Among V(CO)_6,Cr(CO)_5,Cu(CO)_3,Mn(CO)_5,Fe(C | vedclass

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(C) To be isoelectronic with $Ni(CO)_4$,the species must have the same Effective Atomic Number $(EAN)$.$EAN$ of $Ni(CO)_4 = 28 + 4 	imes 2 = 36$.Calc

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$3$

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(C) To be isoelectronic with $Ni(CO)_4$,the species must have the same Effective Atomic Number $(EAN)$.$EAN$ of $Ni(CO)_4 = 28 + 4 	imes 2 = 36$.Calculating $EAN$ for the given species:$1. V(CO)_6 = 23 + 6 	imes 2 = 35$$2. Cr(CO)_5 = 24 + 5 	imes 2 = 34$$3. Cu(CO)_3 = 29 + 3 	imes 2 = 35$$4. Mn(CO)_5 = 25 + 5 	imes 2 = 35$$5. Fe(CO)_5 = 26 + 5 	imes 2 = 36$ (Matches)$6. [Co(CO)_3]^{3-} = 27 + 3 + 3 	imes 2 = 36$ (Matches)$7. [Cr(CO)_4]^{4-} = 24 + 4 + 4 	imes 2 = 36$ (Matches)$8. Ir(CO)_3 = 77 + 3 	imes 2 = 83$The species with $EAN = 36$ are $Fe(CO)_5$,$[Co(CO)_3]^{3-}$,and $[Cr(CO)_4]^{4-}$.Thus,the total number of species isoelectronic with $Ni(CO)_4$ is $3$.

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 : $V=23$,$Cr=24$,$Mn=25$,$Fe=26$,$Co=27$,$Ni=28$,$Cu=29$,$Ir=77$ ]

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