JEE Advanced 2025 — Paper 1 — Question 35

(i) $\mathrm{Ne}_{2} \Rightarrow\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{\mathrm{z}}^{2}\right)\left(\pi 2 p_{\mathrm{x}}^{2}=\pi 2 p_{\mathrm{y}}^{2}\right)\left(\pi^{*} 2 p_{\mathrm{x}}^{2}=\pi^{*} 2 p_{\mathrm{y}}^{2}\right)\left(\sigma^{*} 2 p_{\mathrm{z}}^{2}\right)$

B.O. $=\frac{6-6}{2}=0$

(ii) $\mathrm{F}_{2} \Rightarrow\left(\sigma 1 \mathrm{~s}^{2}\right)\left(\sigma^{*} 1 \mathrm{~s}^{2}\right)\left(\sigma 2 \mathrm{~s}^{2}\right)\left(\sigma^{*} 2 \mathrm{~s}^{2}\right)\left(\sigma 2 \mathrm{p}_{\mathrm{z}}^{2}\right)\left(\pi 2 \mathrm{p}_{\mathrm{x}}^{2}=\pi 2 \mathrm{p}_{\mathrm{y}}^{2}\right)\left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}^{2}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}^{2}\right)$

(iii) $\mathrm{O}_{2}^{\oplus} \Rightarrow\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{z}^{2}\right)\left(\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}\right)\left(\pi^{*} 2 p_{x}^{1}=\pi^{*} 2 p_{y}\right)$

B.O. $=\frac{6-1}{2}=2.5$

$\mathrm{O}_{2} \Rightarrow\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{\mathrm{z}}^{2}\right)\left(\pi 2 p_{\mathrm{x}}^{2}=\pi 2 p_{\mathrm{y}}^{2}\right)\left(\pi^{*} 2 p_{\mathrm{x}}^{1}=\pi^{*} 2 p_{\mathrm{y}}^{1}\right)$

B.O. $\frac{6-2}{2}=2$ (Bond order increases, Bond strength increases)

(iv) Size of atom increases, Bond length increases Size of Li > B

So, Bond length of $\mathrm{Li}_{2}>\mathrm{B}_{2}$