Question: From NCERT NEET - 2021

Chapter|Topic|Subtopic-Group 17 - Preparation,Properties & Uses |

Statement I:

Acid strength increases in the order given as HF << HCl << HBr << HI.

Statement II:

As the size of the elements F, Cl, Br, I increase down the group, the bond strength of HF, HCI, HBr, and HI decreases and so the acid strength increases.

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

1. Statement I is correct but Statement II is false.

2. Statement I is incorrect but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer ▽

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Question: From NCERT NEET - 2021

Topic- Subtopic: Group 18- Preparation,Properties & Uses|

The incorrect statement about noble gases is -

1. Noble gases have weak dispersion forces.

2. Noble gases have large positive values of electron gain enthalpy.

3. Noble gases are sparingly soluble in water.

4. Noble gases have very high melting and boiling points.

Answer ▽

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Question: From NCERT NEET - 2021

Chapter|Topic|Subtopic- Group 15 -Preparation,Properties & Uses|

In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?

1. NH3<PH3<AsH3<SbH3 : Increasing acidic character

2. CO2<SiO2<SnO2<PbO2 : Increasing oxidizing power

3. HF<HCl<HBr<HI : Increasing acidic strength

4. H2O<H2S<H2Se<H2Te : Increasing pKa values

Answer ▽

On going down the group thermal stability order for $H_{2} E$ Hdecreases because H−EH−E bond energy decreases
Hence,order of stability would be
$H_{2} P o < H_{2} T e < H_{2} S e < H_{2} S < H_{2} O$

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Question: From NCERT NEET - 2020

Chapter|Topic|Subtopic-Group 16- Preparation,Properties & Uses |

Which of the following compound of sulphur has $- O - O -$ linkage?

1. $H_{2} {SO}_{4}, sulphuric acid$

2. $H_{2} S_{2} O_{8}, peroxodisulphuric acid$

3. $H_{2} S_{2} O_{7}, pyrosulphuric acid$

4. $H_{2} {SO}_{3}, sulphurous acid$

Answer ▽

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Question: From NCERT NEET - 2020

Chapter|Topic|Subtopic-Group 18- Preparation,Properties & Uses |

Match the compounds of Xe in column I with the molecular structure in column II.

Column-I Column-II

(a) XeF2 (i) Square planar

(b) XeF4 (ii) Linear

(d) XeOF4 (iv) Pyramidal

1. (a)-(ii) , (b)-(i) , (c)-(iii) , (d)-(iv)

2. (a)-(ii) , (b)-(iv) , (c)-(iii) , (d)-(i)

3. (a)-(ii) , (b)-(iii) , (c)-(i) , (d)-(iv)

4. (a)-(ii) , (b)-(i) , (c)-(iv) , (d)-(iii)

Answer ▽

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Question: From NCERT NEET - 2019

Chapter|Topic|Subtopic-Group 18- Preparation,Properties & Uses |

Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option.

Column I Column II
A. XeF6 1. Distorted octahedral
B. XeO3 2. Square planar
C. XeOF4 3. Pyramidal
D. XeF4 4. Square pyramidal

Codes
A B C D
(a) 1 2 4 3
(b) 4 3 1 2
(c) 4 1 2 3
(d) 1 3 4 2

Answer ▽

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Question: From NCERT NEET - 2019

Chapter|Topic|Subtopic- Group 16- Preparation,Properties & Uses|

Which is the correct thermal stability order for $H_{2} E (E = O, S, Se, Te and Po) ?$

1. $H_{2} Se < H_{2} Te < H_{2} Po < H_{2} O < H_{2} S$

2. $H_{2} S < H_{2} O < H_{2} Se < H_{2} Te < H_{2} Po$

3. $H_{2} O < H_{2} S < H_{2} Se < H_{2} Te < H_{2} Po$

4. $H_{2} Po < H_{2} Te < H_{2} Se < H_{2} S < H_{2} O$

Answer ▽

On going down the group thermal stability order for $H_{2} E$ Hdecreases because H−EH−E bond energy decreases
Hence,order of stability would be
$H_{2} P o < H_{2} T e < H_{2} S e < H_{2} S < H_{2} O$

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Question: From NCERT NEET - 2019

Chapter|Topic|Subtopic- Group 15/16/17 -Preparation,Properties & Uses |

Match the following:

