Tag: p-block elements

Questions Related to p-block elements

The following acids have been arranged in order of decreasing acid strength. Identify the correct order.
I. HOCl


II. HOBr 

III. HOI

  1. I > II > III

  2. II > I > III

  3. III > II > I

  4. I > III > II


Correct Option: A
Explanation:

As we move down the group acidic strength decreases as more is the electronegativity of the central atom the shared pair of $e^-$ will be more towards the central atom; resulting in the release of Hydrogen easy.

Identify (a) and (b).

$ Br _2 + \underset {hot  }{ OH^- } \rightarrow (a) + (b) $

$ (a) + (b) + H^+ \rightarrow Br _2 $ :

  1. a- $Br^-$

  2. b- $BrO$

  3. a- $BrO$

  4. b- $BrO _3^-$

  5. None of the above


Correct Option: A,D
Explanation:

Method is for enriching $Br _2$:

$ 3Br _2 + 6 OH^{-} \rightarrow \underset {(A)  }{ 5Br^- }+ \underset {(B)  }{ BrO^- _3} $ +$3H _2O$

$ 5Br^- + BrO^- _3 + 6H^+ \rightarrow 3Br _2 + 3H _2O $

Hence,option A and D is correct.

When $I^-$ is oxidised by $Mn{O}^- _4$ in alkaline medium, $I^-$ converts into :

  1. $\mathrm{I}\mathrm{O} _{3}^{-}$

  2. $\mathrm{I} _{2}$

  3. $\mathrm{I}\mathrm{O} _{4}^{-}$

  4. $\mathrm{I}\mathrm{O}^{-}$


Correct Option: A
Explanation:

$2\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+2\mathrm{K}\mathrm{O}\mathrm{H}\rightarrow 2\mathrm{K} _{2}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+\mathrm{H} _{2}\mathrm{O}+\mathrm{O}$
$\frac{2\mathrm{K} _{2}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+2\mathrm{H} _{2}\mathrm{O}\rightarrow 2\mathrm{M}\mathrm{n}\mathrm{O} _{2}+4\mathrm{K}\mathrm{O}\mathrm{H}+2\mathrm{O}}{2\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+\mathrm{H} _{2}\mathrm{O}\xrightarrow[]{alkalme}2\mathrm{M}\mathrm{n}\mathrm{O} _{2}+2\mathrm{K}\mathrm{O}\mathrm{H}+3[\mathrm{O}]}$
$\frac{KI+[\mathrm{O}] \rightarrow \mathrm{K}\mathrm{I}\mathrm{O} _{3}}{2\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O} _{4}+\mathrm{K}\mathrm{I}+\mathrm{H} _{2}\mathrm{O}\rightarrow 2 KOH +2\mathrm{M}\mathrm{n}\mathrm{O} _{2}+\mathrm{K}\mathrm{I}\mathrm{O} _{3}}$
$KI+3[O]\rightarrow KIO _3^{-}$
Hence option A is correct.

When hypohalites are heated at $300K$, the product obtained are:

  1. halide

  2. halogen

  3. perhalate

  4. halate


Correct Option: A,D
Explanation:

Hypohalites $(XO^-)$ disproportionates in aqueous solution on heating to produce halide ion and halate ion.


$\underset {(Hypohalite)}{3XO^-}\overset {300K}{\rightleftharpoons}\underset {(Halide)}{{2X}^-}+\underset {(Halate)}{{XO _3}^-}$

Iodine solution is prepared by dissolving iodine in

  1. $NaOH$

  2. $Na _{2}CO _{3}$

  3. $H _{2}O$

  4. $KI$


Correct Option: D
Explanation:

Iodine solutions are prepared by dissolving $I _2$ in a concentrated solution of potassium iodide ($KI$).
Hence option D is correct.

The colour of the ${X} _{2}$ molecules of group $17$ elements changes gradually from yellow to violet down the group. This is due to:

  1. the physical state of ${X} _{2}$ at room temperature changes from gas to solid down the group

  2. decrease in ${\pi}^{\ast} - {\sigma}^{\ast}$ gap down the group

  3. decrease in $HOMO-LUMO$ gap down the group

  4. decrease in ionization energy down the group


Correct Option: B,C
Explanation:

The colour of the $\displaystyle X _2$ molecules of group 17 elements changes gradually from yellow to violet down the group. This is due to  (B) Decrease in $\displaystyle {\pi}^{\ast} - {\sigma}^{\ast}$ gap down the group and (C) decrease in HOMO−LUMO gap down the group. Halogens absorb visible light and appear coloured. Outer electrons are excited to higher energy level. Smaller halogen has higher excitation energy and vice versa.

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidizing power.

Ion $Cl{ O } _{4}^{-}$ $I{ O } _{4}^{-}$ $Br{ O } _{4}^{-}$
Reduction potential $E^{\circ}$/V $E^{\circ}$= 1.19V $E^{\circ}$= 1.65V $E^{\circ}$= 1.74V
  1. $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Br{ O } _{4}^{-}$

  2. $I{ O } _{4}^{-}$ > $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$

  3. $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$

  4. $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$


Correct Option: C
Explanation:

The reduction potential of the substance is the ability of the substance to be reduced.So as the reduction potential increases reducing ability of the  substance increases.It means oxidizing power of the substance (The ability of the substance to make the other substance to lose the electrons increases which noting but oxidizing power) increases.

So the decreasing order of oxidizing power is $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$.

Hence option $C$ is correct.

Which one of the following is an amphoteric oxide?

  1. $Na _2O$

  2. $SO _2$

  3. $B _2O _3$

  4. $ZnO$


Correct Option: D
Explanation:

$ZnO$ is an amphoteric oxide. It reacts with both acid and base to produce respective salts. 

A group of acidic oxide is:

  1. $CrO _{3},Mn _{2}O _{7}$

  2. $ZnO,Al _{2}O _{3}$

  3. $CaO,ZnO$

  4. $Na _{2}O,Al _{2}O _{3}$


Correct Option: A
Explanation:

here,${ CrO } _{ 3 }$ are acidic oxides

${ CrO } _{ 3 }+{ H } _{ 2 }O\longrightarrow { H } _{ 2 }{ CrO } _{ 4 }\ { Mn } _{ 2 }{ O } _{ 7 }+{ H } _{ 2 }O\longrightarrow { 2HMnO } _{ 4 }$

Which of the following constitutes a set amphoteric species.?

  1. $H _3O^+, H _2PO^- _4, HCO^- _3$

  2. $H _2O, HPO^{2-} _4, H _2PO^- _2$

  3. $H _2O, H _2PO^- _3, HPO^{2-} _4$

  4. $HC _2O^- _4, H _2PO^- _4, SO^{2-} _4$


Correct Option: C