Consider these complex formation reactions. (i) Fe³⁺ + excess CN⁻ gives P, with secondary valency 6. (ii) Ni²⁺ + excess NH₃ gives Q, octahedral. (iii) Cu²⁺ + excess NH₃ gives R, square planar with its

Question

Consider these complex formation reactions. (i) Fe³⁺ + excess CN⁻ gives P, with secondary valency 6. (ii) Ni²⁺ + excess NH₃ gives Q, octahedral. (iii) Cu²⁺ + excess NH₃ gives R, square planar with its unpaired electron in the d $x^{2} - y^{2}$ orbital. Using crystal field theory, find |X + Y − Z|. X is the number of electrons in the $t_{2 g}$ orbitals of P, Y the number in the $e_{g}$ orbitals of Q, and Z the number in the d $x^{2} - y^{2}$ orbital of R.

TestHub · Accepted Answer

P = [Fe(CN)₆]³⁻ is low-spin d⁵, t 2 g ​ ⁵, so X = 5. Q = [Ni(NH₃)₆]²⁺ is d⁸ with t 2 g ​ ⁶ e g ​ ², so Y = 2. R = [Cu(NH₃)₄]²⁺ has one electron in d x 2 − y 2 , so Z = 1. |5 + 2 − 1| = 6.

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