Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
Bond length of \( \text{Li}_2 \) is larger than the bond length of \( \text{B}_2 \)
(A) Bond order of \( \text{Ne}_2 \) is zero
Electronic configuration of \( \text{Ne}_2 \): 10 bonding and 10 antibonding electrons.
\[ \text{Bond order} = \frac{10 - 10}{2} = 0 \quad \Rightarrow \text{Correct} \]
(B) HOMO of \( \text{F}_2 \) is \( \sigma \)-type
Electronic configuration of \( \text{F}_2 \) based on MO theory:
\[\sigma_{1s}^2 \, \sigma_{1s}^{*2} \, \sigma_{2s}^2 \, \sigma_{2s}^{*2} \, \sigma_{2p_z}^2 \, \pi_{2p_x}^2 = \pi_{2p_y}^2 \, \pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}\]
So, the HOMO is the \( \pi^* \) orbital — not a \( \sigma \)-type orbital.
\[ \Rightarrow \text{Incorrect} \]
(C) Bond energy of \( \text{O}_2^+ \) is smaller than \( \text{O}_2 \)
Bond order of \( \text{O}_2 \) is 2; for \( \text{O}_2^+ \), one electron is removed from an antibonding orbital, increasing bond order to 2.5.
Since bond energy \( \propto \) bond order:
\[ \text{O}_2^+ \text{ has higher bond energy than } \text{O}_2 \Rightarrow \text{Incorrect} \]
(D) Bond length of \( \text{Li}_2 \) is larger than \( \text{B}_2 \)
Bond length generally increases with the size of the atom. Lithium is a larger atom than boron. So:
\[ \text{Bond length of Li}_2 > \text{Bond length of B}_2 \Rightarrow \text{Correct} \]
Final Answer: \( \boxed{\text{(B), (C)}} \)
Which of the following statement is true with respect to H\(_2\)O, NH\(_3\) and CH\(_4\)?
(A) The central atoms of all the molecules are sp\(^3\) hybridized.
(B) The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
(D) Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
(E) A solution of NH\(_3\) in H\(_2\)O is basic. In this solution NH\(_3\) and H\(_2\)O act as Lowry-Bronsted acid and base respectively.
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is