Mn²⁺ - 1860 KJ/mole

Fe²⁺ - 2000 KJ/mole

Co²⁺ - 2078 KJ/mole

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

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

Uses of catalyst is very specific for a particular reaction.