Question 1

Question: Identify the incorrect statement from the following:

Options: 

A) Nitrogen can form $p \pi-p \pi$ multiple bonds with itself. 

B) $\mathrm{P}(\mathrm{C}_2\mathrm{H}_5)_3$ and $\mathrm{As}(\mathrm{C}_6\mathrm{H}_5)_3$ form $d \pi-d \pi$ bond with transition metals. 

C) Phosphorus, arsenic and antimony show catenation property. 

D) Nitrogen can form $d \pi-p \pi$ bond with oxygen.

Correct Answer: D

Year: 2026 

Solution: Nitrogen cannot form $d \pi-p \pi$ bonds because it does not have $d$-orbitals in its valence shell.

Step Solution:

1.  Analyze Nitrogen's electronic configuration: $1s^2 2s^2 2p^3$.

2.  Note that the second energy level ($n=2$) only contains $s$ and $p$ subshells.

3.  Observe that Nitrogen lacks $d$-orbitals entirely.

4.  Conclude that any bonding involving $d$-orbitals (like $d \pi-p \pi$) is physically impossible for Nitrogen.

Difficulty Level: Medium

Concept Name: Atomic Orbitals and Bonding Constraints

Short cut solution: Nitrogen is a second-period element; second-period elements never use $d$-orbitals for bonding.

Question 1

   Question: Given below are two statements: 

       Statement I: Like nitrogen that can form ammonia, arsenic can form arsine.

       Statement II: Antimony cannot form antimony pentoxide.

    In the light of the above statements, choose the most appropriate answer from the options given below.

   Options: 

       A. Statement I is correct but Statement II is incorrect

       B. Statement I is incorrect but Statement II is correct

       C. Both Statement I and Statement II are correct

       D. Both Statement I and Statement II are incorrect

   Correct Answer: A

   Year: NEET 2025

   Solution: All elements in group 15, including nitrogen and arsenic, can form hydrides with the formula $EH_3$. For instance, nitrogen forms ammonia ($NH_3$), and arsenic forms arsine ($AsH_3$). Elements in group 15 are capable of forming two types of oxides: $E_2O_3$ and $E_2O_5$. Antimony, being a group 15 element, can indeed form antimony pentoxide ($Sb_2O_5$).

   Step Solution:

    1.  Analyze Statement I: Nitrogen and arsenic both belong to Group 15.

    2.  Verify Hydrides: Group 15 elements form $EH_3$ hydrides; $N \rightarrow NH_3$ (ammonia) and $As \rightarrow AsH_3$ (arsine). Statement I is correct.

    3.  Analyze Statement II: Check oxide formation for Group 15.

    4.  Verify Antimony: These elements form oxides in +3 and +5 states ($E_2O_3, E_2O_5$). Antimony ($Sb$) forms $Sb_2O_5$.

    5.  Conclusion: Statement II says it cannot form pentoxide, which is incorrect.

   Difficulty Level: Easy

   Concept Name: General Properties of Group 15 Elements (Hydrides and Oxides)

   Shortcut: Group 15 elements generally form $EH_3$ hydrides and $E_2O_5$ oxides.

Question 6

   Question: Which of the following reactions is a part of the large scale industrial preparation of nitric acid?

   Options: 

       A. $Cu(NO_3)_2 + 2NO_2 + 2H_2O \xrightarrow{Pt, 500K, 9 bar} 4HNO_3 + Cu$

       B. $NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3$

       C. $4NH_3 + 5O_2 \text{ (from air)} \xrightarrow{Pt, 500K, 9 bar} 4NO + 6H_2O$

       D. $4HPO_3 + 2N_2O_5 \rightarrow 4HNO_3 + P_4O_{10}$

   Correct Answer: C

   Year: NEET Re-2022

   Solution: On a large scale, nitric acid is prepared by Ostwald's process. The process involves the catalytic oxidation of $NH_3$ by atmospheric oxygen: $4NH_{3(g)} + 5O_{2(g)} \xrightarrow{Pt/Rh \text{ gauge, } 500K, 9 bar} 4NO_{(g)} + 6H_2O_{(g)}$.

   Step Solution:

    1.  Identify Method: Industrial preparation of $HNO_3$ is the Ostwald's process.

    2.  Step 1: Catalytic oxidation of ammonia: $4NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O$.

    3.  Step 2: Oxidation of NO: $2NO + O_2 \rightarrow 2NO_2$.

    4.  Step 3: Dissolution in water: $3NO_2 + H_2O \rightarrow 2HNO_3 + NO$.

    5.  Match: Option C matches the critical catalytic first step.

   Difficulty Level: Medium

   Concept Name: Ostwald’s Process

   Shortcut: Large scale $HNO_3$ = Ostwald's = $NH_3$ oxidation with Pt catalyst.

Question 10

   Question: Urea reacts with water to form A which will decompose to form B. B when passed through $Cu^{2+}$ (aq), deep blue colour solution C is formed. What is the formula of C from the following?

   Options: 

       A. $[Cu(NH_3)_4]^{2+}$

       B. $Cu(OH)_2$

       C. $CuCO_3 \cdot Cu(OH)_2$

       D. $CuSO_4$

   Correct Answer: A

   Year: 2020

   Solution: $NH_2CONH_2 + H_2O \rightarrow (NH_4)_2CO_3 \xrightarrow{\Delta} NH_3(g) + CO_2(g) + H_2O$. The ammonia gas (B) reacts with $Cu^{2+}$ to form the blue coloured complex $[Cu(NH_3)_4]^{2+}$.

   Step Solution:

    1.  Urea Hydrolysis: $NH_2CONH_2 + H_2O \rightarrow (NH_4)_2CO_3$ (A).

    2.  Decomposition: $(NH_4)_2CO_3 \xrightarrow{\Delta} 2NH_3 + CO_2 + H_2O$. So, B is $NH_3$.

    3.  Complex Formation: $Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$.

    4.  Identify Color: This coordination complex provides the characteristic deep blue color.

   Difficulty Level: Medium

   Concept Name: Qualitative Analysis / Coordination Compounds

   Shortcut: $Cu^{2+} + \text{excess } NH_3 \rightarrow \text{Deep Blue } [Cu(NH_3)_4]^{2+}$.

