Na₂B₄O₇ .10H₂O + H₂SO₄ (conc.)\(\rightarrow\) Na₂SO₄ + H₂B₄O₇ + 10H₂O
Borax
H₂B₄O₇ + 5H₂O \(\rightarrow\)  4H₃BO₃
                              Boric acid
2H₃BO₃ \(\xrightarrow[]{Heat}\) B₂O₃ + 3H₂O
                         Boric oxide
B₂O₃ + 3Mg \(\xrightarrow[]{\triangle}\)2B + 3MgO

The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium. 
The white precipitate (compound A) obtained is aluminium hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).
\(2 \mathrm{Al}+3 \mathrm{NaOH} \ \longrightarrow \mathrm{Al}(\mathrm{OH})_{3} \downarrow+3 \mathrm{Na}^{+}\\ Aluminium (X) Sodium hydroxide White ppt.(A) \)
\(\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{NaOH} \longrightarrow \ \mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]\\ (A) \ Sodium tetrahydroxoaluminate (III) ( Soluble complex B)\)
Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.
\(\mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{HCl} \longrightarrow \mathrm{AlCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \)
(A)                             (B)
Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.

Aluminium being amphoteric in nature dissolves both in acids and alkalies evolving H₂ gas which burns with a pop sound.
\(2Al+6HCl\longrightarrow { 2AlCl }_{ 3 }+3{ H }_{ 2 }\)
\(2Al+NaOH+{ 2H }_{ 2 }O\longrightarrow { 2NaAlO }_{ 2 }+3{ H }_{ 2 }\)
But when Al is treated with conc.HNO₃.a thin protective layer of
Al₂O₃ is formed on its surface which prevents further action.
\(2Al+6{ HNO }_{ 3 }\longrightarrow { Al }_{ 2 }{ O }_{ 3 }+{ 6NO }_{ 2 }+{ 3H }_{ 2 }O\)

The hydroxide formed in aqueous solution is Al(OH)₃

Carbon form covalent compounds because of two reasons.
i) Carbon cannot lose electrons to form C⁴⁺ because the ionization energy required is very high
ii) It cannot gain electrons to form C^4- because it is energetically not favorable.
But on moving down the group, the I, E, decrease. Pb, being the last element has so low I.E. that it can lose electrons to form ionic compounds.

Boric acid is sparingly soluble in cold water but fairly soluble in hot water. It acts as a weak monobasic acid. It is not a protonic acid but it acts as aLewis acid by accepting a hydroxide ion of water and releasing a proton into the solution.
H-OH + B(OH)₃\(\rightarrow\) [B(OH)₄]^- +H⁺

\({ SiO }_{ 2 }+2NaOH\longrightarrow { Na }_{ 2 }{ SiO }_{ 3 }+{ H }_{ 2 }O\)

\({ CaCO }_{ 3 }(aq)+{ CO }_{ 2 }+{ H }_{ 2 }O\longrightarrow Ca{ ({ HCO }_{ 3 }) }_{ 2 }\)
                                           Calcium bicarbonate (soluble)

Carbon have the tendency to from  \(p\pi -p\pi \)  bonds.

Lead is the most metallic element (as down the group metallic character increases)