Tag: p- block elements-ii

Questions Related to p- block elements-ii

Which of the following can produce ammonia from ammonium salts?

  1. Quick lime

  2. Caustic soda

  3. slaked lime

  4. All


Correct Option: B,C
Explanation:

Quick lime is $CaO$,caustic soda is$NaOH$, slaked lime is $Ca(OH) _2$.

$NH _4Cl+NaOH \rightarrow NH _3+H _2O+NaCl$
$NH _4Cl+Ca(OH) _2 \rightarrow NH _3+H _2O+CaCl _2$


Haber's process is a/an:

  1. endothermic process

  2. exothermic process

  3. spontaneous with no heat liberation

  4. None


Correct Option: B
Explanation:

Haber's process is an exothermic process.

$N _2+3H _2 \rightarrow 2NH _3$

Urea on reacting with water gives:

  1. $NH _3$

  2. $N _2O _3$

  3. $HNO _2$

  4. $HNO _3$


Correct Option: A
Explanation:

$NH _2CONH _2+H _2O \rightarrow 2NH _3+CO _2$

Urea on reaction with water gives carbondioxide and ammonia.

The complex formed in Brown Ring Test is ___________.

  1. $[Fe(H _2O) _4(NO) _2]^{2+}$

  2. $[Fe(H _2O) _5(NO)]^{2+}$

  3. $[Fe(H _2O) _3(NO) _3]^{2+}$

  4. $[Fe(H _2O) _6(NO)]^{2+}$


Correct Option: B
Explanation:

$[Fe(H _2O) _5(NO)]SO _4$ is the complex formed in brown ring test done to identify the presence of nirates.

Which has a lowest electron affinity among the following?

  1. I

  2. Br

  3. Cl

  4. F


Correct Option: A
Explanation:

Down the group $e^-$ affinity decreases due to increase in size. Here $I$ having large size has least $e^-$ affinity.

The electronic configuration of four atoms are given in brackets:
L : ($1s^2\,2s^2\,2p^1$);                    M : ($1s^2\,2s^2\,2p^5$);

Q : ($1s^2\,2s^2\,2p^6\,3s^1$);             R : ($1s^2\,2s^2\,2p^2$).

The element that would most readily form a diatomic molecule is :

  1. Q

  2. M

  3. R

  4. L


Correct Option: B
Explanation:

From the electronic configuration given it is clear that element M is 1 short of octet configuration. It belongs to group 17. Group 17 elements exist as diatomic molecules.


So, correct option is B.

Which is the correct arrangement of the compounds based on their bond strength?

  1. HF > HCl > HBr > HI

  2. HI>HBr>HCl>HF

  3. HCl>HF>HBr>HI

  4. HF>HBr>HCl>HI


Correct Option: A
Explanation:

Because electronegativity decreases down the group thereby $e^-$ attracting tendency decreases down the group. $F$ with high electronegativity will have strongest bond and $I$ with least electronegativity will have least bond strength.

Mark the correct statements about halogens.

  1. Electron affinity of halogens is in the order F > Cl > Br > I.

  2. HF is the strongest hydrohalic acid.

  3. $F _{2}$ has lower bond dissociation energy than $Cl _{2}$.

  4. All halogens show variable oxidation states.


Correct Option: C
Explanation:

$F _2$ has lower bond dissociation energy than $Cl _2$ due to its small size and high electronegativity.

The property of halogens which is not correctly matched is:

  1. F>Cl>Br>I (Ionisation energy)

  2. F>Cl>Br>I (Electronegativity)

  3. I>Br>Cl>F (Density)

  4. F>Cl>Br>I (Electron affinity)


Correct Option: D
Explanation:

$e^-$ affinity of $F$ is less than that of $Cl$ because of the small size and high electronegativity of $F$ it repels the incoming $e^-$ thereby decreasing its $e^-$ affinity whereas $Cl$ having bigger size than $F$ & can accommodate the new incoming extra $e^-$.

Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy? 

  1. $HF$

  2. $HCl$

  3. $HBr$

  4. $HI$


Correct Option: A
Explanation:
As the size of the halogen atom decreases the bond length between halogen and hydrogen decreases ,So the bond becomes stronger and hence the energy required to break the bond ( bond dissociation enthaphy) increases.

So the compound with highest bond dissociation energy is $HF$

Hence option A is correct.