P Block Group 15 (Nitrogen Family) Notes, Properties & Reactions

Q 1 :

Given below are two statements: [2026]

Statement I: The halogen that makes the longest bond with hydrogen in HX, has the smallest covalent radius in its group.

Statement II: A group 15 element’s hydride ${EH}_{3}$ has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4.

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

Both Statement I and Statement II are true
Statement I is true but Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true

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$HF < HCl < HBr < HI (bond length order)$

$F < Cl < Br < I (radius order)$

${PH}_{3} < {AsH}_{3} < {NH}_{3} < {SbH}_{3} < {BiH}_{3} (Boiling point order)$

$Maximum possible covalency of phosphorus is 6$

Q 2 :

Elements X and Y belong to Group 15. The difference between the electronegativity values of ‘X’ and phosphorus is higher than that of the difference between phosphorus and ‘Y’. ‘X’ & ‘Y’ are respectively [2026]

As & Sb
As & Bi
N & As
Bi & N

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Element EN
N 3.0
P 2.1
As 2.0
Sb 1.9
Bi 1.9

X = Nitrogen (N)

Y = Arsenic (As)

Q 3 :

Regarding the hydrides of group 15 elements $E H_{3} (E = N, P, A s, S b)$ , select the correct statement from the following:

A. The stability of hydrides decreases down the group.
B. The basicity of hydrides decreases down the group.
C. The reducing character increases down the group.
D. The boiling point increases down the group.

Choose the correct answer from the options given below: [2026]

A & D only
A, B, C & D
A, B & C only
B & C only

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${Stability: NH}_{3} > {PH}_{3} > {AsH}_{3} > {SbH}_{3} > {BiH}_{3}$

${Basicity: NH}_{3} > {PH}_{3} > {AsH}_{3} > {SbH}_{3} > {BiH}_{3}$

${Reducing character: NH}_{3} < {PH}_{3} < {AsH}_{3} < {SbH}_{3} < {BiH}_{3}$

${Boiling point: PH}_{3} < {AsH}_{3} < {NH}_{3} < {SbH}_{3} < {BiH}_{3}$

Q 4 :

Statement – I: All elements of group 15 form hydrides of type ${EH}_{3}$ where E = N, P, As, Sb or Bi

Statement – II: All elements of group 15 form halides of type ${EX}_{3}$ where E = N, P, As, Sb or Bi

Statement – III: In case of halides of nitrogen only ${NF}_{3}$ is known to be stable

Statement – IV: The oxides of type $E_{2} O_{3}$ of arsenic and antimony are amphoteric and those of bismuth are predominantly basic

Select the correct combination

I, II, III only
I, II, III, IV
I, III, IV only
I, IV only

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(2)

Q 5 :

Which of the following is a neutral oxide?

$N a_{2} O$
$N_{2} O$
$N a O_{2}$
$N O_{2}$

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$N_{2} O$ , NO and CO are neutral oxides.

Q 6 :

Consider the following orders against the mentioned property:

I. $Mg < Si < Al < S < P$ (2nd ionisation potential)

II. ${Na}_{2} O < MgO < {Al}_{2} O_{3} < P_{2} O_{5}$ (Acidic strength)

III. $C < N < O < F$ (Electron affinity)

The incorrect order(s) is/are

Only I
Only III
Only I and III
I, II and III

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(3)

Hint: On moving left to right, acidic strength of oxide increases.

Correct order of 2nd I.P. : $M g < S i < A l < P < S$

Correct order of electron affinity: $N < C < O < F$

Q 7 :

Assertion: Boiling point of $P H_{3} < A s H_{3} < S b H_{3} < N H_{3}$

Reason: $N H_{3}$ possesses intermolecular H-bonding

A is True, R is true R is correct explanation of A
A is True, R is True R is not correct explanation of A
A is True, R is false
A is false, R is True

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(4)

$B.P. P H_{3} < A s H_{3} < N H_{3} < S b H_{3}$

Q 8 :

$Which of the following is incorrect$

$Acidic order: N_{2} O_{5} > P_{2} O_{5} > A s_{2} O_{5} > S b_{2} O_{5} > B i_{2} O_{5}$
$Stability: Black-P > Red-P > White-P$
$Covalent Radius: P o > T e > S e > S > O$
$Ionisation Energy S > O > S e > T e > P o$

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$The correct order for Ionisation Energy is O > S > S e > T e > P o$

Q 9 :

Assertion (A): The bond angles of the hydrides decreases in the order $N H_{3} > P H_{3} > A s H_{3} > S b H_{3} > B i H_{3}$ .

