The s-Block Elements Test 24

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The s-Block Elements Test 24

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Weekly Quiz Competition

Question 1
1 / -0

Which is the following is not true?

A
Group $$2$$ elements are electropositive strong reducing agents but not as strong as a group $$1$$ elements

B
the reducing power of group $$2$$ elements increases down the group

C
$$Be$$ has the most negative standard reduction potential

D
The magnesium cation is more easily reduced then the heavier members of the group

Solution

Li has the most negative standard reduction potential
Question 2
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The lattice energy of the carbonates of Mg, Ca, Sr and Ba follow the order :

A
$$Mg{CO}_3$$ > $$Ca{CO}_3$$ > $$Sr{CO}_3$$ > $$Ba{CO}_3$$

B
$$Ca{CO}_3$$ > $$Sr{CO}_3$$ > $$Ba{CO}_3$$ > $$Mg{CO}_3$$

C
$$Ba{CO}_3$$ > $$Sr{CO}_3$$ > $$Ca{CO}_3$$ > $$Mg{CO}_3$$

D
$$Sr{CO}_3$$ > $$Mg{CO}_3$$ > $$Ca{CO}_3$$ > $$Ba{CO}_3$$

Solution

Order of lattice energy of carbonates is
$$MgCO_3>CaCO_3>SrCO_3>BaCO_3$$
This is because as the ionic radius increases, the lattice energy decreases.
Question 3
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Which among these does not represent the correct order?

A
$$MgO> Mg{Cl}_{2}> NaCl$$ (Lattice energy)

B
$$Ca{F}_{2}> Sr{F}_{2}> B{F}_{2}$$ (Solubility in water)

C
$$LiF> NaF> KF$$ (solubility in water)

D
Both b and c

Solution
Question 4
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Which of the following shows least melting point?

A
$$Be$$

B
$$Mg$$

C
$$Ca$$

D
$$Sr$$

Solution

Ans : $$B$$ $$\rightarrow$$ $$Mg$$
Generally melting point of the metals decreases down the group. But $$Mg$$ has anomalously low melting point in group $$2$$.
Group $$2$$ metals have metallic bonding. If we look at the crystal structure, $$Be$$ & $$Mg$$ are hexagonal close packed, $$Ca$$ & $$Sr$$ are face centered cubic.
At melting point, $$Be$$, $$Ca$$ and $$Sr$$ are body centered where as $$Mg$$ is hexagonal closed packed. Due to this arrangement, $$Mg$$ shows least melting point
Question 5
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Which of the following compound is not soluble in water?

A
LiF

B
NaCl

C
NaBr

D
LiBr

Solution

Due to its low hydration energy and partial covalent and partial ionic character LiCl is soluble in water as well as acetone. In Lithium fluoride the lattice enthalpy is very high due to small size of fluoride ions. In this case the hydration enthalpy is very less. Hence, LiF is insoluble in water.
Question 6
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The correct order of increasing thermal stability of $$K_2CO_3,MgCO_3 \, CaCO_3$$ and $$BeCO_3$$ is:

A
$$BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$$

B
$$MgCO_3 < BeCO_3 < CaCO_3 < K_2CO_3$$

C
$$K_2CO_3 < MgCO_3 < CaCO_3 < BeCO_3$$

D
$$BeCO_3 < MgCO_3 < K_2CO_3 < .CaCO_3$$
Solution
- In all cases, for a particular set of e.g. Group 1 or Group 2 compounds, the thermal stability increases down the group as the ionic radius of the cation increases, and its polarising power decreases.
- Group 1 compounds tend to be more thermally stable than group 2 compounds because the cation has a smaller charge and a larger ionic radius, and so a lower polarising power. particularly when adjacent metals on the same period are compared.
Hence, the order of increasing thermal stability is $$BeCO_3<MgCO_3<CaCO_3<K_2CO_3$$
Question 7
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Which of the following cannot react with $$NaOH$$?

