Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

Eutrophication of water body results in loss of biodiversity.

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

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 $$ $\propto$ $\frac{1}{Ionicmobility}$ $\frac{}{}$

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

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]

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

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

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 = 3     B.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 p orbital.

The NCERT Exemplar includes analytical, application-based, and reasoning-type questions. It goes beyond the NCERT textbooks and helps students apply theories like VSEPR, hybridization, and MOT. It helps students to prepare for various entrance exams like NEET and JEE by strengthening their conceptual clarity.

Hybridization predicts the geometry and bond angles in complex molecules and it helps in the understanding of how four equivalent bonds in methane (CH? )  can be formed from atoms like carbon. For observed molecular geometries, it explains the formation of the equivalent orbitals (like sp³, sp², sp).