Tag: group 17 elements - properties

Questions Related to group 17 elements - properties

Iodine is oxidised by fuming nitric acid. The major product formed is:

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

  1. $HI$

  2. $HIO _2$

  3. $HIO _3$

  4. $HIO _4$

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Correct Option: C
Explanation:

Reaction of iodine with fuming nitric acid is given by 

${ I } _{ 2 }+{ 10HNO } _{ 3 }\longrightarrow { 2HIO } _{ 3 }+{ 10NO } _{ 2 }+{ 4H } _{ 2 }O$.
So, the major product formed will be ${ HIO } _{ 3 }$.
So, correct answer is option $C$.

Which one of the following orders is not in accord with the property stated against it?

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

  1. $F _{2} > Cl _{2} > Br _{2}> I _{2}$ ; Oxidising power

  2. $HI > HBr > HCl > HF$ ; Acidic property in water

  3. $F _{2} > Cl _{2}> Br _{2} > I _{2}$; Electronegativity

  4. $F _{2}> Cl _{2} > Br _{2} > I _{2}$ ; Bond dissociation energy

Show answer
Correct Option: D
Explanation:

The bond dissociation energies of $I _2$, $Br _2$ and $Cl _2$ increase in the following order:

$I _2$ < $Br _2$ < $Cl _2$ 

In case of $F _2$, the bond dissociation energy is lower than that of $Cl _2$ and $Br _2$ because of the existence of strong inter-electronic repulsions owing to its small size. The overall trend of bond dissociation energies is thus as follows:
$I _2$ < $F _2$ < $Br _2$ < $Cl _2$ 

Which of the following compounds is used as a sedative (sleep-inducing substance)?

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

  1. Sodium Chloride

  2. Potassium bromide

  3. Calcium Chloride

  4. Phosphorous trichloride

Show answer
Correct Option: B
Explanation:

Potassium bromide ($KBr$) is a salt, widely used as an anticonvulsant and a sedative in the late 19th and early 20th centuries.

Its action is due to the bromide ion (sodium bromide is equally effective).
Hence option B is correct answer.

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


II. HOBr 

III. HOI

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

  1. I > II > III

  2. II > I > III

  3. III > II > I

  4. I > III > II

Show answer
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 $ :

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

  1. a- $Br^-$

  2. b- $BrO$

  3. a- $BrO$

  4. b- $BrO _3^-$

  5. None of the above

Show answer
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 :

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

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

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

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

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

Show answer
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:

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

  1. halide

  2. halogen

  3. perhalate

  4. halate

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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

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

  1. $NaOH$

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

  3. $H _{2}O$

  4. $KI$

Show answer
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:

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

  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

Show answer
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
group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

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

Show answer
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.