Tag: group 17 elements - trends in chemical properties

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Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

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
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

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. $\mathrm{I}\mathrm{O}^{-}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

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

  1. halide

  2. halogen

  3. perhalate

  4. halate
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

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


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

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

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
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

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. $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.