NEET Questions Solved


Which of the following is the correct statement?

(1) F2 has higher dissociation energy than Cl2

(2) F has higher electron affinity than Cl

(3) HF is stronger acid than HCl

(4) Boiling point increases down the group in halogens

(4)

(1)-Bond dissociation energy of F2 is less than that of Cl2

(2)-Cl has higher E.A. than fluorine

(3)-HF is weaker acid than HCl, due to higher bond energy

Difficulty Level:

  • 17%
  • 13%
  • 27%
  • 45%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
NEET - 2016

Among the following, which one is a wrong statement?

(a) PH5 and BiCl5 do not exist
(b) ρp-dp bonds are present in SO2
(c) SeF4 and CH4 have same shape
(d) I3+ has bent geometry

(c) PH5 does not exist due to very less electronegativity difference between P and H. Hydrogen is slightly more electronegative than phosphorus, thus could not hold significantly the sharing electrons.

On the other hand, ВіCl5 does not exist due to inert pair effect. On moving down the group, +5 oxidation state becomes less stable while +3 oxidation state becomes more stable.

In SO2p-dp and pp-pp both types of bonds are present.

Thus, SeF4 and CH4 do not have same shape.

Thus, option (c) is incorrect statement.

Difficulty Level:

  • 22%
  • 17%
  • 42%
  • 21%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
NEET - 2015

Which of the statements given below is incorrect?

(a) Cl2O7 is an anhydride of perchloric acid

(b) O3 molecule is bent

(c) ONF is isoelectronic with NO2

(d) OF2 is an oxide of fluorine

(a) Cl2O7 is an anhydride of perchloric acid 
               Δ      
2HClO4 
Cl2O7  
            -H2O

(b) Shape of O3 molecule is bent

(c) Number of electrons in ONF=24
     Number of electrons in NO2=24
     ∴ONF and NO2 both are isoelectronic

(d) OF2 is a fluoride of oxygen because electronegativity of fluorine is more than that of oxygen.
OF2 = Oxygen difluoride

Difficulty Level:

  • 12%
  • 11%
  • 42%
  • 37%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
NEET - 2015

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?

(a) The electronegativity of fluorine is much higher than for other elements in the group

(b) There is strong hydrogen bonding between HF molecules

(c) The bond energy ol HF molecules is greater than in other hydrogen halides

(d) The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule

Since, there is a strong hydrogen bonding between HF molecules. Hence, boiling point is highest for HF.
HF > HI > HBr > HI

Difficulty Level:

  • 17%
  • 69%
  • 11%
  • 4%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
NEET - 2012

When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

(a) zero to +1 and zero to -5
(b) zero to -1 and zero to +5
(c) zero to -1 and zero to +3
(d) zero to +1 and zero to -3

 

(b) When chlorine gas reacts with hot and concentrated NaOH solution, it disproportionates into chloride (Cl-) and chlorate (ClO3-) ions.

In this process, oxidation number of chlorine changes from 0 to -1 and 0 to +5.

Note In disproportionation reactions, the same element undergoes oxidation as well as reduction.

Difficulty Level:

  • 12%
  • 62%
  • 23%
  • 6%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
NEET - 2009

Among the following which is the strongest oxidising agent ?
(a) F2
(b) Br2
(c) I2
(d) Cl2

Key Idea Element having higher tendency to get reduced or to accept an electron, is strong oxidising agent. Fluorine is the most electronegative element because electronegativity decreases on moving down the group. Hence, it gets reduced readily into F- ion and is a strongest oxidising agent.

Difficulty Level:

  • 71%
  • 5%
  • 17%
  • 9%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
BOARD

(i) What happens when                               (5 marks)

(a) chlorine gas reacts with cold and dilute

solution of NaOH ?

(b) XeF2 undergoes hydrolysis ?

(ii) Assign suitable reasons for the following :

(a) SF6 is inert towards hydrolysis.

(b) H3PO3 is diprotic.

(c) Out of noble gases only Xenon is known to form

established chemical compounds.

(i) (a) 2NaOH+Cl2NaCl+H2O                           (1 mark)

(cold and dilute)

(b) 2XeF2(s)+2H2O(l)2Xe(g)+4HF(aq)+O2(g)    (1 mark)

(ii) (a) Sulphur is sterically protected by six F atoms,    (1 mark)

hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are     (1 mark)

present as –OH groups, thus behaves as dibasic acid.

(c) Xe has least ionization energy among the noble gases (1 mark)

and hence it forms chemical compounds particularly with O2 and F2.

Difficulty Level:

  • 70%
  • 18%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
BOARD

(i) Considering the parameters such as bond dissociation     (5 marks)

enthalpy, electron gain enthalpy and hydration enthalpy,

compare the oxidizing power of F2 and Cl2.

(ii) Complete the following reactions :

(a) Cu + HNO3(dilute)

(b) Fe3+ + SO2 + H2O

(c) XeF4 + O2F2

(i) a. Fluorine has less negative electron gain enthalpy than chlorine,  (1/2x4)

b. Fluorine has low enthalpy of dissociation than chlorine

c. Fluorine has very high enthalpy of hydration than chlorine.

d. Fluorine is stronger oxidizing agent than chlorine.

(ii) a) 3Cu+8HNO3(dilute)3Cu(NO3)2+2NO+4H2O       (1 mark)

b) 2Fe3++SO2+2H2O2Fe2++SO42-+4H+                  (1 mark)

c) XeF4+O2F2XeF6+O2                                             (1 mark)

                               (balancing equation may be ignored)

Difficulty Level:

  • 67%
  • 17%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
BOARD

Give reasons :

(i) SO2 is reducing while TeO2 is an oxidizing agent.

(ii) Nitrogen does not form pentahalide.

(iii) ICl is more reactive than I2                                                                         (3)

 

(i) Because stability of higher oxidation state decreases as we move down the group / S is more stable in higher (+6) oxidation state whereas Te is more stable in +4 oxidation state.                                                                    (1)

(ii) Due to absence of d orbital .                                                      (1)

(iii) Because I – Cl bond is weaker than I-I bond.                              (1)

Difficulty Level:

  • 75%
  • 0%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)
BOARD

(a) Account for the following :                                                 (5 marks)

     (i) Acidic character increases from HF to HI.

     (ii) There is large difference between the melting and boiling points of oxygen and sulphur.

     (iii) Nitrogen does not form pentahalide.

(b) Draw the structures of the following :

      (i) ClF3

      (ii) XeF4

(a)

     (i) Due to decrease in bond dissociation enthalpy from HF to HI, there is an increase in acidic

          character observed.             (1 marks)

     (ii) Oxygen exists as diatomic O2 molecule while sulphur as polyatomic S8            (1 marks)

     (iii) Due to non availability of d orbitals                            (1 marks)

 

(b)

                                                 

                                (i) ClF3                                                       (ii) XeF4

                                (1 marks)                                                   (1 marks)

Difficulty Level:

  • 79%
  • 0%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 28, 2019)