(a) Pure nitrogen (i) Chlorine

(b) Haber process (ii) Sulphuric acid

(d) Deacon's process (iv) Sodium azide or Barium azide

The correct matching among the following is -

(a) (b) (c) (d)

1. (iv) (iii) (ii) (i)

2. (i) (ii) (iii) (iv)

3. (ii) (iv) (i) (iii)

4. (iii) (iv) (ii) (i)

Answer ▽

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Question: From NCERT NEET - 2019

Chapter|Topic|Subtopic- |

The incorrect statement among the following related to ${PCl}_{5}$ is:

1. ${PCl}_{5}$ molecules is non-reactive

2. Three equatorial $P - Cl$ bonds make an angle of $120 °$ with each other

3. Two axial $P - Cl$ bonds make an angle of $180 °$ with each other

4. Axial $P - Cl$ bonds are longer than equatorial $P - Cl$ bonds

Subtopic: Group 15 -Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET - 2019

Chapter|Topic|Subtopic- |

A compound 'X' upon reaction with $H_{2} O$ produces a colorless gas 'Y' with a rotten fish smell. Gas 'Y' is absorbed in a solution of ${CuSO}_{4}$ to give ${Cu}_{3} P_{2}$ as one of the products. The compound 'X' is-

1. Ca3P2

2. ${NH}_{4} Cl$

3. ${As}_{2} O_{3}$

4. Ca3(PO4)2

Subtopic: Group 15 -Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET - 2019

Chapter|Topic|Subtopic: Group 15 -Preparation,Properties & Uses |

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

(1) $H_{4} P_{2} O_{7}$

(2) $H_{3} P O_{3}$

(3) $H_{3} P O_{2}$

(4) $H_{3} P O_{4}$

Answer ▽

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Question: From NCERT NEET - 2018

Chapter|Topic|Subtopic- |

Which of the following statements is not true for halogens?

1. All form monobasic oxyacids.
2. All are oxidizing agents.
3. All but fluorine show positive oxidation states.
4. Chlorine has the highest electron-gain enthalpy.

Subtopic: Group 17 - Preparation,Properties & Uses |

Answer ▽

Fluorine shows only $1$ oxidation state and other halogen shows negative and positive oxidation state. Due to high electronegativity and small size only F form one oxoacid that is HOF.

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Question: From NCERT NEET - 2017

Chapter|Topic|Subtopic- Interhalogen Compounds|

Match the interhalogen compounds of Column I with the geometry in Column II and assign the correct code.

Column I Column II
A. XX' (i) T-shape
B. XX'3 (ii) Pentagonal bipyramidal
C. XX'5 (iii) Linear
D. XX'7 (iv) Square-pyramidal
(v) Tetrahedral

The correct match among the following is -
A B C D
1. (iii) (iv) (i) (ii)

2. (iii) (i) (iv) (ii)

3. (v) (iv) (iii) (ii)

4. (iv) (iii) (ii) (i)

Answer ▽

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Question: From NCERT NEET - 2017

Chapter|Topic|Subtopic- Group 17 - Preparation,Properties & Uses ||

hich of the following pairs of the compound is isostructural?

{TeF}_{2}, {XeF}_{2}

IB {r_{2}}^{-}, {XeF}_{2}

{IF}_{3}, {XeF}_{2}

{BeCl}_{2}, {XeF}_{2}

Answer ▽

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Question: From NCERT NEET - 2016

Chapter|Topic|Subtopic-Group 15 -Preparation,Properties & Uses | |

Copper upon heating with conc. HNO3 produces -

1. Cu(NO3)2 and NO

2. Cu(NO3)2, NO and NO2

3. Cu(NO3)2 and N2O

4. Cu(NO3)2 and NO2

Answer ▽

When copper is heated with conc. $H N O_{3}$ it produces cupric nitrate $C u (N O_{3})_{2}$ and nitrogen dioxide $N O_{2}$ .

