The correct structures of trans-[NiBr_{2}(PPh | vedclass

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

Question

(A) $1$. For the square planar complex $trans-[NiBr_{2}(PPh_{3})_{2}]$,the two identical ligands ($Br$ or $PPh_{3}$) must be positioned opposite to ea

Vedclass · Accepted Answer

Vedclass · Answer

(A) $1$. For the square planar complex $trans-[NiBr_{2}(PPh_{3})_{2}]$,the two identical ligands ($Br$ or $PPh_{3}$) must be positioned opposite to each other (at $180^{\circ}$ angle). In the structure shown in option $A$,the $Br$ atoms are trans to each other,and the $PPh_{3}$ groups are trans to each other.$2$. For the octahedral complex $meridional-[Co(NH_{3})_{3}(NO_{2})_{3}]$,the three identical ligands must lie on the same plane (meridian). In the structure shown in option $A$,the three $NH_{3}$ ligands are in a meridional arrangement,and the three $NO_{2}$ ligands are also in a meridional arrangement.$3$. Therefore,the structure in option $A$ correctly represents both the trans-isomer of the nickel complex and the meridional-isomer of the cobalt complex.

The correct structures of trans-$[NiBr_{2}(PPh_{3})_{2}]$ and meridional-$[Co(NH_{3})_{3}(NO_{2})_{3}]$,respectively,are

Similar Questions

Which of the following ions has the highest magnetic moment?

Which of the following compounds is not coloured?

For a $t_{2g}^6 e_g^2$ system,the value of the magnetic moment $(\mu)$ is equal to ..... $B.M.$

The correct order of the spin-only magnetic moments of the following complexes is
$(I) [Cr(H_2O)_6]Br_2$
$(II) Na_4[Fe(CN)_6]$
$(III) Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$
$(IV) (Et_4N)_2[CoCl_4]$

The calculated spin-only magnetic moments $(BM)$ of the species $[Fe(CN)_6]^{4-}$ and $[Fe(H_2O)_6]^{2+}$,respectively,are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module