The p-Block Elements Test

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The p-Block Elements Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Boron exists in two isotopic form $$B^{10}$$ and $$B^{11}$$. If the average atomic weight of $$B$$ is $$10.20\ u$$, select the correct statement.

A
$$\% $$ abundance of $$B^{10}=\%$$ abundance of $$B^{11}$$

B
$$\% $$ abundance of $$B^{10} > \%$$ abundance of $$B^{11}$$

C
$$\% $$ abundance of $$B^{10} < \%$$ abundance of $$B^{11}$$

D
$$\% $$ abundance of $$B^{10}=\dfrac{\%\ abundance\ of\ B^{11}}{2}$$

Solution

Let the percentage existence of $$B^{10}$$ isotope be $$x$$ . Therefore percentage existence of the $$B^{11}$$ isotope is $$(100-x)$$
mass of $$B^{10}$$ isotope is $$10amu$$.
mass of $$B^{11}$$ isotope is $$11 amu$$.

Therefore,

$$\dfrac{10+11(100-x)}{100} = 10.2$$

$$10+11(100-x) = 1020$$

$$x=80

Therefore the percentage of $$Boron-10$$ isotope is $$80$$%
and $$Boron-11$$ isotope is $$(100-x)=20$$%
Hence, option B is the correct answer.
Question 2
1 / -0

The element with lowest electronegativity in 13th group is:

A
B

B
Al

C
Ga

D
In

Solution

Down the $$13^{th}$$ group, electronegativity first decreases from B to Al and then increases marginally. This is because of the discrepancies in the atomic size of the elements.

Hence option B is the correct answer.
Question 3
1 / -0

The electropositive character is the maximum for which of the following elements in group 13?

A
$$Al$$

B
B

C
Ga

D
In

Solution

Electro positive character of elements first increase from boron to aluminium and then decrease from aluminium to Thalium.
So, Aluminium has highest electro positive character in group B.
Question 4
1 / -0

Boric acid is used in carom boads for smooth gliding of pawns because

A
$$H_3BO_3$$ molecules are loosely chemically bonded and hence soft

B
its low density makes it fluffy

C
it can be powdered to a very small grain size

D
it is chemically inert with the plywood

E
H-bonding in $$H_3BO_3$$ GIVES IT A LAYERED STRUCTURE.

Solution

Boric acid is a solid lubricant and is used to reduce friction. It is used as a lubricant to make motion smooter. In carrom boards, boric acid is used for smooth gliding of porous because $$H_3BO_3$$ molecules are loosely chemically bounded and hence soft and thus reduce friction.
Question 5
1 / -0

Teflon polymer is formed by the polymerisation of:

A
$$CH_2CN-CN$$

B
$$F_2C=CF_2$$

C
$$Cl_2C=CH_2$$

D
$$H_2C=CHCl$$

Solution

Teflon is a thermally stable polymer of tetrafluoro ethylene (F2C = CF2).
Question 6
1 / -0

Which of the following compounds of elements in group IV expected to be most ionic?

A
$$PbCl_{2}$$

B
$$PbCl_{4}$$

C
$$CCl_{4}$$

D
$$SiCl_{4}$$

Solution

According to fajans rule greater the charge on cation,greater would be the polarising power and greater covalent character.
$$PbCl_2\rightarrow Pb^{2+}$$
$$PbCl_4\rightarrow Pb^{+4}$$
$$PbCl_2$$ is ionic and $$PbCl_4$$ in covalent.
Similarly $$CCl_4,SiCl_4$$ are also covalent.
Question 7
1 / -0

The electropositive character is the maximum for which of the following elements in group 13?

