Among the following molecules/ions,

\({\rm{C}}_2^{2 - },{\rm{\;N}}_2^{2 - },{\rm{O}}_2^{2 - },{{\rm{O}}_2}\)

which one is diamagnetic and has the shortest bond length?

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JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 2)
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  1. O2
  2. \(N_2^{2 - }\)
  3. \(O_2^{2 - }\)
  4. \(C_2^{2 - }\)

Answer (Detailed Solution Below)

Option 4 : \(C_2^{2 - }\)
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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Concept:

Bond Length:

  • Bond Length is defined as the distance between the centers of two covalently bonded atoms.
  • The length of the bond is determined by the number of bonded electrons. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms.
  • Bond order is \(\propto \frac{1}{{Bond\;length}}\)

Diamagnetic:

  • Whenever two electrons are paired together in an orbital, or their total spin is 0, they are diamagnetic electrons. Atoms with all diamagnetic electrons are called diamagnetic atoms.
  • Diamagnetic is a quantum mechanical effect that occurs in all materials, when it is the only contribution to the magnetism, the material is called diamagnetic.

Paramagnetic:

  • Paramagnetic compounds are attracted to the magnetic field and also have unpaired electrons. So, all atoms with incompletely filled atomic orbitals are paramagnetic.
  • Due to their spin, unpaired electrons have a magnetic dipole moment and act like a tiny magnet.

Calculation:

\(O_2^{2 - }\) is diamagnetic with bond order 1.

  • \(\sigma 1{s^2},{\sigma ^{\rm{*}}}1{s^2},\sigma 2{s^2},{\rm{\;}}{\sigma ^{\rm{*}}}2s_1^{2\;},\sigma 2p_2^2\left[ {\pi 2p_x^2 = \pi 2p_y^2} \right]{\rm{\;}}\sigma 2p_x^1\left[ {\pi *2p_x^1 = \pi x2p_y^1} \right]n\)

\({\rm{Bond\;order}} = \frac{{10 - 8}}{2} = 1\)

  • \(N_2^{2 - } \to \sigma 1{s^2},{\rm{\;}}{\sigma ^{\rm{*}}}1{s^2},{\rm{\;}}\sigma 2{s^2},{\sigma ^{\rm{*}}}2s_1^2\left[ {\pi 2p_x^2 = \pi 2p_y^2} \right]{\rm{\;}}\sigma 2p_2^2\left[ {\pi *2p_x^1 = \pi x2p_y^1} \right]\)

\({\rm{Bond\;order}} = \frac{{10 - 6}}{2} = 2\)

\(N_2^{2 - }\) is paramagnetic with bond order 2.

  • \(C_2^{2 - } \to \sigma 1{s^2},{\sigma ^{\rm{*}}}1{s^2},\sigma 2{s^2},{\sigma ^{\rm{*}}}2{s^2}\left[ {\pi 2p_x^2 = \pi 2p_y^2} \right]\sigma 2{p_x}^2\)

\({\rm{Bond\;order}} = \frac{{10 - 4}}{6} = 3\)

\(C_2^{2 - }\) is diamagnetic with bond order 3.

Hence \(C_2^{2 - }\) has the least bond length and it is diamagnetic.

Hence the O2 is paramagnetic with bond order 2.

  • Then Bond order is \(\propto \frac{1}{{Bond\;length}}\)

 

Symbol

 

No. of unpaired electron

 

Bond Order

 

Magnetic Character

\(O_2^{2 - }\)

0

1

Diamagnetic

\(C_2^{2 - }\)

0

3

Diamagnetic

\(N_2^{2 - }\)

2

2

Paramagnetic

O2

1

2

Paramagnetic

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