The incorrect statement regarding ${15}^{\mathrm{th}}$ group hydrides is-

The correct statement among the following is-

A non-metal M forms ${\mathrm{MCl}}_3, {\mathrm{M}}_2{\mathrm{O}}_5 \mathrm{and} {\mathrm{Mg}}_3{\mathrm{M}}_2$ but does not form ${\mathrm{MI}}_5$. Then incorrect statement regarding non-metal M is:

The correct order of $\mathrm{S}—\mathrm{S}$ bond length in the following oxyanion is :

(I) ${\mathrm{S}}_2{\mathrm{O}}_4^{2-}$                  (II) ${\mathrm{S}}_2{\mathrm{O}}_5^{2-}$                      (III) ${\mathrm{S}}_2{\mathrm{O}}_6^{2-}$

(a) I$>$II$>$III                                      (b) I$>$III$>$II

(c) III$>$II$>$I                                       (d) III$>$I$>$II

In which of the following reaction product does not contain 'Peroxy' linkage?

(a) $2\mathrm{OF}\stackrel{\mathrm{Dimerisation}}{\to }$                               (b) ${\mathrm{H}}_4{\mathrm{P}}_2{\mathrm{O}}_8\stackrel{+{\mathrm{H}}_2\mathrm{O}}{\to }$

(c) $2\mathrm{Na}\underset{∆}{\overset{\mathrm{excess} {\mathrm{O}}_2}{\to }}$                                   (d) None of these 

Consider the following statements in respect of oxides of sulphur.

(1) In gas phase ${\mathrm{SO}}_2$ molecule is V-shape.

(2) In gas phase ${\mathrm{SO}}_3$ molecule is planar.

(3) $\mathrm{y}‐{\mathrm{SO}}_3$ is cyclic trimer.

Which of the above statements are correct ?

Gas that can not be collected over water is :

One gas bleaches the colour of flowers by reduction, while the other gas bleaches by oxidation, the two gases respectively are :

Which of the following halides cannot be hydrolysed at room temperature?

  $\left(\mathrm{I}\right) {\mathrm{TeF}}_6 \left(\mathrm{II}\right) {\mathrm{SF}}_6 \left(\mathrm{III}\right) {\mathrm{NCl}}_3 \left(\mathrm{IV}\right) {\mathrm{NF}}_3$

A parent oxy acid that does not have its hypo acid is -

$$\begin{gathered} 1. {\mathrm{H}}_2{\mathrm{SO}}_{2 } \\ 2. {\mathrm{HNO}}_2 \\ 3. {\mathrm{H}}_3{\mathrm{PO}}_3 \\ 4. {\mathrm{HClO}}_3 \end{gathered}$$

Consider the following reactions :

$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{PCl}}_3+3{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }{\mathrm{H}}_3{\mathrm{PO}}_3+3\mathrm{HCl} \left(\mathrm{ii}\right) {\mathrm{SF}}_4+3{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }{\mathrm{H}}_2{\mathrm{SO}}_3+4\mathrm{HF} \\ \left(\mathrm{iii}\right) {\mathrm{BCl}}_3+3{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }{\mathrm{H}}_3{\mathrm{BO}}_3+3\mathrm{HCl} \left(\mathrm{iv}\right) {\mathrm{XeF}}_6+3{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }{\mathrm{XeO}}_3+6\mathrm{HF} \end{gathered}$$

Then according to given information the incorrect statement is :

Consider the oxy acids ${\mathrm{HClO}}_{\mathrm{n}}$ series, here value of n is 1 to 4. Then incorrect statement regarding these oxy acids is :

The correct statement regarding ${\mathrm{CIO}}_n^{‐}$ molecular ion is :

One of the products of the following reaction $KCNO+{\left(NH\right)}_{4}S{O}_4$ is :

$\mathrm{In}, {\mathrm{Cl}}_2{\mathrm{O}}_6\left(\mathrm{l}\right)+\mathrm{HF} \underset{}{\to } \mathrm{P}+\mathrm{Q}$

