Assume internuclear axis to be z-axis, the correct molecular orbital representation of \(\pi^*\) antibonding molecular orbital formed by overlapping between two \(p\) orbitals will be:

Concept: \(\pi\) and \(\pi^*\) molecular orbitals are formed by sideways overlap of \(p\)-orbitals perpendicular to the internuclear axis. • \(\pi\) bonding MO: constructive overlap of lobes. • \(\pi^*\) antibonding MO: destructive overlap with a nodal plane between nuclei. If the internuclear axis is \(z\)-axis, then overlap occurs between \(p_x\) or \(p_y\) orbitals.
Step 1: Nature of antibonding orbital In \(\pi^*\) antibonding MO: • Opposite phases overlap. • A node exists between the nuclei. • Electron density is outside the internuclear region.
Step 2: Identify correct diagram Among the given options, diagram (3) correctly shows: • sidewise overlap • opposite phases of \(p\)-orbitals • nodal region between nuclei Thus the correct representation is option (3). \[ \boxed{\text{Correct option: (3)}} \]
Two atoms are bonding along the Z-axis. How will the \( \pi^* \) orbital look like (Z is the internuclear axis)? 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
As shown in the figure, the ratio of \(T_1\) and \(T_2\) is 