Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Chemical Bonding and Molecular Structure question

2023 · 24 Jan · Shift 1 · Q11
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Chemical Bonding and Molecular Structure
  5. /2023 · 24 Jan · Shift 1 · Q11

Chemical Bonding and Molecular Structure question

2023 · 24 Jan · Shift 1 · Q11

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Order of Covalent bond; A. KF>KI;LiF>KF\mathrm{KF \gt KI ; LiF \gt KF}KF>KI;LiF>KF B. KF<KI;LiF>KF\mathrm{KF \lt KI ; LiF \gt KF}KF<KI;LiF>KF C. SnCl4>SnCl2;CuCl>NaCl\mathrm{SnCl_4 \gt SnCl_2 ; CuCl \gt NaCl}SnCl4​>SnCl2​;CuCl>NaCl D. LiF>KF;CuCl<NaCl\mathrm{LiF \gt KF ; CuCl \lt NaCl}LiF>KF;CuCl<NaCl E. KF<KI;CuCl>NaCl\mathrm{KF \lt KI ; CuCl \gt NaCl}KF<KI;CuCl>NaCl Choose the correct answer from the options given below :
  1. A
    B, C only
  2. B
    A, B only
  3. C
    C, E only
  4. D
    B, C, E only
View written solutionFree

Correct answer: D

  1. Use Fajans’ rule to compare covalent character

    Covalent character in an ionic compound increases when:

    • the cation is small and highly polarising,
    • the anion is large and highly polarizable.

    So:

    • smaller cation ⇒\Rightarrow⇒ more covalent character,
    • larger anion ⇒\Rightarrow⇒ more covalent character.
  2. Check each statement


    Statement A: KF>KI; LiF>KF\mathrm{KF > KI;\ LiF > KF}KF>KI; LiF>KF

    • Compare KF\mathrm{KF}KF and KI\mathrm{KI}KI:

      • same cation K+\mathrm{K^+}K+,
      • I−\mathrm{I^-}I− is larger than F−\mathrm{F^-}F−,
      • hence KI\mathrm{KI}KI has more covalent character than KF\mathrm{KF}KF.

      Therefore, KF>KI\mathrm{KF > KI}KF>KI is false; correct order is KF<KI\mathrm{KF < KI}KF<KI

    • Compare LiF\mathrm{LiF}LiF and KF\mathrm{KF}KF:

      • same anion F−\mathrm{F^-}F−,
      • Li+\mathrm{Li^+}Li+ is smaller than K+\mathrm{K^+}K+,
      • so Li+\mathrm{Li^+}Li+ polarises more strongly.

      Hence, LiF>KF\mathrm{LiF > KF}LiF>KF is true.

    Since first part is false, Statement A is false.


    Statement B: KF<KI; LiF>KF\mathrm{KF < KI;\ LiF > KF}KF<KI; LiF>KF

    From above:

    • KF<KI\mathrm{KF < KI}KF<KI is true,
    • LiF>KF\mathrm{LiF > KF}LiF>KF is true.

    Hence Statement B is true.


    Statement C: SnCl4>SnCl2; CuCl>NaCl\mathrm{SnCl_4 > SnCl_2;\ CuCl > NaCl}SnCl4​>SnCl2​; CuCl>NaCl

    • Compare SnCl4\mathrm{SnCl_4}SnCl4​ and SnCl2\mathrm{SnCl_2}SnCl2​:

      • cation in SnCl4\mathrm{SnCl_4}SnCl4​ is effectively Sn4+\mathrm{Sn^{4+}}Sn4+,
      • cation in SnCl2\mathrm{SnCl_2}SnCl2​ is effectively Sn2+\mathrm{Sn^{2+}}Sn2+,
      • higher charge means greater polarising power.

      Therefore, SnCl4>SnCl2\mathrm{SnCl_4 > SnCl_2}SnCl4​>SnCl2​ is true.

    • Compare CuCl\mathrm{CuCl}CuCl and NaCl\mathrm{NaCl}NaCl:

      • Cu+\mathrm{Cu^+}Cu+ has greater polarising power than Na+\mathrm{Na^+}Na+ due to smaller size and pseudo-inert gas configuration effect,
      • therefore CuCl\mathrm{CuCl}CuCl is more covalent than NaCl\mathrm{NaCl}NaCl.

      So, CuCl>NaCl\mathrm{CuCl > NaCl}CuCl>NaCl is true.

    Hence Statement C is true.


    Statement D: LiF>KF; CuCl<NaCl\mathrm{LiF > KF;\ CuCl < NaCl}LiF>KF; CuCl<NaCl

    • LiF>KF\mathrm{LiF > KF}LiF>KF is true,
    • but CuCl<NaCl\mathrm{CuCl < NaCl}CuCl<NaCl is false because actually CuCl>NaCl\mathrm{CuCl > NaCl}CuCl>NaCl

    Hence Statement D is false.


    Statement E: KF<KI; CuCl>NaCl\mathrm{KF < KI;\ CuCl > NaCl}KF<KI; CuCl>NaCl

    • KF<KI\mathrm{KF < KI}KF<KI is true,
    • CuCl>NaCl\mathrm{CuCl > NaCl}CuCl>NaCl is true.

    Hence Statement E is true.

  3. Collect the correct statements

    Correct statements are: B, C, and E\boxed{\text{B, C, and E}}B, C, and E​

  4. Match with given options

    The option containing B,C,E only\mathrm{B, C, E\ only}B,C,E only is: D\boxed{\text{D}}D​

  5. Comparison with stored correct answer

    Stored correct answer = D\boxed{\text{D}}D​

    My derived answer also = D\boxed{\text{D}}D​.

    So they agree.

PreviousNext

More from Chemical Bonding and Molecular Structure

  • Decreasing order of the hydrogen bonding in following forms of water is correctly represented by A. Liquid water B. Ice C. Impure water Choose the correct answer from the options given below :2023 · MCQ
  • What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : N2​;N2+​;O2​;O2+​ ?2023 · MCQ
  • The total number of lone pairs of electrons on oxygen atoms of ozone is ​.2023 · Numerical
  • Statement I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre. Statement II : The crossed arrow of the dipole moment…2023 · MCQ
  • The number of molecules or ions from the following, which do not have odd number of electrons are ​. (A) NO 2​ (B) ICl 4−​ (C) BrF 3​ (D) ClO 2​ (E) NO 2+​ (F) NO2023 · Numerical
  • The magnetic behaviour of Li2​O,Na2​O2​ and KO2​, respectively, are :2023 · MCQ
  • The bond dissociation energy is highest for2023 · MCQ
  • According to MO theory the bond orders for O22−​, CO and NO+ respectively, are2023 · MCQ