p-Block Elements Test

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p-Block Elements Test - 2

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Weekly Quiz Competition

Question 1
4 / -1

The hybridisation of N in solid state for N₂O₅ is

A
sp and sp²

B
Only sp²

C
sp² and sp²

D
Only sp³

Solution

In solid state, N₂O₅ exists as nitronium nitrate
In nitrogen is sp and in , nitrogen is sp².
Question 2
4 / -1

PH₃ becomes spontaneously inflammable due to the presence of

A
NH₃

B
N₂H₄

C
P₂H₄

D
All of the above

Solution

Pure phoshine is not inflammable but due to contamination of P₂H₄ and P₄ vapours, it becomes inflammabl.
Question 3
4 / -1

Which is incorrect among the following options?

A
Reducing property of H₃PO₄ > H₃PO₃ > H₃PO₂

B
Oxidation state of nitrogen N₂O < NO < N₂O₃ < N₂O₅

C
Basicity NH₃ > PH₃> AsH₃ > SbH₃

D
Boiling point of SbH₃ > AsH₃ > NH₃ > PH₃

Solution

Reducing property of P oxyacids is due to P—H bonds in the acid. Correct reducing property order is
Question 4
4 / -1

The number of P – O bonds and lone pairs of electron present in P₄O₆ molecule respectively

A
12 and 4

B
8 and 8

C
12 and 16

D
12 and 12

Solution

Number of P – O bonds = 12
Number of pair of electron = 16
Question 5
4 / -1

Which is the correct order of basic strength ?

A
NH₂OH < N₂H₄ < NH₃

B
N₂H₄ < NH₂OH < NH₃

C
NH₃ < N₂H₄ < NH₂OH

D
NH₂OH < NH₃ < N₂H₄

Solution

NH₂OH and NH₂—NH₂ may be considered as NH₃ derivatives in which H is replaced but — OH and NH₂ respectively. Due to their electron withdrawing nature, these groups decreases electron density over nitrogen making them less basic. The effect of — OH group is stronger than —NH₂
Question 6
4 / -1

Solid PCl₅ and solid PBr₅ exist respectively as

A
I and III

B
I and IV

C
II and III

D
II and IV

Solution

PBr does not split in the same fashion. The anion is not possible due to large size of Br-atoms. Hence, it splits differently.
Question 7
4 / -1

Number of moles of NaOH needed to neutralise one mole each of H₃PO₂, H₃PO₃ and H₃PO₄ respectively are

A
3, 3 and 3

B
1, 2 and 3

C
2, 2 and 3

D
2, 1 and 3

Solution

H₃PO₂, H₃PO₃ and H₃PO₄ are mono, di and tribasic acids respectively.
Question 8
4 / -1

Which of the following set of elements have the strongest tendency to form anions?

A
N, O and F

B
P, S and Cl

C
N, P and Cl

D
N, P and S
Question 9
4 / -1

Mono atomic element among the following is:

A
phosphorus

B
oxygen

C
krypton

D
sulphur

Solution

Oxygen is diatomic gas.

Phosphorous is tetra atomic.

Sulphur is polyatomic.

Whereas being an inert gas, Krypton is monoatomic.
Question 10
4 / -1

According to Coulson the high reactivity of fluorine than other halogens is because of its:

A
high electronegativity

B
lesser internuclear repulsions

C
greater repulsions between non bonding electrons

D
All the above

Solution

According to Coulson, since fluorine atoms are small it's intermolecular repulsion is appreciable. The large electron-electron repulsion between the lone pairs of electrons on the fluorine atom weakens the bond.
Question 11
4 / -1

The inert gas present in the second long period is:

A
Kr

B
Xe

C
Ar

D
Rn

Solution

Periods 4 and 5 are called long periods as they contain 18 elements each.

4th period is called the first long period and the inert gas in this period is Kr.

5th period is the second long period where Xe is the inert gas element.
Question 12
4 / -1

Which is correctly matched?

