p-Block Elements Test

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p-Block Elements Test - 14

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Question 1
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The following question has four choices, out of which ONLY ONE is correct.

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P₄O₈ reacts with sufficient water to produce

A
four moles of phosphoric acid + O₂ gas

B
four moles of phosphorous acid + O₂ gas

C
phosphoric acid + phosphorous acid

D
three moles of phosphorous acid + one mole of PH₃ gas

Solution

In P₄O₈, two phosphorus atoms are in + 5 oxidation state and two in + 3 oxidation state. Hence, H₃PO₃ and H₃PO₄ are formed. P₄O₈ + H₂O → H₃PO₃ + H₃PO₄
Question 2
1 / -0

The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence

A
SiX₂ < GeX₂ < SnX₂ < PbX₂

B
PbX₂ < SnX₂ < GeX₂ < SiX₂

C
GeX₂ < SiX₂ < SnX₂ < PbX₂

D
SiX₂ < GeX₂ < PbX₂ < SnX₂

Solution

Due to inert pair effect, the stability of + 2 oxidation state increases as we move down a group.
Question 3
1 / -0

Upon hydrolysis, S₂Cl₂ slowly forms HCl, SO₂ and 'X'. Which of the following is 'X'?

A
SO₃

B
H₂

C
O₂

D
S

Solution

S₂Cl₂ is hydrolysed immediately by H₂O as per the following equation:

2S₂CI₂ + 2H₂O → 3S + SO₂+ 4HCI
Question 4
1 / -0

Fluoride ions are added to toothpaste to

A
maintain brightness of teeth

B
harden enamel of teeth

C
improve digestion

D
remove bacteria from mouth

Solution

Fluoride ions make enamel of teeth much harder by converting hydroxy apatite [3Ca₃(PO₄)₂, Ca(OH)₂] to much harder fluorapatite [3Ca₃(PO₄)₂, CaF₂].
Question 5
1 / -0

Fluorine cannot show oxidation state other than - 1 because

A
it is the most electronegative element and has no vacant d-orbital

B
it is small in size

C
it cannot form covalent bonds because of its highest electronegativity

D
its atomic volume is very small

Solution

Since it is the most electronegative element, it cannot donate electrons to other elements to form positive oxidation states. Also, because of the non-availability of vacant d-orbitals, it cannot promote electrons to produce more than one unpaired electron.
Question 6
1 / -0

Which of the following species is not a pseudohalide?

A
N₃^–

B
CN^–

C
OCN^–

D
IF₈^–

Solution

It is not a pseudohalogen. It is a polyhalide ion, which behaves differently.
Question 7
1 / -0

Xenon can form XeF₄, but He, Ne or Ar cannot form tetrafluorides because

A
ionisation energy of He, Ne and Ar is much higher than that of Xe

B
sizes of He, Ne and Ar are comparable to the size of fluorine, but that of Xe is much bigger

C
electropositivity of Xe is much higher than that of He, Ne and Ar

D
only Xe amongst these has a vacant d-orbital

Solution

This is true as the ionisation energy of Xe is much lower than that of He, Ne and Ar.

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p-Block Elements Test - 14

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p-Block Elements Test - 14

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Question 1

four moles of phosphoric acid + O2 gas

four moles of phosphorous acid + O2 gas

phosphoric acid + phosphorous acid

three moles of phosphorous acid + one mole of PH3 gas

Solution

Question 2

SiX2 < GeX2 < SnX2 < PbX2

PbX2 < SnX2 < GeX2 < SiX2

GeX2 < SiX2 < SnX2 < PbX2

SiX2 < GeX2 < PbX2 < SnX2

Solution

Question 3

SO3

H2

O2

S

Solution

Question 4

maintain brightness of teeth

harden enamel of teeth

improve digestion

remove bacteria from mouth

Solution

Question 5

it is the most electronegative element and has no vacant d-orbital

it is small in size

it cannot form covalent bonds because of its highest electronegativity

its atomic volume is very small

Solution

Question 6

N3–

CN–

OCN–

IF8–

Solution

Question 7

ionisation energy of He, Ne and Ar is much higher than that of Xe

sizes of He, Ne and Ar are comparable to the size of fluorine, but that of Xe is much bigger

electropositivity of Xe is much higher than that of He, Ne and Ar

only Xe amongst these has a vacant d-orbital

Solution

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four moles of phosphoric acid + O₂ gas

four moles of phosphorous acid + O₂ gas

three moles of phosphorous acid + one mole of PH₃ gas

SiX₂ < GeX₂ < SnX₂ < PbX₂

PbX₂ < SnX₂ < GeX₂ < SiX₂

GeX₂ < SiX₂ < SnX₂ < PbX₂

SiX₂ < GeX₂ < PbX₂ < SnX₂

SO₃

H₂

O₂

N₃^–

CN^–

OCN^–

IF₈^–