p-Block Elements Test

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p-Block Elements Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

Hot conc. H₂SO₄ acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H₂SO₄ into two gaseous products?

A
Cu

B
S

C
C

D
zn

Solution

Option A: Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
Option B: S + 2H₂SO₄ → 2SO₂ +2H₂O
Option C: C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O
Option D: Zn + 2H₂SO₄ → ZnSO₄ + SO₂ + 2H₂O
Hence, option C is correct.
Question 2
4 / -1

Correct order of 2nd ionisation energy of C, N, O and F is

A
N > F > O > C

B
O > F > N > C

C
F > O > N > C

D
O > F > C > N

Solution

The second ionization energy refers to the energy required to remove the electron from the corresponding monovalent cation of the respective atom.
It is expected to increase from left to right in the periodic table with the decrease in atomic size.
Since the Oxygen atom gets a stable electronic configuration, 2s²2p³ after removing one electron, the O⁺ shows greater ionization energy than F⁺ as well as N⁺.
Thus, correct order will be: O > F > N > C
Question 3
4 / -1

Which trend enthalpy is correct ?

A
N — O > P — O

B
N— F > P — F

C
N ≡ N > P ≡ P

D
N—N > P—P

Solution

Bonds → Bond enrgy
N-O → 201 kJ/mol
P-O → 340 kJ/mol
N-F → 272 kJ/mol
P-F → 490 kJ/mol
N≡N → 946 kJ/mol
P≡P → 490 kJ/mol
N-N → 160 kJ/mol
P-P → 209 kJ/mol
Hence option C is correct.
Question 4
4 / -1

Boiling Point of liquid nitrogen is

A
4 K

B
20 K

C
77 K

D
87 K

Solution

Liquid nitrogen is a cryogenic fluid that can cause rapid freezing on contact with living tissue.
The temperature is held constant at 77 K by slow boiling of the liquid, resulting in the evolution of nitrogen gas.
Question 5
4 / -1

Oxidation of ammonia with CuO produce a gaseous chemical which can also be obtained by:

A
Heating ammonium nitrate.

B
Reacting excess ammonia with chlorine.

C
Heating potassium dichromate.

D
Catalytic oxidation of ammonia.

Solution

Actually when ammonia is passed through a solution of calcium hypochlorite (bleaching powder), bromide water or passed over heated Cu oxide, it is oxidized to dinitrogen gas.
2NH₃+ 3CuO → 3Cu + 3H₂O + N₂
8NH₃(excess) + 3Cl₂ → 6NH₄Cl + N₂
Question 6
4 / -1

Which of the following molecular species has unpaired electron(s) ?

A
N₂

B
O₂

C
NO⁺

D
CN^-

Solution

O₂ contains two unpaired electrons and is paramagnetic in nature. On the other hand, N₂, F₂ and contains all paired electrons and are diamagnetic in nature.
Question 7
4 / -1

Which is the correct order w.r.t the given property?

A
N > P > As > Sb > Bi (Atomic mass)

B
N > P > As > Sb = Bi (Electronegativity)

C
N > P > As > Sb = Bi (Covalent radii)

D
N = P > As > Sb > Bi (Density)

Solution

Only electronegativity of the given 4 decreases down the group whilst all the others increase or stay the same down the group.
Question 8
4 / -1

The ratio of bond pairs and lone pairs in a P₄ molecule is

A
1:1

B
2:3

C
3:2

D
4:1

Solution

P₄ has six P-P single bonds and four lone pair of electrons.
Ratio: 6/4 = 3/2 i.e. 3:2
Question 9
4 / -1

Fluorine does not show variable oxidation states because_________

A
it is a halogen

B
of absence of d-orbitals

C
absence of s-orbital

D
absence of p-orbitals

Solution

Fluorine cannot expand its octet due to absence of d-orbitals.

Therefore it cannot show variable oxidation states.

Option B is correct.
Question 10
4 / -1

The halogen which can form both cations and anions is:

A
F

B
Cl

C
Br

D
I

Solution

Iodine can form both cations, as well as anions. Iodine can form cations due to the poor shielding effect of d orbitals where electrons can be removed from the valence shell and the cation is formed.

The cations and anions of Iodine is represented as follow:

I⁻ and I⁺
Question 11
4 / -1

In the third period of the periodic table the element having smallest size is

A
Na

B
CI

C
Ar

D
Si

Solution

Atomic size decreases across the period.
Cl has a smaller size than Ar due to the inert nature atoms exist as single atoms.
Question 12
4 / -1

The correct statements among the given are :

A
Antimony belongs to 15th group and 5th period.

B
Electron gain enthalpy of P > N > S > O.

C
Minimum and maximum oxidation number of phosphorus is -3 and +6.

D
Fluoroapatite, formula is Ca₆(PO₄)₆ CaF₂.

