Ncert Solutions Chemistry Class 12th

Blue cupric metaborate is reduced to cuprous metaborate in a luminous flame.

$2Cu{\left(B{O}_2\right)}_2+2NaB{O}_2+C\underset{}{\overset{Luminous}{\to }}2CuB{O}_2+N{a}_2{B}_4{O}_7+CO$            

Cupric metaborate is obtained by heating boric anhydride & copper sulphate in a non-luminous flame as

$CuS{O}_4+B_2{O}_3\underset{}{\overset{Non-LuminiousFlame}{\to }}\underset{Blue}{Cu{\left(B{O}_2\right)}_2}+S{O}_3.$            

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

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.

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 → 2B
t = 0: 1 mole 0
after 100 min: (1 - 0.1) mol 0.2 mol
K = (1/t)ln [a/ (a-n)]
K = (2.303/100)log (1/0.9) min? ¹
0.693/t? /? = (2.303/100) [log10 - log9] = (2.303/100) x 0.046
t? /? = 69.3 / (2.303 x 0.046) = 654.15 min
Ans. = 654 (the nearest integer)

Percentage packing fraction of diamond is 34% and number of particle per unit cell is 8.

Coordination no.

X 6
Y 12

Trans product is more stable than Cis. ( Trans product feel less repulsion due to opposite direction)

BOD is the amount of oxygen required by bacteria to break down the organic matter present in a certain volume of sample of water.