A
$$Be$$

B
$$Fe$$

C
$$Al$$

D
$$Zn$$

Solution

Be is an alkaline earth metal.
It has only 2 valance electrons that is, it has fully filled orbitals.
It forms mostly covalent compounds.
Question 8
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Which of the following compounds on thermal decomposition yields a basic as well as an acidic oxide?

A
$$KClO_3$$

B
$$NaNO_3$$

C
$$K_2CO_3$$

D
$$MgCO_3$$

Solution

$$MgCO_3\xrightarrow{\Delta} MgO+CO_2$$
Since oxides of metals are basic in nature, $$MgO$$ is basic, and $$CO_2$$ is acidic. Hence decoposition of $$MgCO_3$$ yields both acidic and basic oxides
Question 9
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Beryllium and aluminium show many same characters but in which way these differ to each other:

A
Show maximum covalency in their compounds

B
Nature of oxides are amphoteric

C
Form covalent halide

D
Form polymeric hydride
Solution
- Due to diagonal relationship Be and Al exhibit many similar properties but covalency of Be is 4 because it has an absence of d-orbital whereas, covalency of Al is up to 6 due to the presence of vacant d-orbital.
- Hence option A is correct answer.
Question 10
1 / -0

Which one of the following chlorides has maximum tendency to form hydrate?

A
$$BaCl_2$$

B
$$NaCl$$

C
$$MgCl_2$$

D
$$LiCl$$

Solution

A correct option would be Lithium.
Lithium is the smallest in size amongst all. Hence $$Li^+$$ ion polarises water molecule easily than the others. Water molecules get attached to $$Li^+$$ as the water of crystallization. The common example is ($$LiCl.2H_2O$$). As the size increases the polarising power of cation decreases, therefore, higher member of the group generally form anhydrous salt.

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The s-Block Elements Test 24

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Question 1

Group $$2$$ elements are electropositive strong reducing agents but not as strong as a group $$1$$ elements

the reducing power of group $$2$$ elements increases down the group

$$Be$$ has the most negative standard reduction potential

The magnesium cation is more easily reduced then the heavier members of the group

Solution

Question 2

$$Mg{CO}_3$$ > $$Ca{CO}_3$$ > $$Sr{CO}_3$$ > $$Ba{CO}_3$$

$$Ca{CO}_3$$ > $$Sr{CO}_3$$ > $$Ba{CO}_3$$ > $$Mg{CO}_3$$

$$Ba{CO}_3$$ > $$Sr{CO}_3$$ > $$Ca{CO}_3$$ > $$Mg{CO}_3$$

$$Sr{CO}_3$$ > $$Mg{CO}_3$$ > $$Ca{CO}_3$$ > $$Ba{CO}_3$$

Solution

Question 3

$$MgO> Mg{Cl}_{2}> NaCl$$ (Lattice energy)

$$Ca{F}_{2}> Sr{F}_{2}> B{F}_{2}$$ (Solubility in water)

$$LiF> NaF> KF$$ (solubility in water)

Both b and c

Solution

Question 4

$$Be$$

$$Mg$$

$$Ca$$

$$Sr$$

Solution

Question 5

LiF

NaCl

NaBr

LiBr

Solution

Question 6

$$BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$$

$$MgCO_3 < BeCO_3 < CaCO_3 < K_2CO_3$$

$$K_2CO_3 < MgCO_3 < CaCO_3 < BeCO_3$$

$$BeCO_3 < MgCO_3 < K_2CO_3 < .CaCO_3$$

Solution

Question 7

$$Be$$

$$Fe$$

$$Al$$

$$Zn$$

Solution

Question 8

$$KClO_3$$

$$NaNO_3$$

$$K_2CO_3$$

$$MgCO_3$$

Solution

Question 9

Show maximum covalency in their compounds

Nature of oxides are amphoteric

Form covalent halide

Form polymeric hydride

Solution

Question 10

$$BaCl_2$$

$$NaCl$$

$$MgCl_2$$

$$LiCl$$

Solution

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