JEE Advanced · Chemistry · 4. Chemical Bonding
Among \(H_2, H e_2^{+}, L i_2, B e_2, B_2, C_2, N_2, O_2^{-}\), and \(F_2\), the number of diamagnetic species is -
(Atomic number : \(H=1, H e=2, L i=3, B e=4, B=5,\) \(C=6, N =7, O=8, F=9\) )
- A 5
- B 10
- C 15
- D 20
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
| \(H_2 \Rightarrow \sigma 1 s^2\) | (Dia magnetic) |
| \(H e_2^{+} \Rightarrow \sigma 1 s^2 \sigma^* 1 s^1\) | (Para magnetic) |
| \(L i_2 \Rightarrow \sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2\) | (Dia magnetic) |
| \(B e_2 \Rightarrow \sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2 \quad\)\(\sigma^* 2 s^2\) | (Does not exist) |
| \(B_2 \Rightarrow \sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2 \quad\)\( \sigma^* 2 s^2 \quad \pi 2 p_x^1=\pi 2 p_y^1\) | (Para magnetic) |
| \(C_2 \Rightarrow \sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2 \quad\)\( \sigma^* 2 s^2 \quad \pi 2 p_x^2=\pi 2 p_y^2\) | (Dia magnetic) |
| \(N_2 \Rightarrow \sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2 \quad\)\( \sigma^* 2 s^2\quad \pi 2 p_x^2=\pi 2 p_y^2 \quad \sigma 2 p_z^2\) | (Dia magnetic) |
| \(O_2^{\Theta} \Rightarrow\sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2 \quad\)\(\sigma^* 2 s^2 \quad\sigma 2 p_z^2 \quad \pi 2 p_x^2=\pi 2 p_y^2 \quad\)\(\pi^* 2 p_x^2=\pi^* 2 p_y^1\) | (Para magnetic) |
| \(F_2 \Rightarrow \sigma 1 s^2 \quad \sigma^* 1 s^2 \quad \sigma 2 s^2 \quad\)\( \sigma^* 2 s^2 \quad\sigma 2 o_z^2 \quad \pi 2 p_x^2=\pi 2 p_y^2 \quad\)\(\pi^* 2 p_x^2=\pi^* 2 p_y^2\) | (Dia magnetic) |
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