What is the hybridization of the central metal ion in [Cr(NH3)6]3+? (2020) 2020
Practice Questions
Q1
What is the hybridization of the central metal ion in [Cr(NH3)6]3+? (2020) 2020
sp
sp2
d2sp3
dsp2
Questions & Step-by-Step Solutions
What is the hybridization of the central metal ion in [Cr(NH3)6]3+? (2020) 2020
Step 1: Identify the central metal ion in the complex [Cr(NH3)6]3+. The central metal ion is chromium (Cr).
Step 2: Determine the oxidation state of chromium in the complex. Since the overall charge is +3 and ammonia (NH3) is a neutral ligand, the oxidation state of Cr is +3.
Step 3: Write the electron configuration of chromium in its elemental form. Chromium (Cr) has an atomic number of 24, so its electron configuration is [Ar] 3d5 4s1.
Step 4: Adjust the electron configuration for the +3 oxidation state. Remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d3.
Step 5: Determine the coordination number. In [Cr(NH3)6]3+, there are six ammonia ligands coordinating to the chromium ion, giving a coordination number of 6.
Step 6: Identify the geometry based on the coordination number. A coordination number of 6 typically results in an octahedral geometry.
Step 7: Determine the hybridization based on the octahedral geometry. For octahedral complexes, the hybridization is d2sp3.