Question 15

   Question: Identify the incorrect statement related to $PCl_5$ from the following:

   Options: 

       A. $PCl_5$ molecule is non-reactive.

       B. Three equatorial P−Cl bonds make an angle of $120^\circ$ with each other.

       C. Two axial P−Cl bonds make an angle of $180^\circ$ with each other.

       D. Axial P−Cl bonds are longer than equatorial P−Cl bonds.

   Correct Answer: A

   Year: NEET 2019

   Solution: Due to longer and hence weaker axial bonds, $PCl_5$ is a reactive molecule.

   Step Solution:

    1.  Structure of $PCl_5$: It has trigonal bipyramidal geometry ($sp^3d$).

    2.  Check Angles: 3 equatorial bonds are at $120^\circ$ (Statement B is true).

    3.  Check Axial Angles: 2 axial bonds are at $180^\circ$ to each other (Statement C is true).

    4.  Check Bond Lengths: Axial bonds are longer due to greater repulsion (Statement D is true).

    5.  Reactivity: Longer bonds are weaker; therefore, $PCl_5$ is highly reactive, making A incorrect.

   Difficulty Level: Easy

   Concept Name: Chemical Bonding (Structure and Reactivity of $PCl_5$)

   Shortcut: $PCl_5$ is always reactive because its axial bonds are weak and easily broken.

Question 16

   Question: A compound 'X' upon reaction with $H_2O$ produces a colourless gas 'Y' with rotten fish smell. Gas 'Y' is absorbed in a solution of $CuSO_4$ to give $Cu_3P_2$ as one of the products. Predict the compound 'X'.

   Options: 

       A. $Ca_3P_2$

       B. $NH_4Cl$

       C. $As_2O_3$

       D. $Ca_3(PO_4)_2$

   Correct Answer: A

   Year: Odisha NEET 2019

   Solution: $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_{3(g)}$. $PH_3$ is gas 'Y' which has a rotten fish smell. Reaction with copper sulphate: $3CuSO_4 + 2PH_3 \rightarrow Cu_3P_2 + 3H_2SO_4$.

   Step Solution:

    1.  Identify Gas Y: "Rotten fish smell" is the signature of Phosphine ($PH_3$).

    2.  Confirm Y: Reaction with $CuSO_4$ yielding $Cu_3P_2$ confirms Y is $PH_3$.

    3.  Find X: Compound X must be a phosphide that releases $PH_3$ with water.

    4.  Reaction: $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$.

    5.  Conclusion: X is $Ca_3P_2$.

   Difficulty Level: Medium

   Concept Name: Properties of Phosphine and Phosphides

   Shortcut: Rotten fish smell = Phosphine ($PH_3$); Calcium phosphide + water $\rightarrow$ Phosphine.

Question 17

   Question: Which of the following oxoacids of phosphorus has strongest reducing property?

   Options: 

       A. $H_4P_2O_7$

       B. $H_3PO_3$

       C. $H_3PO_2$

       D. $H_3PO_4$

   Correct Answer: C

   Year: Odisha NEET 2019

   Solution: Acids which contain P−H bonds have strong reducing properties. Among the given compounds, $H_3PO_2$ is the strongest reducing agent as it contains two P−H bonds.

   Step Solution:

    1.  Principle: Reducing power in phosphorus oxoacids depends on the number of P−H bonds.

    2.  Evaluate $H_3PO_4$: Zero P−H bonds.

    3.  Evaluate $H_3PO_3$: One P−H bond.

    4.  Evaluate $H_3PO_2$: Two P−H bonds.

    5.  Conclusion: $H_3PO_2$ has the most P−H bonds and is the strongest reducing agent.

   Difficulty Level: Easy

   Concept Name: Oxoacids of Phosphorus (Reducing Character)

   Shortcut: Reducing power $\propto$ number of P−H bonds. $H_3PO_2$ (hypophosphorous acid) has 2.

Question 23

   Question: Which is the correct statement for the given acids?

   Options: 

       A. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.

       B. Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid.

       C. Both are diprotic acids.

       D. Both are triprotic acids.

   Correct Answer: A

   Year: NEET-I 2016

   Solution: Hypophosphorous acid or Phosphinic acid ($H_3PO_2$) is Monobasic. Orthophosphorous acid or Phosphonic acid ($H_3PO_3$) is Dibasic.

   Step Solution:

    1.  Identify Formulae: Phosphinic acid is $H_3PO_2$; Phosphonic acid is $H_3PO_3$.

    2.  Determine Basicity Rule: Basicity (proticity) is determined by the number of P—OH bonds.

    3.  Analyze Phosphinic Acid: $H_3PO_2$ contains one P—OH group and two P—H bonds; thus, it is monoprotic.

    4.  Analyze Phosphonic Acid: $H_3PO_3$ contains two P—OH groups and one P—H bond; thus, it is diprotic.

    5.  Conclusion: Phosphinic is monoprotic and phosphonic is diprotic.

   Difficulty Level: Medium

   Concept Name: Basicity of Phosphorus Oxoacids

   Shortcut: Phosphinic ($H_3PO_2$) has 1 OH bond $\rightarrow$ monoprotic. Phosphonic ($H_3PO_3$) has 2 OH bonds $\rightarrow$ diprotic.

Question 25

   Question: When copper is heated with conc. $HNO_3$ it produces

   Options: 

       A. $Cu(NO_3)_2$, $NO$ and $NO_2$

       B. $Cu(NO_3)_2$ and $N_2O$

       C. $Cu(NO_3)_2$ and $NO_2$

       D. $Cu(NO_3)_2$ and $NO$

   Correct Answer: C

   Year: NEET-I 2016

   Solution: $Cu + 4HNO_3 (conc.) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$

   Step Solution:

    1.  Reactants: Copper ($Cu$) and concentrated Nitric Acid ($HNO_3$).

    2.  Oxidation State: Conc. $HNO_3$ is a strong oxidizing agent.

    3.  Nitrogen Product: Concentrated $HNO_3$ always reduces to Nitrogen dioxide ($NO_2$) when reacting with metals like copper.

    4.  Metal Product: Copper is oxidized to Copper(II) nitrate ($Cu(NO_3)_2$).

    5.  Balanced Equation: $Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$.

   Difficulty Level: Easy

   Concept Name: Reactions of Nitric Acid with Metals

   Shortcut: Conc. $HNO_3$ + Metal $\rightarrow$ $NO_2$ gas.

