Nickel (Z = 28) combines with a uninegative m | vedclass

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

Question

(D) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.In the complex $[NiX_4]^{-2}$,let the oxidation state

Vedclass · Accepted Answer

$2$,tetrahedral

Vedclass · Answer

(D) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.In the complex $[NiX_4]^{-2}$,let the oxidation state of $Ni$ be $y$.$y + 4(-1) = -2 \implies y = +2$.So,$Ni^{+2}$ has the configuration $[Ar] 3d^8$.In the $3d$ orbital,there are $8$ electrons. According to Hund's rule,these are distributed as $3$ paired and $2$ unpaired electrons.Since the complex is paramagnetic and the ligand $X^-$ is a weak field ligand (as it forms a tetrahedral complex),no pairing of electrons occurs in the $3d$ orbitals.Thus,there are $2$ unpaired electrons.The hybridization involved is $sp^3$,which corresponds to a tetrahedral geometry.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex $[NiX_4]^{-2}.$ The number of unpaired electron$(s)$ in the nickel and geometry of this complex ion are,respectively $:$

Similar Questions

The example of $dsp^2$ hybridisation is

Choose the correct statement for the $[Ni(CN)_4]^{2-}$ complex ion (Atomic number of $Ni=28$).

In Wilkinson's catalyst,the hybridization of the central metal ion and its shape are respectively:

The total number of unpaired electrons present in $[Co(NH_3)_6]Cl_2$ and $[Co(NH_3)_6]Cl_3$ is $......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module