List - I [Species] List - II [Stretching frequency of CO (cm -1 )] (P) CO (1) 2143 (Q) Cr ( CO ) 6 (2) 2000 (R) [ V ( CO ) 6 ] − (3) 1860 (S) [ Mn ( CO ) 6 ] + (4) 2090

Question

List - I [Species] List - II [Stretching frequency of CO (cm -1 )] (P) CO (1) 2143 (Q) Cr ( CO ) 6 ​ (2) 2000 (R) [ V ( CO ) 6 ​ ] − (3) 1860 (S) [ Mn ( CO ) 6 ​ ] + (4) 2090

TestHub · Accepted Answer

The metal atom donates electrons to the $π^{*}$ orbital of CO; the weaker the C–O bond becomes, the lower the stretching frequency.

P (CO): Free CO has the strongest bond and highest frequency ( $2143 c m^{- 1}$ ).

S $[M n (CO)_{6}]$ ⁺: The positive charge on Mn reduces back-donation, keeping the frequency high ( $2090 c m^{- 1}$ ).

Q $C r (CO)_{6}$ : A neutral complex with moderate back-donation results in a frequency of $2000 c m^{- 1}$ .

R $[V (CO)_{6}]$ ⁻: The negative charge on V increases back-donation significantly, weakening the C–O bond and giving the lowest frequency ( $1860 c m^{- 1}$ ).

Therefore, the correct matching is (A) P $\to$ 1; Q $\to$ 2; R $\to$ 3; S $\to$ 4.

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Doubts & Discussion

List - I [Species]	List - II [Stretching frequency of CO (cm^-1 )]
(P) CO	(1) 2143
(Q) $Cr (CO)_{6}$	(2) 2000
(R) $[V (CO)_{6}]^{-}$	(3) 1860
(S) $[Mn (CO)_{6}]^{+}$	(4) 2090