p-Block Elements Test

Result

p-Block Elements Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). (B) reacts with organic compounds containing -OH group converting them into chloro derivatives. (B) on hydrolysis gives (C) and is finally converted to phosphoric acid. (A), (B) and (C) are?

A
P₄, PCl₃, H₃PO₄

B
P₄,PCl₅,H₃PO₃

C
P₄, PCl₅, POCl₃

D
P₄,PCl₃, POCl₃

Solution

A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). Here B is PCl₅.PCl₅ reacts with organic compounds containing -OH group converting them into chloro derivative POCl₃..PCl₅ on hydrolysis gives (C).C is POCl₃ and is finally POCl₃ converted to phosphoric acid.
So (A), (B) and (C) are P₄,PCl₅,POCl₃.
Question 2
1 / -0

Which of the following is not correctly matched?

A
PCl₅ - sp³d hybridisation

B
PCI₃ - sp³ hybridisation

C
PCl₅ (solid) - [PtCl₄]⁺ (PtCI₆]^-

D
PCI₅ - brownish powder

Solution

Colour of PCl₅ is not brownish but a greenish yellow crystalline solid with an irritating odour.
Question 3
1 / -0

On reaction with Cl₂, phosphorus forms two types of halides 'A' and 'B'. Halide 'A' is yellowish-white powder but halide 'B' is colourless oily liquid. What would be the hydrolysis products of 'A' and 'B' respectively?

A
H₃PO₄, H₃PO₃

B
HOPO₃, H₂PO₂

C
H₃PO₃, H₃PO₄

D
HPO₃, H₃PO₃

Solution
Question 4
1 / -0

Which of the following statements is not correct about the structure of PCl₅?

A
PCl₅ has a trigonal bipyramidal structure

B
Three equatorial P-Cl bonds are equivalent

C
The two axial bonds are different and longer than equatorial bonds

D
Equatorial bond pairs suffer more repulsion than that of the axial bond pairs

Solution

The axial bond pairs suffer more repulsion as compared to equatorial bond pairs.
Question 5
1 / -0

Why are all P-F bonds in PF₅ are not equivalent?

A
PF₅ has sp³d hybridisation, out of five P−F bonds three are equatorial which have different lengths

B
PF₅ has sp³ hybridisation, out of P−F bonds two are equatorial which have different lengths

C
Out of five P−F bonds two are axial and three equatorial. All five bonds have different bond lengths

D
PF₅ is made up of two types of bond namely covalent and coordinate, hence are not equivalent

Solution

Structure of PF₅
PF₅ has three axial bonds and two equitorial bonds and has sp₃d hybridisation PCl₅ has trigonal bipyramidal structure .
Due to this two Cl atoms lie along axial line and other three Cl atoms lie along the equatorial plane.
Hence Cl atoms which lie along axis have different bond length than that of Cl atoms lying on equatorial plane.
Question 6
1 / -0

Match the column I with column II and mark the appropriate choice.

A
(A) → (i), (B) → (ii), (C) → (iii), (D) → (iv)

B
(A) → (iv), (B) → (i), (C) → (ii), (D) → (iii)

C
(A) →(ii), (B) →(iii), (C) → (iv), (D) → (i)

D
(A) → (iii), (B) →(iv), (C) → (i), (D) → (ii)

Solution

(A) Laughing Gas ⟶ Nitrous oxide.
As this gas sometimes generates a feeling of euphoria due to which a person wants to laugh after smelling it.
(B) Anhydride of HNO₃ ⟶ anhydride means a compound obtained after water has been removed from other compounds.
From HNO₃,H₂O can't be removed so, if we take 2(HNO₃)
(C) Anhydride of HPO₃
Similarly as above if we take
(D) Acid Anhydride of nitrogen.
Acid Anhydride is formed by the combination of two acid structures with an ion of a water molecule.
Hydrazine reacts with nitrous acid giving acid anhydride of nitrogen, i.e, hydrazoic acid with loss of 2 molecules of water.
N₂H₄ + HNO₂ ⟶ HN₃ + 2H₂O
Question 7
1 / -0

How many P - O - P bonds appear in cyclic metaphosphoric acid?

A
Four

B
Three

C
Two

D
One

Solution

The number of P-O-P bond in cyclic metaphosphoric acid is three.
Question 8
1 / -0

Which of the following is a tetrabasic acid?

A
Hypophosphorous acid

B
Metaphosphoric acid

C
Pyrophosphoric acid

D
Orthophosphoric acid

Solution

Pyrophosphoric acid (H₄P₂O₇) is tetrabasic acid.
Question 9
1 / -0

Match the column I and column II and mark the appropriate choice.

A
(A) → (i), (B) →(iii), (C) →(ii), (D) →(iv)

B
(A) →(ii), (B) →(iv), (C) →(iii), (D) →(i)

C
(A) →(iii), (B) → (ii), (C) →(iv), (D) →(i)

D
(A) → (iv), (B) →(i), ((C)) →(ii), (D) →(iii)

Solution

H₃PO₂⟶ 3 + x − 4 = 0
x = +1
H₃P₃O₉ ⟶ 3 + 3x − 18 = 0
⇒ 3x = 15
⇒ x = +5
H₃PO₄ ⟶ 3 + x − 8 = 0
⇒ x = +5
H₃PO₃ ⟶ 3 + x − 6 = 0
⇒ x = +3
(A) H₃PO₂⟶ monobasic acid having 1 removable H⁺ ion.
(B) H₃P₃O₉ ⟶ cyclic oxoacid
(Refer to Image)
(C) H₃PO₄⟶ Tribasic ⟶ having 3 removable H⁺ ion.
Question 10
1 / -0

Phosphorous acid on heating gives the following products:

The above reaction is an example of

A
Oxidation

B
Thermal decomposition

C
Disproportionation

D
Reduction

Solution

It is disproportional where Phosphorus changes from +3 to −3 and +5. So, this is an example of disproportion reaction.