A
$$Al$$

B
$$Ti$$

C
$$B$$

D
$$Tl$$

Solution

Amongst the elements of group 13, the sum of the first three ionization enthalpy for Boron is the highest and hence it is the least electropositive element in the group.
As we move down, due to increase in atomic size, the sum of the ionization enthalpies decreases. Hence the electropositive nature increases from Boron to Aluminium.
However, going down further increasesthe ionization enthalpies due to poor shielding of the inner d and f orbittals. Hence electropositive nature again decreases from Aluminium to Thallium. Therefore, Aluminium is the most electropositive element in group 13
Question 8
1 / -0

Select the incorrect statement among the following:

A
$${BF}_{3}$$ and $$B{Cl}_{3}$$ both have equal bond angles

B
$$Xe{O}_{2}{F}_{2}$$ has two $$p\pi-p\pi$$ bonds

C
$${SF}_{6}$$ has all equal bond lengths

D
$${CH}_{2}{Cl}_{2}$$ is less polar than $${CH}_{3}Cl$$

Solution

1) Lesser the electronegativity of the attached atoms more is the bond angle.
2) Size of attached atoms (eg $$Br, \; Cl$$ or $$F$$ in this case)
The bond order will be-
$$BBr_3>BCl_3>BF_3$$
Question 9
1 / -0

The electropositive character is the maximum for which of the following elements in group 13?

A
$$Al$$

B
$$In$$

C
$$B$$

D
$$Ga$$

Solution

Amongst the elements of group 13, the sum of the first three ionization enthalpy for Boron is the highest and hence it is the least electropositive element in the group. As we move down, due to an increase in atomic size, the sum of the ionization enthalpies decreases. Hence the electropositive nature increases from Boron to Aluminium. However, going down further increases the ionization enthalpies due to poor shielding of the inner d and f orbitals. Hence electropositive nature again decreases from Aluminium to Thallium. Therefore, Aluminium is the most electropositive element in group 13
Question 10
1 / -0

Aluminium does not react with:

A
NaOH

B
HCI

C
$$N_2$$

D
$$HNO_3$$

Solution

Before reaction with $$HNO_3$$ it forms an oxide layer of $$Al_2O_3$$.

oxide layer does not allow reaction of $$HNO_3$$ with Aluminium.

Because of this reason $$Al$$ vessel is used to store $$HNO_3. HNO_3$$ does'nt react with $$Al$$ Or any other metal (except manganese and magnesium) because $$HNO_3$$ is a strong oxidising agent.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 48

Result

The p-Block Elements Test - 48

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Question 1

$$\% $$ abundance of $$B^{10}=\%$$ abundance of $$B^{11}$$

$$\% $$ abundance of $$B^{10} > \%$$ abundance of $$B^{11}$$

$$\% $$ abundance of $$B^{10} < \%$$ abundance of $$B^{11}$$

$$\% $$ abundance of $$B^{10}=\dfrac{\%\ abundance\ of\ B^{11}}{2}$$

Solution

Question 2

B

Al

Ga

In

Solution

Question 3

$$Al$$

B

Ga

In

Solution

Question 4

$$H_3BO_3$$ molecules are loosely chemically bonded and hence soft

its low density makes it fluffy

it can be powdered to a very small grain size

it is chemically inert with the plywood

H-bonding in $$H_3BO_3$$ GIVES IT A LAYERED STRUCTURE.

Solution

Question 5

$$CH_2CN-CN$$

$$F_2C=CF_2$$

$$Cl_2C=CH_2$$

$$H_2C=CHCl$$

Solution

Question 6

$$PbCl_{2}$$

$$PbCl_{4}$$

$$CCl_{4}$$

$$SiCl_{4}$$

Solution

Question 7

$$Al$$

$$Ti$$

$$B$$

$$Tl$$

Solution

Question 8

$${BF}_{3}$$ and $$B{Cl}_{3}$$ both have equal bond angles

$$Xe{O}_{2}{F}_{2}$$ has two $$p\pi-p\pi$$ bonds

$${SF}_{6}$$ has all equal bond lengths

$${CH}_{2}{Cl}_{2}$$ is less polar than $${CH}_{3}Cl$$

Solution

Question 9

$$Al$$

$$In$$

$$B$$

$$Ga$$

Solution

Question 10

NaOH

HCI

$$N_2$$

$$HNO_3$$

Solution

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