P-block Elements Group 15 Questions

The p-block elements are found in Periodic Table Groups 13 to 18. Their electronic valence shell configuration is ns² np^1–6 (except He which has 1s² configuration). For Group 15, the configuration is ns² np³. Atomic sizes, ionisation enthalpy, electron gain enthalpy, and electronegativity all have a big impact on the properties of p-block elements. Group 15 elements, also known as the pnictogens which include nitrogen, phosphorus, arsenic, antimony, and bismuth.

P-block Elements Group 15 Chemistry Questions with Solutions

Q-1: Illustrate with the help of an equation, why +5 oxidation state of Bi is less stable than +3 state.

Answer:

Bi shows +3 and +5 oxidation states in BiCl₃ and BiCl₅ respectively. But due to the inert pair effect, the +5 oxidation state is less stable than +3 oxidation state. This is evident from the observation that BiCl₃ even on prolonged heating with Cl₂ does not form BiCl₅.

The chemical reaction shown below does not proceed due to BiCl₃ being more stable than BiCl₅.

BiCl₃ + Cl₂ → BiCl₅

Q-2: Which of the following is not known?

1. NCl₅
2. NCl₃
3. NI₃
4. SbCl₃

Answer: a) NCl₅

Explanation: N is from the second period. There is no d-orbital in the second subshell. As a result, arranging five pairs of bonding electrons around a nitrogen atom is impossible. As a result, NCl₅ does not exist.

Q-3: Which of the Group 15 elements does not exhibit allotropy?

1. Nitrogen
2. Phosphorous
3. Antimony
4. Bismuth

Answer: d) Bismuth

Explanation: Allotropy, also known as allotropes of the elements, is the property of some chemical elements to exist in two or more different forms in the same physical state.

Bismuth does not exhibit allotropy, but other elements do.

Nitrogen→ α-nitrogen and β-nitrogen (solid crystalline forms)

Phosphorus →White, Red and Black forms

Arsenic →Yellow and Grey forms

Antimony →White,Yellow and Black forms.

Q-4: The number of P-O-P bonds in cyclic tri metaphosphoric acid is

1. Zero
2. Two
3. Three
4. four

Answer: c) Three

Explanation: The general formula for cyclic metaphosphoric acid is (HPO₃)_n, where the value of ‘n’ gives the number of P-O-P bonds.

The formula of cyclic tri meta phosphoric acid is (HPO₃)₃.This shows ‘n’ is equal to 3. Hence, the number of P-O-P bonds are 3.

The structure for the cyclic tri metaphosphoric acid is shown below:

Q-5: Select the incorrect statement:

1. Black phosphorus has layered graphite like structure and is thermodynamically most stable.
2. White phosphorus is less reactive than red phosphorus.
3. Red phosphorus has a polymeric structure with P-P linkages.
4. Angle strain in red phosphorus is less than white phosphorus.

Answer: b) White phosphorus is less reactive than red phosphorus.

Explanation:

White phosphorus: The angles in the P₄ molecule are only 60 degrees, making it less stable and thus more reactive than the other solid phases under typical conditions. It easily catches fire in the air, producing dense white P₄O₁₀ vapours. The structure of white phosphorous is shown below:

Red phosphorus It has an iron grey lustre. It has no odour, is non-poisonous, and is insoluble in both water and carbon disulphide. Red phosphorus is less reactive chemically than white phosphorus. It is not visible in the dark. It’s polymeric, made up of chains of P₄ tetrahedra connected by P-P connections as shown below

Black phosphorus There are two types of black phosphorus: α-black phosphorus and β–black phosphorus. When red phosphorus is heated to 803K in a sealed tube, black phosphorus is created. At normal temperature and pressure, black phosphorus is thermodynamically stable. It has opaque monoclinic or rhombohedral crystals and can be sublimed in air. It does not oxidise when exposed to air. Black phosphorus creates layered structures just like graphene as shown in the below figure.White phosphorus is heated to 473 K under high pressure to produce black phosphorus. Up to 673 K, it does not burn in air.

Q-6: Complete the following reactions:

i) PbO₂ + HNO₃ →

ii)

Answer:

i) PbO₂ + HNO₃ → Pb(NO₃)₂ + H₂O + O₂

ii)

Q-7: What happens when

i) Solid XeF₂ reacts with AsF₅ in 1:1 ratio

ii) White phosphorus reacts with NaOH.

Answer:

i) Strong Lewis acids like AsF₅, SbF₅, IF₅, PtF₅ act as F^– acceptor with XeF₄ or XeF₂.

The following reaction takes place:

XeF₂ + AsF₅ → [XeF]⁺ + [AsF₆]^–

ii) Heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂ gives Phosphine (PH₃).

The reaction is shown below:

P₄ + 3NaOH+ 3H₂O→ PH₃ + 3NaH₂PO₂

Q-8: What are the Pseudohalogens and Pseudohalides? Give examples

Answer: A few ions are known consisting of two or more atoms of which at least one is N, that have properties similar to those of the halide ions. They are therefore called pseudohalides ions. These ions are univalent and form salts resembling the halide salts.

