Tag: p-block elements

Questions Related to p-block elements

Iodine reacts with hot NaOH solution giving the products as-___?

  1.  $NaI$

  2. $ \displaystyle NaIO _{3}$.

  3. A and B

  4. non of these


Correct Option: A
Explanation:
Iodine reacts with hot concentrated NaOH solution giving the products sodium iodide (NaI) and  sodium iodate $\displaystyle  (NaIO _3)$.
With cold and dil NaOH, iodine gives sodium hypo iodite $\displaystyle  (NaOI)$

Observe the following statements.
I .Bleaching powder is used in the preparation of chloroform.
II. Bleaching powder decomposes in the presence of $\mathrm{C}\mathrm{o}\mathrm{C}1 _{2}$ to liberate $\mathrm{O} _{2}$.
III. Aqueous $\mathrm{K}\mathrm{H}\mathrm{F} _{2}$ is used in the preparation of fluorine.

  1. I,II and III are correct

  2. Only II is correct.

  3. Only I and III are correct

  4. Only I and II are correct.


Correct Option: D
Explanation:

Molten $KHF _{2}$ is used in the preparation of fluorine.

Rest two statements are true.

Tincture of iodine contains _____?

  1. $ \displaystyle I _{2}$

  2.  $KI$

  3.  Rectified spirit.

  4. All of these


Correct Option: D
Explanation:

Tincture of iodine contains 0.5 ounce iodine,  0.25 ounce KI and 1 pint of rectified spirit.

Iodine deficiency in diet is ____.


  1. Solutions of iodine in KI are used in treatment of Goiter.

  2. Goiter is enlargement of thyroid gland.


  3. Iodine deficiency in diet is known to cause Goiter.

  4. Iodine is necessary for normal functioning of the thyroid gland.


Correct Option: A,B,C,D
Explanation:
Iodine is necessary for normal functioning of the thyroid gland.
Iodine deficiency in diet is known to cause Goiter.
Goiter is enlargement of thyroid gland.
Solutions of iodine in KI are used in treatment of Goiter.

Which of the following will displace the halogen from the solution of of halide
a $) Br _{2}$ added to an $NaCl$ solution
$\mathrm{b}) Cl _{2}$ added to a $KBr$ solution
$\mathrm{c}) Cl _{2}$ added to an $NaF$ solution
$\mathrm{d}) Br _{2}$ added to to a $KI$ solution
Find the correct answer :

  1. a, b, c

  2. b, c

  3. b, d

  4. all are correct


Correct Option: C
Explanation:

$Br^{-}$ can reduce $Cl _{2}$ to $Cl^{-}$ $(E _{cell}=+ve)$

$I^{-}$ can reduce $Br _{2}$ to $Br^{-}$ $(E _{cell}=+ve)$

$E _{cell}=+ve$ in both case

$\therefore$ Reaction is feasible

Iodine can be obtained from NaI solution by the action of:

  1. chlorine

  2. bromine

  3. soluble chloride

  4. soluble bromine


Correct Option: A,B
Explanation:

The iodine can be obtained by reacting chlorine or bromine by NaI. This is because the electronegativity of chlorine and bromine is more than the iodine and also the reactivity of chlorine and bromine is more than the iodine.

Which of the following is incorrect?

  1. iodine is a solid

  2. chlorine is insoluble in water

  3. iodine is more reactive than bromine

  4. bromine is more reactive than chlorine


Correct Option: B,C,D
Explanation:

Reactivity order:-
$F _2>Cl _2> Br _2>I _2$
Thus, chlorine is more reactive than bromine and bromine is more reactive than iodine.

Chlorine is soluble in water.

In KI solution, $ I _2 $ readily dissolves and forms :

  1. $ I^{-}$

  2. $ KI _2 $

  3. $ KI _2^- $

  4. $ KI _3 $


Correct Option: D
Explanation:

In KI solution, $I _2$ readily dissolves and forms $KI _3$ which contains polyhalide ion $I _3^-$


$KI + I _2  \rightarrow KI _3$

Due to this, iodine is partly soluble in water but highly soluble in the $KI$ solution.

So, the correct option is $D$

Correct order of reactivity of halogens is:

  1. $I _2 > Br _2 > Cl _2 > F _2$

  2. $Br _2 > I _2 > Cl _2 > F _2$

  3. $Cl _2 > Br _2 > I _2 > F _2$

  4. $F _2 > Cl _2 > Br _2 > I _2$


Correct Option: D
Explanation:

The reactivity of halogens decreases with increase in atomic number of halogens. So the correct order of reactivity of halogens is $F _2 > Cl _2 > Br _2 > I _2.$

Iodide of MilIon's base is :

  1. $HIO _3$

  2. $K _2Hgl _4$

  3. $NH _2HgO.HgI $

  4. $Hg(NH _2)I$


Correct Option: C
Explanation:

Iodide of MilIon's base is $\displaystyle NH _2HgO.HgI$
It is formed when Nessler's reagent (alkaline solution of $\displaystyle K _2[HgI _4]$) reacts with $NH _3$. 

Iodide of MilIon's base is a reddish brown ppt.