Solution

Option A: Group 5A (or VA) of the periodic table are the pnictogens: the nonmetals nitrogen (N), and phosphorus (P), the metalloids arsenic (As) and antimony (Sb), and the metal bismuth (Bi).
Option B: The electron gain enthalpy of P< N< S< O.
Option C: Minimum and maximum oxidation number of phosphorus are -3 and +5 respectively.
Option D: Fluorapatite is a phosphate mineral with the formula Ca₅(PO₄)₃F .
Hence, option A is correct.
Question 13
4 / -1

The halogen which can displace three halogens from their compounds is:

A
F

B
Cl

C
Br

D
I

Solution

F can displace all other halogens from their compounds i.e three halogens this is because of small size and high electronegativity of fluorine.
Question 14
4 / -1

Passage
Phosphorus was discovered by Brand (1669), Scheele isolated from bone ash and Lavoisier proved its elemental nature (1777). The principal minerals are phosphate rock, fluoroacetate, and chloroacetate. Phosphorus is prepared by the direct reduction of phosphorite by carbon in the presence of silica. It exists in different allotropic forms such as yellow or white, red, a-black,f3-black, etc. White P is most reactive, poisonous, glows in dark, and readily catches fire due to unstable discrete P4 molecules. Red P is inert, non-poisonous, does not glow, etc., due to its polymeric structure. a-black, f3 -black allotropes are also chemically inert, do not ignite at normal temperature. It has a layer structure like graphite and acts as a conductor.
Q. The allotrope of phosphorus with low ignition temperature is:

A
White P

B
Red P

C
α - Black

D
β - Black

Solution

White P is an allotrope of phosphorus with low ignition temperature.
Question 15
4 / -1

Passage
Phosphorus was discovered by Brand (1669), Scheele isolated from bone ash and Lavoisier proved its elemental nature (1777). The principal minerals are phosphate rock, fluoroacetate, and chloroacetate. Phosphorus is prepared by the direct reduction of phosphorite by carbon in the presence of silica. It exists in different allotropic forms such as yellow or white, red, a-black,f3-black, etc. White P is most reactive, poisonous, glows in dark, and readily catches fire due to unstable discrete P4 molecules. Red P is inert, non-poisonous, does not glow, etc., due to its polymeric structure. a-black, f3 -black allotropes are also chemically inert, do not ignite at normal temperature. It has a layer structure like graphite and acts as a conductor.
Q. The allotrope of phosphorus that has layer lattice-like graphite is:

A
White P

B
Yellow P

C
Red P

D
Black P

Solution

Black P has layer lattice like graphite.
Question 16
4 / -1

Matching List Type
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
Match column I with Column II and mark the correct option from the codes given below:

A

B

C

D

Solution

(i) Nitrogen is a non-metal.
(ii) Phosphorus is a non-metal.
(iii) Arsenic is a metalloid and shows Sublimation.
(iv) Bismuth is metal and shows the Inert pair effect.
Hence, option A is correct.
Question 17
4 / -1

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

A

B

C

D

Solution

(i) Phosphine Oxide, R₃P=O, presents an important example of the participation of d-atomic orbitals of nonmetallic elements in π bonding. The presence of π bonding is detected with the help of evidence, such as reduction in the bond length, increase in the bond strength, and the stabilization of charge distribution. On these grounds, compared to ammine oxide phosphine oxide presents strong evidence of the presence of dπ - pπ bond, in very high stability of P = O.
(ii) Nitrogen has a unique ability to form Pπ−Pπ multiple bonds with itself and with other elements due to its small size and high electronegativity.
(iii) N₂O₃ doesn't contain protons, so it is not a Brønsted acid. If N₂O₃ is dissolved in water, you get HNO₂ which is a weak acid, so it is an anhydride of an acid.
(iv) Ca₃N₂: The oxidation state is -3.
Hence, option B is correct.
Question 18
4 / -1

A brown ring is formed in the ring test for NO₃^– ion. It is due to the formation of

A
[Fe(H₂O)₅ (NO)]²⁺

B
FeSO₄.NO₂

C
[Fe(H₂O)₄(NO)₂]²⁺

D
FeSO₄.HNO₃

Solution

When freshly prepared solution of FeSO₄is added in a solution containing NO₃^–ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.
Question 19
4 / -1

The word Argon represents the term_______

A
Sun

B
Hidden

C
Lazy

D
Stranger

Solution

The word Argon represents the term Lazy.

Option C is correct.
Question 20
4 / -1

Which of the following halogens does not form oxyacids at room temperature?

A
F₂

B
Cl₂

C
Br₂

D
I₂

Solution

Fluorine does not hove d-sub shell so it can only have one oxidation state. Therefore, it cann't form oxy acids. Fluorine is the most electronegative element and always show −1 oxiadation state. In oxyacids. reast of the halogens have positive oxidation state.
Question 21
4 / -1

Statement Type
This section is based on Statement I and Statement II. Select the correct answer from the codes given below.
Statement I: N₂ is less reactive than P₄.
Statement II: Nitrogen has more electron gain enthalpy than phosphorus.

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

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

C
Statement I is correct but Statement II is incorrect.

D
Statement II is correct but Statement I is incorrect.

Solution

P — P single bond (213 kJ mol^-1) in P₄molecule is much weaker than N ≡ N triple bond (941.4 kJ mol^-1) in N₂.
Thus, Statement I is correct and Statement II is incorrect.