Consider the following combination of atomic orbitals :<br>Combinding orbitals (internuclear axis)<b — Chemical Bonding Chemistry Question
Question
Consider the following combination of atomic orbitals :<br>Combinding orbitals (internuclear axis)<br>(i) $s + p_x (x)$<br>(ii) $d_{xy} + d_{xy} (x)$<br>(iii) $d_{yz} + p_z (z)$<br>(iv) $p_y + p_y (y)$<br>(v) $d_{yz} + d_{yz} (x)$<br>(vi) $s + s (z)$<br>(vii) $d_{z^2} + d_{z^2} (z)$<br>(viii) $d_{xy} + d_{xy} (z)$<br>(ix) $p_x + p_x (y)$<br>(x) $s + p_z (x)$<br>Then calculate value of "$a^2 + b^2 + 2cd$". (where a = $\sigma$ M.O.; b = $\pi$ M.O.; c = $\delta$ M.O.; d = non-bonding M.O.)
💡 Solution & Explanation
Step 1: Determine the geometry of ethyne (C2H2): each carbon is sp hybridized, yielding a symmetric linear structure (H-C≡C-H). Step 2: Analyze the structures of the options: H2O2 is open-book (non-planar), NO2 is bent (sp2 with an odd electron), SnCl2 is bent (sp2 with one lone pair). Step 3: Carbon dioxide (CO2) is sp hybridized and linear (O=C=O). Therefore, C2H2 is isostructural with CO2, matching option (d).