Reason (R): The Lewis basic strength of the hydrides increases in the order $N H_{3} < P H_{3} < A s H_{3} < S b H_{3}$ .

Both A and R are correct and R is the correct explanation of A
Both A and R are correct and R is not the correct explanation of A
A is correct but R is incorrect
A is incorrect but R is correct

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$N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$ : Lewis basic structure

Q 10 :

Correct statements from the following are:

A. Nitrogen in oxidation states from +1 to +4 disproportionates in acid medium.

B. Nitrogen has the ability to form $d π - p π$ multiple bonds with itself and other elements with small size and high electronegativity.

C. N-N single bond is stronger than P-P single bond.

D. Nitrogen has highest density in its group due to small size.

E. The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.

Choose the correct answer from the options given below: [2026]

B, C and D only
C, D and E only
A, C and E only
A and E only

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A. For nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution. For example,

$3 H \overset{+ 3}{N} O_{2} \to H \overset{+ 5}{N} O_{3} + H_{2} O + 2 \overset{+ 2}{N} O$

B. Nitrogen has unique ability to form $p π - p π$ multiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O).

C. The single N–N bond is weaker than the single P–P bond because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length.

D. Dinitrogen is a diatomic gas while all others are solids. Thus dinitrogen has lowest density in the group.

E. Nitrogen is restricted to a maximum covalency of 4 since only four (one $s$ and three $p$ ) orbitals are available for bonding

Q 11 :

Find the correct statements related to group 15 hydrides.

A. Reducing nature increases from ${NH}_{3}$ to ${BiH}_{3}$ .

B. Tendency to donate lone pair of electrons decreases from ${NH}_{3}$ to ${BiH}_{3}$ .

C. The stability of hydrides decreases from ${NH}_{3}$ to ${BiH}_{3}$ .

D. $HEH$ bond angle decreases from ${NH}_{3}$ to ${SbH}_{3}$ ( $E = Elements of group 15$ )

Choose the correct answer from the options given below: [2026]

A and B only
B and C only
A, B, C and D
A, C and D only

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A. C. Bond length of hydrides increases down the group. Thus, bond dissociation energy decreases down the group. This results in decrease of stability down the group. As bond length increases down the group, abstraction of $H^{-}$ becomes easier and reducing nature increases down the group.

B. Down the group, lone pair of electrons gets diffused as size of orbital containing lone pair increases. Thus effectiveness of electron donation decreases down the group.

D. Down the group, electronegativity of central atom decreases. Thus bond pair of electrons is shifted farther away from the central atom. This reduces bond pair- bond pair repulsion and hence bond angle reduces down the group.

Q 12 :

A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with $K_{2} {CrO}_{4}$ to give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. ${HNO}_{3}$ and ammonium molybdate, it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively: [2026]

${Ca}^{2 +}$ and ${SO}_{4}^{2 -}$
${Ba}^{2 +}$ and ${PO}_{4}^{3 -}$
${Mn}^{2 +}$ and ${PO}_{4}^{3 -}$
${Ba}^{2 +}$ and ${SO}_{4}^{2 -}$

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(i) ${BaCl}_{2}$ gives apple green colour of the flame observed by naked eye.

(ii) ${BaCO}_{3} + 2 {CH}_{3} COOH \to {({CH}_{3} COO)}_{2} Ba + H_{2} O + {CO}_{2}$

${(CH}_{3} {COO)}_{2} Ba + K_{2} {CrO}_{4} \to \underset{(yellow ppt.)}{{BaCrO}_{4}} + 2 {CH}_{3} COOK$

(iii) ${Na}_{2} {HPO}_{4} + 12 {({NH}_{4})}_{2} {MoO}_{4} + 23 {HNO}_{3} \to \underset{Canary yellow ppt.}{{({NH}_{4})}_{3} [P {({Mo}_{3} O_{10})}_{4}]} + 2 {NaNO}_{3} + 21 {NH}_{4} {NO}_{3} + 12 H_{2} O$

Q 13 :

Treatment of a gas ' $X$ ' with a freshly prepared ferrous sulphate solution gives a compound ' $Y$ ' as a brown ring. The compounds $X$ and $Y$ are [2026]

