Tag: chemistry

Questions Related to chemistry

A one litre flask is full of brown bromine vapour. Intensity of brown colour vapour will not decrease a appreciable amount on adding to flask some of:

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

  1. Piece of marble

  2. Carbon disulphide

  3. Carbon tetrachloride

  4. Animal charcoal powder

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Correct Option: A
Explanation:

A one litre flask is full of brown bromine vapour. Intensity of brown colour vapour will not decrease a appreciable amount on adding to flask some of Piece of marble.

Hence option A is correct.

Fluorine is stronger oxidising agent than chlorine in water. The factor which is not responsible for that:

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

  1. Heat of dissociation

  2. e-affinity

  3. Heat of hydration

  4. Ionisation potential

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Correct Option: B
Explanation:

Affinity do not have any relation with the efficiency of oxidizing ability.

Hence option A is correct.

$Br _2$ turns starch iodide paper :

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

  1. blue

  2. red

  3. colourless

  4. yellow

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Correct Option: A
Explanation:

The bromine releases iodine from the potassium iodide Iodine reacts with starch to produce a blue color.

$Br _2+KI\rightarrow KBr+I _2$

Hence option A is correct

Which of the following reactions are not possible?

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

  1. $2NaCl+F _2$ $\longrightarrow \ $ $2NaF+Cl _2$

  2. $2NaCl+Br _2$ $\longrightarrow \ $ $2NaBr+Cl _2$

  3. $2NaF+Cl _2$ $\longrightarrow \ $ $2NaCl+F _2$

  4. $2NaBr+Cl _2$ $\longrightarrow \ $ $2NaCl+Br _2$

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Correct Option: B,C
Explanation:

Option B and C reactions are not possible because the bromine do not displace chlorine and chlorine do not displace florine as chlorine and florine are most stable than bromine and chorine in their compounds respectively.

Hence option B,C are correct.

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

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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

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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

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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

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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.