12th Grade > Chemistry
S BLOCK ELEMENTS MCQs
Total Questions : 30
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Answer: Option B. -> Common ion effect and high solubility of MgCl2 and CaCl2 when compared to NaCl
:
B
As Cl−concentration increases precipitation takes place.
:
B
As Cl−concentration increases precipitation takes place.
Answer: Option B. -> Li+(aq.)>Na+(aq.)>K+(aq.)>Rb+(aq.)>Cs+(aq.)
:
B
If we just look at only the ionic radii - then clearly Li+ 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 - Li+ has the highest charge density and is thus hydrated to the largest extent. Thus Li+ has the highest hydrated ion radius and least conductivity in water.
:
B
If we just look at only the ionic radii - then clearly Li+ 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 - Li+ has the highest charge density and is thus hydrated to the largest extent. Thus Li+ has the highest hydrated ion radius and least conductivity in water.
Answer: Option B. -> I, II, IV correct
:
B
Options I, II and IV are all very accurate. These are all characteristics of the corresponding compounds. As for option III - the compounds LiCl and MgCl2 both form hydrates. Only - LiCl forms LiCl.2H2O whereas MgCl2 forms MgCl2.8H2O
:
B
Options I, II and IV are all very accurate. These are all characteristics of the corresponding compounds. As for option III - the compounds LiCl and MgCl2 both form hydrates. Only - LiCl forms LiCl.2H2O whereas MgCl2 forms MgCl2.8H2O
Answer: Option B. -> Li2O Na2O2 KO2
:
B
Li forms monoxide, Na forms peroxide, K forms super oxide on burning with oxygen
:
B
Li forms monoxide, Na forms peroxide, K forms super oxide on burning with oxygen
Question 15. The correct statements among the following are:
I. Due to greater polarizing power of Li+, Li2CO3 is most stable
II. LiHCO3 does not exist as solid
III. NaNO3 on strong heating gives NaNO2 whereas LiNO3 will give Li2O
IV. Sodium & Potassium ions are involved in the transport of sugars and amino acids into the cells
V. LiF is less soluble due to smaller hydration energy of Li+ ion
I. Due to greater polarizing power of Li+, Li2CO3 is most stable
II. LiHCO3 does not exist as solid
III. NaNO3 on strong heating gives NaNO2 whereas LiNO3 will give Li2O
IV. Sodium & Potassium ions are involved in the transport of sugars and amino acids into the cells
V. LiF is less soluble due to smaller hydration energy of Li+ ion
Answer: Option C. -> II, III & IV
:
C
Li2CO3 is unstable and Li+ has high hydration energy.
:
C
Li2CO3 is unstable and Li+ has high hydration energy.
Answer: Option A. -> High energy
:
A
Violet belongs to the high-energy (low wavelength) section in the visible part of the Electromagnetic spectrum.
:
A
Violet belongs to the high-energy (low wavelength) section in the visible part of the Electromagnetic spectrum.
Answer: Option D. -> Solvated electrons
:
D
The (strongly) reducing nature is due to solvated or ammoniated electrons.
:
D
The (strongly) reducing nature is due to solvated or ammoniated electrons.
Answer: Option B. -> A and R are true but the R is not the correct explanation of A
:
B
Sodium is stored under kerosene because it does not react with kerosene. On the other hand, Sodium violently reacts with water.
Taken one by one - both the assertion and reason are true but the reason does not explainthe assertion.
:
B
Sodium is stored under kerosene because it does not react with kerosene. On the other hand, Sodium violently reacts with water.
Taken one by one - both the assertion and reason are true but the reason does not explainthe assertion.
Answer: Option C. -> CaCO3 + NaCl
:
C
The overall reaction for the Solvay or Ammonia-soda processis:
2NaCl+CaCO3⟶Na2CO3+CaCl2
:
C
The overall reaction for the Solvay or Ammonia-soda processis:
2NaCl+CaCO3⟶Na2CO3+CaCl2
Answer: Option B. -> Na2CO3.H2O
:
B
Na2CO3.10H2O is efflorescent and looses 9 water molecules on exposure to air to give Na2CO3.H2O.
:
B
Na2CO3.10H2O is efflorescent and looses 9 water molecules on exposure to air to give Na2CO3.H2O.