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

$CO and {Cl}_{2}$

$H_{2} S and {Br}_{2}$

${NH}_{3} and {SO}_{3}$

${SO}_{2} and {Cl}_{2}$

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Explanation

(d) Aqueous solution of ${SO}_{2}$ acts as a reducing agent

${SO}_{2} + 2 H_{2} O \overset{}{\to} H_{2} {SO}_{4} + \underset{nascent hydrogen}{2 H}$

Thus, ${SO}_{2}$ in presence of moisture is used as bleaching agent. This is due

to the reducing nature of ${SO}_{2}$ .

For delicate articles

$Coloured matter + H \overset{}{\to} Colourless matter$

Similarly, ${Cl}_{2}$ acts as bleaching agent in presence of moisture

${Cl}_{2} + H_{2} O \overset{}{\to} 2 HCl + [O] Coloured matter + [O] \overset{}{\to} Colourless matter$

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

$(I) {TeF}_{6} (II) {SF}_{6} (III) {NCl}_{3} (IV) {NF}_{3}$

$III and IV$

$I, II and III$

$I, II and IV$

$II and IV$

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Explanation

(d) In ${SF}_{6}, S$ sterically hindered by six fluorine atoms hence, attack of $H_{2} O$

molecule will not occur. ${NF}_{3}$ is not hydrolysed due to absence of vacant

orbital either on $N$ and $F$ atom.

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

Acidic character of oxy acids increases with increasing value of n.

oxidising power of oxy acids increases with decreasing value of n.

Thermal stability of oxy acids decreases with increasing value of n.

$' C — O'$ bond order decreases with decreasing value of n

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Explanation

(c)

$Acidic character : HOCl < {HClO}_{2} < {HCIO}_{3} < {HClO}_{4} Oxidising power : HOCl > {HClO}_{2} > {HCIO}_{3} > {HClO}_{4} Thermal stability : HOCl < {HClO}_{2} < {HCIO}_{3} < {HClO}_{4}' Cl — O' bond order : HOCl < {HClO}_{2} < {HCIO}_{3} < {HClO}_{4}$

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The correct statement regarding ${CIO}_{n}^{‐}$ molecular ion is :

On decreasing value of $' n',' Cl — O'$ bond order increases

On the increasing value of $' n',' Cl — O'$ bond length increases

On the increasing value of $n,$ oxidation number of central atom increases

On the increasing value of $n,$ hybrid orbitals on the central atom increase

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Explanation

(c)

$\Rightarrow {ClO}^{-}_{\underset{Hyb . of Cl : {sp}^{3}}{B . O . = 1.0}}^{+ I} {ClO}_{2}^{-}_{\underset{Hyb . of Cl : {sp}^{3}}{B . O . = 1.5}}^{+ III} {ClO}_{3}^{-}_{\underset{Hyb . of Cl : {sp}^{3}}{B . O . = 1.67}}^{+ V} {ClO}_{4}^{-}_{\underset{Hyb . of Cl : {sp}^{3}}{B . O . = 1.75}}^{+ VII}$

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Auto-oxidation of bleaching powder gives :

only calcium chlorate

only calcium chloride

only calcium hypochlorite

both (a) and (b)

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Explanation

(d) $6 {CaOCl}_{2} \to_{oxidation}^{Auto} Ca ({CIO}_{3})_{2} + 5 {CaCl}_{2}$

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

$Element Electronic configuration X 1 s^{2} 2 s^{2} 2 p^{2} Y 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} Z 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{5}$

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

Hydride of X Hydride of Y Hydride of Z

Colourless gas insoluble in Silver/grey solid reacts with $H_{2} O$ Colourless gas forms a strong

$H_{2} O$ to form an alkali acid in $H_{2} O$

Colourless liquid, no reaction Silver/grey solid forms $H_{2}$ and Ionic solid with formula $ZH$

with $H_{2} O$ $H_{2} O$

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

with ${Cl}_{2}$ in light formula ${YH}_{2}$ to phosphorus and element Z

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

with air violently with acids at STP

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Explanation

(a) $X = {CH}_{4}, Y = NaH, Z = HBr$

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Which of the following is not true about helium?

It has the lowest boiling point

It has the highest first ionization energy

It can diffuse through rubber and plastic material

It can form clathrate compounds

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Explanation

(d) Due to small size of $He$ , it escapes from interstitial spaces/voids of

molecular lattice of quinols.

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${SbF}_{5}$ reacts with ${XeF}_{4}$ to form an adduct. The shapes of cation and anion in the adduct are respectively:

square planar, trigonal bipyramidal

T-shaped, Octahedral

square pyramidal, octahedral

square planar, octahedral

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Explanation

(b)

${XeF}_{4} + {SbF}_{5} \overset{}{\to} {[{XeF}_{3}]}^{+} {[{SbF}_{6}]}^{-} \overset{}{\to} \underset{\overset{{sp}^{3} d}{bent T shape}}{{[{XeF}_{3}]}^{+}} + \underset{\overset{{sp}^{3} d^{2}}{octahedral}}{{[{SbF}_{6}]}^{-}}$

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Select the correct statement about elements of group $15^{t h}$ .

The order of stability of oxidation state for +3 is $B i^{3 +} > S b^{3 +} > A s^{3 +}$ and for +5 is $B i^{5 +} < S b^{5 +} < A s^{5 +}$

In the case of nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution.

There is considerable increase in covalent radius from N to P but from As to Bi only a small increase in covalent radius is observed.

All of these.

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Explanation

(D) (A) This is due to the inert pair effect in the heavier p-block elements.

(B) $3 H N O_{2} \to H N O_{3} + H_{2} O + 2 N O$

(C) This is due to the presence of completely filled d and l or f orbitals in heavier elements. Covalent radius l pm; N =70, P =110, As =120, Bi =150.

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Select the incorrect statement

Nitrous oxide supports combustion more vigorously than air.

Phosphorous pentaoxide dehydrates nitric acid forming nitrogen pentaoxide.

Reactivity order of various allotropic forms of phosphorous is white > red > black.

Red phosphorous changes to white phosphorus on heating in the atmosphere of $C O_{2}$ or coal gas at 573 K.

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Explanation

(D) (A) $2 N_{2} O \to 2 N_{2} + O_{2}$

It produces one third oxygen by volume of total gas produced while air has one fifth volume of oxygen.

(B) $2 H N O_{3} \overset{P_{2} O_{5}}{\to} H_{2} O + N_{2} O_{5}$

(D) White phosphorus turns to red phosphorus on heating in the atmosphere of $C O_{2}$ or coal gas
at 573 K.
So (A), (B), (C) are correct and (D) is correct.

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