JEE Advanced 2024 Paper 1, Chemistry Section 3 Q4: Bonding Of Co-ordination Compounds

JEE Advanced2024Paper 1Chemistry Section 3

Among V(CO) $_{6}$ , Cr(CO) $_{5}$ , Cu(CO) $_{3}$ , Mn(CO) $_{5}$ , Fe(CO) $_{5}$ , [Co(CO) $_{3}$ ] $^{3 -}$ , [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]

Your answer

Solution

Total electrons in Ni(CO) $_{4}$ = $28 + 4 \times 14 = 28 + 56 = 84$ .

Compute total electron count for each species (atomic number of metal + $14 \times$ number of COs, adjusted for charge):

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) $_{3}$ ] $^{3 -}$ : $27 + 42 + 3 = 72$ .

[Cr(CO) $_{4}$ ] $^{4 -}$ : $24 + 56 + 4 = 84$ . $✓$ (matches Ni(CO) $_{4}$ )

Ir(CO) $_{3}$ : $77 + 42 = 119$ .

Only [Cr(CO) $_{4}$ ] $^{4 -}$ has 84 electrons. Count = 1.

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JEE Advanced 2024 Paper 1, Chemistry Section 3 Q4: Bonding Of Co-ordination Compounds

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