Among [ Ni(CO)4],[ Ni(CN) 4]2- ,[ NiCl4]2- species, the hybridization states of the Ni atom are, respectively(At. No. of Ni = 28) [2004]a)sp3, dsp2, dsp2b)sp3, dsp2, sp3c)sp3, sp3, dsp2d)dsp2 , sp3 , sp3Correct answer is option 'B'. Can you explain this answer?

Question

Answer

In Ni(CO)4, nickel is sp3 hybridised because in it
oxidation state of Ni is zero. So, configuration of
28Ni=1s2,2s22p6,3s23p63d8,4s2
(CO is a strong field ligand, hence does pairing of
electrons)
In [Ni(CN)4]2−nickel is present as Ni2+, so its
configuration = 1s2,2s22p6,3s23p63d8
CN− is strong field ligand, hence it makes
Ni2+ electrons to be paired up.
In [NiCl4]2−, nickel is present as Ni2+ , so its
configuration = 1s2,2s22p6,3s23p6,3d8
Cl− is a weak field ligand, hence in Ni2+ electrons
are not paired....

Answer

₂₈ Ni⁰ :

Remember CO is a strong ligand

Remember CN^– is also a strong ligand

Cl^– is a weak ligand, hence no pairing of electrons takes place.

Answer

The hybridization states of the Ni atom in the given species are as follows:

1. [Ni(CO)4]
In this species, the nickel atom is bonded to four carbon monoxide (CO) ligands. Carbon monoxide is a strong field ligand, which means it forms strong bonds with the metal atom and causes the d orbitals to split into a lower energy set and a higher energy set.
The electron configuration of the Ni atom is 3d8 4s2. In the presence of strong field ligands, the 3d orbitals are involved in bonding. The four CO ligands will occupy the lower energy d orbitals, leaving the 4s and higher energy d orbitals empty.
Thus, the hybridization state of the Ni atom in [Ni(CO)4] is sp3.

2. [Ni(CN)4]2-
In this species, the nickel atom is bonded to four cyanide (CN) ligands. Cyanide is a weak field ligand, which means it forms weak bonds with the metal atom and causes little splitting of the d orbitals.
The electron configuration of the Ni atom is 3d8 4s2. In the presence of weak field ligands, the 3d orbitals are not involved in bonding. The four CN ligands will occupy the 4s and 4p orbitals, leaving the 3d orbitals empty.
Thus, the hybridization state of the Ni atom in [Ni(CN)4]2- is sp3.

3. [NiCl4]2-
In this species, the nickel atom is bonded to four chloride (Cl) ligands. Chloride is an intermediate field ligand, which means it forms moderate strength bonds with the metal atom and causes partial splitting of the d orbitals.
The electron configuration of the Ni atom is 3d8 4s2. In the presence of intermediate field ligands, the 3d orbitals are partially involved in bonding. The four Cl ligands will occupy the lower energy d orbitals, leaving the 4s and higher energy d orbitals empty.
Thus, the hybridization state of the Ni atom in [NiCl4]2- is sp3.

In summary, the hybridization states of the Ni atom in the given species are sp3, sp3, and dsp2 for [Ni(CO)4], [Ni(CN)4]2-, and [NiCl4]2- respectively.

Similar Class 12 Doubts

Among [Ni(CO)4],[Ni(CN)4]2- and [NiCI4]2- species the hybridisation states of Ni atom are respectively

Among [NiCl2 (PPh3)2], [Ni(HDMG)2] and [Ni(CO)4] complexes the hybridisation statesof Ni atom are respectively.

Qut of TiF62–, CoF63–, Cu2Cl2 and NiCl24– (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28), the colourless species are: [2009]

Some of the properties of the two species,NO-3 and H3O+ are describe below. Which one of them is correct?

Number of chemical species having negative oxidation state for nitrogen amongNF3, NCI3, NH2OH, NH3,CH3NH2, NH-2, L13N, N20, HCN, HNC, NO-2 Correct answer is '8'. Can you explain this answer?

Top Courses for Class 12

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams