Chemistry NCERT Exemplar Solutions Class 12th Chapter Thirteen

Ge (Z = 32)

^{$=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^2$}         

$m_{\mathcal{l}}=0$       (for 4s, 3s, 2s, 1s) 4 orbital

$m_{\mathcal{l}}$       = 0 (one p orbital of 2p and 3p)

$m_{\mathcal{l}}$ = 0 (one d orbital)

Total orbitals = 7

Ans. = 7

One water molecule is associated with hydrogen bond.

Ans. = 1

Cl, Br & I form halic (V) acid i.e. HClO₃, HBrO₃ & HlO₃.

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

HgS, PbS, CuS, Sb₂S₃, As₂S₃, & CdS are given sulphides

              CdS, PbS, As₂S₃ & CuS are soluble in 50% HNO₃ but Sb₂S₃ & HgS are not soluble.

              Ans. = 4

Change in oxidation number = 2

$\Delta G°=-n{E}_{cell}^{o}F$

$\Delta G°=-n{E}_{cell}^{o}F$

=-2 * 4.315 * 96487 Jmol^-1

^{$\therefore \Delta G°=\Delta H°-T\Delta S°$}

^{$\Delta S°=\frac{\Delta H°-\Delta G°}{T}$}

$=\frac{-825.2*1000-\left(-2*4.315*96487\right)}{298.15}J{K}^{-1}$

$=\frac{7482.81}{298.15}=25.09J{K}^{-1}$

$\approx$ 25.1 JK^-1

Ans. = 25

Molarity (M) = $\frac{w*1000}{molecularmass*volumeofsolution\left(ml\right)}$

= $\frac{6.3*1000}{126*250}=\frac{4}{20}=0.2M$

Molecular mass of oxalic acid $\left(H_2{C}_2{O}_4.2H_{2}O\right)$

= 1 * 2 + 12 * 2 + 16 * 4 + 2 * 18

              = 26 + 64 + 36 = 126

              M = 2 * 10^-1 M

              = 20 * 10^-2M

$\therefore x*10^{-2}=20*10^{-2}$

$\therefore x=20$

Ans. = 20

$A_3{B}_{2\left(s\right)}?3A_{\left(aq\right)}^{+2}+2B_{\left(aq\right)}^{3-}$

Solubility $\frac{x}{M}$ $\frac{3x}{M}$ $\frac{x}{M}$

$\therefore Ksp={\left[A^{+2}\right]}^3{\left[B^{-3}\right]}^2$

$={\left(\frac{3x}{M}\right)}^3{\left(\frac{2x}{M}\right)}^2$

$=108{\left(\frac{x}{M}\right)}^5$

$Ksp=a{\left(\frac{x}{M}\right)}^5=108{\left(\frac{x}{M}\right)}^5$

Ans. $\therefore$ a = 108