Chemistry

$E_{M{n}^{+3}/M{n}^{+2}}^o=1.51V$

the strongest oxidizing agent have the highest reduction potential. So Mn³⁺ is the strongest oxidizing agent.

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l}C{e}^{+4}+e^{-}\to C{e}^{+3}{E}^0=1.61V\\ C{e}^{+3}+3e^{-}\to Ce{E}^0=-2.336V\end{array}$

It exist as Ce+4 and acts like a strong oxidizing agent by gaining electrons

$AgC\mathcal{l}+2N{H}_3\to \underset{\left(Solublecomplex\right)}{\left[Ag{\left(N{H}_3\right)}_2\right]C\mathcal{l}}$

Size of hydrated ion $\propto$ $\frac{1}{Ionicmobility}$

Size of hydrated ions $\propto \frac{1}{Ionicsize}$

Size of hydrated ions ; $B{e}^{+2}>M{g}^{+2}>C{a}^{+2}>S{r}^{+2}$

Ionic mobility ; $B{e}^{+2}<M{g}^{+2}<C{a}^{+2}<S{r}^{+2}$

Number of electrons in

$P{H}_3=15+3=18$

$B_2{H}_6=5*2+6=16$

$\begin{array}{l}CC{l}_4=6+17*4=6+68=74\\ N{H}_3=7+1*3=10\\ LiH=3+1=4\\ BC{l}_3=5+17*3=56\end{array}$

$B_2{H}_6\&BC{l}_3$ are e^- deficient molecules. B₂H₆ is dimer of BH₃, both compound has 6e^- only.

Au + NaCN + O₂ → Na [Au (CN)₂]

$Zn+Na\left[Au{\left(CN\right)}_2\right]\to N{a}_2\left[Zn{\left(CN\right)}_4\right]+Au$

$Ais{\left[Au{\left(CN\right)}_2\right]}^{-}andBis{\left[Zn{\left(CN\right)}_4\right]}^{-2}$

Au + NaCN + O₂ → Na [Au (CN)₂]

$Zn+Na\left[Au{\left(CN\right)}_2\right]\to N{a}_2\left[Zn{\left(CN\right)}_4\right]+Au$

$Ais{\left[Au{\left(CN\right)}_2\right]}^{-}andBis{\left[Zn{\left(CN\right)}_4\right]}^{-2}$

 $A{\mathcal{l}}^{+3}$ → number of electrons = 10

Br → number of electrons = 36

$B{e}^{+2}$ → number of electrons = 2

$C{\mathcal{l}}^{-}$ → number of electrons = 18

$L{i}^{+}$ → number of electrons = 2

$S^{-2}$ → number of electrons = 18

K⁺ → number of electrons = 18

O^-2 → number of electrons = 10

Mg⁺² → number of electrons = 10

O^-2 & Mg⁺² are isoelectronic with Al⁺³

Bond strength $\propto$ Bond order

$N_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2$

$O_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*1}$

$C_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2$

$B_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1\equiv {\pi }_{2py}^1$

NO Number of electron = 7 + 8 = 15

B.O. Similar to $N_2^{-}$

$N_2^{-}\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*}$

B.O. of N₂ = 3B.O of C₂ = $\frac{8-4}{2}=2$

Removal of e form antibonding molecular orbital increases bond order.

In NO & O₂ has valance e in orbital.