JEE Advanced · Chemistry · 4. Chemical Bonding
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)
- A Bond order of \(\text{Ne} _2\) is zero.
- B The highest occupied molecular orbital (HOMO) of \(\text F_2\) is \(\sigma\)-type.
- C Bond energy of \(\text O _2^{+}\)is smaller than the bond energy of \(\text O _2\).
- D Bond length of \(\text{Li} _2\) is larger than the bond length of \(\text B _2\)
Answer & Solution
Correct Answer
(C) Bond energy of \(\text O _2^{+}\)is smaller than the bond energy of \(\text O _2\).
Step-by-step Solution
Detailed explanation
(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 \(\mathrm{Li}_2>\mathrm{B}_2\)
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 \(\mathrm{Li}_2>\mathrm{B}_2\)
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