Chemistry

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.

MnO? + H? C? O?2H? O - (H? )-> Mn²? + CO?
nf = (5) nf = (2)
Milli equivalent of C? O? ²? = mili equivalent of MnO?
2 x M x 10 = 5 x 0.05 x 10
M = 0.125 M
M = Strength / M.M of H? C? O?2H? O
Strength = 0.125 x 126g/L
= 15.75g/L

H? SO? + 2NaOH → Na? SO? + 2H? O
Milli mole: 80 60 (limiting reagent)
H? + OH? → H? O
Milli mole: 160 60
Milli mole: 160-60 0 60
Total heat produced = (60/1000) x 57.1 x 1000 = 3426 J
Total heat absorbed = msΔT
ΔT = Totalheat / (m x s)
= 3426J / (1000 x 4.18) = 0.819K
= 81.9 x 10? ² K
Ans. = 82 (the nearest)

3 [Co (en)? Cl? ]Cl + 3AgNO? (excess) → 3AgCl↓ (ppt) + 3 [Co (en)? Cl? ] NO?
Secondary valency of complex = 6

E.C = σ1s² σ1s² σ2s² σ2s² σ2p? ² π2p? ² = π2p? ² π2p? ² = π2p? ²
Total electrons in BMO = 10

E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398

Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

Kp = (Po? )¹/² = 4
Po? = 16 atm

1.     The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers? 2? 16.