Question 27

   Question: Strong reducing behaviour of $H_3PO_2$ is due to

   Options: 

       A. high electron gain enthalpy of phosphorus

       B. high oxidation state of phosphorus

       C. presence of two -OH groups and one P—H bond

       D. presence of one -OH group and two P—H bonds.

   Correct Answer: D

   Year: 2015

   Solution: All oxyacids of phosphorus which have P—H bonds act as strong reducing agents. $H_3PO_2$ has two P—H bonds hence, it acts as a strong reducing agent 

   Step Solution:

    1.  Formula: Hypophosphorous acid is $H_3PO_2$.

    2.  Structural Principle: Reducing power is proportional to the number of P—H bonds.

    3.  Analyze Bonds: $H_3PO_2$ has a central P atom with 1 =O, 1 —OH, and 2 —H atoms.

    4.  Basicity vs. Reducing Power: It is monobasic (1 —OH) but a strong reductant (2 P—H).

    5.  Match: Option D correctly identifies the bond distribution.

   Difficulty Level: Easy

   Concept Name: Reducing Property of Phosphorus Oxoacids

   Shortcut: Reducing agent = P—H bonds. $H_3PO_2$ has 2 P—H bonds.

Question 31

   Question: Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?

   Options: 

       A. Is soluble in water.

       B. Is used as a food preservative

       C. Forms ‘acid-rain’

       D. Is a reducing agent.

   Correct Answer: B

   Year: 2015 Cancelled

   Solution: $NO_2$ is not used as a food preservative.

   Step Solution:

    1.  Water Solubility: Both $NO_2$ and $SO_2$ react with water to form acids.

    2.  Acid Rain: Both gases are major contributors to acid rain ($HNO_3$ and $H_2SO_4$).

    3.  Redox Property: Both act as reducing agents in specific reactions.

    4.  Preservation: $SO_2$ is widely used in food preservation; $NO_2$ is not.

    5.  Differentiator: The industrial use as a preservative applies only to $SO_2$.

   Difficulty Level: Medium

   Concept Name: Chemical Properties of $p$-Block Oxides

   Shortcut: $SO_2$ is a common preservative; $NO_2$ (toxic brown gas) is not.

Question 37

   Question: In which of the following compounds, nitrogen exhibits highest oxidation state?

   Options: 

       A. $N_2H_4$

       B. $NH_3$

       C. $N_3H$

       D. $NH_2OH$

   Correct Answer: C

   Year: NEET 2013

   Solution: Oxidation states: $N_2H_4$ (-2), $NH_3$ (-3), $N_3H$ (-1/3), $NH_2OH$ (-1).

   Step Solution:

    1.  Calculate $N_2H_4$: $2x + 4(+1) = 0 \Rightarrow x = -2$.

    2.  Calculate $NH_3$: $x + 3(+1) = 0 \Rightarrow x = -3$.

    3.  Calculate $N_3H$: $3x + 1(+1) = 0 \Rightarrow x = -1/3$.

    4.  Calculate $NH_2OH$: $x + 2(H) + 1(O) + 1(H) \rightarrow x + 2 - 2 + 1 = 0 \Rightarrow x = -1$.

    5.  Compare: $-1/3$ is the largest (closest to zero) value among the negative states.

   Difficulty Level: Easy

   Concept Name: Oxidation State Calculation

   Shortcut: In $N_3H$, nitrogen's state is fractional (-0.33), which is higher than -1, -2, or -3.

Question 38

   Question: Which of the following statements is not valid for oxoacids of phosphorus?

   Options: 

       A. Orthophosphoric acid is used in the manufacture of triple superphosphate

       B. Hypophosphorous acid is a diprotic acid

       C. All oxoacids contain tetrahedral four coordinated phosphorus.

       D. All oxoacids contain atleast one P=O unit and one P-OH group

   Correct Answer: B

   Year: 2012

   Solution: Hypophosphorous acid is a monoprotic acid. 

   Step Solution:

    1.  Recall Formula: Hypophosphorous acid is $H_3PO_2$.

    2.  Analyze Structure: It contains one P=O, two P—H, and one P—OH group.

    3.  Determine Proticity: Proticity depends on the number of replaceable hydrogens (P—OH bonds).

    4.  Identify Error: Since it has only one P—OH bond, it is monoprotic, not diprotic.

    5.  Confirming others: Most oxoacids are tetrahedral and contain P=O and P—OH units.

   Difficulty Level: Medium

   Concept Name: Structure and Basicity of Phosphorus Oxoacids

   Shortcut: Hypophosphorous ($H_3PO_2$) = 1 OH bond = monoprotic. Statement B is false.

Question 41

   Question: Oxidation states of P in $\mathbf{H}_{4}\mathbf{P}_{2}\mathbf{O}_{5}, \mathbf{H}_{4}\mathbf{P}_{2}\mathbf{O}_{6}, \mathbf{H}_{4}\mathbf{P}_{2}\mathbf{O}_{7}$ are respectively

   Options: 

       A. +3, +5, +4

       B. +5, +3, +4

       C. +5, +4, +3

       D. +3, +4, +5

   Correct Answer: D

   Year: 2010

   Solution: The oxidation state can be calculated as:

       For $H_4P_2O_5$: $+4 + 2x + 5(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow x = +3$.

       For $H_4P_2O_6$: $+4 + 2x + 6(-2) = 0 \Rightarrow 2x - 8 = 0 \Rightarrow x = +4$.

       For $H_4P_2O_7$: $+4 + 2x + 7(-2) = 0 \Rightarrow 2x - 10 = 0 \Rightarrow x = +5$.

   Step Solution: 

    1.  Assign oxidation states: $H = +1$ and $O = -2$.

    2.  For $H_4P_2O_5$: $4(+1) + 2(P) + 5(-2) = 0 \Rightarrow 4 + 2P - 10 = 0 \Rightarrow 2P = 6 \Rightarrow \mathbf{P = +3}$.

    3.  For $H_4P_2O_6$: $4(+1) + 2(P) + 6(-2) = 0 \Rightarrow 4 + 2P - 12 = 0 \Rightarrow 2P = 8 \Rightarrow \mathbf{P = +4}$.

    4.  For $H_4P_2O_7$: $4(+1) + 2(P) + 7(-2) = 0 \Rightarrow 4 + 2P - 14 = 0 \Rightarrow 2P = 10 \Rightarrow \mathbf{P = +5}$.

    5.  Match the sequence: +3, +4, +5.

   Difficulty Level: Easy

   Concept Name: Oxidation State Calculation

   Short cut solution: Use the formula: $\text{Oxidation State} = \frac{((\text{Oxygen atoms} \times 2) - (\text{Hydrogen atoms}))}{\text{Phosphorus atoms}}$. 

       For $O_5$: $(10-4)/2 = 3$. 

