It is an experiment fact that Cs 2[ CuCl 4] is orange coloured but ( NH 4)2[ CuCl 4] is yellow. It is further known that total paramagnetic moment of a unpaired electron is due to spin as well as due to nature of orbital, 'd' orbital contributing more than 's' or ' p '. Thus the total paramagnetic moment of orange compound is found to be more than that of yellow compound. Then which of the following is correct :

Question

It is an experiment fact that Cs 2[ CuCl 4] is orange coloured but ( NH 4)2[ CuCl 4] is yellow. It is further known that total paramagnetic moment of a unpaired electron is due to spin as well as due to nature of orbital, 'd' orbital contributing more than 's' or ' p '. Thus the total paramagnetic moment of orange compound is found to be more than that of yellow compound. Then which of the following is correct : (A) Anion of orange compound is tetrahedral and that of yellow is square planar (B) Anion of orange compound is square planar and that of yellow is tetrahedral (C) Both the anions are tetrahedral (D) Both the anions are square planar

Tardigrade Admin · Accepted Answer

Cs 2[ CuCl 4] (orange) is tetrahedral because in Cu ( II ) the unpaired electron is in 3 d. But ( NH 4)2[ CuCl 4] yellow is square planar because the unpaired electron is not in 3 d rather in some promoted state 's' or ' p '.

Anion of orange compound is tetrahedral and that of yellow is square planar

Anion of orange compound is square planar and that of yellow is tetrahedral

Both the anions are tetrahedral

Both the anions are square planar

Cs2​[CuCl4​] (orange) is tetrahedral because in Cu(II) the unpaired electron is in 3d. But (NH4​)2​[CuCl4​] yellow is square planar because the unpaired electron is not in 3d rather in some promoted state 's' or ' p '.

$C s_{2} [C u C l_{4}]$ (orange) is tetrahedral because in $C u (II)$ the unpaired electron is in $3 d$ . But $(N H_{4})_{2} [C u C l_{4}]$ yellow is square planar because the unpaired electron is not in $3 d$ rather in some promoted state 's' or ' $p$ '.