$\mathrm{If} {\mathrm{H}}^{+} \mathrm{of} \mathrm{acid} \mathrm{HF} \mathrm{attaches} \mathrm{with} \mathrm{Q} \mathrm{then} \mathrm{correct} \mathrm{option} \mathrm{for} \mathrm{hybrization} \mathrm{of} \mathrm{Cl}-\mathrm{atom}$

$\mathrm{and} \angle \mathrm{OCIO} \mathrm{in} \mathrm{the} \mathrm{P} \mathrm{and} \mathrm{Q} \mathrm{ions} :$

$\left(\mathrm{a}\right) P\mathbf{:}s{p}^{\mathit{2}};>120°$

$\left(\mathrm{b}\right) Q\mathbf{:}s{p}^3; 109°28\text{'}$

$\left(\mathrm{c}\right) P\mathbf{:}s{p}^3;<109°28\text{'}$

$\left(\mathrm{d}\right) Q\mathbf{:}s{p}^3;>109°28\text{'}$

Bromine is commercially prepared from sea water by displacement reaction 

                       ${\mathrm{Cl}}_2+{\mathrm{Br}}^{‐}\left(\mathrm{aq}\right) \stackrel{}{\to } 2{\mathrm{Cl}}^{‐}\left(\mathrm{aq}\right)+{\mathrm{Br}}_2$

${\mathrm{Br}}_2$ gas thus formed is dissolved into solution of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ and then pure ${\mathrm{Br}}_2$ is obtained by treatment of the solution with :

$\left(\mathrm{a}\right) \mathrm{Ca}(\mathrm{OH}{)}_2$

$\left(\mathrm{b}\right) \mathrm{NaOH}$

$\left(\mathrm{c}\right) {\mathrm{H}}_2{\mathrm{SO}}_4$

$\left(\mathrm{d}\right) \mathrm{HI}$

$$\begin{gathered} {\mathrm{Cl}}_2\left(\mathrm{g}\right)+\mathrm{Ba}{\left(\mathrm{OH}\right)}_2 \stackrel{}{\to } \mathrm{X}(\mathrm{aq}.)+{\mathrm{BaCl}}_2+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{X}+{\mathrm{H}}_2{\mathrm{SO}}_4 \stackrel{ }{\to } \mathrm{Y}+{\mathrm{BaSO}}_4 \\ \mathrm{Y} \underset{∆>365\mathrm{K}}{\overset{∆}{\to }} \mathrm{Z}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2 \end{gathered}$$

Y and Z are respectively :

Auto-oxidation of bleaching powder gives :

The three elements X, Y and Z with electronic configuration shown below all form hydrides :

            $$\begin{gathered} \mathbf{Element}\mathbf{ } \mathbf{Electronic}\mathbf{ }\mathbf{configuration} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathrm{X} 1s^{\mathit{2}}2s^{\mathit{2}}2p^{\mathit{2}} \\ \mathrm{Y} 1s^{\mathit{2}}2s^{\mathit{2}}2p^{6}3s^{\mathit{1}} \\ \mathrm{Z} 1s^{\mathit{2}}2s^{\mathit{2}}2p^{\mathit{6}}3s^{\mathit{2}}3p^{\mathit{6}}3d^{\mathit{10}}4s^{\mathit{2}}4p^{\mathit{5}} \end{gathered}$$

Which set of properties match correctly with properties of the hydrides of these elements :

     Hydride of X                                      Hydride of Y                                   Hydride of Z

(a) Colourless gas insoluble in        Silver/grey solid reacts with ${\mathrm{H}}_2\mathrm{O}$      Colourless gas forms a strong

        ${\mathrm{H}}_2\mathrm{O}$                                        to form an alkali                                 acid in ${\mathrm{H}}_2\mathrm{O}$

(b) Colourless liquid, no reaction    Silver/grey solid forms ${\mathrm{H}}_2$ and          Ionic solid with formula $\mathrm{ZH}$

        with ${\mathrm{H}}_2\mathrm{O}$                                  ${\mathrm{H}}_2\mathrm{O}$                                