       For $O_6$: $(12-4)/2 = 4$. 

       For $O_7$: $(14-4)/2 = 5$.

Question 43

   Question: How many bridging oxygen atoms are present in $\mathbf{P}_{4}\mathbf{O}_{10}$

   Options: 

       A. 6

       B. 4

       C. 2

       D. 3

   Correct Answer: A

   Year: 2010 Mains

   Solution: Bridging oxygen atoms are those that connect two phosphorus atoms in the structure. In $P_4O_{10}$, there are 6 such P—O—P bonds. 

   Step Solution: 

    1.  Visualize the $P_4$ skeleton as a tetrahedron.

    2.  Place one phosphorus atom at each of the 4 vertices.

    3.  Identify the edges of the tetrahedron; each edge represents a bridging oxygen atom location (P—O—P).

    4.  Count the number of edges in a tetrahedron, which is 6.

    5.  Each Phosphorus also has one terminal oxygen (P=O), but these are not bridging.

   Difficulty Level: Medium

   Concept Name: Molecular Structure of Phosphorus Oxides

   Short cut solution: In $P_4O_6$ and $P_4O_{10}$, the number of bridging oxygens is always the number of edges in a tetrahedron, which is 6.

Question 46

   Question: The electronegativity difference between N and F is greater than that between $\mathbf{N}$ and $\mathbf{H}$ yet the dipole moment of $\mathbf{NH}_{3}$ (1.5D) is larger than that of $\mathbf{NF}_{3} (0.2D)$. This is because

   Options: 

       A. in $NH_3$ the atomic dipole and bond dipole are in the opposite directions whereas in $NF_3$ these are in the same direction

       B. in $NH_3$ as well as in $NF_3$ the atomic dipole and bond dipole are in the same direction

       C. in $NH_3$ the atomic dipole and bond dipole are in the same direction whereas in $NF_3$ these are in opposite directions

       D. in $NH_3$ as well as in $NF_3$ the atomic dipole and bond dipole are in opposite directions.

   Correct Answer: C

   Year: 2006

   Solution: The dipole moment of $NF_3$ is 0.24 D and of $NH_3$ is 1.48 D. The difference is due to the fact that while the dipole moment due to N—F bonds in $NF_3$ are in the opposite direction to the lone pair dipole (partly cancelling out), the N—H bond dipoles in $NH_3$ are in the same direction as the lone pair dipole (adding up). 

   Step Solution: 

    1.  Identify the lone pair dipole direction: always directed away from the Nitrogen atom.

    2.  Analyze $NH_3$: Nitrogen is more electronegative than Hydrogen, so bond dipoles point towards Nitrogen.

    3.  Combine: In $NH_3$, bond dipoles and lone pair dipoles point in the same direction, increasing the net moment.

    4.  Analyze $NF_3$: Fluorine is more electronegative than Nitrogen, so bond dipoles point away from Nitrogen (towards Fluorine).

    5.  Combine: In $NF_3$, bond dipoles oppose the lone pair dipole, resulting in a very low net moment.

   Difficulty Level: Hard

   Concept Name: Resultant Dipole Moment / Vector Addition of Dipoles

   Short cut solution: In $NH_3$, lone pair + bond dipoles add up. In $NF_3$, they subtract.

Question 54

   Question: Which of the following oxides is most acidic?

   Options: 

       A. $As_2O_5$

       B. $P_2O_5$

       C. $N_2O_5$

       D. $Sb_2O_5$

   Correct Answer: C

   Year: 1999

   Solution: As among N, P, As and Sb, the former has highest electronegativity, so its oxide is most acidic. As the electronegativity of the element increases, the acidic character of the oxide also increases.

   Step Solution: 

    1.  Identify the group: All central atoms (N, P, As, Sb) belong to Group 15.

    2.  Check Oxidation States: All elements are in the +5 oxidation state.

    3.  Apply Trend: For oxides in the same oxidation state, acidity increases with the electronegativity of the central atom.

    4.  Compare Electronegativity: $N > P > As > Sb$.

    5.  Conclusion: $N_2O_5$ is the most acidic.

   Difficulty Level: Easy

   Concept Name: Periodic Trends in Acidity of Oxides

   Short cut solution: Acidity $\propto$ Electronegativity. Nitrogen is the most electronegative in the list.

Question 55

   Question: Which of the following phosphorus is the most reactive?

   Options: 

       A. Scarlet phosphorus

       B. White phosphorus

       C. Red phosphorus

       D. Violet phosphorus

   Correct Answer: B

   Year: 1999

   Solution: White phosphorus is the most reactive allotrope. (Note: Outside of the sources, this is due to the angular strain in the $P_4$ molecule where angles are $60^\circ$).

   Step Solution: 

    1.  List phosphorus allotropes: White, Red, Black (and others like Scarlet/Violet).

    2.  Compare stability: White phosphorus consists of discrete $P_4$ units.

    3.  Identify Strain: The bond angles in $P_4$ are small ($60^\circ$), creating significant instability.

    4.  Relate to Reactivity: Higher instability leads to higher reactivity.

    5.  Conclusion: White phosphorus is the most reactive and catches fire spontaneously in air.

   Difficulty Level: Easy

   Concept Name: Allotropes of Phosphorus

   Short cut solution: White P = Most strained = Most reactive.

Question 57

   Question: Repeated use of which one of the following fertilizers would increase the acidity of the soil?

   Options: 

       A. Ammonium sulphate

       B. Superphosphate of lime

       C. Urea

       D. Potassium nitrate

   Correct Answer: A

   Year: 1998

   Solution: Ammonium sulphate is a salt of a strong acid ($H_2SO_4$) and a weak base ($NH_4OH$). Repeated use increases the concentration of sulphuric acid as ammonia is used up by plants, increasing soil acidity.

   Step Solution: 

    1.  Identify the chemical formula: Ammonium sulphate is $(NH_4)_2SO_4$.

    2.  Determine salt type: It is formed from $NH_4OH$ (weak base) and $H_2SO_4$ (strong acid).

    3.  Analyze plant uptake: Plants consume the $NH_4^+$ (cationic) part as a nitrogen source.

    4.  Identify residue: The $SO_4^{2-}$ (sulfate) anion remains in the soil.

    5.  Effect: The sulfate ion associates with $H^+$ from water/biological processes to form sulphuric acid, lowering the soil pH.

   Difficulty Level: Medium

   Concept Name: Salt Hydrolysis and Soil Chemistry

   Short cut solution: Ammonium + Strong Acid Anion (like sulfate) always leaves an acidic residue in soil.

