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, $1^2 + 2^2 + 6^2 = 41$.. Step 1: Ammonia (NH3) is a Lewis base that coordinates to copper ions through the lone pair of electrons on its nitrogen atom. Step 2: In acidic solutions, the high concentration of protons (H^+) causes ammonia to undergo protonation to form the ammonium ion (NH4^+). Step 3: Because NH4^+ lacks a lone pair of electrons, it cannot coordinate to Cu^2+ ions. Therefore, ammonia coordinates with protons instead of forming the complex, which corresponds to option (d).

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