Chemistry - Coordination Chemistry Question with Solution | TestHub

Bond order of ${\mathrm{O}}_2$ and ${\mathrm{NO}}^{-}$have unpaired electrons & both are 16-electron system hence B.O. $=\frac{10-6}{2}=2$

The question asks to identify the correct statement regarding an isoelectronic pair and a given property.

Option A: O₂ and NO⁻; same bond order

O₂ has 16 electrons. Its electronic configuration is $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^2$. Bond order = (8-4)/2 = 2.

NO⁻ also has 16 electrons. Its electronic configuration is $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^2$. Bond order = (8-4)/2 = 2.

Therefore, O₂ and NO⁻ are isoelectronic and have the same bond order.

Option B: Ni(CO)₄ and [Fe(CO)₄]²⁻; same metal-carbon bond length

Ni(CO)₄: Ni is in the 0 oxidation state. The electronic configuration of Ni is [Ar] 3d⁸ 4s². So, Ni⁰ has 10 d-electrons.

[Fe(CO)₄]²⁻: Fe is in the -2 oxidation state. The electronic configuration of Fe is [Ar] 3d⁶ 4s². So, Fe²⁻ has 8+2 = 10 d-electrons.

Both are isoelectronic. However, the charge on the metal center affects the synergic-bonding to CO. Higher negative charge on the metal center in [Fe(CO)₄]²⁻ will result in increased synergic-bonding, which will increase the M-C bond length and decrease the C-O bond length. Therefore, the M-C bond lengths are NOT the same.

Option C: CO₃²⁻ and NO₃⁻; same bond length

CO₃²⁻ has 24 valence electrons. The Lewis structure has three resonance structures with C-O bond orders of 4/3 = 1.33.

NO₃⁻ has 24 valence electrons. The Lewis structure has three resonance structures with N-O bond orders of 4/3 = 1.33.

the bond order is the same but bond length cannot be same due to different atoms.

Option D: Fe(CO)₅ and Cr(CO)₆; different magnetic behavior

Fe(CO)₅: Fe is in the 0 oxidation state. The electronic configuration of Fe is [Ar] 3d⁶ 4s². CO is a strong field ligand, so it will cause pairing of electrons. Fe(CO)₅ has a trigonal bipyramidal structure. It is diamagnetic.

Cr(CO)₆: Cr is in the 0 oxidation state. The electronic configuration of Cr is [Ar] 3d⁵ 4s¹. CO is a strong field ligand, so it will cause pairing of electrons. Cr(CO)₆ has an octahedral structure. It is diamagnetic.

Both are diamagnetic, so they have the SAME magnetic behavior.

Final Answer: The correct answer is A.