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NEET CHEMISTRYOrganic Chemistry – Some Basic Principles and Techniques Easy

Question

A paramagnetic molecule among the following is:

A

O₂⁻

B

CN⁻

C

NO⁺

D

CO

Step-by-Step Solution

  1. Define Paramagnetism: Paramagnetism arises due to the presence of unpaired electrons in an atom, ion, or molecule [Class 12 Chemistry, Unit 4, Sec 4.3.9].
  2. Analyze Electron Counts:
  • O2−O_2^- (Superoxide Ion): Total electrons = 8(O)+8(O)+1(charge)=178 (O) + 8 (O) + 1 (charge) = 17. An odd number of electrons ensures at least one is unpaired. According to MOT, the configuration is σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗2=π2py∗1)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}). It has 1 unpaired electron. Thus, it is Paramagnetic.
  • CN−CN^- (Cyanide Ion): Total electrons = 6(C)+7(N)+1=146 (C) + 7 (N) + 1 = 14. All electrons are paired (sigma1s2...sigma2pz2\\sigma_{1s}^2 ... \\sigma_{2p_z}^2). Diamagnetic.
  • NO+NO^+ (Nitrosonium Ion): Total electrons = 7(N)+8(O)−1=147 (N) + 8 (O) - 1 = 14. Isoelectronic with CN−CN^- and N2N_2. All electrons paired. Diamagnetic.
  • COCO (Carbon Monoxide): Total electrons = 6(C)+8(O)=146 (C) + 8 (O) = 14. Isoelectronic with N2N_2. All electrons paired. Diamagnetic.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Organic Chemistry – Some Basic Principles and Techniques . Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYOrganic Chemistry – Some Basic Principles and Techniques paramagneticmoleculefollowing

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Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?

A.$\text{BCl}_3$ and $\text{BrCl}_3$
B.$\text{NH}_3$ and $\text{NO}_3^-$
C.$\text{NF}_3$ and $\text{BF}_3$
D.$\text{BF}_4^-$ and $\text{NH}_4^+$
MediumSolve

Among the following, isoelectronic species are:

A.$CO_{2}, NO_{2}$
B.$NO_{2}^{-}, CO_{2}$
C.$CN^{-}, CO$
D.$SO_{2}, CO_{2}$
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Which one of the following pairs of species have the same bond order?

A.$\text{CO}$, $\text{NO}$
B.$\text{O}_2^-$, $\text{CN}^-$
C.$\text{CN}^-$, $\text{NO}^+$
D.$\text{N}^+$, $\text{CN}^+$
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In a regular octahedral molecule, $MX_{6}$, the number of $X–M–X$ bonds at 180º is:

A.Two
B.Six
C.Four
D.Three
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Which of the following pairs of ions are isoelectronic and isostructural?

A.CO₃²⁻, SO₃²⁻
B.ClO₃⁻, CO₃²⁻
C.SO₃²⁻, NO₃⁻
D.ClO₃⁻, SO₃²⁻
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Which orbital does the additional electron occupy when $O_{2}$ is converted to the $O_{2}^{-}$ ion?

A.$\pi$ orbital
B.$\sigma^{*}$ orbital
C.$\sigma$ orbital
D.$\pi^{*}$ orbital
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A species that does not exhibit paramagnetism is:

A.$N_{2}^{+}$
B.$O_{2}^{-}$
C.$CO$
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MediumSolve

The total number of $\pi$-bond electrons in the following structure is:

A.4
B.8
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D.16
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