The pair of metal carbonyl complexes that are isoelectronic is :

Question

The pair of metal carbonyl complexes that are isoelectronic is : (A) [Co(CO)4]- and Ni(CO)4  (B) Ni(CO)4 and V(CO)6 (C) [Cr(CO)6] and V(CO)6 (D)  [Fe(CO)4]- and Cr(CO)6

Tardigrade Admin · Accepted Answer

As Co (d7) and Ni (d0) elements differ by one electron and both contain four (CO) groups thus [ Co ( CO )4]- and [ Ni ( CO )4] have same number of electrons. Hence, both are isoelectronic.

Q. The pair of metal carbonyl complexes that are isoelectronic is :

$[C o (CO)_{4}]^{-}$ and $N i (CO)_{4}$

$N i (CO)_{4}$ and $V (CO)_{6}$

$[C r (CO)_{6}]$ and $V (CO)_{6}$

$[F e (CO)_{4}]^{-}$ and $C r (CO)_{6}$

As $C o (d^{7})$ and $N i (d^{0})$ elements differ by one electron and both contain four (CO) groups thus $[C o (CO)_{4}]^{-}$ and $[N i (CO)_{4}]$ have same number of electrons.

Hence, both are isoelectronic.

Q. The pair of metal carbonyl complexes that are isoelectronic is :

[Co(CO)4​]− and Ni(CO)4​

Ni(CO)4​ and V(CO)6​

[Cr(CO)6​] and V(CO)6​

[Fe(CO)4​]− and Cr(CO)6​

As Co(d7) and Ni(d0) elements differ by one electron and both contain four (CO) groups thus [Co(CO)4​]− and [Ni(CO)4​] have same number of electrons. Hence, both are isoelectronic.

$[C o (CO)_{4}]^{-}$ and $N i (CO)_{4}$

$N i (CO)_{4}$ and $V (CO)_{6}$

$[C r (CO)_{6}]$ and $V (CO)_{6}$

$[F e (CO)_{4}]^{-}$ and $C r (CO)_{6}$

As $C o (d^{7})$ and $N i (d^{0})$ elements differ by one electron and both contain four (CO) groups thus $[C o (CO)_{4}]^{-}$ and $[N i (CO)_{4}]$ have same number of electrons.

Hence, both are isoelectronic.