In the coordination compound [Cr(NH₃)(en)₂Cl]Br₂, the coordination number (C.N.) and oxidation number (O.N.) of the metal atom are, respectively, A) C.N. = 6; O.N. = +4. B) C.N. = 6; O.N. = +3. C) C.N. = 5; O.N. = +2. D) C.N. = 4; O.N. = +2. E) C.N. = 4; O.N. = +3.

Question

Quizplus · Accepted Answer

The compound contains one chromium (Cr) atom, which is surrounded by one en molecule, one chloride (Cl) ion, one bromide (Br) ion, one thiolate ligand (NHSBU13SB0) and one thiocyanate ligand (en) which means it has a total of six ligands.
Thus, the coordination number (C.N.) of Cr is 6.
To determine the oxidation number (O.N.) of Cr, we can use the sum of the oxidation numbers of all the ligands and the charge on the complex ion.
The Br ion has a charge of -1, and the Cl ion has a charge of -1. The thiocyanate ligand has a charge of -1, and the thiolate ligand has a charge of -1. The en molecule is neutral.
Therefore, the total charge on the complex ion is -3 (-1 from Br and -1 from Cl and -1 from thiolate or en).
To balance this charge, the oxidation number of Cr must be +3. Thus, the correct choice is B.

In the Coordination Compound [Cr(NH3)(en)2Cl]Br2, the Coordination Number (C

In the Coordination Compound [Cr(NH₃)(en)₂Cl]Br₂, the Coordination Number (C