Question

% s-character of bonding orbital of sulphur in $H_{2}S$ is (Bond angle $H-S-H={92}^{\circ };\cos {92}^{\circ }=-0.035)$

• A. 25%
• B. 20%
• C. 3.37%
• D. 33.33%

Answer 1

The correct option is C 3.37%

The formula to calculate the % s-character is :$\frac{\cos \alpha }{\cos \alpha -1}\times 100$
$\therefore \frac{\cos {92}^{\circ }}{\cos {92}^{\circ }-1}\times 100=3.37\%$