Question 58

   Question: Which of the following has the highest dipole moment?

   Options: 

       A. $SbH_3$

       B. $AsH_3$

       C. $NH_3$

       D. $PH_3$

   Correct Answer: C

   Year: 1997

   Solution: Due to greater electronegativity of nitrogen, dipole moment for $NH_3$ is greater.

   Step Solution:

    1.  Identify Central Atoms: The elements (N, P, As, Sb) all belong to Group 15.

    2.  Evaluate Electronegativity (EN): Nitrogen has the highest electronegativity in this group.

    3.  Relate EN to Bond Dipole: A higher electronegativity difference between the central atom and Hydrogen ($2.1$) results in a more polar bond.

    4.  Consider Geometry: All these hydrides are pyramidal with a lone pair. 

    5.  Conclusion: The high polarity of the $N-H$ bond, combined with the lone pair contribution, gives $NH_3$ the highest net dipole moment.

   Difficulty Level: Easy

   Concept Name: Dipole Moment and Electronegativity Trends

   Short cut solution: In a group of similar hydrides, the one with the most electronegative central atom (Nitrogen) has the highest dipole moment.

Question 59

   Question: The structural formula of hypophosphorous acid is

   Options: 

       A.

       B.

       C.

       D. None of these.

   Correct Answer: C

   Year: 1997

   Solution: The formula of hypophosphorous acid is $H_3PO_2$ as shown in (c). It is a monobasic acid.

   Step Solution:

    1.  Identify Formula: Hypophosphorous acid is $H_3PO_2$.

    2.  Determine Structure: It consists of a central Phosphorus atom.

    3.  Identify Bonds: It has one $P=O$ bond, one $P-OH$ bond, and two $P-H$ bonds.

    4.  Determine Basicity: Only the Hydrogen in the $-OH$ group is ionizable.

    5.  Conclusion: The structure with one $-OH$ group (monobasic) corresponds to $H_3PO_2$.

   Difficulty Level: Easy

   Concept Name: Structure of Phosphorus Oxoacids

   Short cut solution: Hypophosphorous acid = $H_3PO_2$ = 1 OH group (Monobasic).

Question 61

   Question: The basic character of hydrides of the V group elements decreases in the order

   Options: 

       A. $NH_3 > PH_3 > AsH_3 > SbH_3$

       B. $SbH_3 > AsH_3 > PH_3 > NH_3$

       C. $SbH_3 > PH_3 > AsH_3 > NH_3$

       D. $NH_3 > SbH_3 > PH_3 > AsH_3$

   Correct Answer: A

   Year: 1996

   Solution: All the hydrides of group V elements have one lone pair of electrons on their central atom. Therefore, they can act as Lewis bases. The basic character of these hydrides decreases down the [group].

   Step Solution:

    1.  Identify Basicity Type: These are Lewis bases due to a lone pair on the central atom.

    2.  Analyze Atomic Size: Moving from $N$ to $Sb$, the atomic size increases.

    3.  Determine Electron Density: The lone pair is concentrated in a small volume in $N$ but spread over a larger volume in $Sb$.

    4.  Evaluate Donation Tendency: Higher electron density makes it easier to donate the lone pair.

    5.  Establish Order: $NH_3$ (highest density) is the strongest base, followed by $PH_3$, $AsH_3$, and $SbH_3$.

   Difficulty Level: Easy

   Concept Name: Basic Strength of Group 15 Hydrides

   Short cut solution: Basic strength $\propto$ 1 / Size of central atom. Smallest atom (Nitrogen) = Strongest base.

Question 62

   Question: Among the following oxides, the lowest acidic is

   Options: 

       A. $As_4O_6$

       B. $As_4O_{10}$

       C. $P_4O_6$

       D. $P_4O_{10}$

   Correct Answer: A

   Year: 1996

   Solution: The acidic character of the oxides decreases with the decrease in the oxidation state and also decreases down the group.

   Step Solution:

    1.  Identify Trends: Acidity increases with higher oxidation states and decreases down a group.

    2.  Compare Oxidation States: $E_4O_6$ has a $+3$ state, while $E_4O_{10}$ has a $+5$ state.

    3.  Apply State Rule: $+3$ oxides are less acidic than $+5$ oxides.

    4.  Compare Elements: Phosphorus ($P$) is above Arsenic ($As$) in the group.

    5.  Final Comparison: $As_4O_6$ has both the lower oxidation state (+3) and the lower element position, making it the least acidic.

   Difficulty Level: Medium

   Concept Name: Periodic Trends in Oxide Acidity

   Short cut solution: Lowest acidity = Lower oxidation state (+3) + element further down the group ($As$).

Question 67

   Question: The electronic configuration of an element is $1s^2 2s^2 2p^6 3s^2 3p^3$. What is the atomic number of the element, which is just below the above element in the periodic table?

   Options: 

       A. 36

       B. 49

       C. 33

       D. 34

   Correct Answer: C

   Year: 1995

   Solution: Atomic number of the given element is 15 and it belongs to 5th group. Therefore atomic number of the element below the above element $= 15 + 18 = 33$.

   Step Solution:

    1.  Calculate Atomic Number ($Z$): Sum the electrons: $2+2+6+2+3 = \mathbf{15}$.

    2.  Identify Group: The configuration $ns^2 np^3$ indicates it is in Group 15 (or 5th group in old notation).

    3.  Identify Period: The highest principal quantum number is $n=3$, so it is in the 3rd period.

    4.  Determine Shift: To find the element in the next period (4th) of the same group, add 18.

    5.  Final Calculation: $15 + 18 = \mathbf{33}$.

   Difficulty Level: Easy

   Concept Name: Atomic Number and Group Trends

   Short cut solution: Atomic number 15 is Phosphorus ($P$). Adding the magic number 18 gives 33 (Arsenic).

Question 68

   Question: Which of the following oxides of nitrogen is paramagnetic?

   Options: 

       A. $NO_2$

       B. $N_2O_3$

       C. $N_2O$

       D. $N_2O_5$

   Correct Answer: A

   Year: 1994

   Solution: $NO_2$ is paramagnetic due to the presence of unpaired electrons.

   Step Solution:

    1.  Identify Paramagnetism: Paramagnetism occurs in molecules with unpaired (odd) electrons.

    2.  Calculate Valence Electrons ($NO_2$): $5 (\text{from } N) + 2 \times 6 (\text{from } O) = \mathbf{17}$. (Odd, so paramagnetic).

