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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Chemistry
Chapter
Coordination Compounds

JEE Advanced 2024 · Paper 1 · Chemistry · Coordination Compounds

Among V(CO)_6, Cr(CO)_5, Cu(CO)_3, Mn(CO)_5, Fe(CO)_5,[Co(CO)_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−\mathrm{V}(\mathrm{CO})_6, \mathrm{Cr}(\mathrm{CO})_5, \mathrm{Cu}(\mathrm{CO})_3, \operatorname{Mn}(\mathrm{CO})_5, \mathrm{Fe}(\mathrm{CO})_5,\left[\mathrm{Co}(\mathrm{CO})_3\right]^{3-},\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-}, and Ir⁡(CO)3\operatorname{Ir}(\mathrm{CO})_3, the total number of species isoelectronic with Ni(CO)4\mathrm{Ni}(\mathrm{CO})_4 is ________. [Given, atomic number: V=23,Cr=24,Mn=25,Fe=26,Co=27,Ni=28,Cu=29,Ir=77\mathrm{V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27, \mathrm{Ni}=28, \mathrm{Cu}=29, \mathrm{Ir}=77]

Correct answer

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Step-by-step solution

In case of complexes, isoelectronic species should be those having same effective atomic number (EAN) Ni(CO)4⇒28+4×2=36\mathrm{Ni}(\mathrm{CO})_4 \Rightarrow 28+4 \times 2=36 (i) V(CO)6⇒23+2×6=35\mathrm{V}(\mathrm{CO})_6 \Rightarrow 23+2 \times 6=35 (ii) Cr⁡(CO)5⇒24+2×5=34\operatorname{Cr}(\mathrm{CO})_5 \Rightarrow 24+2 \times 5=34 (iii) Cu(CO)3⇒29+2×3=35\mathrm{Cu}(\mathrm{CO})_3 \Rightarrow 29+2 \times 3=35 (iv) Mn(CO)5⇒25+2×5=35\mathrm{Mn}(\mathrm{CO})_5 \Rightarrow 25+2 \times 5=35 (v) Fe(CO)5⇒26+2×5=36\mathrm{Fe}(\mathrm{CO})_5 \Rightarrow 26+2 \times 5=36 (vi) [Co(CO)3]3−⇒27+3+2×3=36\left[\mathrm{Co}(\mathrm{CO})_3\right]^{3-} \Rightarrow 27+3+2 \times 3=36 (vii) [Cr(CO)4]4−⇒24+4+2×4=36\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-} \Rightarrow 24+4+2 \times 4=36 (viii) [Ir⁡(CO)3]⇒77+2×3=83\left[\operatorname{Ir}(\mathrm{CO})_3\right] \Rightarrow 77+2 \times 3=83
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