S Block Elements(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

S BLOCK ELEMENTS MCQs

Total Questions : 30 | Page 1 of 3 pages

The correct statements among the following are:
I. Due to greater polarizing power of $L i^{+}$ , $L i_{2} C O_{3}$ is most stable
II. $L i H C O_{3}$ does not exist as solid
III. $N a N O_{3}$ on strong heating gives $N a N O_{2}$ whereas $L i N O_{3}$ will give $L i_{2} O$
IV. Sodium & Potassium ions are involved in the transport of sugars and amino acids into the cells
V. $L i F$ is less soluble due to smaller hydration energy of $L i^{+}$ ion

I, II & V
III, IV & V
II, III & IV
I, III & V

Discuss Question

Answer: Option C. -> II, III & IV
:
C

$L i_{2} C O_{3}$ is unstable and $L i^{+}$ has high hydration energy.

Question 2.

The correct order of thermal stability of alkali metal hydrides is:

$C s H > R b H > K H > N a H > L i H$
$L i H > N a H > K H > R b H > C s H$
$C s H > N a H > K H > R b H > L i H$
$L i H > K H > N a H > R b H > C s H$

Discuss Question

Answer: Option B. ->

L i H > N a H > K H > R b H > C s H

:
B

The stability of alkali metal hydrides decreases with increase in the size of the metal cation. In other words, as we go down the group - the stability of the hydrides decreases.

Question 3.

The reaction of sodium is highly exothermic with H₂O. The rate of reaction is lowered by

Lowering the temperature
Mixing with alcohol
Mixing with acetic acid
Making an amalgam

Discuss Question

Answer: Option D. -> Making an amalgam
:
D

Sodium amalgam, commonly denoted Na(Hg), is an alloy of mercury and sodium.
Na(Hg) is less reactive than Na. It will decrease the rate of reaction.
Sodium is highly reactive with water and decreasing the temperature does not decrease the rate of the reaction if we consider Le Chatelier's principle.
Sodium reacts with alcohol and acetic acid also. They will not affect the rate of reaction of sodium with water.

Question 4.

Statement-1 : Alkali metals dissolve in liquid ammonia to give blue solutions.

Statement-2 : Alkali metals in liquid ammonia give solvated species of the type $[M (N H_{3})_{n}]^{+}$ (M =alkali metals).

Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

Discuss Question

Answer: Option B. -> Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
:
B

Alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature. The blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light. The second statement describes the solvated cations (positively charged species) whereas the colour is due to the solvated electrons. Do note however that both the statements are accurate.

Question 5.

$N a H$ reacts with water to give:

Alkaline solution
$N H_{3}$ gas
Acidic solution
Neutral solution

Discuss Question

Answer: Option A. -> Alkaline solution
:
A

$N a H + H_{2} O \to N a O H + H_{2} (g)$
As you can see - we get $N a O H$ which is very strongly basic.

Question 6.

Look at the following reaction:
$L i H + A l H_{3} ⟶ L i A l H_{4}$
Here $A l H_{3}$ and $L i H$ act as:

Lewis acid and Lewis base
Lewis base and Lewis acid
Bronsted base and Bronsted acid
None of these

Discuss Question

Answer: Option A. -> Lewis acid and Lewis base
:
A

As shown in the figure, the central metal atom in $A l H_{3}$ has only six valence electrons. Hence, it could accept a pair of electrons (Lewis acid) to complete its octet. On the other hand, $H^{-}$ is a Lewis base because it has two valence electrons thus having a completely filled shell. Further, since the ion readily shares or donates electrons - it acts as a Lewis base.

Question 7.

The ionic radii of alkali metal ions in water (hydrated radii) are in the order

$L i^{+} > N a^{+} > K^{+} > R b^{+} < C s^{+}$
$L i^{+} (a q .) > N a^{+} (a q .) > K^{+} (a q .) > R b^{+} (a q .) > C s^{+} (a q .)$
$L i^{+} < N a^{+} > K^{+} > R b^{+} > C s^{+}$
$L i^{+} (a q .) > N a^{+} (a q .) < K^{+} (a q .) < R b^{+} (a q .) < C s^{+} (a q .)$

Discuss Question

Answer: Option B. ->

L i^{+} (a q .) > N a^{+} (a q .) > K^{+} (a q .) > R b^{+} (a q .) > C s^{+} (a q .)

:
B

If we just look at only the ionic radii - then clearly $L i^{+}$ should be the smallest and common sense might compel us to think that the small ion should exhibit the best conductivity. However, among alkali metal cations - $L i^{+}$ has the highest charge density and is thus hydrated to the largest extent. Thus $L i^{+}$ has the highest hydrated ion radius and least conductivity in water.

Question 8.

Cs⁺ ions impart violet colour to Bunsen flame. This is due to the fact that the emitted radiations are of

High energy
Lower frequencies
Longer wavelength
Zero wave number

Discuss Question

Answer: Option A. -> High energy
:
A

Violet belongs to the high-energy (low wavelength) section in the visible part of the Electromagnetic spectrum.

Question 9.

The correct order of density of alkali metals is:

Discuss Question

Answer: Option A. ->

:
A

As we go down a group - density increases generally. However, there is a notable exception in potassium, which is less dense than sodium. Simply put, in the case of potassium the increase in shell size outweighs the pull of the core on the outer shell electron and so potassium is less dense than sodium.

Question 10.

Paramagnetic, paramagnetic
Paramagnetic, diamagnetic
Diamagnetic, paramagnetic
Diamagnetic ,diamagnetic

Discuss Question

Answer: Option A. -> Paramagnetic, paramagnetic
:
A

X is NO₂& Y is O₂ . Both the species have unpaired electrons in their electronic configurations (For $O_{2}$ you will have to use MOT).
Due to the presence of unpaired electrons, both $N O_{2}$ and $O_{2}$ are paramagnetic.