$C u + 4 H N O_{3} \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O$

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Question: From NCERT NEET - Derived

Chapter|Topic|Subtopic-Oxides of Nitrogen |

Addition of $c o n c . H N O_{3}$ to $c o n c . H_{2} S O_{4}$ gives:

$S O_{4}^{2 -}$

$S O_{3}$

$N O_{3}^{-}$

$N O_{2}^{+}$

Answer ▽

Correct option is

$N O_{2}^{+}$

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Question: From NCERT NEET - Derived

Chapter|Topic|Subtopic- Oxides of Nitrogen|

Nitrogen dioxide is not produced on heating :

$K N O_{3}$

$P b (N O_{3})_{2}$

$C u (N O_{3})_{2}$

$A g N O_{3}$

Medium

Answer ▽

Correct option is

$K N O_{3}$

Nitrogen dioxide is not produced on heating $K N O_{3}$
Heavy metal nitrates liberate $N O_{2}$ on heating.

$2 K N O_{3} Δ 2 K N O_{2} + O_{2}$

$2 P b (N O_{3})_{2} Δ 2 P b O + 4 N O_{2} + O_{2}$

$2 C u (N O_{3})_{2} Δ 2 C u O + 4 N O_{2} + O_{2}$

$2 A g N O_{3} Δ 2 A g + 2 N O_{2} + O_{2}$

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Question: From NCERT NEET - 2016

Chapter|Topic|Subtopic-Group 15 -Preparation,Properties & Uses | |

The correct statement among the following regarding acids is -

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

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

3. Both are triprotic acids

4. Both are diprotic acids

Answer ▽

Phosphinic acid is a monoprotic acid as it contains one $P - O H$ bond while phosphonic acid is a diprotic acid as it contains two $P - O H$ bonds.

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Question: From NCERT NEET - Derived

Chapter|Topic|Subtopic-Oxoacids of Phosphorus |

Assertion(A): The basicity of orthophosphours acid is two.
Reason (R): In orthophosphorous acids, two re-placeable hydrogen atoms are present.

Both (A) and (R) are true and (R) is the correct

explanation of (A)

Both (A) and (R) are true and (R) is not the

correct explanation of (A)

(A) is true and (R) is false

(A) is false but (R) is true

3

Medium

Answer ▽

Correct option is

Both (A) and (R) are true and (R) is the correct

explanation of (A)

Basicity = 2
Reason is false only one replaeeable hydrogen atom

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Question: From NCERT NEET - Derived

Chapter|Topic|Subtopic-Oxoacids of Phosphorus |

The number of hydrogen atoms attached to phosphorus atom in hypophosphorous acid is :

zero

two

one

three

Medium

Answer ▽

Molecular formula of Hypophosphorous Acid is $H_{3} P O_{2}$ .
In this, two hydrogen atoms are attached to phosphorus atom with single bond, one -OH group is attached to phosphorus atom with single bond and one oxygen atom is attached to phosphorus atom with double bond.

Hence,option B is correct.

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Question: From NCERT NEET -

Chapter|Topic|Subtopic- Oxidation Number|

The correct order of $N -$ compounds in its decreasing order of oxidation states is:

$H N O_{3}, N H_{4} C l, N O, N_{2}$

$H N O_{3}, N O, N_{2}, N H_{4} C l$

$N H_{4} C l, N_{2}, N O, H N O_{3}$

$H N O_{3}, N O, N H_{4} C l, N_{2}$

Medium

NEET

Answer ▽

Correct option is

$H N O_{3}, N O, N_{2}, N H_{4} C l$

Oxidation state of $N$ in $H N O_{3} = + 5$
Oxidation state of $N$ in $N O = + 2$
Oxidation state of $N$ in $N_{2} = 0$
Oxidation state of $N$ in $N H_{4} C l = - 3$

Hence, the correct option is $B$

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Question: From NCERT NEET - 2016

Chapter|Topic|Subtopic-Group 16- Preparation,Properties & Uses |

The reaction among the following that does not show oxidising behaviour of H2SO4 is -

1. $Cu + 2 H_{2} {SO}_{4} \to {CuSO}_{4} + {SO}_{2} + 2 H_{2} O$

2. $3 S + 2 H_{2} {SO}_{4} \to 3 {SO}_{2} + 2 H_{2} O$

3. $C + 2 H_{2} {SO}_{4} \to {CO}_{2} + 2 {SO}_{2} + 2 H_{2} O$

4. ${CaF}_{2} + H_{2} {SO}_{4} \to {CaSO}_{4} + 2 HF$

Answer ▽

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Question: From NCERT NEET - 2016

Chapter|Topic|Subtopic-Group 16- Preparation,Properties & Uses |

Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?