    3.  Calculate Valence Electrons ($N_2O$): $2 \times 5 + 6 = 16$. (Even).

    4.  Calculate Valence Electrons ($N_2O_3$): $2 \times 5 + 3 \times 6 = 28$. (Even).

    5.  Calculate Valence Electrons ($N_2O_5$): $2 \times 5 + 5 \times 6 = 40$. (Even).

   Difficulty Level: Easy

   Concept Name: Magnetic Properties of Molecules (Odd-electron Species)

   Short cut solution: $NO$ and $NO_2$ are the only common nitrogen oxides that are paramagnetic because they have an odd number of valence electrons.

Question 70

   Question: Which of the following fluorides does not exist?

   Options: 

       A. $NF_5$

       B. $PF_5$

       C. $AsF_5$

       D. $SbF_5$

   Correct Answer: A

   Year: 1993

   Solution: Nitrogen cannot form pentahalides because it cannot expand its octet due to non-availability of d-orbitals.

   Step Solution:

    1.  Identify Group: All central atoms belong to Group 15.

    2.  Determine Valence Shell: Nitrogen ($N$) is in the 2nd period, meaning its valence shell is $n=2$.

    3.  Check Orbitals: For $n=2$, only $2s$ and $2p$ orbitals are available; there are no $2d$ orbitals.

    4.  Analyze Bonding: To form 5 bonds (as in $NF_5$), an element must expand its octet using d-orbitals.

    5.  Conclusion: Since Nitrogen lacks d-orbitals, it cannot form $NF_5$, making option A the correct answer.

   Difficulty Level: Easy

   Concept Name: Octet Rule and Availability of d-orbitals

   Short cut solution: Nitrogen is in period 2; no d-orbitals = no pentahalides.

Question 75

   Question: Number of electrons shared in the formation of nitrogen molecule is

   Options: 

       A. 6

       B. 10

       C. 2

       D. 8

   Correct Answer: A

   Year: 1992

   Solution: Nitrogen molecule is diatomic containing a triple bond between two N atoms, $\ddot{N} \equiv \ddot{N}$ therefore, nitrogen molecule is formed by sharing six electrons.

   Step Solution:

    1.  Determine Valence Electrons: Each Nitrogen atom has 5 valence electrons ($2s^2 2p^3$).

    2.  Identify Bonding Requirement: To achieve an octet, each atom needs 3 more electrons.

    3.  Form Bonds: Two Nitrogen atoms share 3 pairs of electrons to fulfill the octet.

    4.  Count Shared Electrons: 3 pairs $\times$ 2 electrons/pair = 6 shared electrons.

    5.  Verify Structure: This creates a triple bond ($\equiv$).

   Difficulty Level: Easy

   Concept Name: Lewis Dot Structure / Covalent Bonding

   Short cut solution: Nitrogen forms a triple bond; 1 bond = 2 electrons, so 3 bonds = 6 electrons.

Question 76

   Question: Sugarcane on reaction with nitric acid gives

   Options: 

       A. $CO_2$ and $SO_2$

       B. $(COOH)_2$

       C. 2 $HCOOH$ (two moles)

       D. no reaction

   Correct Answer: B

   Year: 1992

   Solution: Sugarcane (sucrose) reacts with concentrated nitric acid to produce oxalic acid. (Note: While the chemical solution text is sparse in the source, the correct option B corresponds to the formula for oxalic acid).

   Step Solution:

    1.  Identify Reactants: Sugarcane (Sucrose, $C_{12}H_{22}O_{11}$) and Nitric acid ($HNO_3$).

    2.  Determine Reaction Type: Concentrated $HNO_3$ acts as a powerful oxidizing agent.

    3.  Analyze Oxidation: Sucrose is oxidized by the acid.

    4.  Identify Product: The carbon chain is broken down and oxidized to oxalic acid.

    5.  Match Formula: Oxalic acid is represented as $(COOH)_2$, which is option B.

   Difficulty Level: Medium

   Concept Name: Oxidizing Nature of Nitric Acid

   Short cut solution: Oxidation of sugar by $HNO_3$ $\rightarrow$ Oxalic acid.

Question 77

   Question: Nitrogen is relatively inactive element because

   Options: 

       A. its atom has a stable electronic configuration

       B. it has low atomic radius

       C. its electronegativity is fairly high

       D. dissociation energy of its molecule is fairly high

   Correct Answer: D

   Year: 1992

   Solution: $N_2$ molecule contains triple bond between N atoms having very high dissociation energy ($946\text{ kJ mol}^{-1}$) due to which it is relatively inactive.

   Step Solution:

    1.  Analyze $N_2$ Structure: Nitrogen exists as $N \equiv N$ with a triple bond.

    2.  Define Reactivity: Chemical reactivity often requires the breaking of existing bonds to form new ones.

    3.  Evaluate Bond Strength: A triple bond is extremely strong and difficult to break.

    4.  Quantify Energy: The bond dissociation energy is very high at $946\text{ kJ/mol}$.

    5.  Conclusion: High energy requirement for bond cleavage makes $N_2$ chemically inert at room temperature.

   Difficulty Level: Easy

   Concept Name: Bond Dissociation Enthalpy

   Short cut solution: Triple bond = Very high energy to break = Chemical Inactivity.

Question 78

   Question: $H_3PO_2$ is the molecular formula of an acid of phosphorus. Its name and basicity respectively are

   Options: 

       A. phosphorous acid and two

       B. hypophosphorous acid and two

       C. hypophosphorous acid and one

       D. hypophosphoric acid and two

   Correct Answer: C

   Year: 1992

   Solution: $H_3PO_2$ is named as hypophosphorous acid. As it contains only one P—OH group, its basicity is one. 

   Step Solution:

    1.  Name the Acid: $H_3PO_2$ is known as hypophosphorous acid.

    2.  Determine Structure: The central Phosphorus atom is bonded to one $=O$, two $-H$, and one $-OH$ group.

    3.  Define Basicity: Basicity is the number of ionizable hydrogen atoms (those bonded to Oxygen in oxoacids).

    4.  Count Ionizable Hydrogens: There is only one P—OH bond in $H_3PO_2$.

    5.  Conclusion: The basicity is one (monobasic/monoprotic).

   Difficulty Level: Easy

   Concept Name: Basicity of Oxoacids

   Short cut solution: $H_3PO_2$ has 1 OH group $\rightarrow$ Basicity = 1.

