Table of Contents
- 1 How is dsp2 hybridization identified?
- 2 What is the geometrical shape of PT Cl 4 2?
- 3 Which of the following has dsp2 hybridization?
- 4 What hybridization is the in PT ptcl42?
- 5 What is the shape of dsp2 hybridization?
- 6 Why PtCl4 2 is square planar?
- 7 Why is the hybridisation of PtCl4^2- is square planar?
- 8 What is the hybridization of electrons of 5s in 4D series?
- 9 Why is [PdCl4]2 a low spin inner orbital complex?
How is dsp2 hybridization identified?
dsp2 type of hybridization is seen specially in case of transition metal ions. The orbitals involved in this type of Hybridization are dx2- y2, s and two p. If the compound undergoes reaction with strong ligand than dsp2, dsp3,d2sp3 hybridization occurs.
What is the geometrical shape of PT Cl 4 2?
square planar
The hybridization of [PtCl4]2- is square planar. Platinum has six valence electrons. (2-) indicates the addition of two more electrons to platinum.
Which of the following has dsp2 hybridization?
NiCl42−
How can you distinguish between Sp3 and dsp2 hybridization?
Sp3 molecules are regular tetrahedral with 4bp only. dsp2 is a special case of octahedral geometry. In dsp2 – inner d-orbital and outer s,p orbitals are used.
What is the bond angle of dsp2 hybridization?
dsp2−90o.
What hybridization is the in PT ptcl42?
Therefore, it undergoes sp3 hybridization. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.
What is the shape of dsp2 hybridization?
The orbitals participating are 3d, 4s, and 4p resulting in hybridization dsp2. The shape is square planar.
Why PtCl4 2 is square planar?
Platinum has 6 valence electrons. A strong repulsion with the electron and ligand take place in between the platinum which leads to a strong crystal field splitting. Thus, the splitting breaks the degeneracy of dx2-y2 and dz2. This degeneracy stabilizes more to the square planar arrangement than the tetrahedral.
Why is dsp2 not sp2d?
The latest empty orbital out of 5 from 3d can participate in hybridization with one 4s and two 4p orbitals. This will lead to formation of dsp2 hybridization. This would never happen in case of weak field ligands.
Is dsp2 possible?
There is no vacant d orbital as 3d orbitals are fulfilled. Thus, dsp2 hybridisation is not possible for Ni(CO)4.
Why is the hybridisation of PtCl4^2- is square planar?
Even though Cl is weak field ligand hybridisation of PtCl4 ^2- have dsp2 (square planar) because of high nuclear charge of Pt (being a d-block element).
What is the hybridization of electrons of 5s in 4D series?
So electrons of 5s becomes zero. For 4d and 5d series all ligands act as strong ligands hence pairing occur. The 2 unpaired electron in 4d of Pt get paired and hence there is one vacant orbital. -> resulting in dsp2 type of hybridization and square planar geometry. Hope it helps.
Why is [PdCl4]2 a low spin inner orbital complex?
Firstly, [PdCL4]-2 is a low spin inner orbital complex. Secondly, this happens even though it carries a weak field ligand because of its high effective nuclear charge. The unexpectedly high nuclear charge occurs due to poor shielding of d electrons (10 greater than that in the case of Ni).
What is the valency of 4D and 5D series of PT?
Here valency of Pt is +2. So electrons of 5s becomes zero. For 4d and 5d series all ligands act as strong ligands hence pairing occur. The 2 unpaired electron in 4d of Pt get paired and hence there is one vacant orbital.