(c) Non-polar compound reacts       Silver/grey ionic solid with               Forms when water is added

     with ${\mathrm{Cl}}_2$ in light                            formula ${\mathrm{YH}}_2$                                 to phosphorus and element Z

(d) Colourless gas which burns        Silver/grey solid which reacts           Colourless, corrosive liquid

          with air                                     violently with acids                             at STP

The incorrect order is :

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{Hl} \mathbf{:}\mathbf{ }\mathrm{Acidic} \mathrm{strength} \\ \left(\mathrm{b}\right) \mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{Hl} \mathbf{:}\mathbf{ }\mathrm{Thermal} \mathrm{stability} \\ \left(\mathrm{c}\right) \mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{Hl} \mathbf{:}\mathbf{ }\mathrm{Boiling} \mathrm{point} \\ \left(\mathrm{d}\right) \mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{Hl} \mathbf{:}\mathbf{ }\mathrm{Bond} \mathrm{dissociation} \mathrm{enthalpy} \end{gathered}$$

$N{H}_{4}Cl{O}_4+HN{O}_3\left(dilute\right)\to HCl{O}_4+\left[X\right]$

The correct order of acidic strength of oxyacid of phosphorous is

Peroxoacids of sulphur among the following is -

The correct statement among the following  is when ${\mathrm{XeF}}_2 \mathrm{and} {\mathrm{XeF}}_6$ are separately hydrolyzed;

$1. \mathrm{Both} \mathrm{give} \mathrm{out} {\mathrm{O}}_2$

$2. {\mathrm{XeF}}_6 \mathrm{gives} {\mathrm{O}}_2 \mathrm{and} {\mathrm{XeF}}_2 \mathrm{does} \mathrm{not}$

$3. {\mathrm{XeF}}_2 \mathrm{alone} \mathrm{gives} {\mathrm{O}}_2$

$4. \mathrm{Neither} \mathrm{of} \mathrm{them} \mathrm{gives} \mathrm{HF}$

$\mathrm{MF}+{\mathrm{XeF}}_4 \stackrel{}{\to }\text{'}\mathrm{A}\text{'} ({\mathrm{M}}^{+}=\mathrm{Alkali} \mathrm{metal} \mathrm{cation})$

The state of hybridisation of the central atom in 'A' and shape of the species are:

$\left(\mathrm{a}\right) s{p}^{3}d, \mathrm{TBP}$

$\left(\mathrm{b}\right) s{p}^{\mathit{3}}{d}^3, \mathrm{pentagonal} \mathrm{planar}$

$\left(\mathrm{c}\right) s{p}^{\mathit{3}}{d}^2, \mathrm{distorted} \mathrm{octahedral}$

$\left(\mathrm{d}\right) \mathrm{No} \mathrm{compound} \mathrm{formed} \mathrm{at} \mathrm{all}$

${\mathrm{XeF}}_6$ dissolves in anhydrous $\mathrm{HF}$ to give a good conducting solution that contains-

The incorrect statement among the following regarding helium is -

Which of the following is an uncommon hydrolysis product of ${\mathrm{XeF}}_2 \mathrm{and} {\mathrm{XeF}}_4$?

$\left(\mathrm{a}\right) \mathrm{Xe}$

$\left(\mathrm{b}\right) {\mathrm{XeO}}_3$

$\left(\mathrm{c}\right) \mathrm{HF}$

$\left(\mathrm{d}\right) {\mathrm{O}}_2$

${\mathrm{SbF}}_5$ reacts with ${\mathrm{XeF}}_4$ to form an adduct. The shapes of cation and anion in the adduct are respectively:

Incorrect statement regarding following reactions is :

(a) 'X' is explosive

(b) 'Y' is an oxyacid of xenon

(c) Both are example of non-redox reaction

(d) ${\mathrm{XeF}}_6$ can undergo partial hydrolysis