$C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$

$C a F_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H F$

$C u + 2 H_{2} S O_{4} \to C u S O_{4} + S O_{2} + 2 H_{2} O$

$S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$

Answer ▽

Correct option is

$C a F_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H F$

$C a F_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H F$

In this reaction, the oxidation number of $C a, H, S$ and $O$ are not changed (oxidation number not changed means it is not a redox reaction).

Hence, $H_{2} S O_{4}$ is not acting as an oxidising agent and reducing agent.

Hence, option $B$ is the correct answer.

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Question: From NCERT NEET -Derived

Chapter|Topic|Subtopic-Group 16- Preparation,Properties & Uses|

In which of the following reactions does sulphurous acid function as an oxidising agent?

$H_{2} S O_{3} + B r_{2} + H_{2} O \to H_{2} S O_{4} + 2 H B r$

$5 H_{2} S O_{3} + 2 K I O_{3} \to I_{2} + 4 H_{2} S O_{4} + H_{2} O + K_{2} S O_{4}$

$H_{2} S O_{3} + 2 H_{2} S \to 3 H_{2} O + 3 S$

$H_{2} S O_{3} + 2 F e C l_{3} + H_{2} O \to H_{2} S O_{4} + 2 F e C l_{2} + 2 H C l$

Answer ▽

Correct option is

$H_{2} S O_{3} + 2 H_{2} S \to 3 H_{2} O + 3 S$

In $H_{2} S O_{3} + 2 H_{2} S \to 3 H_{2} O + 3 S$ reaction, $H_{2} S O_{3}$ acts as oxidising agent. In this reaction $H_{2} S$ is oxidised and converted into sulphur by using $H_{2} S O_{3}$ .
So, answer is C.

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Question: From NCERT NEET - 2016

Chapter|Topic|Subtopic- Group 17 - Preparation,Properties & Uses | |

The correct order of the bond dissociation enthalpy of halogen molecules is -

1. Cl2 > Br2 > F2 > l2

2. Br2 > l2 > F2 > Cl2

3. F2 > Cl2 > Br2 > l2

4. l2 > Br2 > Cl2 > F2

Answer ▽

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Question: From NCERT NEET - 2016

Chapter|Topic|Subtopic- Group 17 - Preparation,Properties & Uses ||

The variation of the boiling point of the hydrogen halides is in the order HF > HI > HBr >

HCl. What explains the higher boiling point of hydrogen fluoride?

1. The electronegativity of fluorine is much higher than for other elements in the group

2. There is strong hydrogen bonding between HF molecules

3. The bond energy ol HF molecules is greater than in other hydrogen halides

4. The effect of nuclear shielding is much reduced in fluorine which polarises the HF

molecule

Answer ▽

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Question: From NCERT NEET - 2015

Chapter|Topic|Subtopic- Group 15 -Preparation,Properties & Uses|

Strong reducing behaviour of H₃PO2 is due to -

1. Presence of one-OH group and two P-H bonds

2. High electron gain enthalpy of phosphorus

3. High oxidation state of phosphorus

4. Presence of two-OH groups and one P-H bond

Answer ▽

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Question: From NCERT NEET - 2015

Chapter|Topic|Subtopic- Group 17/16 - Preparation,Properties & Uses|

Which of the statements given below is incorrect?

1. $C l_{2} O_{7}$ is an anhydride of perchloric acid

2. $O_{3}$ molecule is bent

3. ONF is isoelectronic with $N O_{2}^{-}$

4. $O F_{2}$ is an oxide of fluorine

Answer ▽

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Question: From NCERT NEET - 2014

Chapter|Topic|Subtopic- |

Acidity of diprotic acids in aqueous solutions increases in the order:

1. H2S<H2Se<H2Te

2. H2Se<H2S<H2Te

3. H2Te<H2S<H2Se

4. H2Se<H2Te<H2S

Subtopic: Group 16- Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET - 2013

Chapter|Topic|Subtopic-Group 15 -Preparation,Properties & Uses |

A compound among the following that does not give oxygen on heating is -

1. Zn(ClO3)2

2. K2Cr2O7

3. (NH4)2Cr2O7

4. KClO3

Answer ▽

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Question: From NCERT NEET - 2012

Chapter|Topic|Subtopic- |

When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation

number of chlorine changes from

1. zero to +1 and zero to -5

2. zero to -1 and zero to +5

3. zero to -1 and zero to +3

4. zero to +1 and zero to -3

Subtopic: Group 17 - Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET - 2012

