Among $\mathrm{V}(\mathrm{CO})_{6}$, $\mathrm{Cr}(\mathrm{CO})_{5}$, $\mathrm{Cu}(\mathrm{CO})_{3}$, $\mathrm{Mn}(\mathrm{CO})_{5}$, $\mathrm{Fe}(\mathrm{CO})_{5}$, $[\mathrm{Co}(\mathrm{CO})_{3}]^{3-}$, $[\mathrm{Cr}(\mathrm{CO})_{4}]^{4-}$, and $\mathrm{Ir}(\mathrm{CO})_{3}$, the total number of species isoelectronic with $\mathrm{Ni}(\mathrm{CO})_{4}$ is ____. [Given, atomic number: $\mathrm{V}=23$, $\mathrm{Cr}=24$, $\mathrm{Mn}=25$, $\mathrm{Fe}=26$, $\mathrm{Co}=27$, $\mathrm{Ni}=28$, $\mathrm{Cu}=29$, $\mathrm{Ir}=77$]

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Coordination Compounds

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

]

Answer: 3

For metal complexes, isoelectronic species have the same effective atomic number (EAN). For

Ni (CO)_{4}

: EAN

= 28 + 4 \times 2 = 36

. Checking each: (i)

V (CO)_{6} = 23 + 12 = 35

; (ii)

Cr (CO)_{5} = 24 + 10 = 34

; (iii)

Cu (CO)_{3} = 29 + 6 = 35

; (iv)

Mn (CO)_{5} = 25 + 10 = 35

; (v)

Fe (CO)_{5} = 26 + 10 = 36

✓; (vi)

[Co (CO)_{3}]^{3 -} = 27 + 3 + 6 = 36

✓; (vii)

[Cr (CO)_{4}]^{4 -} = 24 + 4 + 8 = 36

✓; (viii)

Ir (CO)_{3} = 77 + 6 = 83

. Three species have EAN = 36.

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Integer answer