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

P Block Elements question

2015 · 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 Advanced
  3. /Chemistry
  4. /P Block Elements
  5. /2015 · Shift 1 · Q11

P Block Elements question

2015 · Shift 1 · Q11

JEE AdvancedChemistryP Block ElementsNumerical+3 / −1
The number of lone pairs of electrons in N2{}_22​O3{}_33​ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Find the total number of valence electrons

For N2O3\mathrm{N_2O_3}N2​O3​:

  • Each nitrogen has 555 valence electrons, so 2×5=102 \times 5 = 102×5=10
  • Each oxygen has 666 valence electrons, so 3×6=183 \times 6 = 183×6=18

Thus, total valence electrons: 10+18=2810 + 18 = 2810+18=28

  1. Write the likely structure of N2O3\mathrm{N_2O_3}N2​O3​

N2O3\mathrm{N_2O_3}N2​O3​ is the anhydride of nitrous acid and has the structure: O=N−O−N=O\mathrm{O=N-O-N=O}O=N−O−N=O

That is:

  • two terminal oxygens are double bonded to nitrogen
  • one oxygen bridges the two nitrogens by single bonds
  1. Count bonding electrons

In O=N−O−N=O\mathrm{O=N-O-N=O}O=N−O−N=O:

  • Two N=O\mathrm{N=O}N=O double bonds contribute 2×4=82 \times 4 = 82×4=8 electrons
  • Two N−O\mathrm{N-O}N−O single bonds contribute 2×2=42 \times 2 = 42×2=4 electrons

Total bonding electrons: 8+4=128 + 4 = 128+4=12

  1. Find nonbonding electrons

Total valence electrons = 282828

So lone-pair electrons: 28−12=1628 - 12 = 1628−12=16

Since each lone pair contains 222 electrons, number of lone pairs is: 162=8\frac{16}{2} = 8216​=8

  1. Check atom-wise
  • Each terminal double-bonded oxygen has 222 lone pairs: 2×2=42 \times 2 = 42×2=4
  • Bridging oxygen has 222 lone pairs: 222
  • Each nitrogen has 111 lone pair: 2×1=22 \times 1 = 22×1=2

Total lone pairs: 4+2+2=84 + 2 + 2 = 84+2+2=8

So, the number of lone pairs in N2O3\mathrm{N_2O_3}N2​O3​ is: 8\boxed{8}8​

PreviousNext

More from P Block Elements

  • Three moles of B2​H6​ are completely reacted with methanol. The number of moles of boron containing product formed is ​.2015 · Numerical
  • The correct statements regarding (i) HClO, (ii) HClO2​, (iii) HClO3​ and (iv) HClO4​ is (are)2015 · Multiple correct
  • The correct statement(s) for orthoboric acid is/are2014 · Multiple correct
  • Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is Includes diagram2014 · MCQ
  • The product formed in the reaction of SOCl2​ with white phosphorus is2014 · MCQ
  • Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of2013 · MCQ
  • The correct statement(s) about O3​ is(are)2013 · Multiple correct
  • The reaction of Cl2​ gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2​ gas reacts with SO2​ gas, in presence of charcoal, to give a…2013 · MCQ