Chapter|Topic|Subtopic- |

The incorrect statement among the following regarding oxoacids of phosphorus is -

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

2. Hypophoshphorous acid is a diprotic acid

3. All oxoacids contain tetrahedral four coordinated phosphorus

4. All oxoacids contain at least one P=O unit and one P-OH group

Subtopic: Group 15 -Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET - 2010

Chapter|Topic|Subtopic- |

Oxidation states of P in H4P2O5, H4P2O6, H4P2O7, are respectively

1. +3, +5, +4

2. +5, +3, +4

3. +5, +4, +3

4. +3, +4, +5

Answer ▽

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Question: From NCERT NEET - 2009

Chapter|Topic|Subtopic- |

Among the following which is the strongest oxidising agent ?

1. F2

2. Br2

3. I2

4. Cl2

Subtopic: Group 17 - Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET - 2008

Chapter|Topic|Subtopic-Group 17 - Preparation,Properties & Uses | |

The incorrect statement among the following is -

1. F2 > Cl2 > Br2 > I2 : Oxidising power

2. F < Cl > Br > I : Electron gain enthalpy

3. F2 > Cl2 > Br2 > I2 : Bond dissociation energy

4. F> Cl > Br > I : Electronegativity

Answer ▽

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Question: From NCERT NEET - 2006

Chapter|Topic|Subtopic- |

The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is :

1. 2

2. 3

3. 4

4. 1

Subtopic: Group 16- Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET -

Chapter|Topic|Subtopic-Chlorine |

In the structure of $C l F_{3}$ , the number of lone pairs of electrons on central atom $C l$ is:

four

one

three

two

Answer ▽

Correct option is

two

In the structure of $C l F_{3}$ , the number of lone pairs of electrons on central atom Cl is two. Cl has 7 valence electrons out of which 3 are involved in bond formation with 3 F atoms. $7 - 3 = 4$ valence electrons remains in the form of 2 lone pairs of electrons.

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Question: From NCERT NEET

Chapter|Topic|Subtopic-Halogen |

$H g C l_{2}$ and $I_{2}$ both when dissolved in water containing $I^{-}$ ions, the pair of species formed is:

$H g_{2} I_{2}, I^{-}$

$H g I_{2}, I_{3}^{-}$

$H g I_{2}, I^{-}$

$H g I_{4}^{2 -}, I_{3}^{-}$

Medium

NEET

Answer ▽

Correct option is

$H g I_{4}^{2 -}, I_{3}^{-}$

$H g C l_{2}$ and $I_{2}$ both when dissolved in water containing $I^{-}$ ions the pair of species formed is $H g I_{4}^{2 -}, I_{3}^{-}$ .

$I_{2} + I^{-} \to I_{3}^{-}$

$H g C l_{2} + 4 I^{-} \to H g I_{4}^{2 -} + 2 C l^{-}$ .

Hence, the correct option is $D$ .

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Question: From NCERT NEET Derived

Chapter|Topic|Subtopic- |

Bleaching action of SO2 is due to its-

1. Acidic property

2. Basic property

3. Reducing property

4. Oxidising property

Subtopic: Group 16- Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET Derived

Chapter|Topic|Subtopic- |

Incorrect statement regarding following reactions is :

(a) 'X' is explosive

(b) 'Y' is an oxyacid of xenon

(d) ${XeF}_{6}$ can undergo partial hydrolysis

Subtopic: Group 18- Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET -Derived

Chapter|Topic|Subtopic- |

The incorrect order is:

$(a) N > P > As : Strength of π ‐ bond with oxygen atom$

$(b) {SiF}_{4} > {SiCl}_{4} > {SiBr}_{4} > {Sil}_{4} : Thermal stability$

$(c) H_{2} S > H_{2} Se > H_{2} Te : Arrhenious acid character$

$(d) {CaBr}_{2} > {MgBr}_{2} > {BeBr}_{2} : Melting point$

Subtopic: Group 16- Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET Derived

Chapter|Topic|Subtopic- |

Select the correct statement about elements of group $15^{t h}$ .

(A) The order of stability of oxidation state for +3 is $B i^{3 +} > S b^{3 +} > A s^{3 +}$ and for +5 is $B i^{5 +} < S b^{5 +} < A s^{5 +}$

(B) In the case of nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution.

