Chemistry M.C.Q (1 Marks)

$\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}$ .

Statement $II$ due to intermolecular $\mathrm{H}$-bonding $\mathrm{H}_2 \mathrm{O}$ shows higher boiling point than respective hydrides of group $16$.

(Both Statement are true)

Order from $\mathrm{H}_2 \mathrm{Te}$ to $\mathrm{H}_2 \mathrm{~S}$ is due to decreasing molar mass.

$M +(x+y) NH _3 \rightarrow\left[ M \left( NH _3\right)\right]^{+}+\left[e\left( NH _3\right) y\right]^{-}$

So, assertion statement is correct but reason is incorrect.

$H _{2} O \,\,\,\,\,\,\,\,\,\,\,\,\,\, \underbrace{ H _{2} S \,\,\, H _{2} Se \,\,\, H _{2} Te}$

$\,\,$H-Bond $\,\,\,\,\,\,\,\,\,\,$ VMA $\propto$ mol.wt.

$\text { B.P. } \rightarrow H _{2} S < H _{2} Se < H _{2} Te < H _{2} O$

$ICl$ is more reactive due to polar bonds.

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

$\mathrm{CH}_{3}-\mathrm{F}>\mathrm{CH}_{3}-\mathrm{Cl}>\mathrm{CH}_{3}-\mathrm{Br}>\mathrm{CH}_{3}-\mathrm{I}$

Down the group with the increase in size of halogen in C-X bond, energy decreases.

$(B)$ $2 KO _2+2 H _2 O \rightarrow 2 KOH + H _2 O _2+ O _{2( g )} \uparrow$

$(C)$ $Na _2 O _2+2 H _2 O \rightarrow 2 NaOH + H _2 O _2$

$(D)$ $Li _2 O _2+2 H _2 O \rightarrow 2 LiOH + H _2 O _2$

$\mathrm{XeF}_{6}-\mathrm{sp}^{3} \mathrm{d}^{3}, \quad$ Distorted octahedral

$\mathrm{XeOF}_{4}-\mathrm{sp}^{3} \mathrm{d}^{2},\quad$ Square pyramidal

$\mathrm{XeO}_{3}-\mathrm{sp}^{3}, \quad$ Pyramidal

${\left(2 \mathrm{NaN}_{3} \stackrel{4}{\longrightarrow} 2 \mathrm{Na}+3 \mathrm{N}_{2}\right)}$

${\left(\mathrm{Ba}\left(\mathrm{N}_{3}\right)_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{Ba}+3 \mathrm{N}_{2}\right)}$

Haber process $\Rightarrow$ Formation of Ammonia

$\left(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\right)$

Contact process $\Rightarrow$ manufacture of $\mathrm{H}_{2} \mathrm{SO}_{4}$

Deacon's process $\Rightarrow$ Formation of $\mathrm{Cl}_{2}$ gas $\left(\mathrm{HCl}+\mathrm{O}_{2(\mathrm{Atmosphere})} \stackrel{\mathrm{cucl_2}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}\right)$

$2\; mole \;\mathrm{NH}_{3}(\mathrm{g})$ requires $3\; mole \;\mathrm{H}_{2}(\mathrm{g})$

$20 \;mole\; \mathrm{NH}_{3}(\mathrm{g})$ requires

$=\frac{3}{2} \times 20 \text { mole } \mathrm{H}_{2}(\mathrm{g})$

$=30\;mole$

$3 XeF _4+6 H _2 O \rightarrow 2 Xe + XeO _3+\frac{3}{2} O _2+12 HF$

$\therefore$ One mole of $XeF _4$ gives 4 moles of $HF$ on hydrolysis.

$\mathrm{NO} \to +2$

$\mathrm{N}_{2} \to 0$

$\mathrm{NH}_{4} \mathrm{Cl}\to -3$

$\mathrm{NO} \quad \quad \quad \quad \quad 15 \quad \quad \quad \quad \quad \quad 2.5$

$\mathrm{CN}^{-} \quad \quad \quad \quad \;\;14 \quad \quad \quad \quad \quad \quad 3$

${\mathrm{CN}^{+}} \quad \quad \quad \quad \;\; {12} \quad \quad \quad \quad \quad \quad {2}$

${\mathrm{CN}} \quad \quad \quad \quad \quad {13}\quad \quad \quad \quad \quad \quad {2.5}$

$XX$ ' has linear geometry as only two atoms are present.

$(b)-(i)$

$XX _3^{\prime}$ has T-shaped geometry as $3$ bond pairs and $2$ lone pairs of electrons are present. The electron pair geometry is trigonal bipyramidal.

$(c)-(iv),$

$XX _5^{\prime}$ has Square - pyramidal geometry as $5$ bond pairs and $1$ lone pair of electrons are present. The electron pair geometry is octahedral.

$(d)-(ii)$

$XX _7^{\prime}$ has pentagonal bipyramidal geometry as $7$ bond pairs and $0$ lone pairs of electrons are present.

In $XeO _3$, the central $Xe$ atom undergoes $sp ^3$ hybridisation and the molecular geometry is pyramidal with $1$ lone pair and $3$ bonding domains of electrons.

In $XeOF _4$, the central $Xe$ atom undergoes $sp ^3 d ^2$ hybridization and the molecular geometry is square pyramidal with $1$ lone pair and $5$ bond pairs of electrons.

In $XeF _4$, the central $Xe$ atom undergoes $sp ^3 d ^2$ hybridization and the molecular geometry is square planar with $2$ lone pairs and $4$ bond pairs of electrons.

$H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}$

$+1\quad \quad +3\quad\quad\quad+5\quad\quad\quad+7$

Bond dissociation energy of $F_{2}$ less than $C l_{2}$ and $B r_{2}$ because small size of atom result in strong lone pair repulsion.

$\mathrm{Cu}+4 \mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NO}_{2}$

All oxy-acid of phosphorus which contain $\mathrm{P}-\mathrm{H}$ bond act as reductant.

presence of one -OH group and two $P-H$ bonds

$N O_{2}$ is not used as food preservative.

$\mathrm{HF} > \mathrm{Hl} > \mathrm{HBr} > \mathrm{Hl}$

of central atom increases which weakens the

$\mathrm{M}-\mathrm{H}$ bond. Since, the size increases from $\mathrm{S}$ to $\mathrm{Te}$

thus acidic strength follows the order.

$H_{2} S < H_{2} S e < H_{2} T e$

Acidic nature $\propto \frac{1}{\text { Bond disspciation enthalpy }}$

$S$ to $Te$ size increases, bond dissociation enthalpy

decreases and acidic nature increases.

He has the lowest critical temperature $(5.1 \,K$ ) while $Rn$ has the highest critical temperature $(373.5\, K$ ). It increases as we move down the group from helium to radon.

The decrease of the critical temperature follows the trend $Xe \,>\, Kr \,>\, Ar \,>\,Ne\, >\, He$ and hence, the same trend is followed for liquefaction.

$(b)$ $2 \mathrm{XeF}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Xe}+4 \mathrm{HF}+\mathrm{O}_{2}$

$(c)$ $6 \mathrm{XeF}_{4}+12 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_{3}+24 \mathrm{HF}+3 \mathrm{O}_{2}$

$(d)$ $\mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{XeO}_{3}+6 \mathrm{HF}$