The p-Block Elements Test

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The p-Block Elements Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The electronic configuration of few elements is given below. Mark the statement which is not correct about these elements.
(i) $$1s^{2} 2s^{2} 2p^{6} 3s^{1}$$
(ii) $$1s^{2} 2s^{2} 2p^{5}$$
(iii) $$1s^{2}2s^{2}2p^{6}$$
(iv) $$1s^{2}2s^{2}2p^{3}$$

A
(i) is an alkali metal.

B
(iii) is a noble metal.

C
(i) and (ii) form ionic compounds.

D
(iv) has high ionisation enthalpy.

Solution

(i) is an alkali metal. It readily loses an electron to attain a noble gas configuration.
(iii) is a noble gas. It has completed its octet.
Highly electropositive (i) and highly electronegative (ii) form ionic compounds.
(iv) has high ionization enthalpy as removal of an electron will break half-filled 2p subshell.
Question 2
1 / -0

The electronic configuration of four atoms are given in brackets:
L : ($$1s^2\,2s^2\,2p^1$$); M : ($$1s^2\,2s^2\,2p^5$$);
Q : ($$1s^2\,2s^2\,2p^6\,3s^1$$); R : ($$1s^2\,2s^2\,2p^2$$).

The element that would most readily form a diatomic molecule is :

A
Q

B
M

C
R

D
L

Solution

From the electronic configuration given it is clear that element M is 1 short of octet configuration. It belongs to group 17. Group 17 elements exist as diatomic molecules.

So, correct option is B.
Question 3
1 / -0

Which among the group-$$15$$ elements does not exists as tetra atomic molecule?

A
Nitrogen

B
Phosphorus

C
Arsenic

D
Antimony

Solution

Nitrogen does not exist as tetra atomic molecule. It exists as diatomic molecule $$N_2$$ that can be represented as $$N \equiv N$$.
Question 4
1 / -0

Which of the following gases cannot be kept in a glass bottle because it chemically reacts with glass?

A
Sulphur

B
Chlorine

C
Fluorine

D
Methane

Solution

Sealing fluorine in regular glass will do more than fog the surface. It will fairly quickly etch the glass and escape. Hence Fluorine gas can't be stored in glass bottle.
Hence option C is correct answer.
Question 5
1 / -0

Which compound is prepared by the following reaction?
Xe +$$F_{2}$$ (2:1 volume ratio)$$\overset{Ni}\rightarrow$$at 673 K

A
$$XeF_{4}$$

B
$$XeF_{2}$$

C
$$XeF_{6}$$

D
None of these

Solution

$$Xe+F_2\overset {673K}{\underset {Ni}{\longrightarrow}} XeF_2\quad (1:3$$ $$Ratio)$$
$$Xe+2F_2\overset {873K}{\underset {Ni}{\longrightarrow}} XeF_4\quad (1:5$$ $$Ratio)$$
$$Xe+3F_2\overset {673K}{\underset {Ni}{\longrightarrow}} XeF_6\quad (1:20$$ $$Ratio)$$
Question 6
1 / -0

Xenon has closed shell configuration but is known to give compounds with fluorine because:

A
Xe atom has large size and lower ionisation potential as compared to other noble gases

B
Xe has unpaired electrons which can form covalent bonds

C
Xe has highest boiling point hence it can form compounds with fluorine

D
fluorine is the smallest element hence it can react with all noble gases

Solution

$$Xe$$ is inert though it gives compounds with Fluorine due to its large size and lower ionization potential as compared to other noble gases. Also, it is more reactive than other noble gases.
Question 7
1 / -0

The structure of white phosphorus is:

A
square planar

B
pyramidal

C
tetrahedral

D
trigonal planar

Solution

The structure of white phosphorus is tetrahedral.Phosphorus or s tetraphosphorus ($$P_4$$) exists as molecules made up of four atoms in a tetrahedral structure. The molecule is described as consisting of six single P–P bonds.
Hence option C is correct.
Question 8
1 / -0

The halogen that is most easily reduced is:

A
$$F_{2}$$

B
$$Cl_{2}$$

C
$$Br_{2}$$

D
$$I_{2}$$

Solution

$$F_2$$ is easily reduced as $$F^{-1}$$ is the strongest oxidising agent. Oxidising power can be explained by the fact that the ease with which a halogen $$[X]$$ forms the halide ion $$H^-$$ decreases down the group.
Question 9
1 / -0

In the preparation of $$HNO_{3}$$, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of $$NH_{3}$$ will be :

A
2

B
3

C
4

D
6

Solution

In the preparation of $$HNO_{3}$$, we get NO gas by catalytic oxidation of ammonia.In the below equation 4 moles of $$NH_3$$ produced 4 moles of $$NO$$.So the moles of $$NO$$ produced by the oxidation of two moles of $$NH_{3}$$ will be2 moles.
$$4NH_3+5O_2\rightarrow 4NO+6H_2O$$.
Hence option A is correct.
Question 10
1 / -0

Which of the following statements is wrong?

A
Single N-N bond is stronger than the single P-P bond.

B
$$PH_{3}$$ can act as a ligand in the formation of coordination compound with transition elements.

C
$$NO_{2}$$ is paramagnetic in nature.

D
Covalency of nitrogen in $$N_{2}O_{5}$$ is four.

Solution

N-N sigma bond (single bod) is weaker than P-P sigma bond (single bond) due to the small bond length between the nitrogen atoms. The lone pair of electrons of both the atoms nitrogen repel each other making the single bon between $$N-N$$ weaker than P-P sigma bond.In $$P-P$$ bond the repulsion between sigma bond and lone pair electrons is less as the size of phosphorous is large.
So statement A is wrong.
Hence option A is correct.

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The p-Block Elements Test - 52

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The p-Block Elements Test - 52

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Question 1

(i) is an alkali metal.

(iii) is a noble metal.

(i) and (ii) form ionic compounds.

(iv) has high ionisation enthalpy.

Solution

Question 2

Q

M

R

L

Solution

Question 3

Nitrogen

Phosphorus

Arsenic

Antimony

Solution

Question 4

Sulphur

Chlorine

Fluorine

Methane

Solution

Question 5

$$XeF_{4}$$

$$XeF_{2}$$

$$XeF_{6}$$

None of these

Solution

Question 6

Xe atom has large size and lower ionisation potential as compared to other noble gases

Xe has unpaired electrons which can form covalent bonds

Xe has highest boiling point hence it can form compounds with fluorine

fluorine is the smallest element hence it can react with all noble gases

Solution

Question 7

square planar

pyramidal

tetrahedral

trigonal planar

Solution

Question 8

$$F_{2}$$

$$Cl_{2}$$

$$Br_{2}$$

$$I_{2}$$

Solution

Question 9

2

3

4

6

Solution

Question 10

Single N-N bond is stronger than the single P-P bond.

$$PH_{3}$$ can act as a ligand in the formation of coordination compound with transition elements.

$$NO_{2}$$ is paramagnetic in nature.

Covalency of nitrogen in $$N_{2}O_{5}$$ is four.

Solution

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