Question 86

   Question: Aqueous solution of ammonia consists of

   Options: 

       A. $H^+$

       B. $OH^−$

       C. $NH_4^+$

       D. $NH_4^+$ and $OH^−$

   Correct Answer: D

   Year: 1991

   Solution: Aqueous solution of ammonia contains $NH_4^+$ and $OH^−$ ions. $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^−$.

   Step Solution:

    1.  Analyze Reactants: Ammonia ($NH_3$) and Water ($H_2O$).

    2.  Identify Chemical Nature: $NH_3$ acts as a Lewis base because it has a lone pair of electrons to donate.

    3.  Analyze Reaction: $NH_3$ accepts a proton ($H^+$) from water.

    4.  Form Products: This results in the formation of the ammonium ion ($NH_4^+$) and the hydroxide ion ($OH^-$).

    5.  Equation: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$.

   Difficulty Level: Easy

   Concept Name: Lewis Base / Brønsted-Lowry Theory

   Short cut solution: Ammonia + Water $\rightarrow$ $NH_4^+$ and $OH^−$.

Question 87

   Question: $P_2O_5$ is heated with water to give

   Options: 

       A. hypophosphorous acid

       B. phosphorous acid

       C. hypophosphoric acid

       D. orthophosphoric acid

   Correct Answer: D

   Year: 1991

   Solution: $P_2O_5 + 3H_2O \xrightarrow{\Delta} 2H_3PO_4$

   Step Solution: 

    1.  Identify the reactants: Phosphorus pentoxide ($P_2O_5$) and Water ($H_2O$).

    2.  Identify the nature of $P_2O_5$: It is the acid anhydride of phosphoric acid.

    3.  Apply the hydration reaction: $P_2O_5 + 3H_2O \rightarrow 2H_3PO_4$.

    4.  Name the product: $H_3PO_4$ is known as orthophosphoric acid.

   Difficulty Level: Easy

   Concept Name: Hydration of Oxides

   Short cut solution: $P_2O_5$ (+5 oxidation state) reacts with water to form the corresponding +5 oxoacid, orthophosphoric acid.

Question 88

   Question: Basicity of orthophosphoric acid is

   Options: 

       A. 2

       B. 3

       C. 4

       D. 5

   Correct Answer: B

   Year: 1991

   Solution: Orthophosphoric acid, $H_3PO_4$ contains three P—OH groups and is therefore, tribasic. 

   Step Solution: 

    1.  Recall the formula for orthophosphoric acid: $H_3PO_4$.

    2.  Determine its structure: The central Phosphorus atom is bonded to one $=O$ and three $-OH$ groups.

    3.  Identify ionizable hydrogens: Only hydrogens attached to oxygen (in —OH groups) are ionizable.

    4.  Count the groups: There are three P—OH groups.

    5.  Conclusion: The basicity is 3 (tribasic).

   Difficulty Level: Easy

   Concept Name: Basicity of Oxoacids

   Short cut solution: In $H_3PO_4$, all three hydrogens are part of P—OH groups, making it tribasic.

Question 89

   Question: $PCl_3$ reacts with water to form

   Options: 

       A. $PH_3$

       B. $H_3PO_3$, $HCl$

       C. $POCl_3$

       D. $H_3PO_4$

   Correct Answer: B

   Year: 1991

   Solution: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$

   Step Solution: 

    1.  Analyze the reaction type: This is a hydrolysis reaction of a phosphorus halide.

    2.  Check the oxidation state of P in $PCl_3$: It is +3.

    3.  Form the products: Hydrolysis results in the corresponding oxoacid with the same oxidation state (+3).

    4.  Identify the acid: $H_3PO_3$ (phosphorous acid) is the +3 oxoacid.

    5.  Balance the equation: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$.

   Difficulty Level: Easy

   Concept Name: Hydrolysis of Phosphorus Halides

   Short cut solution: $PCl_3$ (+3 state) hydrolyzed $\rightarrow$ $H_3PO_3$ (+3 state).

Question 90

   Question: $PH_4I + NaOH$ forms

   Options: 

       A. $PH_3$

       B. $NH_3$

       C. $P_4O_6$

       D. $P_4O_{10}$

   Correct Answer: A

   Year: 1991

   Solution: $PH_4I + NaOH \rightarrow NaI + PH_3 + H_2O$

   Step Solution: 

    1.  Identify the reactants: Phosphonium iodide ($PH_4I$) and Sodium hydroxide ($NaOH$).

    2.  Analyze the reaction: The phosphonium ion ($PH_4^+$) reacts with the hydroxide base ($OH^-$).

    3.  De-protonation: $OH^-$ removes a proton ($H^+$) from $PH_4^+$ to form water ($H_2O$) and phosphine ($PH_3$).

    4.  Form side product: The remaining $Na^+$ and $I^-$ ions form $NaI$.

    5.  Net result: The reaction produces $PH_3$ gas.

   Difficulty Level: Medium

   Concept Name: Preparation of Phosphine

   Short cut solution: Phosphonium salt + Strong base always liberates Phosphine ($PH_3$).

Question 91

   Question: Pure nitrogen is prepared in the laboratory by heating a mixture of

   Options: 

       A. $NH_4OH + NaCl$

       B. $NH_4NO_3 + NaCl$

       C. $NH_4Cl + NaOH$

       D. $NH_4Cl + NaNO_2$

   Correct Answer: D

   Year: 1991

   Solution: $NH_4Cl + NaNO_2 \xrightarrow{Heat} NH_4NO_2 + NaCl$, followed by $NH_4NO_2 \rightarrow N_2 + 2H_2O$ 

   Step Solution: 

    1.  Identify the starting materials: Ammonium chloride and sodium nitrite.

    2.  Analyze the first step: Heating the mixture produces ammonium nitrite ($NH_4NO_2$) via double displacement.

    3.  Analyze the second step: $NH_4NO_2$ is thermally unstable.

    4.  Decomposition: $NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$.

    5.  Conclusion: This reaction sequence produces Pure Nitrogen.

   Difficulty Level: Medium

   Concept Name: Laboratory Preparation of Nitrogen

   Short cut solution: Heating Ammonium salt + Nitrite $\rightarrow$ $N_2$ gas.

Question 93

   Question: Which of the following statement is not correct for nitrogen?

