The number of unpaired electrons in Ni(CO)_4 | vedclass

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

Question

(A) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes the pairing of electrons in the $3d$ and $4s$ orbitals. As a result,the $4s$ electrons are promoted to the $3d$ orbital,leading to a fully filled $3d$ subshell $(3d^{10})$ and an empty $4s$ orbital. Therefore,there are $0$ unpaired electrons in $Ni(CO)_4$.

List-$I$	List-$II$
$A. Ni(CO)_4$	$I. sp^3$
$B. [Ni(CN)_4]^{2-}$	$II. sp^3d^2$
$C. [Co(CN)_6]^{3-}$	$III. d^2sp^3$
$D. [CoF_6]^{3-}$	$IV. dsp^2$

The number of unpaired electrons in $Ni(CO)_4$ is

Similar Questions

The species having tetrahedral shape is

The tetracyanonickelate $(II)$ complex ion has a ..... geometry.

Which among the following statements about $[Ni(CN)_4]^{2-}$ is $NOT$ true?

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A. Ni(CO)_4$ $I. sp^3$

$B. [Ni(CN)_4]^{2-}$ $II. sp^3d^2$

$C. [Co(CN)_6]^{3-}$ $III. d^2sp^3$

$D. [CoF_6]^{3-}$ $IV. dsp^2$

Choose the correct answer from the options given below.

Which of the following pairs of structure and magnetic properties is correct for the tetracyanonickelate$(II)$ ion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module