JEE Advanced 2024 Paper 1 · Q11 · Metal Carbonyls — Isoelectronic Counting
Among $V(CO)_6, Cr(CO)_5, Cu(CO)_3, Mn(CO)_5, Fe(CO)_5, [Co(CO)_4]^{3-}, [Cr(CO)_4]^{4-}$, and $Ir(CO)_3$, the total number of species isoelectronic with $Ni(CO)_4$ is ______. [Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]
Reveal answer + step-by-step solution
Correct answer:1
Solution
Total electron count of $Ni(CO)_4$ = $28 + 4 \times 14 = 84$ (Ni provides 28; each CO provides 14, i.e., $6_C + 8_O$). Compute total electrons for each species: $V(CO)_6: 23 + 84 = 107$; $Cr(CO)_5: 24 + 70 = 94$; $Cu(CO)_3: 29 + 42 = 71$; $Mn(CO)_5: 25 + 70 = 95$; $Fe(CO)_5: 26 + 70 = 96$; $[Co(CO)_4]^{3-}: 27 + 56 + 3 = 86$; $[Cr(CO)_4]^{4-}: 24 + 56 + 4 = \mathbf{84}$ ✓ ; $Ir(CO)_3: 77 + 42 = 119$. Only $[Cr(CO)_4]^{4-}$ has the same total electron count (84) as $Ni(CO)_4$. Answer = $\mathbf{1}$.
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