(C) There is considerable increase in covalent radius from N to P but from As to Bi only a small increase in covalent radius is observed.

(D) All of these.

Subtopic: Group 15 -Preparation,Properties & Uses |

Answer ▽

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Question: From NCERT NEET Derived

Chapter|Topic|Subtopic- Group 15 -Preparation,Properties & Uses|

Which of the following compounds does not give nitrogen on heating?

(1) $N a N_{3}$

(2) ${(N H_{4})}_{2} S O_{4}$

(3) $N H_{4} N O_{2}$

(4) $(N H_{4}) 2 C r_{2} O_{7}$

Answer ▽

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Question: From NCERT NEET Derived

Chapter|Topic|Subtopic- Group 15 -Preparation,Properties & Uses ||

The incorrect statement among the following is -

1. Nitrous oxide supports combustion more vigorously than air.

2. Phosphorous pentaoxide dehydrates nitric acid-forming nitrogen pentaoxide.

3. Reactivity order of various allotropic forms of phosphorous is white > red > black.

4. Red phosphorous changes to white phosphorus on heating in the atmosphere of ${CO}_{2}$ or coal gas at 573 K.

Answer ▽

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Question: From NCERT NEET -Derived

Chapter|Topic|Subtopic- |

The incorrect statement among the following is -

1. $H_{3} {PO}_{2}, H_{3} {PO}_{3} and H_{3} {PO}_{4}$ all are tribasic and reducing in nature.

2. Amongst anions ${NO}_{3}^{-}, {SO}_{3}^{2 -}, {CO}_{3}^{2 -} and {BO}_{3}^{3 -} only {SO}_{3}^{2 -}$ $have pπ - π d bonding .$

3. Amongst anions ${SO}_{3}^{2 -}, {SO}_{4}^{2 -}, S_{2} O_{4}^{2 -} and {HSO}_{4}^{-}, {SO}_{3}^{2 -}$ is basic and reducing in nature.

4. Number of lone pair(s) of electrons on Xe atoms in ${XeF}_{2}, {XeF}_{4} and {XeF}_{6}$ are 3, 2 and 1 respectively.

Subtopic: Group 18- Preparation,Properties & Uses | Group 16- Preparation,Properties & Uses | Group 15 -Preparation,Properties & Uses |

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The p-Block Elements MCQs*

On going down the group thermal stability order for H2​EHdecreases because H−EH−E bond energy decreasesHence,order of stability would be H2​Po<H2​Te<H2​Se<H2​S<H2​O

On going down the group thermal stability order for H2​EHdecreases because H−EH−E bond energy decreasesHence,order of stability would be H2​Po<H2​Te<H2​Se<H2​S<H2​O

When copper is heated with conc. HNO3​ it produces cupric nitrate Cu(NO3​)2​ and nitrogen dioxide NO2​. Cu+4HNO3​→Cu(NO3​)2​+2NO2​+2H2​O

Addition of conc.HNO3​ to conc.H2​SO4​ gives:

SO42−​

SO3​

NO3−​

NO2+​

NO2+​

Nitrogen dioxide is not produced on heating :

KNO3​

Pb(NO3​)2​

Cu(NO3​)2​

AgNO3​

KNO3​

Nitrogen dioxide is not produced on heating KNO3​Heavy metal nitrates liberate NO2​ on heating.2KNO3​Δ​2KNO2​+O2​2Pb(NO3​)2​Δ​2PbO+4NO2​+O2​2Cu(NO3​)2​Δ​2CuO+4NO2​+O2​2AgNO3​Δ​2Ag+2NO2​+O2​

Phosphinic acid is a monoprotic acid as it contains one P−OH bond while phosphonic acid is a diprotic acid as it contains two P−OH bonds.

Assertion(A): The basicity of orthophosphours acid is two.Reason (R): In orthophosphorous acids, two re-placeable hydrogen atoms are present.

