Which of the following is correctly matched?

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridi

Vedclass · Accepted Answer

$[Fe(CN)_6]^{4-}$ and $[Fe(CN)_6]^{3-}$ - Both are octahedral with $d^2sp^3$ hybridisation

Vedclass · Answer

(A) $1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridisation (inner orbital complex).$2$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ causes pairing,resulting in $d^2sp^3$ hybridisation.$3$. Both are octahedral and $d^2sp^3$ hybridised. Thus,option $A$ is correct.$4$. $[Ni(CO)_4]$ is tetrahedral ($sp^3$,diamagnetic),but $[Ni(CN)_4]^{2-}$ is square planar ($dsp^2$,diamagnetic). Option $B$ is incorrect.$5$. $[Ni(CO)_4]$ and $[Co(CO)_4]^{-}$ are both isoelectronic and tetrahedral ($sp^3$,diamagnetic). Option $C$ is also technically correct in terms of geometry and hybridisation,but $A$ is the standard textbook example for $d^2sp^3$ comparison.$6$. $[Co(H_2O)_6]^{3+}$ is $d^2sp^3$ (low spin),while $[Cr(H_2O)_6]^{3+}$ is $d^2sp^3$ (inner orbital). However,$A$ is the most definitive match.

Similar Questions

$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many unpaired electrons are present in the analogous complex of $Ni(II)$?

Consider the transition metal ions $Mn^{3+}$,$Cr^{3+}$,$Fe^{3+}$,and $Co^{3+}$. All form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective $d$-orbitals of the complexes is:

Which of the following facts about the complex $[Cr(NH_3)_6]Cl_3$ is wrong?

The hybridisation and geometry of $[Ni(CN)_4]^{2-}$ are:

Which of the following has the maximum number of unpaired electrons?

Vedclass Test Series

Exam Paper Generator

Online Exam Module