Tag: group 17 elements

Questions Related to group 17 elements

Select the correct statement (s)

  1. HCI and HF do not from free radical easily due to larger bond-energy

  2. HI with a minimum bond energy is every reactive and instead from ${{\text{I}} _{\text{2}}}\,\,\,$

  3. Both (a) and (b)

  4. None of the above


Correct Option: C
Explanation:

Bond dissociation energy with hydrogen decreased down the group so HI is will be the most reactive . HI is more reative because the bond dissociation energy is low due to difference in size of hydrogen and iodine so it can be easily breaked.

Addition of halogen $(C{l _2} or B{r _2})$ to alkenes goes through formation of :

  1. Carbocation

  2. Carbanion

  3. Cyclic halonium ion

  4. Carbon free radical


Correct Option: C

In which of the following, oxidation number of chlorine is $ + 5 ?$

  1. $Cl _{2}O _{7}$

  2. $\mathrm { ClO } _ { 3 } ^ { - }$

  3. $\mathrm { CHO } ^ { - }$

  4. $\mathrm { ClO } _ { 4 } ^ { - }$


Correct Option: B
Explanation:

Oxidation number of $Cl _2O _7$:-

   $2x-14=0$
$x=7$

Oxidation number of ${ClO _3}^{-}$:-
$x-6=-1$
$x=5$
Oxidation number of $ClO^-$:-
$x-2=-1$
$x=1$
Oxidation number of ${ClO _4}^-$:-

$x-8=-1$
$x=7$
hence option B is correct.

Which of the following has highest chlorine content ?

  1. Pyrene

  2. DDT

  3. Chloral

  4. Gammaxene


Correct Option: A

Compare oxygen and flourine in Ionic radius$\quad O^{2-} , F^-$:

  1. $\space O^{2-} = F^-$

  2. $\space O^{2-} < F^-$

  3. $\space O^{2-}> F^-$

  4. $\space O^{2-} \geq F^-$


Correct Option: C
Explanation:

Among the elements of same period the ion with more number of electrons and less number of protons is having more radius.If the number of electrons are same the ion with less number of protons is having more radius.

Here $\quad O^{2-} $and $ F^-$ both are having same 10 electrons but oxygen is having 8 protons(which can atrract outer electrons with less force than florine) and florine is having 9 protons.So the ionic radius of $\space O^{2-}>  F^-$

Hence option C is correct.

Oxidation of hydrogen halide, HX affords a method for the industrial and laboratory preparation of the halogen, $X _2$, in the free state in respect of all of the following except :

  1. fluorine.

  2. chlorine.

  3. bromine.

  4. iodine.


Correct Option: A
Explanation:

Due to higher electronegativity, HF doesn't follow such reaction and fluorine doesn't form.

Which f the following atoms has the highest ionisation energy ?

  1. F

  2. Cl

  3. Br

  4. I


Correct Option: A
Explanation:

Halogens have little tendency to lose electron. Thus they have very high ionisation enthalpy. Due to increase in atomic size, ionisation enthalpy decreases down the group.

Which of the following is incorrect statement? 

  1. Bond dissociation energy of Bromine is more than that of chlorine.

  2. The standard electrode potential of chlorine is more than that of fluorine.

  3. The percentage of chlorine in bleaching powder compound is 35-38%.

  4. All


Correct Option: D
Explanation:

Bond dissociation:
$Cl _{2} > Br _{2}$
Standard electrode potential:
$F>Cl$
Bleaching powder $(Ca(ClO) _{2})$             

Percentage of $Cl = \frac{35.5\times 2}{(35.5+16)\times 2 +40}$
$=\frac{71}{143}\times 100 \simeq 50$%

The correct order of reactivity is:

  1. $\mathrm{F}>\mathrm{Cl}>\mathrm{B}\mathrm{r}> \mathrm{I} $

  2. $\mathrm{F}<\mathrm{Cl}<\mathrm{B}\mathrm{r}<\mathrm{I}$

  3. $\mathrm{F}<\mathrm{Cl}>\mathrm{B}\mathrm{r}>\mathrm{I}$

  4. $\mathrm{F}>\mathrm{Cl}<\mathrm{B}\mathrm{r}<\mathrm{I}$


Correct Option: A
Explanation:

As we move down the group chemical reactivity decreases due to decrease in electronegativity

The outer most electronic configuration of the most electronegative element is:

  1. $\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{3}$

  2. $\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{4}$

  3. $\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{5}$

  4. $\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{6}$


Correct Option: C
Explanation:

Halogens are most electronegative.


Therefore outer electronic configuration is $ns^{2}\ np^{5}$

Option C is correct.