The p-Block Elements Test

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The p-Block Elements Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in $$[B(OH_4)]^-$$ and the geometry of the complex are respectively:

A
$$sp^3$$, tetrahedral

B
$$sp^3$$, square planar

C
$$sp^3d^2$$, octahedral

D
$$dsp^2$$, square planar

Solution

Boron has electronic configuration:
$$1s^22s^22p_x^1 2p^0_y 2p^0_z$$
In the excited state,2s orbital electrons are impaired and one electron is shifted to a p-orbital. Now, hybridisation occurs between one s-and three p-orbitals to give $$sp^3$$ hybridisation and tetrahedral geometry.
Question 2
1 / -0

The exhibition of the highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as a central atom in $$MF_6^{3-}$$?

A
$$B$$

B
$$Al$$

C
$$Ga$$

D
$$In$$

Solution

The element $$M$$ in the complex ion $${M{F}_{6}}^{3-}$$ has a coordination number of six. Since boron has only s and p-orbitals and no d-orbitals, therefore, boron is restricted to a maximum covalence of four (using 2s and three 2p orbitals). Thus, $$B$$ cannot form a complex of the type $${M{F}_{6}}^{3-}$$.
Question 3
1 / -0

Clay is used for making bricks and flower pots. If the clay article is not dried before 'firing' in a kiln, there will be evaporation of water from the surface and the :

A
kiln fire will be put out by steam

B
article will becomes soft

C
article will become hard

D
article will develop cracks

Solution

A kiln is a thermally insulated chamber, a type of oven, that produces temperatures sufficient to complete some process, such as hardening, drying, or chemical changes.
While manufacturing clay article if the clay is not dried before 'firing' in a kiln then the water content present on the surface will evaporate and articles will develop cracks on it which makes article non-durable.
Hence option D is correct answer.
Question 4
1 / -0

Elements of group 14 exhibit oxidation sate of:

A
+4 only

B
+2 and +4 only

C
+1 and +3 only

D
+2 only

Solution

The element of group 14 has 4 valence electrons. Therefore, the oxidation state of the group is +4. However, as a result of the inert pair effect, the lower oxidation state becomes more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.
Question 5
1 / -0

Which is not correct?

A
$$GeO$$ is acidic

B
$$Ge{ Cl }_{ 2 }$$ is more stable than $$Ge{ Cl }_{ 4 }$$

C
$$Ge{ Cl }_{ 2 }$$ is acidic

D
$$Ge{ Cl }_{ 4 }$$ in $$HCl$$ forms $${ \left[ Ge{ Cl }_{ 6 } \right] }^{ 2- }$$ ion
Solution
- The phenomena of the shielding effect of electrons explains this. Since there are fewer electrons in $$Ge ^{4+}$$, the nucleus has a stronger pull on those electrons than in $$Ge ^{+2}$$.
- Both $$Ge ^{4+}$$ and $$Ge ^{2+}$$ have the same stabilizing attractive force and $$Ge ^{2+}$$ has more electrons and thus more of the de-stabilizing repulsive force.So, $$Ge^{4+}$$ is more stable than $$Ge ^{2+}$$
- Hence option B is not correct.
Question 6
1 / -0

Element $$M+{ N }_{ 2 }\xrightarrow [ ]{ \Delta } \xrightarrow [ ]{ { H }_{ 2 }O } { NH }_{ 3 }$$
Element $$M$$ belonging to group $$13$$ can be ?

A
$$B$$ or $$Al$$

B
$$Ga$$ or $$Al$$

C
$$B$$ or $$Ga$$

D
$$In$$ or $$TI$$

Solution

$$2B + N_2 \longrightarrow 2BN$$
$$2Al +N_2 + \longrightarrow 2AlN$$
These 2 nitrites on heating with presence of $$H_2O$$ releases $$NH_3$$. So, the group $$13$$ elements are $$B$$ and $$Al$$
Question 7
1 / -0

Which of the following statements is correct with respect to the property of elements with increase in atomic number in the carbon family (group 14)?

A
Their metallic character decreases.

B
The stability of +2 oxidation state increases.

C
Their ionization energy increases.

D
Their atomic size decreases.

Solution

(B) The stability of +2 oxidation state increases.
Due to the increase in inert pair effect as we move down the group, s-orbital electrons stop participating in bonding in an atom which decreases the stability of +4 oxidation state and increases the stability of +2 oxidation state.
Question 8
1 / -0

Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding:

A
$$MeSiCl_3$$

B
$$Me_2SiCl_2$$

C
$$Me_3SiCl$$

D
$$Me_4Si$$

Solution

$${Me}_{3}SiCl$$
The chain length of Silicone polymer can be controlled by adding $${(C{H}_{3})}_{3}SiCl$$, which blocks the ends as shown below:
$${(C{H}_{3})}_{3}SiCl \xrightarrow[-HCl]{{H}_{2}O}{(C{H}_{3})}_{3}SiOH$$
Question 9
1 / -0

The element with atomic number $$33$$ belongs to:

A
group $$13$$

B
group $$14$$

C
group $$15$$

D
group $$16$$

Solution

Electronic configuration of atomic number 33 will be
$${1s}^22s^22p^63s^23p^64s^2{3d}^{10}4p^3$$
So, its configuration is similar to 15th group and it has 5 valence electrons .
Hence it belongs to 15th group of the periodic table.
Question 10
1 / -0

Assertion : The heavier p-block elements do not form strong $$\pi$$ bonds.
Reason : The heavier elements of p-block form $$d\pi-d\pi$$ bonds.

A
Both A and R are true and R is the correct explanation of A.

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true.

Solution

The presence of the d-orbitals influences the chemistry of the heavier elements in a number of other ways. The combined effect of size and availability of d orbitals considerably influences the ability of these elements to form π bonds.

The heavier elements do form π bonds but this involves d orbitals (dπ – pπ or dπ –dπ ). As the d orbitals are of higher energy than the p orbitals, they contribute less to the overall stability of molecules than does pπ - pπ bonding of the second row elements.

Option A is correct.

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The p-Block Elements Test - 46

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The p-Block Elements Test - 46

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Question 1

$$sp^3$$, tetrahedral

$$sp^3$$, square planar

$$sp^3d^2$$, octahedral

$$dsp^2$$, square planar

Solution

Question 2

$$B$$

$$Al$$

$$Ga$$

$$In$$

Solution

Question 3

kiln fire will be put out by steam

article will becomes soft

article will become hard

article will develop cracks

Solution

Question 4

+4 only

+2 and +4 only

+1 and +3 only

+2 only

Solution

Question 5

$$GeO$$ is acidic

$$Ge{ Cl }_{ 2 }$$ is more stable than $$Ge{ Cl }_{ 4 }$$

$$Ge{ Cl }_{ 2 }$$ is acidic

$$Ge{ Cl }_{ 4 }$$ in $$HCl$$ forms $${ \left[ Ge{ Cl }_{ 6 } \right] }^{ 2- }$$ ion

Solution

Question 6

$$B$$ or $$Al$$

$$Ga$$ or $$Al$$

$$B$$ or $$Ga$$

$$In$$ or $$TI$$

Solution

Question 7

Their metallic character decreases.

The stability of +2 oxidation state increases.

Their ionization energy increases.

Their atomic size decreases.

Solution

Question 8

$$MeSiCl_3$$

$$Me_2SiCl_2$$

$$Me_3SiCl$$

$$Me_4Si$$

Solution

Question 9

group $$13$$

group $$14$$

group $$15$$

group $$16$$

Solution

Question 10

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

A is true but R is false.

A is false but R is true.

Solution

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