Consider the following complex compounds
(i) [Cu(NH3)4][Cu(NO2)4]
(ii) [Cr(py)6][Cr(SCN)6]
(iii) [Co(NH3)5(NO2)][Pt(SCN)4]
If are more possible coordination isomers of given complex compou

Question

Consider the following complex compounds
(i) [Cu(NH3)4][Cu(NO2)4]
(ii) [Cr(py)6][Cr(SCN)6]
(iii) [Co(NH3)5(NO2)][Pt(SCN)4]
If $x_1, x_2, x_3$ are more possible coordination isomers of given complex compounds respectively then calculate value of $x_1^2 + x_2^2 + x_3^2$.

Prashant Kotian · Accepted Answer

Correct answer: **ANSWER: 41**<BR>SOLUTION: THE NUMBER OF MORE POSSIBLE COORDINATION ISOMERS ARE $X_1 = 1$, $X_2 = 2$, AND $X_3 = 6$. SO, $1^2 + 2^2 + 6^2 = 41$.

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