Both (A) and (R) are true and (R) is the correctexplanation of (A)

Both (A) and (R) are true and (R) is not thecorrect explanation of (A)

(A) is true and (R) is false

(A) is false but (R) is true3

Both (A) and (R) are true and (R) is the correctexplanation of (A)

Basicity = 2Reason is false only one replaeeable hydrogen atom

The number of hydrogen atoms attached to phosphorus atom in hypophosphorous acid is :

zero

two

one

three

Molecular formula of Hypophosphorous Acid is H3​PO2​. In this, two hydrogen atoms are attached to phosphorus atom with single bond, one -OH group is attached to phosphorus atom with single bond and one oxygen atom is attached to phosphorus atom with double bond.Hence,option B is correct.

The correct order of N− compounds in its decreasing order of oxidation states is:

HNO3​,NH4​Cl,NO,N2​

HNO3​,NO,N2​,NH4​Cl

NH4​Cl,N2​,NO,HNO3​

HNO3​,NO,NH4​Cl,N2​

HNO3​,NO,N2​,NH4​Cl

Oxidation state of N in HNO3​=+5Oxidation state of N in NO=+2Oxidation state of N in N2​=0Oxidation state of N in NH4​Cl=−3Hence, the correct option is B

Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?

C+2H2​SO4​→CO2​+2SO2​+2H2​O

CaF2​+H2​SO4​→CaSO4​+2HF

Cu+2H2​SO4​→CuSO4​+SO2​+2H2​O

S+2H2​SO4​→3SO2​+2H2​O

CaF2​+H2​SO4​→CaSO4​+2HF

CaF2​+H2​SO4​→CaSO4​+2HFIn this reaction, the oxidation number of Ca,H,S and O are not changed (oxidation number not changed means it is not a redox reaction). Hence, H2​SO4​ is not acting as an oxidising agent and reducing agent.Hence, option B is the correct answer.

In which of the following reactions does sulphurous acid function as an oxidising agent?

H2​SO3​+Br2​+H2​O→H2​SO4​+2HBr

5H2​SO3​+2KIO3​→I2​+4H2​SO4​+H2​O+K2​SO4​

H2​SO3​+2H2​S→3H2​O+3S

H2​SO3​+2FeCl3​+H2​O→H2​SO4​+2FeCl2​+2HCl

H2​SO3​+2H2​S→3H2​O+3S

In H2​SO3​+2H2​S→3H2​O+3S reaction, H2​SO3​ acts as oxidising agent. In this reaction H2​S is oxidised and converted into sulphur by using H2​SO3​.So, answer is C.

In the structure of ClF3​, the number of lone pairs of electrons on central atom Cl is:

four

one

three

two

two

In the structure of ClF3​, the number of lone pairs of electrons on central atom Cl is two. Cl has 7 valence electrons out of which 3 are involved in bond formation with 3 F atoms. 7−3=4 valence electrons remains in the form of 2 lone pairs of electrons.

HgCl2​ and I2​ both when dissolved in water containing I− ions, the pair of species formed is:

Hg2​I2​,I−

HgI2​,I3−​

HgI2​,I−

HgI42−​,I3−​

HgI42−​,I3−​

HgCl2​ and I2​ both when dissolved in water containing I−ions the pair of species formed is HgI42−​,I3−​.I2​+I−→I3−​HgCl2​+4I−→HgI42−​+2Cl−.Hence, the correct option is D.

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On going down the group thermal stability order for $H_{2} E$ Hdecreases because H−EH−E bond energy decreases
Hence,order of stability would be
$H_{2} P o < H_{2} T e < H_{2} S e < H_{2} S < H_{2} O$

On going down the group thermal stability order for $H_{2} E$ Hdecreases because H−EH−E bond energy decreases
Hence,order of stability would be
$H_{2} P o < H_{2} T e < H_{2} S e < H_{2} S < H_{2} O$

When copper is heated with conc. $H N O_{3}$ it produces cupric nitrate $C u (N O_{3})_{2}$ and nitrogen dioxide $N O_{2}$ .

$C u + 4 H N O_{3} \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O$

Addition of $c o n c . H N O_{3}$ to $c o n c . H_{2} S O_{4}$ gives:

$S O_{4}^{2 -}$

$S O_{3}$

$N O_{3}^{-}$

$N O_{2}^{+}$

$N O_{2}^{+}$

$K N O_{3}$

$P b (N O_{3})_{2}$

$C u (N O_{3})_{2}$

$A g N O_{3}$

$K N O_{3}$

Nitrogen dioxide is not produced on heating $K N O_{3}$
Heavy metal nitrates liberate $N O_{2}$ on heating.