   Options: 

       A. Its electronegativity is very high.

       B. d-orbitals are available for bonding.

       C. It is a typical non-metal.

       D. Its molecular size is small.

   Correct Answer: B

   Year: 1990

   Solution: In case of nitrogen, d-orbitals are not available for bonding. $N: 1s^2 2s^2 2p^3$

   Step Solution: 

    1.  Check Period: Nitrogen belongs to the 2nd period of the periodic table.

    2.  Check Valence Shell: For the 2nd period, the principal quantum number $n=2$.

    3.  Available Orbitals: The $n=2$ shell only contains $2s$ and $2p$ subshells.

    4.  Analyze d-orbitals: d-orbitals only begin appearing at $n=3$ ($3d$ orbitals).

    5.  Conclusion: Nitrogen lacks d-orbitals in its valence shell, making statement B incorrect.

   Difficulty Level: Easy

   Concept Name: Anomalous Properties of Nitrogen / Atomic Structure

   Short cut solution: Nitrogen is in the 2nd period; 2nd period elements never have available d-orbitals.

Question 94

   Question: Which of the following compound does not exist?

   Options: 

       A. $NCl_5$

       B. $AsF_5$

       C. $SbCl_5$

       D. $PF_5$

   Correct Answer: A

   Year: 1989

   Solution: All the elements of group 15 form trihalides and pentahalides of the type $MX_3$ and $MX_5$ except nitrogen which forms only trihalides. Moreover, nitrogen does not form pentahalides due to the absence of d-orbitals in its valence shell.

   Step Solution:

    1.  Identify the central atom's group: All options belong to Group 15.

    2.  Check the valence shell: Nitrogen ($N$) is in the 2nd period.

    3.  Analyze orbital availability: Period 2 elements only have $2s$ and $2p$ orbitals; there are no d-orbitals.

    4.  Apply bonding theory: Forming 5 bonds (pentahalides) requires octet expansion, which is impossible without available d-orbitals.

    5.  Conclusion: Nitrogen cannot form $NCl_5$.

   Difficulty Level: Easy

   Concept Name: Octet Rule and Availability of d-orbitals

   Short cut solution: Nitrogen is in period 2 $\rightarrow$ no d-orbitals $\rightarrow$ no pentahalides.

Question 95

   Question: Each of the following is true for white and red phosphorus except that they

   Options: 

       A. are both soluble in $CS_2$

       B. can be oxidised by heating in air

       C. consist of the same kind of atoms

       D. can be converted into one another.

   Correct Answer: A

   Year: 1989

   Solution: Red phosphorus is insoluble in $CS_2$ and only white P is soluble in $CS_2$.

   Step Solution:

    1.  Verify Allotropy: Both are forms of the same element, Phosphorus (same atoms).

    2.  Check Reactivity: Both allotropes can be oxidized (burned) when heated in air.

    3.  Check Interconversion: Allotropes of phosphorus can be converted into one another under specific conditions.

    4.  Analyze Solubility: Only white phosphorus dissolves in Carbon disulphide ($CS_2$).

    5.  Conclusion: Statement A is the false one as red phosphorus is insoluble.

   Difficulty Level: Easy

   Concept Name: Properties of Phosphorus Allotropes

   Short cut solution: Only White P is soluble in $CS_2$.

Question 96

   Question: When orthophosphoric acid is heated to $600^\circ C$, the product formed is

   Options: 

       A. $PH_3$

       B. $P_2O_5$

       C. $H_3PO_3$

       D. $HPO_3$

   Correct Answer: D

   Year: 1989

   Solution: On heating, it gives pyrophosphoric acid at 525K and metaphosphoric acid at 875K.

   Step Solution:

    1.  Identify the acid: Orthophosphoric acid is $H_3PO_4$.

    2.  Convert units: $600^\circ C \approx 873K$.

    3.  Analyze heating effect: Heating $H_3PO_4$ causes dehydration (loss of $H_2O$).

    4.  Match Temperature: At high temperatures (approx. 875K), orthophosphoric acid dehydrates to metaphosphoric acid.

    5.  Formula: The formula for metaphosphoric acid is $HPO_3$.

   Difficulty Level: Medium

   Concept Name: Thermal Decomposition of Oxoacids

   Short cut solution: $H_3PO_4 + \text{Strong Heat} \xrightarrow{-H_2O} \mathbf{HPO_3}$ (Metaphosphoric acid).

Question 97

   Question: Which one has the lowest boiling point?

   Options: 

       A. $NH_3$

       B. $PH_3$

       C. $AsH_3$

       D. $SbH_3$

   Correct Answer: B

   Year: 1989

   Solution: Boiling point of hydrides increases with increase in atomic number but ammonia has exceptionally high boiling point due to hydrogen bonding. Thus the correct order of boiling point is, $BiH_3 > SbH_3 > NH_3 > AsH_3 > PH_3$.

   Step Solution:

    1.  Determine General Trend: Boiling points of group hydrides generally increase with molar mass due to van der Waals forces.

    2.  Identify Exception: $NH_3$ has intermolecular hydrogen bonding, which gives it an abnormally high boiling point.

    3.  Compare Others: $PH_3, AsH_3, SbH_3$ follow the molar mass trend.

    4.  Establish Order: $SbH_3 > AsH_3 > PH_3$. Adding $NH_3$ puts it above $AsH_3$.

    5.  Identify Minimum: The resulting sequence shows $PH_3$ has the lowest boiling point.

   Difficulty Level: Medium

   Concept Name: Boiling Point Trends in Group 15 Hydrides

   Short cut solution: Lowest B.P. in Group 15 is always $PH_3$.

Question 102

   Question: Which of the following is a nitric acid anhydride?

   Options: 

       A. $NO$

       B. $NO_2$

       C. $N_2O_5$

       D. $N_2O_3$

   Correct Answer: C

   Year: 1988

   Solution: When 2-molecules of nitric acid undergoes heating, loose a water molecule to form an anhydride. Thus, $N_2O_5$ is nitric acid anhydride.

   Step Solution:

    1.  Identify formula for Nitric Acid: $HNO_3$.

    2.  Combine molecules: Take 2 molecules of the acid: $2 \times (HNO_3) = H_2N_2O_6$.

    3.  Remove Water: Subtract one $H_2O$ molecule: $H_2N_2O_6 - H_2O = \mathbf{N_2O_5}$.

    4.  Verify Oxidation State: In $HNO_3$, Nitrogen is in the +5 state.

    5.  Match Oxide: $N_2O_5$ is the oxide where Nitrogen is also in the +5 state.

   Difficulty Level: Easy

   Concept Name: Acid Anhydrides

   Short cut solution: Anhydride of $HNO_3$ (+5 state) is $N_2O_5$ (+5 state).