Table: The important pseudohalogens

Q-9: SbCl₃ on hydrolysis produces a white turbidity due to the formation of

1. Sb(OH)₃
2. SbOCl
3. SbCl₂(OH)
4. Sb(OH)₂Cl

Answer: b) SbOCl

Explanation: The chemical breakdown of compounds by water is known as hydrolysis. So, in terms of science, hydrolysis entails demonstrating the separation of synthetic molecules by adding water. Water’s reaction to the other chemical molecule results in the formation of at least two things.

Hydrolysis of SbCl₃ produces a white Antimony oxychloride precipitate.

The balanced chemical equation taking place is

SbCl₃ + H₂O → SbOCl + 2HCl

Q-10: Ostwald’s process generates nitric acid from ammonia via the formation of intermediate compounds

1. Nitric oxide and nitrogen dioxide
2. Nitrogen and nitric oxide
3. Nitric oxide and dinitrogen pentoxide
4. Nitrogen and nitrous oxide

Answer: a) Nitric oxide and nitrogen dioxide

Explanation: Ostwald’s process is most commonly used to produce nitric acid.

Ostwald’s process is a two-stage chemical reaction that converts ammonia to nitric acid.

Stage 1 involves the oxidation of ammonia to produce nitric oxide and nitric dioxide, followed by the dissolution of nitrogen dioxide in water and the formation of nitric acid.

The reactions involved are:

4NH₃+ O₂ → 4NO + 6H₂O

2NO + O₂ → 2NO₂

3NO₂ +H₂O→ 2HNO₃ + NO

Hence, we can clearly see that the intermediate compounds are nitric oxide(NO) and nitrogen dioxide(NO₂)

Q-11: The oxidation states of P atom in POCl₃ ,H₂PO₃ and H₄P₂O₆, respectively are

1. +5,+4,+4
2. +5,+5,+4
3. +4,+4,+4
4. +3,+4,+5

Answer: a) +5,+4,+4

Explanation: To determine the oxidation state of P, the oxidation states of each atom in a compound are added.

The sum of the oxidation numbers of the atoms in the compound is then made equal to zero for neutral compounds.

Let the oxidation state of P be x in each compound.

Q-12: Which of the following statements is wrong?

1. A single N-N bond is more powerful than a single P-P bond.
2. In the formation of a coordination compound with transition elements, PH₃ can act as a ligand.
3. NO₂ is paramagnetic in nature
4. Covalency of nitrogen in N₂O₅ is four

Answer: a) A single N-N bond is more powerful than a single P-P bond.

Explanation: A small size of N atom results in a shorter N-N bond length.As a result, the lone pair of electrons on both N atoms repel each other, causing the N-N bond to become unstable or weaken.On the other hand, due to the large size of P, there is no electronic repulsion between P atoms.

Q-13: Give the correct order of acidic character of following oxides.

Al₂O₃, MgO, SiO₂, P₄O₁₀

Answer:

Acidic strength is the tendency of a molecule/compound to liberate protons. In case of proton deficient compounds, acidic strength is governed by the positive oxidation state of the central atom. More is the value of positive oxidation state, more is the acidic strength.

The oxidation states of Al, Mg, Si and P are +3,+2,+4 and +5 respectively.

This makes the order of the acidic strength as:

P₄O₁₀ > SiO₂ > Al₂O₃ > MgO

Q-14: N₂O₅ is an anhydride of

1. HNO₂
2. HNO₃
3. H₂N₂O₂
4. HNO₄

Answer: b) HNO₃

Explanation: The term anhydride literally means “without water.” It is the chemical compound formed by removing water from another compound. Anhydride reacts with water to form a base or an acid.

Because anhydride is formed by removing water, adding water to N₂O₅ anhydride produces the corresponding acid/base.

N₂O₅ +H₂O → 2HNO₃

We can clearly see that HNO₃ is obtained from the N₂O₅ anhydride which is an acid.

Q-15: N₂O is a laughing gas which is prepared by heating

1. NH₄Cl
2. (NH₄)₂SO₄
3. NH₄NO₃
4. NH₄Cl + NaNO₃

Answer: c) NH₄NO₃

Explanation: When ammonium nitrate (NH₄NO₃) is heated, the following reaction takes place resulting in the formation of laughing gas.

NH₄NO₃ + heat → N₂O + 2H₂O

Practise Questions on P-block Elements Group 15

Q-1: Bond dissociation enthalpy of A-H (A=element)bond is given below. Which of the following compound will act as the weakest reducing agent

Q-2: Match the compounds in column I with their boiling point in column II

Q-3: Among the following gases, which oxide of N is coloured?

1. Nitrous oxide
2. Nitrogen monoxide
3. Nitrogen dioxide
4. Dinitrogen pentoxide

Q-4: Among the following, a sublimable substance is

1. NH₄Cl
2. BaSO₄
3. CaHPO₃
4. CaCO₃

Q-5: Write a chemical reaction between ammonia and cupric oxide.

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