$NO$ and $[Fe (NO)] {SO}_{4}$
${NO}_{2}$ and $[Fe ({NO}_{2})] {SO}_{4}$
$N_{2} O$ and $[Fe (N_{2} O)] {SO}_{4}$
$N_{2} O_{4}$ and $[Fe (N_{2} O_{4})] {SO}_{4}$

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${FeSO}_{4} + NO \to [Fe(NO)] {SO}_{4} (brown)$

Topic Question Set

Q 2 :

Elements X and Y belong to Group 15. The difference between the electronegativity values of ‘X’ and phosphorus is higher than that of the difference between phosphorus and ‘Y’. ‘X’ & ‘Y’ are respectively [2026]

(2)

Q 5 :

Which of the following is a neutral oxide?

Q 6 :

Consider the following orders against the mentioned property:

I. $Mg < Si < Al < S < P$ (2nd ionisation potential)

II. ${Na}_{2} O < MgO < {Al}_{2} O_{3} < P_{2} O_{5}$ (Acidic strength)

III. $C < N < O < F$ (Electron affinity)

The incorrect order(s) is/are

Q 7 :

Assertion: Boiling point of $P H_{3} < A s H_{3} < S b H_{3} < N H_{3}$

Reason: $N H_{3}$ possesses intermolecular H-bonding

Q 8 :

$Which of the following is incorrect$

Q 9 :

Assertion (A): The bond angles of the hydrides decreases in the order $N H_{3} > P H_{3} > A s H_{3} > S b H_{3} > B i H_{3}$ .

Reason (R): The Lewis basic strength of the hydrides increases in the order $N H_{3} < P H_{3} < A s H_{3} < S b H_{3}$ .

Q 13 :

Treatment of a gas ' $X$ ' with a freshly prepared ferrous sulphate solution gives a compound ' $Y$ ' as a brown ring. The compounds $X$ and $Y$ are [2026]

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Topic Question Set

Q 2 : Elements X and Y belong to Group 15. The difference between the electronegativity values of ‘X’ and phosphorus is higher than that of the difference between phosphorus and ‘Y’. ‘X’ & ‘Y’ are respectively [2026]

(2)

Q 5 : Which of the following is a neutral oxide?

Q 6 : Consider the following orders against the mentioned property: I. Mg<Si<Al<S<P (2nd ionisation potential) II. Na2O<MgO<Al2O3<P2O5 (Acidic strength) III. C<N<O<F (Electron affinity) The incorrect order(s) is/are

Q 7 : Assertion: Boiling point of PH3<AsH3<SbH3<NH3 Reason: NH3 possesses intermolecular H-bonding

Q 8 : Which of the following is incorrect

Q 9 : Assertion (A): The bond angles of the hydrides decreases in the order NH3>PH3>AsH3>SbH3>BiH3. Reason (R): The Lewis basic strength of the hydrides increases in the order NH3<PH3<AsH3<SbH3.

Q 13 : Treatment of a gas 'X' with a freshly prepared ferrous sulphate solution gives a compound 'Y' as a brown ring. The compounds X and Y are [2026]

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Q 2 :

Elements X and Y belong to Group 15. The difference between the electronegativity values of ‘X’ and phosphorus is higher than that of the difference between phosphorus and ‘Y’. ‘X’ & ‘Y’ are respectively [2026]

Q 5 :

Which of the following is a neutral oxide?

Q 6 :

Consider the following orders against the mentioned property:

I. $Mg < Si < Al < S < P$ (2nd ionisation potential)

II. ${Na}_{2} O < MgO < {Al}_{2} O_{3} < P_{2} O_{5}$ (Acidic strength)

III. $C < N < O < F$ (Electron affinity)

The incorrect order(s) is/are

Q 7 :

Assertion: Boiling point of $P H_{3} < A s H_{3} < S b H_{3} < N H_{3}$

Reason: $N H_{3}$ possesses intermolecular H-bonding

Q 8 :

$Which of the following is incorrect$

Q 9 :

Assertion (A): The bond angles of the hydrides decreases in the order $N H_{3} > P H_{3} > A s H_{3} > S b H_{3} > B i H_{3}$ .

Reason (R): The Lewis basic strength of the hydrides increases in the order $N H_{3} < P H_{3} < A s H_{3} < S b H_{3}$ .

Q 13 :

Treatment of a gas ' $X$ ' with a freshly prepared ferrous sulphate solution gives a compound ' $Y$ ' as a brown ring. The compounds $X$ and $Y$ are [2026]