A
Phosphonic acid (H₃PO₂) is a dibasic acid

B
Metaphosphoric acid is called glacial phosphoric acid

C
H₄P₂O₅ (diphosphorus acid) has P— O—P bond

D
None of these

Solution

-

Since, here 1-hydroxyl group is present so, it is monobasic acid. Since, metaphosphoric, when reacts in the presence of heat orthophosphoric acid is form which is in the form of glacial phosphoric acid.

Option (a), (c) and (d) is not correctly matched.
Question 13
4 / -1

Direction (Q. Nos. 13 and 14) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage
In the all oxyacids of phosphorus, each phosphorus atom is in sp³-hybridised state. All these acids contain P—OH bonds, the hydrogen atom of which are ionisable imparting acidic nature to the compound. The ‘ous’ acids (oxidation state of P is + 1 or + 3) also have P—H bonds in which hydrogens are not ionisable.
The presence of P—H bonds in these acids imparts reducing properties. The structure of some oxyacids are drawn below:

Q.
Although metaphosphoric acid is written as a monomer, it exists as a polymer (HPO3)_n. The number of P—O—P bonds in cyclic trimetaphosphoric acid is

A
zero

B
two

C
three

D
four

Solution
Question 14
4 / -1

In the all oxyacids of phosphorus, each phosphorus atom is in sp³-hybridised state. All these acids contain P—OH bonds, the hydrogen atom of which are ionisable imparting acidic nature to the compound. The ‘ous’ acids (oxidation state of P is + 1 or + 3) also have P—H bonds in which hydrogens are not ionisable.
The presence of P—H bonds in these acids imparts reducing properties. The structure of some oxyacids are drawn below:

Q.
Which of the acids show reducing properties?

A
A and C

B
A and B

C
A , B and D

D
C, D, E and F

Solution

Due to the presence of H in A and B , they showre ducing properties.
Question 15
4 / -1

Direction (Q. Nos. 15 and 16) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.
Q.
Match the Column I with Column II and mark the correct option from the codes given below.

A

B

C

D

Solution
Question 16
4 / -1

Match the Column I with Column II and mark the correct option from the codes given below.

A

B

C

D

Solution
Question 17
4 / -1

The steps for the manufacture of nitric acid by Ostwald process are given. Arrange them in sequence.
(a) Recirculation of nitric oxide
(b) Addition of H₂O to nitrogen dioxide
(c) Oxidation of nitric oxide
(d) Catalytic oxidation of ammonia

A
d c b a

B
d c a b

C
c d b a

D
c d a b

Solution

In ostwald process Ammonia is converted to nitric acid in 2 stages. It is oxidized (in a sense "burnt") by heating with oxygen in the presence of a catalyst to form nitric oxide and water. Nitric oxide is oxidized again to yield nitrogen dioxide. This gas is then readily absorbed by the water and forms nitric acid. Thus the steps of process are :
(d) Catalytic oxidation of ammonia
(c) Oxidation of nitric oxide
(b) Addition of H₂O to nitrogen dioxide
(a) Recirculation of nitric oxide
Question 18
4 / -1

The solubility of iodine in carbon tetrachloride is due to

A
Ionic solvent-solute interaction

B
Charge-transfer phenomenon

C
Van der Waals interaction

D
Dissociation of iodine

Solution

Both I₂ and CCl₄ are non polar in nature. The interaction between them is van der waal.
Question 19
4 / -1

Which of the following gases has the greatest density at STP conditions?

A
Ne

B
Ar

C
He

D
Kr

Solution

The highest density among the inert gases is of Neon (Ne). This a factual data. Thus the highest density among given options is of Ne, as all the options are of inert gases.
Question 20
4 / -1

Direction (Q. No. 20) This section is bassed on Statment I and Statment II. II. Select the correct answer from the codes given below.
Q.
Statement I : NF₃ is a weaker ligand than N(CH₃)₃
Statement II : NF₃ ionises to give F^- ions in aqueous solutio

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statemen I

B
Both Statement I and Statement II are correct but Statement II is not the correct. explanation of Statment I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Due to e^- withdrawing capacity of fluorideion, it withdraws. from nitrogen in NF₃ make it weakerlig and while presence of e^- donating methyl group makes the nitrogen in N(CH₃)₃ a strong ligand. In aqueous medium, NF₃ furnishes fluorideion.