Advertisements
Advertisements
प्रश्न
Write the hybridisation and number of unpaired electrons in the complex `[CoF_6]^(3-)`. (Atomic No. of Co = 27)
Advertisements
उत्तर
[CoF6]3– : sp3d2 hybridisation; octahedral shape, 4 unpaired electrons, i.e. paramagnetic.
APPEARS IN
संबंधित प्रश्न
Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic.
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion.
[NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? (Atomic no. Ni = 28)
Using valence bond theory, explain the following in relation to the complexes given below:
\[\ce{[Co(NH3)6]^{3+}}\]
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
In a coordination entity, the electronic configuration of the central metal ion is t2g3 eg1
Is the coordination compound a high spin or low spin complex?
As the s-character of hybridised orbital increases, the bond angle
Which of the following has square planar structures?
Write the hybridisation and magnetic behaviour of [CoF6]3−.
[Given: Atomic number of Co = 27]
[Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2− has square planar, yet both exhibit diamagnetism. Explain.
[Atomic number: Ni = 28]
During chemistry class, a teacher wrote \[\ce{[Ni(CN)4]^2-}\] as a coordination complex ion on the board. The students were asked to find out the magnetic behaviour and shape of the complex. Pari, a student, wrote the answer paramagnetic and tetrahedral whereas another student Suhail wrote diamagnetic and square planer.
Evaluate Pari’s and Suhail’s responses.
