The s-Block Elements Test 37

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The s-Block Elements Test 37

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Weekly Quiz Competition

Question 1
1 / -0

The metal oxides of alkaline earth metal can be formed by heating corresponding carbonates except:

A
$$BeO$$

B
$$MgO$$

C
$$CaO$$

D
none of these

Solution

Option $$(A)$$ is correct.
The metal oxides of alkaline earth metal can be formed by heating corresponding carbonates except $$BeO$$ due to small size of beryllium ion.$$BeO$$ is covalent and has polymeric nature.
Question 2
1 / -0

All alkaline earth metals combine directly with $$H_2$$ and form ionic hydrides as $$MH_2$$ except:

A
$$Mg$$

B
$$Ca$$

C
$$Be$$

D
$$Sr$$

Solution

Option $$(C)$$ is correct.
$$Be$$ does not combine directly with $$H_2$$ but $$BeH_2$$ is prepared from preformed beryllium(II) compounds and $$BeH_2$$ is covalently bonded.
All other alkaline earth metals react directly with hydrogen to produce the binary hydrides ($$MH_2$$). The hydrides of the heavier alkaline earth metals are ionic, but $$BeH_2$$ has polymeric structures that reflect covalent character.
Question 3
1 / -0

Which of the following arrangements is correct in respect to solubility in water?

A
$$CaSO_{4} > BaSO_{4} > BeSO_{4} > MgSO_{4} > SrSO_{4}$$

B
$$BeSO_{4} > MgSO_{4} > CaSO_{4} > SrSO_{4} > BaSO_{4}$$

C
$$BaSO_{4} > SrSO_{4} > CasO_{4} > MgSO_{4} > BeSO_{4}$$

D
$$BeSO_{4} > CaSO_{4} > MgSO_{4} > SrSO_{4} > BaSO_{4}$$

Solution

The correct order of solubility in water is $$BeSO_{4} > MgSO_{4} > CaSO_{4} > SrSO_{4} > BaSO_{4}$$

The increasing order of the ionic size is-
$$Be^{2+}<Mg^{2+}<Ca^{2+}<Sr^{2+}<Ba^{2+} $$.

The hydration energy decreases more rapidly than lattice energy, hence, solubility decreases down the group. (Hydration energy $$\propto \dfrac {charge}{size}$$).
Question 4
1 / -0

Oxidation number of calcium in calcium phosphate :

A
$$1^+$$

B
$$1^-$$

C
$$0$$

D
$$2^+$$

E
$$3^+$$

Solution

As $$Ca$$ is the Alkaline earth metal hence it has electrovalency to be +2, only. Hence, its oxidation state , can not change, from 2.
$$Ca_3(PO_4)_2$$:
As charge of phosphate $$(PO_4^{3-})$$ group is $$-2$$ , Let the oxidation state of $$Ca$$ be $$x$$. and whole unit is neutral.
$$3 \times x +2\times (-3) = 0 $$
$$3x -6 =0$$
$$3x = +6 $$
$$x = \dfrac {+6}{3} = +2$$
So , $$Ca$$ has $$+2$$ oxidation state.
Question 5
1 / -0

Beryllium differs in properties from other elements of its own group but shows resemblance with aluminium because of:

A
relatively bigger ionic radius and high polarising power of $$Be$$

B
relatively smaller ionic radius and high polarising power of $$Be$$

C
relatively bigger ionic radius is the only reason behind this

D
none of the above

Solution

Due to relatively smaller ionic radius and high polarising power, Be differs in properties from other elements of its own group. Polarising powers of $$Be^{2+}$$ and $$Al^{3+}$$. ions are almost the same, thus they show similarities in their properties.
Question 6
1 / -0

Alkaline earth's metals are denser than alkali metals because metallic bonding in alkaline earth's metal is:

A
stronger

B
weaker

C
volatile

D
none of the above

Solution

Alkaline earth metals ($${ns}^{2}$$) are denser than alkali metals ($${ns}^{1}$$) because metallic bonding in alkaline earth metals is stronger than alkali metals due to presence of two electrons in valence shell as compared to one electron in alkali metals.
Question 7
1 / -0

Berly is an important ore of:

A
$$Be$$

B
$$Ba$$

C
$$Mg$$

D
$$Sr$$

Solution

Option $$(A)$$ is correct.
Berly is an important ore of $$Be$$. It is composed of beryllium aluminium cyclosilicate with the chemical formula of $$Be_3Al_2Si_6O_{18}$$. Beryl is a mineral that contains a significant amount of beryllium. Beryllium is a very rare metal, and that limits the occurrence of Beryl to a few geological situations where beryllium is present in sufficient amounts to form minerals. It mainly occurs in igneous and metamorphic rocks
Question 8
1 / -0

