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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 np1–6 (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.

 Definition: The nitrogen family in the modern periodic table with 5 valence electrons in its valence shell is Group 15 .

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.

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

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

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

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)

$$\begin{array}{l}NH_{4}NO_{2} \overset{\Delta }{\rightarrow}\end{array}$$

i) PbO2 + HNO3 → Pb(NO3)2 + H2O + O2

ii)

$$\begin{array}{l}NH_{4}NO_{2} \overset{\Delta }{\rightarrow} N_{2} + H_{2}O\end{array}$$

Q-7: What happens when

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

ii) White phosphorus reacts with NaOH.

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

 Anion Acid Dimer CN– cyanide ion SCN– thiocyanate ion SeCN– selenocyanate ion OCN– cyanate ion ONC– fulminate ion N3– azide ion HCN hydrogen cyanide HSCN thiocyanic acid HOCN cyanic acid HONC fulminic acid HN3 hydrogen azide (CN)2 cyanogen (SCN)2 thiocyanogen (SeCN)2 selenocyanogen

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

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

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.

 Compound Formula Oxidation state of P POCl3 x+(-2)+3(-1)=0 +5 H2PO3 2(+1)+x+3(-2)=0 +4 H4P2O6 4(+1)+2(x)+6(-2)=0 +4

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

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

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

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

 Compound NH3 PH3 AsH3 SbH3 Bond dissociation enthalpy(KJ/mol) 389 322 297 255

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

 Compound Boiling point (K) A) NH3 i) 290 B) PH3 ii) 211 C) AsH3 iii) 186 D) SbH3 iv) 240 E) BiH3 v) 264

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.

Click the PDF to check the answers for Practice Questions.