Get Answers to all your Questions

header-bg qa

Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?

  • Option 1)

    \left [ Ni(NH_{3})_{6} \right ]^{2+}

  • Option 2)

    \left [ Zn(NH_{3})_{6} \right ]^{2+}

  • Option 3)

    \left [ Cr(NH_{3})_{6} \right ]^{3+}

  • Option 4)

    \left [ Co(NH_{3})_{6} \right ]^{3+}

 

Answers (3)

best_answer

Posted by

Plabita

View full answer

As learnt in

Hybridisation -

sp3d2 - square bipyramidal or octahedral 

d2sp3 - octahedral 

sp3 - tetradedral 

dsp2 - square planar

- wherein

sp3d2 - outer complex

d2sp3 - inner complex

sp3 - [Ni(Cl)_{4}]^{2-}

dsp2 - [Pt(CN)_{4}]^{2-}

 

 Electronic configuration of Ni2+ in [Ni (NH3)6]2+ : 3d8 4s° 4p°

NH3 is a strong field ligand. So,

Since, 2 orbitals are not vacant in Ni2+, so it will form an outer orbital complex with sp3d2 hybridization and 2 unpaired electrons giving it paramagnetic nature.


Option 1)

\left [ Ni(NH_{3})_{6} \right ]^{2+}

This option is correct

Option 2)

\left [ Zn(NH_{3})_{6} \right ]^{2+}

This option is incorrect

Option 3)

\left [ Cr(NH_{3})_{6} \right ]^{3+}

This option is incorrect

Option 4)

\left [ Co(NH_{3})_{6} \right ]^{3+}

This option is incorrect

Posted by

Ankit

View full answer

2

 

 

Posted by

Shandhiya kumar

View full answer