The correct increasing order of solubility of the sulphates in water is:

A
$$BeSO_4>MgSO_4>CaSO_4>SrSO_4>BaSO_4$$

B
$$BaSO_4>SrSO_4>CaSO_4>MgSO_4>BeSO_4$$

C
$$BaSO_4>CaSO_4>SrSO_4>BeSO_4MgSO_4$$

D
$$BaSO_4>SrSO_4>MgSO_4>CaSO_4> BeSO_4$$

Solution

The solubility of the sulphates in water decreases down the group:

$$\underset { Soluble }{ \underbrace { BeSO_{ 4 } > MgSO_4 } } > \underset { Sparingly \quad soluble}{ \underbrace { CaSO_{ 4 } } } > \underset { Insoluble }{ \underbrace { SrSO_{ 4 } > BaSO_4 } } $$

The high solubilities of $$BeSO_4$$ and $$MgSO_4$$ are due to the high enthalpy of hydration of the smaller $$Be^{2+}$$ and $$Mg^{2+}$$ ions, which overcomes the lattice energy factor.

Hence, option $$A$$ is correct.
Question 9
1 / -0

Which of the following alkaline earth metal can never have more than four molecules of water of crystallisation?

A
$$Sr$$

B
$$Ca$$

C
$$Be$$

D
$$Mg$$

Solution

Option $$(C)$$ is correct.
$$Be$$ can never have more than four molecules of water of crystallisation as it has only four available orbitals in its valence shell. ($$Be = 2s^2 2p_x^02p_y^0 2p_z^0$$).
The other elements in II A group may have more than four molecules of water of crystallisation as they have $$d$$- orbitals as well.
Question 10
1 / -0

$$BeF_2$$ is soluble in water, whereas the fluorides of other alkaline earth metals are insoluble because of:

A
greater hydration energy of $$Be^{2+}$$ ion as compared to crystal lattice

B
covalent nature of $$ BeF_2$$

C
ionic nature of $$ BeF_2$$

D
none of these

Solution

Option $$(A)$$ is correct.
$$BeF_2$$ is soluble in water due to greater hydration energy of $$Be^{2+}$$ ion as compared to the crystal lattice.The other fluorides are
almost insoluble in water. Since on descending the group lattice energy decreases more rapidly than the hydration energy.

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The s-Block Elements Test 37

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The s-Block Elements Test 37

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Question 1

$$BeO$$

$$MgO$$

$$CaO$$

none of these

Solution

Question 2

$$Mg$$

$$Ca$$

$$Be$$

$$Sr$$

Solution

Question 3

$$CaSO_{4} > BaSO_{4} > BeSO_{4} > MgSO_{4} > SrSO_{4}$$

$$BeSO_{4} > MgSO_{4} > CaSO_{4} > SrSO_{4} > BaSO_{4}$$

$$BaSO_{4} > SrSO_{4} > CasO_{4} > MgSO_{4} > BeSO_{4}$$

$$BeSO_{4} > CaSO_{4} > MgSO_{4} > SrSO_{4} > BaSO_{4}$$

Solution

Question 4

$$1^+$$

$$1^-$$

$$0$$

$$2^+$$

$$3^+$$

Solution

Question 5

relatively bigger ionic radius and high polarising power of $$Be$$

relatively smaller ionic radius and high polarising power of $$Be$$

relatively bigger ionic radius is the only reason behind this

none of the above

Solution

Question 6

stronger

weaker

volatile

none of the above

Solution

Question 7

$$Be$$

$$Ba$$

$$Mg$$

$$Sr$$

Solution

Question 8

$$BeSO_4>MgSO_4>CaSO_4>SrSO_4>BaSO_4$$

$$BaSO_4>SrSO_4>CaSO_4>MgSO_4>BeSO_4$$

$$BaSO_4>CaSO_4>SrSO_4>BeSO_4MgSO_4$$

$$BaSO_4>SrSO_4>MgSO_4>CaSO_4> BeSO_4$$

Solution

Question 9

$$Sr$$

$$Ca$$

$$Be$$

$$Mg$$

Solution

Question 10

greater hydration energy of $$Be^{2+}$$ ion as compared to crystal lattice

covalent nature of $$ BeF_2$$

ionic nature of $$ BeF_2$$

none of these

Solution

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