Step 1: Recall the direction cosine condition
For a line making angles \( \alpha, \beta, \gamma \) with the positive directions of the \( x \)-axis, \( y \)-axis, and \( z \)-axis respectively, the sum of the squares of the direction cosines is: \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. \]
Step 2: Substitute the given angles
The line makes angles \( \alpha = \frac{\pi}{4} \) and \( \gamma = \frac{\pi}{4} \) with the \( x \)-axis and \( z \)-axis, so: \[ \cos^2 \frac{\pi}{4} + \cos^2 \beta + \cos^2 \frac{\pi}{4} = 1. \] Since \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we have: \[ \left( \frac{1}{\sqrt{2}} \right)^2 + \cos^2 \beta + \left( \frac{1}{\sqrt{2}} \right)^2 = 1. \] Simplify: \[ \frac{1}{2} + \cos^2 \beta + \frac{1}{2} = 1. \]
Step 3: Solve for \( \cos^2 \beta \)
Combine terms:
\[ 1 + \cos^2 \beta = 1 \implies \cos^2 \beta = 0. \] Thus: \[ \cos \beta = 0 \implies \beta = \frac{\pi}{2}. \]
Step 4: {Conclude the result}
The angle which the line makes with the positive direction of the \( y \)-axis is \( \frac{\pi}{2} \).
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: