\((i) \quad 4 \mathrm{NaCl}+\mathrm{MnO}_{2}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Cl}_{2}+\mathrm{MnCl}_{2}+4 \mathrm{NaHSO}_{4}+2 \mathrm{H}_{2} \mathrm{O} \)
\((ii) \quad \mathrm{NaNO}_{2}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{HNO}_{2} \)
\((iii) \quad \mathrm{P}_{4}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{NaH}_{2} \mathrm{PO}_{2}+\mathrm{PH}_{3} \uparrow \)
\((iv) \quad 3 \mathrm{AgNO}_{3}+\mathrm{PH}_{3} \rightarrow \mathrm{Ag}_{3} \mathrm{P}+3 \mathrm{HNO}_{3} \)
\((v) \quad 4 \mathrm{Mg}+10 \mathrm{HNO}_{3} \rightarrow 4 \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{N}_{2} \mathrm{O}+6 \mathrm{H}_{2} \mathrm{O} \)
\((vi) \quad 2 \mathrm{KClO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}+3 \mathrm{O}_{2} \uparrow \)
\((vii) \quad \mathrm{Cu}+Con. Hot \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CuSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{SO}_{2} \uparrow \)
\((viii) \quad 2 \mathrm{Sb}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{SbCl}_{3} \)
\((ix) \quad 2 \mathrm{HBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Br}_{2} \)
\((x) \quad \mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{XeO}_{3}+6 \mathrm{HF} \)
\((xi) \quad 5 \mathrm{XeO}_{6}^{4-}+2 \mathrm{Mn}^{2+}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_{4}^{-}+5 \mathrm{XeO}_{5}+7 \mathrm{H}_{2} \mathrm{O} \)
\((xii) \quad 2 \mathrm{XeOF}_{4}+\mathrm{SiO}_{2} \rightarrow 2 \mathrm{XeO}_{2} \mathrm{~F}_{2}+\mathrm{SiF}_{4} \)
\((xiii) \quad Xe+{ 3F }_{ 2 }\overset { Ni/200atm }{ \underset { 400^{ 0 }C }{ \longrightarrow } }XeF_6\)

a) BrF₅

Valence electron of bromine atom 7+ Number of fluorine atom (5) = 12
\(X=\frac{12}{2}=6\) 
Hybridization: sp³d² ;
Geometry: Square Pyramidal
b) BrF₃

Valence electron of bromine atom 7+ Number of fluorine atom (3) = 10 
X = \(\frac{10}{2}=5\)
Hybridization: sp³d²;
Geometry: Triangular bipyramidal (T - shaped)

Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF =x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

HF is a weak acid i.e. 0.1 M solution is only 10% ionised, but in 5M & 15M solution, HF is stronger acid due to chemical equilibrium.
\(\mathrm{HF}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{F}^{-}  \)
\(\mathrm{HF}+\mathrm{F}^{-} \rightleftharpoons \mathrm{HF}_{2}^{-}\)

On heating PCl₅, it decomposes into PCl₃ and chlorine.
\(\mathrm{PCl}_{5(\mathrm{~g})} \rightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\)

HNO₃ reacts with basic oxides to form salts and water
ZnO + 2HNO₃\(\longrightarrow \) Zn(NO₃)₂ + H₂O
3FeO + 10HNO₃ \(\longrightarrow \) 3Fe(NO₃)₃ + NO + 5H₂O

Thermal stability:
\(4 \mathrm{PH}_{3} \stackrel{317 \mathrm{~K}}{\longrightarrow} \mathrm{P}_{4}+6 \mathrm{H}_{2} \)
Combustion:
\(4 \mathrm{PH}_{3}+8 \mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} \underset{\text { Phosphorus pentoxide }}{\mathrm{P}_{4} \mathrm{O}_{10}+6 \mathrm{H}_{2} \mathrm{O}} \)
\(\mathrm{P}_{4} \mathrm{O}_{10}+6 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta}{\longrightarrow} 4 \mathrm{HPO}_{3}+4 \mathrm{H}_{2} \mathrm{O}\\ \quad \quad \quad \quad \quad \quad \quad \text{Meta phosphoric acid}\)

(i) The general electronic configuration of 15 group elements is ns²np³.
(ii) Nitrogen \(\Rightarrow 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}\) (Valence shell)
(iii) Phosphorus \(\Rightarrow 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\)
(iv) Arsenic \(\Rightarrow 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}\)
(v) Antimony \(\Rightarrow 5 s^{2} 5 p^{3}\)
(vi) Bismuth \(\Rightarrow 6 \mathrm{~s}^{2} 6 \mathrm{p}^{3}\)

Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 12 \mathrm{C}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot 11 \mathrm{H}_{2} \mathrm{O} \\ \text { Sucrose } \)
\(\underset{\text { Formic acid }}{\mathrm{HCOOH}}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CO}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)
\(\underset{\text { Oxalic acid }}{\mathrm{(COOH)}_{2}}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CO}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)

Chlorine is prepared by the action of conc. sulphuric acid on chlorides in presence of manganese dioxide
4NaCl + MnO₂ + 4H₂SO₄ \(\longrightarrow \)Cl₂+ MnCl₂ +4NaHSO₄ + 2H₂O

(i) sp³d³ hybridisation
(ii) Pentagonal bipyramidal structure.

(i) Fluorine is more electro negative than other halogens.
(ii) Because except fluorine all the other halogens have positive oxidation state.
(iii) It has high electron affinity character
Example:
\(\mathrm{Cl}_{2}\mathrm{O} \Rightarrow+1 \text { oxidation state }(\mathrm{Cl}) \)
\(\mathrm{OF}_{2} \Rightarrow-1 \text { oxidation state }(\mathrm{F})\)

(a) OF₂
+ 2 + 2(x) = 0 
+2 = -2x 
2 x = -2 ⇒ x = -1
(b) O₂F₂
2(+1) + 2x = 0 
2x = -2 
x = -1
(c) Cl₂O₃
2(x) + 3(-2) = 0 
2x = +6 
x = +3
(d) I₂O₄
2(x) + 4(-2) = 0 
2x = +8 
x = +4

(i) Fluorine is most electronegative atom.
(ii) It has only one unpaired electron.

(i) The Chalcogens belong to group (16).
(ii) The group consists of elements: Oxygen, Sulphur, Selenium, Tellurium and Polonium.
(iii) These are ore forming elements as most of the ores are oxides and sulphides.
(iv) Chalcos meaning 'ore formers'.

(i) In heavier post transition metals, the outers electrons (ns) have a tendency to remain inert and show reluctance to take part in the bonding, which is known as inert pair effect.
(ii) This effect is also observed in groups 14, 15 and 16.