Q5 / 48 · JEE Advanced 2025 (Paper 1) · Chemistry

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

  • A
    $\text { Bond order of } \mathrm{Ne}_2 \text { is zero. }$
  • Correct answer
    B
    $\text { The highest occupied molecular orbital (HOMO) of } F_2 \text { is } \sigma \text {-type. }$
  • Correct answer
    C
    $\text { Bond energy of } \mathrm{O}_2^{+} \text {is smaller than the bond energy of } \mathrm{O}_2 \text {. }$
  • D
    $\text { Bond length of } \mathrm{Li}_2 \text { is larger than the bond length of } \mathrm{B}_2$

Solution

(i) $\mathrm{Ne}_2 \Rightarrow\left(\sigma 1 \mathrm{~s}^2\right)\left(\sigma^* 1 \mathrm{~s}^2\right)\left(\sigma 2 \mathrm{~s}^2\right)\left(\sigma^* 2 \mathrm{~s}^2\right)\left(\sigma 2 \mathrm{p}_{\mathrm{z}}^2\right)\left(\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right)\left(\pi^* 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^2\right)\left(\sigma^* 2 \mathrm{p}_{\mathrm{z}}^2\right)$ 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 $\mathrm{Li}_2>\mathrm{B}_2$

Hence, the correct answer is options: 'B', 'C'.

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