$2 K N O_{3} Δ 2 K N O_{2} + O_{2}$

$2 P b (N O_{3})_{2} Δ 2 P b O + 4 N O_{2} + O_{2}$

$2 C u (N O_{3})_{2} Δ 2 C u O + 4 N O_{2} + O_{2}$

$2 A g N O_{3} Δ 2 A g + 2 N O_{2} + O_{2}$

Phosphinic acid is a monoprotic acid as it contains one $P - O H$ bond while phosphonic acid is a diprotic acid as it contains two $P - O H$ bonds.

Assertion(A): The basicity of orthophosphours acid is two.
Reason (R): In orthophosphorous acids, two re-placeable hydrogen atoms are present.

Both (A) and (R) are true and (R) is the correct

explanation of (A)

Both (A) and (R) are true and (R) is not the

correct explanation of (A)

(A) is false but (R) is true

3

Both (A) and (R) are true and (R) is the correct

explanation of (A)

Basicity = 2
Reason is false only one replaeeable hydrogen atom

Molecular formula of Hypophosphorous Acid is $H_{3} P O_{2}$ .
In this, two hydrogen atoms are attached to phosphorus atom with single bond, one -OH group is attached to phosphorus atom with single bond and one oxygen atom is attached to phosphorus atom with double bond.

Hence,option B is correct.

The correct order of $N -$ compounds in its decreasing order of oxidation states is:

$H N O_{3}, N H_{4} C l, N O, N_{2}$

$H N O_{3}, N O, N_{2}, N H_{4} C l$

$N H_{4} C l, N_{2}, N O, H N O_{3}$

$H N O_{3}, N O, N H_{4} C l, N_{2}$

$H N O_{3}, N O, N_{2}, N H_{4} C l$

Oxidation state of $N$ in $H N O_{3} = + 5$
Oxidation state of $N$ in $N O = + 2$
Oxidation state of $N$ in $N_{2} = 0$
Oxidation state of $N$ in $N H_{4} C l = - 3$

Hence, the correct option is $B$

$C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$

$C a F_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H F$

$C u + 2 H_{2} S O_{4} \to C u S O_{4} + S O_{2} + 2 H_{2} O$

$S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$

$C a F_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H F$

$H_{2} S O_{3} + B r_{2} + H_{2} O \to H_{2} S O_{4} + 2 H B r$

$5 H_{2} S O_{3} + 2 K I O_{3} \to I_{2} + 4 H_{2} S O_{4} + H_{2} O + K_{2} S O_{4}$

$H_{2} S O_{3} + 2 H_{2} S \to 3 H_{2} O + 3 S$

$H_{2} S O_{3} + 2 F e C l_{3} + H_{2} O \to H_{2} S O_{4} + 2 F e C l_{2} + 2 H C l$

$H_{2} S O_{3} + 2 H_{2} S \to 3 H_{2} O + 3 S$

In $H_{2} S O_{3} + 2 H_{2} S \to 3 H_{2} O + 3 S$ reaction, $H_{2} S O_{3}$ acts as oxidising agent. In this reaction $H_{2} S$ is oxidised and converted into sulphur by using $H_{2} S O_{3}$ .
So, answer is C.

In the structure of $C l F_{3}$ , the number of lone pairs of electrons on central atom $C l$ is:

In the structure of $C l F_{3}$ , the number of lone pairs of electrons on central atom Cl is two. Cl has 7 valence electrons out of which 3 are involved in bond formation with 3 F atoms. $7 - 3 = 4$ valence electrons remains in the form of 2 lone pairs of electrons.

$H g C l_{2}$ and $I_{2}$ both when dissolved in water containing $I^{-}$ ions, the pair of species formed is:

$H g_{2} I_{2}, I^{-}$

$H g I_{2}, I_{3}^{-}$

$H g I_{2}, I^{-}$

$H g I_{4}^{2 -}, I_{3}^{-}$

$H g I_{4}^{2 -}, I_{3}^{-}$

$H g C l_{2}$ and $I_{2}$ both when dissolved in water containing $I^{-}$ ions the pair of species formed is $H g I_{4}^{2 -}, I_{3}^{-}$ .

$I_{2} + I^{-} \to I_{3}^{-}$

$H g C l_{2} + 4 I^{-} \to H g I_{4}^{2 -} + 2 C l^{-}$ .

Hence, the correct option is $D$ .