$\frac{1}{\mathrm{\lambda }}$ = R$\left(\frac{1}{{{\mathrm{n}}_1}^2}-\frac{1}{{{\mathrm{n}}_2}^2}\right)$ Z²

= R$\left(\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2}\right)$ Z²

$\frac{1}{\mathrm{\lambda }}$ = $\frac{3}{4}$RZ²

$\therefore \mathrm{\lambda } \propto \frac{1}{{\mathrm{Z}}^2}$

$\therefore$For shortest $\mathrm{\lambda }$, Z must be maximum, which is for Li²⁺

Pauli's exclusion principle states, "An orbital can have a maximum of two electrons and these must have opposite spins."

E_n = $\frac{{\mathrm{E}}_1}{{\mathrm{n}}^2}$ x Z²

For Li²⁺, the excited state, n = 2 and Z = 3

$\therefore$ E_n = $\frac{- 13.6}{2^2}$ x (3)²

= $\frac{- 13.6}{4}$ x 9

= - 30.6 eV

Millimoles (or milliequivalents) of NaOH = 20 x 0.5 = 10

Millimoles (or milliequivalents) of HCl = 100 x 0.1 = 10

$\therefore$ $\underset{\underset{\mathrm{milliequivalents}}{10}}{\mathrm{NaOH}}$ + $\underset{\underset{\mathrm{milliequivalents}}{10}}{\mathrm{HCl}}$ $\to$ NaCl + H₂O

Thus, heat released when 10 milliequivalents of HCl are. neutralised by 10 millimoles of NaOH = x kJ. But the heat of neutralisation is heat released when 1 equivalent of HCl is neutralised by 1 equivalent of NaOH.

$\therefore$ ${∆}_{\mathrm{neu}}$H = - $\frac{\mathrm{x}}{10 \times {10}^{-3}}$ or - 100 x kJ/mol

Formal charge = total number of valence electrons - total number of nonbonding electrons - 1/2 x total number of bonding electrons

As for structure, $\stackrel{..}{\underset{..}{\mathrm{N}}}-\mathrm{N}-\underset{..}{\stackrel{..}{\mathrm{N}}}$ ;

For Ist and 3rd nitrogen atom

Formal charge = 5 - 4 - 1/2 x 4 = -1

For middle nitrogen atom

Formal charge = 5 - 0 - 1/2 x 8 = +1

As 2 g of metal carbonate is neutralised by 100 mL of 0.1 N HCl.

$\therefore$Number of gram-equivalents of HCl = $\frac{100 \times 0.1}{1000}$

=0.01

$\therefore$ Number of gram-equivalents of metal carbonate= 0.01

Mass of 0.01 gram equivalent metal carbonate = 2 g

Mass of 1 gram equivalent metal carbonate = $\frac{2}{0.01}$

= 200 g

$\therefore$Equivalent mass of metal carbonate = 200

Amino acids exist as Zwitter ions which contain both positive and negative charge but overall electrically neutral. At their the isoelectric point, amino acids do not migrate to any electrode and remain uniformly distributed in the solution. Thus, alanine (isoelectric point 6.0) does not move to any electrode.

At a pH lower than their isoelectric point, they exist as cation and move towards the cathode. Thus, arginine (isoelectric point 10.7 exists as the cation.

At a pH higher than their isoelectric point, they exist as anion and move towards the anode. Thus, glutamic acid (isoelectric point 3.2) at pH 6.0 moves towards the anode.

In the Periodic Table, the minimum ionisation energy is found at alkali metals. Hence sodium (electronic configuration = 1s² 2s² 2p⁶ 3s¹) has the lowest ionisation energy.

Note: Ionisation energy is the amount of energy required to remove an electron from the outermost shell an isolated atom.

CH₃COONa + H₂O $\rightleftharpoons$ $\underset{\mathrm{Weak} \mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}}$ + $\underset{\mathrm{Strong} \mathrm{base}}{\mathrm{NaOH}}$

As sodium acetate is a salt of weak acid and strong base, in the hydrolysis of sodium acetate the pH is given by;

pH = 1/2 pK_w + 1/2 pK_a + 1/2 log C

SO₂ + 1/2 O₂ $\to$SO₃

K₁ = $\frac{\left[{\mathrm{SO}}_3\right]}{\left[{\mathrm{SO}}_2\right][{\mathrm{O}}_2{]}^{1/2}}$

2SO₃ $\to$2SO₂ + O₂

K₂ = $\frac{\left[{\mathrm{SO}}_2{]}^2\right[{\mathrm{O}}_2]}{[{\mathrm{SO}}_3{]}^2}$

Obviously, K₂ = (1/K₁)²