Question

N₂ and O₂ are converted into monoanions, N₂⁻ and O₂⁻
respectively. Which of the following statements is wrong ?
(a) In N₂⁻ , N – N bond weakens
(b) In O₂⁻ , O - O bond order increases
(c) In O₂⁻ , O - O bond order decreases
(d) N₂⁻ becomes paramagnetic

Answer 1

The N₂⁻ becomes paramagnetic due to the presence of unpaired electron.

Option (D) in wrong.

Explanation:

Correct statement:

N₂ and O₂ are converted into monoanions, N₂⁻ and O₂⁻  respectively. Which of the following statements is wrong ?

(a) In N₂⁻ , N – N bond weakens

(b) In O₂⁻ , O - O bond order increases

(c) In O₂⁻ , O - O bond order decreases

(d) N₂⁻ becomes diamagnetic

In N−2 total number of electrons are = 14 + 1  = 15

Electronic configuration of N−2 is

σ1s2,∗σ1s2,σ2s2,∗σ2s2,σ2p2x,π2p2y≈π2p2z,∗π2p1y

The configursation shows that N-2 has only one electron in py orbital.

Due to presence of one unpaired electron, it shows paramagnetic character.

Thus the N₂⁻ becomes paramagnetic due to the presence of unpaired electron.

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Explain the anomalous behaviour of oxygen in elements of group 16 with reference to : 1) atomicity 2) oxidation and write magnetic properties of a) N2O4 b) NO2  ?

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Answer 2

Answer is (b)

Explanation:

Bond order = $\frac{N_{B}- N_{NB}}{2}$

Where $N_{B}$= Bonding electrons and

$N_{NB}$= Non-bonding electrons

$N_{2}$ has 2×7= 14 electrons

$\sigma 1s^{2}\sigma ^{\ast }1s^{2}\sigma 2s^{2}\sigma ^{\ast } 2s^{2}\sigma 2p_{x}^{2}\sigma 2p_{y}^{2}\sigma 2p_{z}^{2}$

here $\ast$ showing anti-bonding orbitals

then for $N_{2}$  Bond order= $\frac{10-4}{2}= 3$

for $N_{2}^{-}$ Bond order $= \frac{10-5}{2}= 2.5$

for $O_{2}$ Bond order is $= \frac{10-6}{2}= 2$

for $O_{2}^{-}$ Bond order is $= \frac{10-7}{2}= 1.5$

for$N_{2}^{-}$ and  $O_{2}^{-}$ bond order decreases

And if bond order is in fraction it will show paramagnetic behaviour.

Hence option (b) is wrong