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Bonding in some hetero nuclear di-atomic molecules: Molecular orbital diagram of Carbon monoxide molecule (CO) Electronic configuration of C atom: 1s2 2s2 2p2 Electronic configuration of O atom: 1s2 2s2 2p4 Electronic configuration of CO molecule : σ1s2, σ1s"2, σ2s2, σ2s"2, π2py2, π2pz2 σ2px2 Bond order = \ (\frac {N_b - N_a}2\) \ (=\frac {10 - 4}2\) \ (= 3\) The molecule has no unpaired electrons hence it is diamagnetic. Learn how to calculate the bond order of CO group using the formula and electronic configuration. The correct answer is 3 and CO has a dative bond. Hint: The molecular orbital theory helps us find out the bond order on the basis of molecular order diagrams. The Atomic no, of C = 6 and O = 8 Complete answer: The electronic configuration of C is 1 s 2 2 s 2 2 p 2 and that of O + is 1 s 2 2 s 2 2 p 4. Steps for drawing the Molecular Orbital Diagram Since C O is a heteronuclear atom, the atom with more electronegativity will be placed lower in the energy level. So For the 1s degenerate orbital, 2 electrons will go to σ 1 s and 2 electrons ... What is Bond Order Bond order refers to the number of chemical bonds formed between two atoms in a molecule. For example, two atoms of hydrogen combine to form a hydrogen molecule (H-H or H 2). Therefore, the bond order of H 2 is one. Similarly, the bond order of oxygen (O=O or O 2) is two and that of nitrogen (N ≡ N or N 2) is three. The value need not be an integer, especially for polyatomic molecules having resonance structures [1-4].
