Chemistry M.C.Q (1 Marks)

$ICl$ is more reactive due to polar bonds.

$X-X'$ bond is weaker than $X - X$ bond except $F _{2}$

Due to inter electronic repulsions $F - F$ bond becomes weak and easily broken.

$(b)$ $1 s ^{2} 2 s ^{2} 2 p ^{4} \rightarrow O$

$(c)$ $1 s ^{2} 2 s ^{2} 2 p ^{3} \rightarrow N$

$(d)$ $1 s ^{2} 2 s ^{2} 2 p ^{1} \rightarrow B$

The ionization enthalpy order is $B < Be < O < N$

Be has more IE compared to $B$ due to extra stability and $N$ has more $IE$ compared to oxygen due to extra stability

Hence, $N \rightarrow 1402 kJ / mol$

$O \rightarrow 1314 kJ / mol$

$B \rightarrow 801 kJ / mol$

$Be \rightarrow 899 kJ / mol$

Due to strong reducing nature of $I^{-}$

$2 Fe ^{3+}+2 I ^{-} \longrightarrow 2 Fe ^{2+}+ I _{2}$

$Cl _{2}+ H _{2} O \rightarrow HCl + HClO$

$HClO \rightarrow HCl +[ O ]$

$Cl _{2}> Br _{2}> F _{2}> I _{2}$

Fluorine has very small size due to which inter-electronic repulsion is there between two fluorine atoms this results in the weaken bond.

(Excess)

$NH _{3}+3 Cl _{2} \rightarrow NCl _{3}+3 HCl$

(Excess)

Hence. $X=6 N H_{4} C l+N_{2}$

$Y=N C l_{3}+3 H C l$

$(H F $ to $HI $ $ ( HF > HCl > HBr > HI )$

Due to successive decrease in the strength of $H X$ bonds, bonds dissociation enthalpy decreases from $H F$ to $H I$ $H X$ Bond dissociation $\underset{574.0}{H F}>\underset{428.1}{H C l}>\underset{362.5}{H B r}>\underset{294}{H I}$ enthalpy ( $k J / m o l)$

$(hot \;and \;conc.)\quad \quad (A)\quad \quad \quad side \;product$

$2 \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{Cl}_{2} \rightarrow \mathrm{Ca}(\mathrm{OCl})_{2}+\mathrm{CaCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

$(dry)\quad \quad \quad \quad \quad \quad \quad \quad (B)\quad \quad \quad side \;product$

$E.A.$ order : $F < Cl$
(ii) Down the group electron affinity decreases due to size increases.

$EA.$ order : $S > Se$       $Li > Na$

         $2NaI + Br_2 \to 2NaBr + I_2$

$C{l_2} + {H_2}O \to HCl + HOCl$

$B{r_2} + {H_2}O \to HBr + HOBr$

${I_2} + {H_2}O + No\,reaction\,[\Delta G > 0\,at\,R.T.]$

Chlorine reacts with water to form hydrochloric and hypochlorous acids.

$Cl_2+ H_2 O \longrightarrow HCl + HClO$

Hypochlorous acid is unstable, and it easily dissociates to form nascent oxygen.

$HClO \longrightarrow HCl +[ O ]$

Nascent oxygen is a more powerful oxidising agent. Its formation is responsible for the bleaching nature of chlorine in the presence of moisture.

When bromine gas is passed through solutions of $NaF , NaCl , NaI$, bromine cannot displace fluorine and chlorine compound but it can displace iodine compound

$2 NaI + Br _2 \rightarrow 2 NaBr + I _2$

Iodine gas is released.

so the reaction $SF_6$ and $BF_3$ is not complete

Hybridization of $\mathrm{Si}-\mathrm{s} p^{3},$     bond angle $=109\,^o28'$

$\angle Cl-\text{N}-\text{Cl}$ in $NCl_3$ is greater than $\angle \text{Cl}-\text{P}-\text{Cl}$ in $\mathrm{PCl}_{3}$ because partidpation of $s-$ orbital in the hybridization decreases from $NCl_3$ to $PCl_3$

Figure $(2)$

$\angle \text{Cl}-\text{P}-\text{Cl}$ in $\text{PC}{{\text{l}}_{3}}>\angle \text{H}-\text{Sb}-\text{H}$ in $\mathrm{SbH}_{3},$ as in $\mathrm{SbH}_{3}$, bond pairs are formed by overlapping of almost pure $p-$ orbitals.

$\angle \text{H}-\text{Sb}-\text{H}$ in $\text{Sb}{{\text{H}}_{3}}<\angle \text{H}-\text{Te}-\text{H}$ in $\mathrm{H}_{2}Te$ because two lone pairs are present on $Te$ while at $\mathrm{Sb}$ there is one lone pair.

$\underbrace{HF\,}_{H\,-\,Bonding}\,>\,\,\underbrace{HI\,\,>\,\,HBr\,\,>\,\,HCl}_{\begin{smallmatrix} 
 Dipole\,\,dipole \\ 
 attraction\,\,(V.W.forces) 
\end{smallmatrix}}$

In case of same type of van der waal's forces of attraction b.pt. $\propto $ molecular mass.

${I_2} + 2N{a_2}{S_2}{O_3} \to 2NaI + N{a_2}{S_4}{O_6}$

Chlorine reacts with water and produces a single oxygen atom also called nascent oxygen. This nascent oxygen when combines with any colour make it colourless.

$Cl _2+ H _2 O \rightarrow 2 HCl +( O )$ ( $O$ is oxidizing)

Sulphur dioxide when reacts with coloured substances it releases oxygen from the substance resulting in loss of colour.

$SO _2+2 H _2 O \rightarrow H _2 SO _4+2( H )$ ( $H$ is reducing)

$I _2\,<\, Br _2\,<\, Cl _2\,<\, F _2$

$F _2$ is better oxidizing agent in water than chlorine due to greater hydration energy.

$X_2Cl_3\Rightarrow X^{3+}Cl^{2-}$

$X_2(SO_4)_3\Rightarrow X^{3+}SO_4^{2-}$

$XPO_4\Rightarrow X^{3+}PO_4^{3-}$

Because $Cl^{2-}$ does not exist. So, $X_2Cl_3$ is incorrect. The correct formula should be $XCl_3$.

Therefore, the product of the reaction of $I _2 I _2$ with $H _2 O _2. H _2 O _2$ in the basic medium is Iodine.

Acidic nature $ \propto  EN \propto  + O.S.$

For compounds of an element